IQBAL LIBRARY THE UNIVERSITY OF KASHMIR C<3ll ACC. No Overdue charges will be levied under rules for each day if the book is kept beyond the date stamed above* ELEMENTARY QUANTITATIVE ANALYSIS BY CARL J. ENGELDER FUNDAMENTALS OF SEMI-MICRO QUALITATIVE ANALYSIS Cloth; 6 by 9K; 389 pages. A TEXTBOOK OF ELEMENTARY QUALITATIVE ANALYSIS Third Edition. Cioth; 6 by 9M; 344 pages. LABORATORY RECORD BOOK OF QUALITATIVE ANALYSIS Second Edition. Paper; 6 by 9^; 91 pages. CALCULATIONS OF QUALITATIVE ANALYSIS Second Edition. Cloth; 6 by 9 ^; 174 pages. A TEXTBOOK OF ELEMENTARY QUANTITATIVE ANALYSIS Third Edition. Cloth; 6 by 9 }^; 283 pages. LABORATORY RECORD BOOK OF QUANTITATIVE ANALYSIS Second Edition. Tables. Paper; 0 by 9M; 96 pages. CALCULATIONS OF QUANTITATIVE ANALYSIS Cloth; 6 by 93^; 174 pages. A LABORATORY MANUAL OF GAS. OIL AND FUEL ANALYSIS Cloth; 6 by 9 ^; 236 pages. BY CARL J. ENGELDER TOBIAS H. DUNKELBERGER AND WILLIAM J. SCHILLER SEMI-MICRO QUALITATIVE ANALYSIS Second Edition. Cloth; 6 by QKl 305 pages. PUBUSHED B7 JOHN WILEY & SONS, IKC. NEW YORK A Textbook of ELEMENTARY QUANTITATIVE ANALYSIS BY CARL J. ENGELDER, Pll.D. PROFESSOR OF ANALYTICAL CHEMISTRY UNrV’ERSITY OF PITTSBORCIl THIRD EDITION New York • JOHN WILEY & SONS, Inc London ■ CHAPMAN & HALL, Limited 10768 Copyright, 1929, 1936, 1943 BY Carl J. Engelder AU Rights Reserved This hook or any part (hereof must not be reproduced in any form without the toriiten permission of the publisher. THIRD EDITION Fifth Printing, September, 1947 PBINTBD IN TBB UNXTBD BTATBB OF AMBBIOA PREFACE This book of Elomcntaiy Quantitative Analysis appeared first in 1929. It was revised in 1936, the most important change being the reversal of the gravimetric with the volumetric technique. Now, again, another ^e^^sion has been made, the third edition being herewith presented. Attention is called to the more noteworthy alterations involved in this revision. The most important shift has been to place volumetric precipitation methods directly preceding gra\'imetric precipitation meth- ods; this has the advantage of not only placing both types of precipita- tion methods close together but of also introducing the concepts of equivalent weights and normal solution with acids and bases. The volu- metric chloride determination is an added procedure in this revision. In general, much of the text material has been rewritten and brought up to '^jite by the adoption of better indicators, less omphasi.s on titers, etc. F*iu.»^.’durcs for limestone and brass have been merely outlined. A chapter Ou tsj ^^matic quantitative analysis has been added, which in- corporates two useful tabulations. Problem work has been enlarged and arranged to cover fifteen weekly assignments, each problem set containing ten problems with answers and ten without answers. This should greatly facilitate the work of instruction in stoichiometry. Acknowledgment is made to Lundell, Hoffmann and Bright, authors of Outlines of Methods of Chemical Analysis^ for use of the table of elec- trode potentials and for tabulations upon which “The Distribution of the Elements in the General Procedure” and “Summary of Analytical Methods” were based. Carl J. Engelder University of Pittsburgh Januai'y^ 19/^3 V CONTENTS PART I FUNDAlVrENTAL PRINCIPLES OF QUANTITATIVE ANALYSIS CHAPTER I. General Considerations of Quantitative Analysis. . . . The Scope of Quantitative Analysis. The Theoretical Basis of Quantitative Analysis. General Operations in Preparing Samples for Analysis. Errors and Precision of Quantitative Methods. Mathematical Operations. The Analytical Balance. Reagents. PART II VOLUMETRIC ANALYSIS II. The Theory, Technique and Calculations of Volu- metric Analysis The Theory of Volumetric Analysis. Volumetric Apparatus and Technique. The Calculations of Volumetric Analysis. III. Neutralization Methods General Laboratory Instructions. General Considerations of Acidimetry and AlkaUmetry. Determinations of Alkalinity and Acidity. Stoichiometric Calculations of Neutralization. The Theory of Neutralization. IV. Oxidation and Reduction Methods General Considerations. Representative Oxidation-Reduction Procedures: A, Dichromate Methods; B, Permanganate Meth- ods; C, Ceric Sulfate Methods; D, Iodine Methods. Volumetric Precipitation Methods Theory of Volumetric Precipitation. « • VII PAGE 1 29 45 99 160 CONTENTS • « « VIU PART III , GRAVIMETRIC ANALYSIS ' X CHAPTER PAGE VI. The Theory, Technique and Calculations op Gravi- metric Precipitation Methods 169 Fundamental Requirements of Gravimetric Precipitation Methods. The Technique of Gravimetric Analysis. The Cal- culations of Gravimetric Analysis. VII. Representative Gravimetric Precipitation Procedures 190 The Precipitation of Hahdes. The Precipitation of Hy- droxides. The Precipitation of Sulfates. The Precipitation of Oxalates. The Precipitation of Phosphates. Calculations of Factor Weight Samples. PART IV SYSTEMATIC QUANTITATIVE ANALYSIS VIII. Systematic Analysis 213 Evolution Methods, Descriptive Outline of Limestone Anal- ysis. Descriptive Outline of the Analysis of Brass. Calcula- tions of Indirect Analysis. Summary of Analytical Methods. APPENDIX The Literature of Analytical Chemistry 241 The Plan of the Course — Suggestions to the Instructor 243 Reagents and Sup-plies 245 Tables 248 Index 281 PART I FUNDAMENTAL PRINCIPLES OF QUANTITATIVE ANALYSIS CHAPTER I GENERAL CONSIDERATIONS OF QUANTITATIVE ANALYSIS Chemistry became an exact science when quantitative methods of experimentation were introduced. The progress of modem chemistry is dependent upon exact quantitative measurements and rests upon the foundation of quantitative analysis first laid down by Lavoisier, Ber- zelius, Stas and others. The composition of matter and its transforma- tions are determined by precise analysis. Back of the discovery of a new clement, the synthesis of a new compound, the perfection of a new industrial process lies the story of careful, painstaking chemical analysis. Synthesis and analysis must go hand in hand in the advancement of chemical science. In practically every field of human activity, chemistry, of which quantitative anal3'’sis is a fundamental and essential part, plays an im- portant role. The application of chemistiy to all phases of industry, to our well-being and to our cverj^day needs involves the work of the analyst. Much of the advance in medical science is the result of a careful chemical investigation of the factors concerned. The control of chemical manufacturing processes is directly dependent upon the anal- ysis of the raw materials, often the intermediate products and usually the finished product. A research problem cannot be successfully pursued without a careful check by analysis at each step. A newly synthesized compound must be analyzed before its composition is definitely kno\vn. Even the determination of atomic weights depends upon the most precise quantitative methods of analysis. The influence of chemistrj^ on our health, our happiness and our material advancement needs no emphasis. The student who undertakes the studj" of quantitative analysis as part of his preparation for the profession of medicine, dentistry, phar- macy, engineering, mining, chemistry or the natural sciences cannot help 1 2 GENERAL CONSIDERATIONS finding an enthusiastic appeal in the subject. Its application to his chosen profession is direct and fundamental. Through the training it offers in sound, logical reasoning, his chemical ideas are placed on a more secure and definite basis and his knowledge of chemical pnnciples and re- lationships is enhanced. His training in the careful, exact mampulations of quantitative technique will always be of co^iderable v^ue to mm. Efints to Students. The beginner's success in quantitative analysis depends largely on his approach to the subject and the attitude he maintains throughout the course. Too frequently at the outset, begin- ners have mistaken ideas that the subject is beyond their ability, that they cannot master the technique, that the calculations will prove too difficult. These ideas are unfounded and are quickly dispelled by the conscientious student who by dihgent study keeps abreast of the labora- tory and classroom assignments. A few students, on the other hand, do not give the subject the serious thought and best effort it deserves and soon find themselves hopelessly submerged. The student s efforts should be directed toward a clear understanding of the principles involved, the development of a reasonable dexterity in manipulation and a satisfactory facility in solving problems which express quantitative relationships. Quantitative analysis imposes a severe discipline, and discouragements are apt to arise. The beginner must remember, however, that accidents sometimes occur in the best conducted experiments and ruin the anal- ysis, and that perfect checks cannot always be hoped for in all deter- minations. Self-confidence and reliance in his own results are effective cures for discouragements. Confidence can be developed only by patient and persistent efforts to acquire the necessary skill in techmque. When confidence in a determination is lost through the introduction of an error which is known to affect seriously the accuracy of the result, the analysis had better be abandoned and a fresh start made. Absolute honesty is demanded of the analyst. The temptation to falsify results or to use the results of others must be resolutely resisted. No matter how discordant the results of an analysis are they should nevertheless be reported. Although closely agreeing results indicate, as a rule, reliable accuracy and reflect good ability in technique, they do not necessarily imply correct results because a similar large error may have been introduced in each determination. The habit of recording all data and calculations in inky in a bound notebook is a good safeguard against any tendency deliberately to forge the results. A laboratory notebook must be kept into which all observations, data and calculations are entered.* Data such as the date, all weighings, * The author has prepared a Laboratory Record Book of Quantitative Analysis, pubhshed by John Wiley & Sons, Inc., New York. THE SCOPE OF QITANTITATIVE ANALYSIS 3 buret readings and oilier observatioas should bo recorded on the right- hand page and all calculations made on the left-hand page. The habit of recording data and making calculations oii loose scraps of pajx'r cannot be tolerated. Valuable data are sometimes lost in this way, errors are made in transcribing and, above all, the reliability of the re.sults may be questioned. A neat, well-kept notebook n'flects the owner’s care and maintains his self-confidence. Writing should be done in ink and no erasures should be made; necessaiy corrections should be made without destroying the original entries. Re.sults, usually computed as percentages, are to be reported on forms supplied by the instructor; samples are showai in the Appcuidix. The procedures in this book call for making all ck'terminations in triplicate, so that if one determination is ruined or the re.sult lacks satisfaotoiy agreement, at least two other results can be considered. Wheth(*r all determinations should be made in duplicate or in triplicate is left to the discretion of the instructor. The results will be judged by their devi- ation from the true composition of the sample. To make intelligent progress the student must know what he is doing. He must thoroughly understand the principle involved and th(‘ reason for each step. Tliis means that the student must study the proc(‘dure before he comes to the laboratory, so that valuable time is not wasted during the laboratorj’- ix^riod and blunders may be avoided, ('ommon sense must be used and the work must be carefully planiie% one atom of oxygen unites with two atoms of hydrogen to form water. Atomic weights are directly proportional to the actual weights. This is, in brief, the Law of Combining Weights, which can be stated thus: Reactions take place between weights of chemical substances which are propor- tional to the atomic or molecular weights of the reacting elements or compounds. Thus, if we know the actual weight (in grams, pounds, grains or any other units) of one of the components, we can by a simple direct proportion between the actual weights and the atomic weights calculate the actual weight of the other components of the reaction. 10 GENERAL CONSIDERATIONS GENERAL OPERATIONS IN PREPARING SAMPLES FOR ANALYSIS Sampling. The securing of a representative sample is an important consideration too frequently ignored by the analyst. No general rules can be laid down in regard to sampling, for much depends upon the nature of the material and the quantity from which a representative sample is to be taken. Liquids are best sampled by taking small samples at various levels from the tank or other container. Samples of alloys and metals may best be secured by drilling or chipping from several sections of the sample bar or ladle. Dry powders should be sampled by taking a small scoopful from each sack. Coarse material such as ores, rocks, coal, etc., may be sampled from the car, bin or dump, about a ton for each carload. This coarse material must then be crushed, placed in a cone-shaped heap, divided into four quarters and two opposite quarters retained. Further crushing and quartering are necessary, the operations being repeated until about five pounds of material of about 60-mesh remain. From this reduced sample the laboratory sample is secured, and sometimes ground to 100-mesh. In grinding it is important to grind all the sample and not discard the more resistant parts. The grinding is best done by a ball mill. Students' samples, as a rule, are supplied all ready for analysis. Certain samples which require diying should be spread on a large watch glass, thoroughly mixed, and dried in an oven at about 105°C. They should then be placed in a weighing bottle and kept in a desiccator. The moisture content of certain industrial samples must be determined in order that the results, obtained from the dried sample, can be recalcu- lated for the sample “as received.” The procedure for a moisture deter- mination is given on page 48. Further reference is made to moisture determinations on page 217. Weighing the Sample. The most important instrument for the analyst is the balance, since all weighing operations are conducted by means of it. The proper weight of sample to be taken for an analysis must be decided upon for each type of material. The rule which governs this is: Enough sample should be taken to give sufficient constituent so that the operations of volumetric or gravimetric analysis can conven- iently be carried out with a single set of apparatus, but not so much that the operations consume too much time. The amount of sample to take is determined by the content of the constituent sought. If the constituent is present in small amount a relatively larger sample must be taken; on the other hand, if the sample contains large amounts, a smaller portion of sample should be taken. The approximate content in the sample can be determined by a well-conducted qualitative analysis. In ERRORS AND PRECISION OF QUANTITATIVE METHODS 11 students* samples the procedures provide for the proper weight to [>0 taken. Dissolving the Sample. Wator-soluhle .samples ordinarily will present no difficulty in dissoUdng. In some cases the wal(T must b(* acidified. Dilute nitric acid is the usual solvent for water-insoluble samj>Ies, Init in some cases dilute hydrochloric acid is ])r(‘fernl»le. Materials which resist acid treatment must be fus('d to g(‘t them info solution. The commonest flux for materials of an acid nature, such as silicates, is NaoCOa. For basic materials an acid flux such as KIIS ()4 is used. ERRORS AND PRECISION OF QUANTITATIVE METHODS Errors. The accuracy of a quantitative' method dejxauls upon how completely the errors introduced in the detenniiiation can Ix' reduced. No method is absolutely free from errors no matter how carefully one carries out the procedure. Tiic analy.st must know the sources of error in the method used and how certain of those <‘iTors may lx* minimizc'd, in order to determine the relative care witli which he must work as w( 11 as the reliance he can place on his results. The errors of an analytical method are of several difTerent kinds, some of wliich can be reduced to a negligible amount and othem wliieli cannot. Errors arising from careless technique sucli as faulty sampling, n:i.'qak('.s in weighing, incomplete precipitation, wasliing and ignition, lo.sst's of material or inclusion of dust and foreign matter during manijxdations, failure to match end points during titrations, di.sregard of temperatun's and mistakes in recording and computing data — all tliose can be prac- tically eliminated by pajdng the closest attention to the details of manip- ulation, by the use of clean apparatus and by developing a certain amount of skill and dexterity in technique. Errors arising from the use of faulty measuring apparatus such as inaccurate balances, weights and burets can be greatly minimized by making sure that the balance is properly adjusted, and by the calibra- tion of the set of weights and the calibration of burets, flasks and pipets. Certain personal errors, which tend to introduce a constant error in the observations, can also in most cases be corrected. Errors inherent in the reaction itself can be only partially eliminated. A reaction never runs absolutely to completion ; if equilibrium is reached while quantities measurable by ordinary analytical methods still remain, an appreciable error is present in the method. To a certain extent this effect can be counterbalanced by using an excess of precipitating agent, by judicious choice of the indicator in titrations or by some other mean.s, but a certain constant error still remains. 12 general Errors, over which the experimenter has no control, are present m every series of measurements and represent the variations m obserya- readings even where every precaution is taken to n_e aU other errors. The magnitude of these errors follows the laws of chanL and can be shown by a probability curve. It is important that the analyst know how closely a number of similar me^uremente m&y L expected to agree with each other and especiaUy how rehable the Twemle result may be. This will give him a measure of the precision °^SL^Sn°iid Accuracy. By precision is meant the agreement in value of two or more similar measurements or observations; precision « most simply indicated by the deviations between two or morynividual Tlues. Such deviations can readily be computed by the student dunng the course of a determination and will show him how consistently he h^ dupUcated the various operations by which the data were deiived. The precision does not, however, tell how accurately he has worked. ^ By accuracy is meant the agreement of individual resists (or their average) with the true value. The accuracy of the student s res^t can only be ascertained by comparing this result with tbe tme content of the sample. The student may obtain a high precision, but by the intro- duction of a large error, carried through two or more duphcate steps, the accuracy of the final result may be poor. -*1, ulti In quantitative analysis we are just as much concerned ^th the ifiti- mate effect of aU errors on the final result as we are with the deviations of the various measurements which are made dunng the cour^ of an analysis. In other words, we want to know witlim what hmts check results may be expected to lie. For this purpose the precision is best expressed in parts per thousand. For example, if two detei^aUoM give 24.95 and 24.90 per cent, respectively, for the amount of a certam constituent contained in a given sample, and the correct res^t is 25.00 per cent, the deviations are 0.05 and 0.10, respectively, and the accuracy is 2 parts and 4 parts per thousand. It is preferable to express the precision in parts per thousand rather than as a percentage of error because the amount of co^tituent is usually reported as a percentage and confusion is thus avoided. The accuracy depends, among other things, upon the amount of con- stituent present; the smaUer the amount the poorer the accuracy. For every analytical method the accuracy is known or can usually be deter- mined. In student analyses, precision is a,pparent from duphcate runs, whereas accuracy is ascertained by the deviation of results from the true value of the sample. The best agreement we can hope for in any ordi- nary volumetric or gravimetric method is about 1 part in a thousand. MATHEMATICAL OPERATIONS 13 MATHEMATICAL OPERATIONS The calculations of quantitative analysis arc of several kinds and are dependent upon the nature of the method used. The most important type of calculation is that based on the Law of Definite Composition, and such a calculation must be made with the; data of every quantitative determination; fundamentally the calculation involves a direct proiwr- tion between actual weights and molecular or atomic weiglits. Second in importance are the calculations involving applications of the Law of Chemical Equilibrium and dealing with precipitation, regulation of ion concentrations, buffer solutions, etc. Certain electrochemical methods require computations based on the laws of electrochemistry and certain gas analytical methods involve the gas laws. Examples of these various types arc given later in problem sets and applied to the data of the procedures. In the present section attention is directed to computation rules with special reference to significant figures and to tlie use of exponents and logarithms. _ Significant Figures. Significant figures are the digits which when placed in order give the value to a number. Thus in the quantity repre- sented by the number 635, the digits 6, 3 and 5 are significant figures. Zeros are used in some cases to locate the decimal point; in others they may be significant figures. In the number 30.25, the zero is a significant figure; in 0.015 the zeros arc not significant but show the order of magni- tude represented by the other digits. Zeros following other digits may or may not be significant. The ordinary analytical balance weighs to one-tenth of a milligram, that is, to the fourth decimal place; hence only four digits to the right of the decimal point, in the weight of an object, are significant. A buret can be estimated to one one-hundredth of a milliliter; hence two digits to the right of the decimal point are significant. In general, percentages are usually calculated to the nearest 0.01 per cent. Computation Rules. In making computations from the data of an analysis, it is important that the proper number of significant figures shall be retained and the superfluous ones discarded, in order that the result may properly represent the degree of precision of the determina- tion. In general each quantity should contain only such a number of sig- nificant figures as to have the last one uncertain. Thus, in a buret reading of 30.25 ml. the last digit, 5, is estimated and hence uncertain, but should be retained. ^Mien superfluous digits are discarded, increase by 1 the last figure retained if the discarded figure is 5 or more. 14 GENERAL CONSIDERATIONS In adding or subtracting numbers, retain in each number only those corresponding to the uncertain figure in the number having the least number of digits to the right of the decimal point. For example, when adding together 30.25, 1.062 and 6.7685, the quantity which deter- t- * les the number of significant figures to retain is 30.25, modif 3 dng the above quantities to read 30.25 + 1.06 + 6.77, the sum becomes 38.08. In multiplying or dividing retain as many significant figures are contained in the quantity having the least number of significant figures. Use of Exponents. In certain types of chemical calculations, involv- ing extremely large or extremely small numbers, it is very cumbersome to carry out the arithmetical operations in the ordinary way. Solubility data, ionization and other equilibrium constants and certain other data of physics and chemistry are frequently obtained with a precision good only to one, two, three or at most four significant figures, yet the actual magnitudes may be extremely small or large. Such quantities are best expressed in the exponential form, the arithmetical work which follows being then much simplified. It is assumed that the student is already familiar with exponents. The rules in their use are briefly summarized below. When multiplying two or more quantities, express each in the expo- nential form and add the expments. The significant figures must of course be treated in the ordinary way. In multiplications involving negative exponents, the algebraic sum is taken. When dividing one quantity by another subtract the expouenis alge- braically. When an operation involving both multiplication and division is necessary, express each quantity in the exponential form, transfer the exponents from the denominator to the numerator with change of sign and take the algebraic sum of the exponents. In raising numbers to a specified power, the exponents are doubled for squaring, trebled for cubing, etc. In extracting roots, the exponent is divided by 2 to obtain the square root, by 3 for the cube root, etc. Use of Logarithms. The solving of problems and of the results of determinations from analytical data is greatly facilitated by the use of logarithms, and it is strongly advised that students use this method. To those not familiar with logarithms, it may be said that many hoxirs of tedious long-hand computation may be saved by acquiring a work- ing knowledge of this very helpful mathematical device. A logarithm is an exponent which must be applied to a number, known as the base, in order to produce any given number. In the common or Briggsian system of logarithms, the base is 10. Thus in the equation 10 =* = 100 MATHEMATICAL OPERATIONS 15 the exponent 2 is the logarithm of 100, when the base is 10. The above equation, in terms of logarithms, can then be written: logio 100 = 2 which states that the logarithm of 100 to the base 10 is 2. The relation of exponential to logarithmic forms for simple, whole numbers as expo- nents is shown in Table I. TABLE I Number Exponential Form ' Logarit 1,000,000 10® 6 1,000 10® 3 100 102 2 10 10^ 1 1 10® 0 0.1 10"* -1 0.01 10"2 -2 0.001 10'® -3 0.000001 10-® -6 Note that the logarithm of 10 is 1, that of 1 is 0 and that for negative exponents there result corresponding negative logarithms. A logarithm is made up of two parts: the jnanlissa, which is found in logarithm tables and which is placed to the right of the decimal point; and the characUristiCf which is placed to the left of the decimal point. In four-place log tables, the mantissas are given to four decimal places. The log table in this book is a. five-place table and is to be used in making the calculations based on the laboratory data. Illustrative cal- culations as well as answers to problems in this book are based on four-place logs. In making computations by logarithms, the logs of the numbers must first be found. Then, if an operation of multiplication is to be made, the logs are added ; if a division is called for, the logs are subtracted ; and to raise a number to a certain power, the log is multiplied by that power. Finally, the number corresponding to the final logarithm is ascertained, as illustrated below, from log tables; a table of antilogarithms is some- times used for this purpose. To write the logarithm of a number, first supply the characteristic. For numbers greater than unity, the characteristic is one less than the number of digits in the given number. Thus, log of 10 = 2 - 1 = 1.0 . . . log of 100 = 3 - 1 = 2.0 . . . log of 1000 = 4 - 1 == 3.0 . . . IQ general considerations - For numbers such as 5, 55, 555, 5555 and 55555, the characteristics are respectively, 0, 1, 2, 3 and 4; the mantissas must be supplied from log'tables. For the first four examples cited, the mantissas are given directly in the five-place log tables in the Appendix and are tabulated below. Interpolation is necessary for numbers such as 55555. Here we must interpolate between the logs of 5555 and 5556. The difference is 0.00005. The difference between the mantissas 74476 and 74468 is .00008. In the “Proportional Parts” column of the log table, under “8” it is found that a difference of 0.00005 or (5) in the original number corresponds to “4.0” or 0.00004. This is then added to the mantissa 74468, giving a mantissa of 74472 for the number 55555. The logarithms of these illustrative numbers are then log 5 = 0.69897 log 55 = 1.74036 log 555 = 2.74429 log 5555 = 3.74468 log 55555 = 4.74472 Numbers less than unity have negative characteristics. The char- acteristic for decimal numbers, having no zeros immediately follow- ing the decimal point, is designated as 1 or 9 (plus mantissa) — 10. For example, the logarithm of 0.50000 would be written 1.69897, or 9.69897 - 10. ’ If in a given number there are no digits before the decimal point and n zeros before the significant figures after the decimal point, the char- acteristic is (1 + n) or (9 — n) plus mantissa — 10. For example, the log of 0.05500 is 2.74036 or 8.74036 — 10 and the log of 0.000555 is 5.74429 or 6.74429 — 10. The use of the system involving (9 — n) — 10 is recommended to students. ' The logarithms of some decimal fractions are here tabulated. log of 0.50000 = T.69897 or 9.69897 - 10 log of 0.05500 = 2.74036 or 8.74036 - 10 log of 0.00555 - 3.74429 or 7.74429 - 10 log of 0.0005555 = 4.74468 or 6.74468 - 10 log of 0.055555 = 2.74472 or 8.74472 - 10 MATHEMATICAL OPERATIONS 17 It should be noted that the logarithm of a decimal fraction such as 0.05500, as expressed above, consists of a negative characteristic, written 2, or 8 — 10, and a positive mantissa (.74036). This might also bo written as a negative logaritlim by adding the two parts, algebraically, ^ving —1.12563. The finding of the number (antilogarithm) corresponding to a given logarithm is, in general, the reverse of finding a logarithm. For logs having positive characteristics, look up in the log table the number corresponding to the given mantissa and adjust the decimal point. For example, in finding the antilogarithm of log 3.74468, wc find that 74468 (in the log table) is the mantissa of 5555. Since tlic given log has a characteristic of 3, the desired number has 4 digits to the left of the decimal point and the number is 5555. Likewise, the antilog of 1.74468 is 55.55, For logarithms having negative characteristics, such as 7.74429 — 10 or 3.74429, look up the number in the log table conesponding to the mantissa; this will be found to be 555, Since the characteristic is 3, the number sought is 0.00555. The rule for locating the decimal point, for numbers having positive characteristics, is to place n + 1 digits before the decimal point, where n is the characteristic. For negative characteristics, place n — 1 zeros after the decimal point. A logarithmic calculation, typical of many similar ones encountered in the laboratory and problem work, is illustrated here. Find the result of the following indicated operations: 25.62 X 0.1234 X 0.05300 0.2013 X 100 log 25.62 log 0.1234 log 0.05300 = log 100 Adding : log 0.2013 Subtracting: or antilog = or 1.40858 9.09132 - 10 8.72428 - 10 2.00000 21.22418 - 20 9.30384 - 10 12.92034 - 10 1.92034 83.242 83.24 18 GENERAL CONSIDERATIONS THE ANALYTICAL BALANCE The analytical balance, by which all weighings are made, is the most important piece of apparatus in the quantitative laboratory. Since the a^^^. fiTacy of an analysis depends so largely on correct weighings, a proper uiKJ iri ending of the construction, use and care of the balance is essen- tial. Balances differ a great deal in details of construction and in sensi- tiveness, depending upon the use to which they are to be put. The analytical balance in common use weighs to the fourth decimal place, that is, to 0.0001 of a gram. The assayer^s balance weighs to the fifth decimal and a microchemical balance to the sixth decimal place. The Mechanical Features ot the Ordinary Balance. The balance operates on the principle of the lever. A beam of light truss work, to the center of which is attached an agate knife-edge, rests upon an agate plate. The beam carries at each end a stirrup with a knife-edge, the two stirrups supporting the pans upon which the weights and object are placed. The combination as a whole acts as a lever and fulcrum. The essential parts of the balance and their fimction are shown in the accompanying sketch. A good balance in order to give accurate and satisfactory service must have the central knife edge and the terminal knife edges in the same horizontal plane. The knife edges must remain true and not become blunted or injured by rough treatment. The arms must be of equal length. The center of gravity must lie slightly below the central knife edge. The balance must oscillate fairly rapidly. The sensitiveness should be such that an extra load of one miUigram on one pan displaces the pointer about tliree pointer-scale divisions. Before attempting any weighings examine the balance assigned to you and study its construction and then check up the following points. 1. See that the balance is level as indicated by the plumb bob or spirit level. If it is not, adjust it by means of the screw supports on the base. 2. Test the beam support and the pan arrests for smooth operation. 3. Check over the weights in the weight box and familiarize yourself w'ith the units. See that the rider is of proper weight for the balance. 4. With balance door closed, release the beam and pans and start the pointer to swinging. If the balance is properly adjusted the pointer should swing equal distances on cither side of the zero mark on the pointer scale. The point of ultimate rest must be determined as de- scribed below. Point of Rest. When the balance is in equilibrium the pointer will come to rest at some point on the pointer scale. In perfect adjustment THE ANALYTICAL BALANCE 19 this point of rest will coincide with the zero position of the scale. In practice a displacement of one small scale division between the actual and the marked zero is permissible. The point of rest must be deter- mined each day the balance is used. This is done as follows: Set the pans in motion, with no load on either pan, so that the pointer traverses at least five scale divisions on either side of the zero. When the pointer is swinging freely take a reading of the pointer position at the end of the oscillation to the left and a reading to the right; take the difference and divide by 2. For instance, suppose the pointer swings to 5.9 on the left and 5.3 on the right. The difference is 0.6 and therefore the point of rest is 0.3 scale divisions to the left of zero. 20 GENERAL CONSIDERATIONS For the case here illustrated the balance is in satisfactory adjustment. If it is off by more than one whole scale division, ask the instructor to adjust the balance. This is done by moving the small thimib nut on the extremity of the beam, thereby adding or subtracting weight from one of the pans. Sensitivity. The sensitivity depends largely upon the length of the beam arms and is indicated by the amount of displacement of the pointer when a small excess load is placed on one of the pans. It is usually defined as the number of scale divisions the pointer is displaced from its point of rest when an additional weight of 1 milligram is placed on the pan or the rider is moved one whole division (1 mg.) to the right. Further instructions are given on page 46. Rules for Weighing. 1. See that the inside of the balance case and the pans are clean and free from dust and, if not, use the large camePs hair brush supplied for this purpose. See that the release is working properly. 2. Determine the point of rest ; have the instructor make any needed adjustments. 3. Place the object to be weighed on the left-hand pan. Samples of dry material may be weighed on a weighed watch glass, kept for this purpose, or from weighing tubes. Never place material directly on the pan. 4. Always use the forceps for handling objects and weights and never touch them with the fingers. 5. Whole and fractional weights should be added to the right-hand pan, starting with the largest weight. Never change weights or remove the object being weighed without first arresting the pans and placing the beam on its support. 6. Close the door and adjust the rider until the swings of the pointer are equal on both sides of the zero position. Never move the rider while the beam is suspended on the knife-edges. 7. Add up the weights missing from the box and check this against the weights as they are replaced after the object is weighed. 8. Do not weigh hot, wet or recently rubbed objects. Hot air cur- rents, varying humidity and electrical effects will introduce large errors. 9. Make sure the beam and pan arrests are restored before leaving the balance. 10. Keep the balance clean and in the best of condition. The weights require the same care. 11. Remember that the balance is a delicate instrument but will give you satisfactory service if not abused. Regard it as your personal property even though others may be assigned to it. THE ANALYTICAL BALANCE 21 Calibration of Weights. For work demanding reasonable precision, the weights should be calibrated. It is not necessary in ordinary work that all the weights be absolutely correct with the international standard of weight, but they must be accurate among themselves. The units of an ordinary inexpensive set are usually in slight error and corrections should be applied to them. A calibration should be made by comparing them with each other, using as the starting point one of the centigram weights and assuming that this weight is correct. In a one-semester course of instruction, the calibration of weights must be dispensed with. To calibrate a set of weights, first mark the duplicate weights of like denominations with punch marks so that they may be icientified. Then place a centigram weight assumed to be correct on the left-hand pan and balance it against a suitable tare. The most convenient tare to use is the weights from another set. To balance the first centigram weight, which is taken as the temporary standard (assumed correct) weight place a tare of somewhat less than 0.0100 gram on the right-hand pan and use the rider to establish equilibrium. Then replace the first centi- gram weight by the second centigram weight of the set, which is marked 0.01*, and move the rider until the point of rest is again established. The amount by which the rider was moved is the difference in weight between the two 1-centigram weights. Suppose that in balancing the second 0.01 gram weight against the tare the rider was moved from 3.2 to 3.5, a distance corresponding to 0.3 of a milligram. Assuming the first centigram weight to be correct, the second centigram weight there- fore weighs 0.0103 gram, that is, it is 0.0003 gram too heavy. Next place both centigram weights on the left-hand pan and weights from the auxiliary set on the right-hand pan until the load is almost balanced and then balance with the rider. The mass now on the left- hand pan is 0.0100 -|- 0.0103 gram. Now exchange the two l-centigram weights for the 2-centigram weight, and move the rider until the loads are balanced. Suppose the rider must be moved one small division (0.1 milligram) to the right; then the weight of the 2-centigram weight in terms of the temporary standard is 0.0203 -f 0.0001 or 0.0204 gram. In the same way determine the relative mass of the second 2-centigram weight (marked 0.02*). Suppose it is found to be 0.0203 gram. To standardize the 0.05-gram weight against the temporary standard, place the standard 0.01, the 0.02 weight which weighs 0.0204 gram and the 0.02* weight which weighs 0.0203 gram, on the left-hand pan, mak- ing a total weight of 0.0507 gram, and balance with a tare and the rider, ^place these weights by the 5-centigram weight and balance with the rider. If the rider must be moved, say two small divisions to the left, the value of the 5-centigram weight is 0.0505 gram. 22 GENERAL CONSIDERATIONS Proceed in this way with the comparison of each weight, building up combinations as the larger weights are determined. Tabulate the value of each weight as compared with the first centigram weight as the assumed standard. Since the unit chosen as the standard is small, it is better to calculate the coiTection which must be applied to each weight, in terms of a weight of a larger denomination, for which a 10-gram weight is usually used. If it is desired, moreover, to have the set of weights agree, not only among themselves but with the standard of the Bureau of Standards, include in the comparison a 10-gram weight from a set furnished by the Bureau of Standards. The 10-gram weight which is now selected as the permanent standard, either the weight from the set or a standard 10-gram weight, when compared with the combination of smaller weights, by the method of substitution used above, will have an apparent mass greater or less than exactly 10 grams. Suppose the summation of the smaller weights gives to the selected standard a mass of 10.0150 grams. To find the true value of each weight in the set, compute the aliquot part of 10.0150 corresponding to each weight. Thus, in the present case, the aliquot part for each 10-gram weight is 10,0150; for the 5-gram weight is 5,0075; for the l-gram weights, 1.0015; for the 0.1-gram weights 0.10015; and so for all other weights in the set. The correction which must be applied to each weight whenever it is used in subsequent weighings is then found as the difference between the apparent weights as previously determined and the proper aliquot part of the standard. To illustrate this, in the data above the 0.05-gram weight was found to have an apparent value of 0.0505 gram and the aliquot part of the standard is 0,050075; the difference is 0.000425. The correction to apply is therefore +0.0004 gram, and this value must be added to the face value (0.0500) of the 0.05-gram weight. If the difference in any case is a negative value, it must be subtracted from the face value of the weight, A table of corrections should be placed on a card and kept in the balance case. Correction for Buoyancy. For very precise work, weighings must be corrected for the buoyancy of the air, that is, weights must be reduced to vacuo. The apparent weight of an object in air is less than its true weight by an amount equal to the weight of the volume of air displaced. A milliliter of air, under ordinary laboratory conditions, weighs 0.0012 gram. The volume of air displaced by an object is equal to its weight (hvided by its density. Since the object being weighed is counter- balanced by brass weights, which also displace a certain volume (and weight) of air, the true weight of the object, that is, the weight it would REAGENTS have in a vacuum, is equal to its weight in air plus the difference in the weights of air displaced by the object and by the weights. If, for example, a porcelain crucible weighs 7.6252 grams in air, what is its weight in vacuo? The density of porcelain is 2.4 and that of brass weights approximately 8.0. The volume of air displaced by the crucible is 7.6252/2.4 = 3.17 ml., and the weight of air is 3.17 X 0.0012 = 0.0038 gram. The volume of air displaced by the weights is 7.6252/8.0 = 0.95 ml., and its weight is 0.95 X 0.0012 = 0.0011 gram. The difference, 0.0027, is the buoyant effect of the air. The true weight of the crucible is therefore 7.6252 -f- 0.0027 = 7.6279 grams. The above steps can be combined into the single equation Weight in vacuo Weight in + air Weight in air Density of object Weight in air Density of weights 0.0012 It should be noted that if the volume, and hence the weight, of air displaced by the weight is greater than that displaced by the object, the object will weigh less in a vacuum than in air. REAGENTS An important part of every chemist's work is the preparation of reagents. Reagent solutions are made in some cases from the solid chemicals by dissolving the required amount of the solute in water; in other cases, as with the common acids, the reagent is made by dilution of the concentrated solution. The strength of the reagent is an impor- tant consideration in every procedure, and as a general rule the strength is so adjusted to the amount of material with which it is to react that the volume of the resulting solution is not abnormally large. Since in many procedures limitations must be placed on the volume of the solu- tion, the reagent must contain enough active reacting substance so that not more than 75 or 100 ml. may be used. For the procedures given in this book reagents of suitable strength are specified, but in devising new methods of analysis the analyst must decide for himself the proper strength to use. The strength of the reagent is most simply expressed in terms of the grams of solute actually contained in 1 ml. of the reagent or the number of grams of material with which 1 ml. of the reagent will react. The strength in grams per milliliter is known as the titer of the solution. Reagents Made from SoUd Chemicals. This type of reagent is easily made by dissolving the required weight of pure, previously tested, salt in water (or acids if demanded) and diluting to the desired volume with 24 GENERAL CONSIDERATIONS distilled water. Calculations involving the strength of reagents of this type are illustrated in the following problem. What is the strength of a AgNOa solution containing 20 grams of AgNOa per liter, in terms of the amount of chloride ion with which. 1 ml. will react? Since the solution contains 20 grams of AgNOa per liter, 1 ml. contains 0.020 gram. This quantity will react with x grams of Cl“ ion. We then have the relation AgNOa : Cl = 0.02 : a: 169.9 35.46 from which x = 0.0048 gram. Therefore 1 ml. of AgNOa will bring into, reaction and precipitate as AgCl 0.0048 gram of chloride ion. This is the chloride-titer of the AgNOa solution. Reagents Made by Dilution. Dilute solutions of hydrochloric, rdtric, sulfuric and certain other acids and of ammonium hydroxide, so often employed in analytical procedures, are prepared by diluting the concen- trated reagent. The proper amount of water to add to the concentrated stock reagent in order to give a dilute solution of the desired strength must be calculated for each individual case. Concentrated hydrochloric acid is manufactured by dissolving hy- drogen chloride (hydrochloric acid gas) in water imtil the solution is saturated. The saturated solution contains 39.11 per cent by weight of HCl and has a specific gravity of 1.20. The ordinary “concentrated*’ acid as supplied to the laboratory is somewhat weaker, and for purposes of calculation may be assumed to have a specific gravity of 1.19 corre- sponding to 37.23 per cent by weight of pure HCl. Concentrated sulfuric acid is the result of the reaction between sulfur trioxide and water. The liquid may be theoretically 100 per cent pure H2SO4, and may even dissolve additional amoimts of SO3, forming fuming sulfuric acid. The “concentrated” reagent as supplied is some- what impure, owing to the fact that it readily absorbs water. It may be assumed to be 95.60 per cent pure H2SO4 and to have a specific gravity of 1.84. The concentrated nitric acid supplied to the laboratory usually has a specific gravity of 1.42 and contains 69.77 per cent by weight of pure HNO3. Ammonium hydroxide, as supplied commerciaUy, contains about 28.33 per cent NH3 by weight and has a specific gravity of 0.900. The ‘ strength of ammonium hydroxide and of acid reagents is usually indi- cated by the specific gravity of the solution. The specific gravity is the ratio of the weight of a given volume of the liquid to the wdght of an REAGENTS 25 equal volume of water. If the weight of a unit volume of water, 1 ml., is taken as 1 gram, the specific gravity, or rather the density of a liquid, is expressed directly as the weight in grams per milliliter of the liquid. The relation between specific gravity and the percentage by weight of the active constituent and the way these quantities vary with dilution must be ascertained from specific gravity tables such as those found in the Appendix. If the strength is given in terms of specific gravity, refer to tables to find the percentage of solute or of active constituent. To find the weight of constituent in 1 ml., multiply the specific gravity by the percentage by weight. To illustrate, the grams of IICl in 1 ml. of hydrochloric acid of specific gravity 1.100 is found by reference to the acid table in the Appendix. This is found to be 20.01 per cent. Wc have then 1.100 X 0.2001 = 0.2201 gram per ml. Problems involving data in which the strength of the solution is given in terms of specific gravity are tlien readily solved by finding the weight present or required per milliliter of the reacting solution. Problems of Dilution. When a concentrated solution is mixed \Wth water the resulting volume will not be equal to the sum of the original volume of the solution plus the volume of the added water. A con traction or expansion takes place upon mixing, making the final volume of the diluted reagent either less or greater than the sum of the volumes of the two original liquids. For example, if 100 ml. of concentrated hydroclfioric acid are mixed with 100 ml. of water or dilute acid, the final volume is not 200 ml. as might be expected, but is somewhat less, depending upon the strength of the concentrated acid. The fact that a change in volume results must be kept in mind in making calculations dealing with dilution. A calculation based on the assumption that no volume change occurs is wTong. Weights, however, do not change, and calculations must be made on a weight basis. The correct way of solving dilution problems is shown by the follow- ing example. How much water must be added to 50 ml. of concentrated hydro- chloric acid having a specific gravity of 1.190 and containing 37.23 per cent of HCl by weight in order to make a solution of specific gravity 1.12 containing 23.82 per cent by weight of HCl? To solve this we must make an equation between weights) since no loss or gain of weight takes place during the mixing of the two liquids, and since the total weight of pure HCl does not change on dilution, it is best to equate the amount of pure HCl in the original acid and in the diluted solution. 26 GENERAL CONSIDERATIONS One miUiliter of the original acid weighs 1.19 grams and 60 ml. weigh 50 X 1.19 = 59.50 grams. The weight of pure HCl is therefore 69.60 X 0.3723 = 22.152 grams. Let X be the amount of water to add. Then the final solution will weigh (59.50 + x) grams. It should contain 23.82 per cent of pure HCl. Combining the above steps, and solving for x: 50 X 1.19 X 0.3723 = (50 X 1.19 + a:)0.2382 22.152 = 14.173 + 0.2382a: X — 33.5 grams (or ml.) of water The calculation is somewhat more complicated when the second liquid is a solution of acid instead of water. Suppose, for example, that 50 ml. of hydrochloric acid of specific gravity 1.19 were mixed with hydro- chloric acid of specific gravity 1.10 so that the resulting solution had a specific gravity of 1.14. How much of the second acid was used? The percentages by weight corresponding to the specific gravities of these three solutions, are, from the acid tables (see Appendix): 1.19 sp. gr. = 0 = 37.23 per cent 1.10 sp. gr. =c= 20.01 per cent 1.14 sp. gr. o 27.66 per cent Let X equal the volume of the second acid solution. Then equating the weights of pure HCl we have (50 X 1.19 X 0.3723) + (x X 1.10 X 0.2001) = [(50 X 1.19) + (x X 1.10)]0.2766 and solving, we find x = 67.7 ml., the volume used. PROBLEM SET 1 1. With due regard to significant figures, indicate the proper recording of the following data: (a) 32.523 ml. when referred to a buret reading. Ans. 32.52 ml. (5) 2.5 grams weighed on the analytical balance. Ans. 2.6000 grams (c) 50 ml. when measured in a measuring cylinder. Ans. 60 ml; id) 0.00255 gram when referred to an analytical weighing. Ans. 0.0026 gram 2. Find the sum of the following numbers, and record the result with the correct number of significant figures. 23.382 + 0.26 + 1.2346 + 1.0 -h 11.28 - 3. A student carrying out a determination for the percentage of a constituent uses the usual apparatus and weighs to the fourth decimal place; he reports the result as 42.937 per cent. How should the result bo reported? Ans. 42.94 per cent PROBLEMS 27 4 . The following results were found in a student's notebook: Reported 42.65 per cent and 42.92 per cent on duplicate determinations of a material known to contain 42.81 per cent. What was the accuracy in parts per thousand for each determina- tion? What was the precision? 6. (a) If in measuring 10.00 ml., an error of 0.10 ml. is made, what is the per- centage error in the measurement? (b) If the same error is made in measuring 25.00 ml.? (c) If in measuring 40.00 ml.? Ans. (a) 1 per cent (6) 0.4 jKjr cent (c) 0.25 per cent 6. Rewrite the following quantities in the exponential form, retaining the proper number of signiBcant figures. (a) 0.000246 (5) 40 million (c) 400 thousand (d) 0.0180 (c) 3250 form. Retain 124.0 0.000376 96540 573 thousand 0.00376 8. Find the common logarithms of the following numbers: (a) 3654 (5) 0.004231 (c) 1.569 (d) 0.9659 (e) 22.469 7. Convert the following exponential quantities into the ordinary the proper number of significant figures. (a) 1.240 X 10^ Ans. (a) (5) 3.76 X 10-^ (6) (c) 9.6540 X 10^ (c) (d) 573 X 10» (d) (e) 37.6 X IQ-^ (c) 9. Find the antilogarithms corresponding to the following logarithms: (а) 4.8692 (б) 0.9681 (c) 1.6322 (d) 7.8643 - 10 (c) 27.8312 - 30 Ans. (a) 73990 (5) 9.292 (c) 42.87 (d) 0.007316 (c) 0.006779 10. Use logarithms to perform the following operations which are arranged as typical analytical data. Retain the proper number of significant figures. 32.30 X 0.1142 X 0.05300 X 100 0.2500 0.2845 X 107.88 169.89 33.03 ? 0.2253 35.45 X 0.1699 28 GENERAL CONSIDERATIONS 11. How much (a) CNS is contained in 100 grams of KCNS; ( 6 ) Cr in 35.0 grams of KaCrzO?; (c) Cl in 10 grams of pxire hydrogen chloride? Ans. (a) 69.8 grams (&) 12.4 grams (c) 9.7 grams 12. What weight of hydrogen chloride is contained in 20 ml. of hydrochloric acid which has a specific gravity of 1.175 and contains 34.42 per cent HCl by weight? 13. What weight of NH 3 is contained in 20 ml. of ammonium hydroxide solution which has a specific gravity of 0.942 and contains 15.04 per cent of NH 3 by weight? Ans. 2.83 grams 14. What volume of concentrated NH 4 OH is required to precipitate the iron fts Fe(OH )3 in a sample of pure Fe 203 weighing 0.4500 gram? 16. Concentrated HCl has a specific gravity of 1.190 and contains 37.23 per cent hydrogen chloride by weight. What volume of concentrated HCl is required to furnish 3.646 grams of hydrogen chloride? Ans. 8.230 ml. 16. It is desired to prepare a solution of nitric acid of specific gravity 1.18 contain- ing 29.37 per cent HNO3 by weight. How much water must be added to 100 ml. of nitric acid of specific gravity 1.42 containing 69.77 per cent HNO3 by weight to give the required strength? 17. If it required 85.0 ml. of dilute HCl to precipitate as AgCl the silver in 0.3000 gram of AgNOs, how much concentrated HCl was used in the preparation of the dilute acid? An«. 0.145 ml. 18. How much water must be added to 50 ml. of aqueous ammonia of specific gravity 0-900 containing 28.33 per cent by weight of NH 3 to give a solution of specific gravity 0.960 containing 9.91 per cent NH 3 by weight? 19. If 20 ml. of nitric acid (sp. gr. 1.20) are mixed with 20 ml. of aqueous ammonia (sp. gr. 0 . 95 ) what is the weight of the excess basic or acidic constituent present? Atis. 0.33 gram NHj 20. What volume of concentrated hj^drochloric acid should be taken to prepare a liter of solution 36.00 ml. of which will neutralize 0.2500 gram of pure Na 2 C 03 ? PART II VOLUMF.TRIC ANALYSIS CHAPTER II THK THEORY, TECHNIQUE AND CALCULATIONS OF VOLUMETRIC ANALYSIS In the volumetric methods of analysis the weight of reagent necessary to complete a chemical reaction is ascertained by determining the volume of solution required to react with the material being analyzed, and from this weight the amount of desired constituent in the sample can be calculated. The method, in short, involves adding measured amounts of the reagent to a suitable weight of the sample until the reaction is complete. The strength of the reagent must be accurately known and its total volume carefully measured. Some \'isible means must be pro- vided for determining when the reaction is completed. The process of determining the exact volume required is called titration. Volumetric methods are also known as titration methods. THE THEORY OF VOLUMETRIC ANALYSIS Classification of Volumetric Methods. Reactions which arc used as the basis of volumetric methods of analysis must go practically to com- pletion. A number of reactions in which precipitates or complex ions are formed, reactions between acids and bases and a great many reac- tions between oxidizing and reducing substances are among those which readily adapt themselves to volumetric technique. Most of the volu- metric methods are of these three tj'pes: 1. Neutralization methods. 2. Oxidation-reduction methods. 3. Volumetric precipitation methods. Neutralization methods have for their object the determination of the acidity or basicity of materials. 29 30 VOLUMETRIC ANALYSIS Oxidation-reduction methods involve reactions between oxidizing and reducing substances and constitute the most important class of volu- metric methods. In volumetric precipitation methods the exact volume and hence the exact weight of precipitating agent required to precipitate the desired ion or to cause a precipitate to form a complex ion is made the basis of the method. Equilibria in Volumetric Reactions. The equilibria set up in a react- ing system determine whether the reaction is sufficiently complete to serve the purpose of a quantitative determination. If equilibrium is reached when appreciable amounts of the initial reactants still remain, the reaction must of course be discarded as a means of determination and other methods of analysis sought. The first requirement of every volumetric method is that the reaction must be practically irreversible. In neutralization methods, the fundamental theory to be considered is the equilibrium between ions and OH“ ions and their possible reaction with the ions of weakly ionized acids and bases. We are pri- marily concerned in this respect with the ionization constants of water, acids and bases and to a certain extent with hydrolysis constants. Application of ionic theory to the determination of acids and bases is taken up in Chapter III. In reactions between oxidizing and reducing agents equilibria are like- wise set up which determine to what extent such reactions are complete. The Law of Chemical Equilibrium, when applied to precipitation methods, has to deal with the Solubility Product Principle. In a later chapter it will be shown how this principle is applied to the volumetric precipitation of a salt of the ion whose determination is under consider- ation. Suffice it to say here that, if the Ks.p. of the precipitated salt is too large or, in other words, if the precipitate is too soluble, equilib- rium is reached too soon, the error is too large and the method is dis- carded at the outset. Standard Solutions. Having selected a satisfactory reaction, the next consideration is to provide the reagent with which the constituent is to react. The strength of the reagent must be accurately known and not be subject to variation. If then a measured volume of the reagent is added, the exact strength of which is known, in sufficient amount to complete the reaction with the desired constituent and a means is pro- vided of telling when the reaction is completed, we have a method of determining the weight o constituent which has reacted with the known amount of reagent. Here the Law of Definite Composition is involved since reactions always take place between weights (here expressed in grams) which are proportional to the atomic or molecular weights of THE THEORY OF VOLUMETRIC ANALYSIS 31 the reacting substances. In applying the Law of Definite Composition the fact that theoretically equilibrium is reached before the reaction has run absolutely to completion is ignored and the calculation is based on the assumption that the reaction is complete. We are justified in doing this because only those reactions which are substantially complete are of any use in quantitative analysis. Volumetric methods are merely indirect weighing methods in wliich the weight of the added reagent is ascertained by finding the volume required to supply the needed weight. The theory of volumetric analysis and its application to calculations of data and problems of volumetric analysis will be much better under- stood if this type of technique is regarded as a simple method of indirect weighing. The strength of the reagent used must be carefully determined prior to its use as the titrating solution. It is quite impossible and entirely too inaccurate to prepare the reagent from stock chemicals except in certain cases, because they are not chemically pure. The actual strength must usually be ascertained by causing an accurately kno\\Ti amount of a pure substance to react with the reagent in a separate titration. This procedure is knowm as standardization and, after the strength of the solution is thus determined, the reagent is spoken of as a standard solution. A general rule in regard to standardization is: Standardize by a method which willy as far as possible, involve the same reactions and amditions as those used later in actual determinations. The standard solution must also be stable and not change its strength on standing as, otherwise, if we are not sure of the strength of the titrating solution the results of the analysis are worthless. The fact that some solutions cannot be properly standardized and others change strength from day to day places limitations on the availability of re- agents which otherwise might be suitable. The second general requirement, therefore, is the employment of stable, accurately standardized solutions which, by reacting quantita- tively with the sample, give a measure of the amount of the constituent being determined. Indicators. When adding the standard solution to the solution which is being titrated we must have some means of knowing definitely when an equivalent amount has been added. In the usual titration methods this is accomplished by the use of a third substance, called the indicator, placed either in the solution being titrated or used outside the solution, the indicator by a marked change in color or in some other property si^fying that the main reaction is complete. An indicator must possess the property of not reacting with the titrating solution until the prin- cipal reaction is complete and then, on the addition of a small drop, to 32 VOLUMETRIC ANALYSIS manifest a sharp color change, form a precipitate or m some other way indicate that the main reaction has been satisfied. In the potentio- metric titration methods, the completion of the reaction is marked by a sudden change in electrical potential. The indicators used for precipitation methods are usually inorganic substances which form precipitates or color the solution. The indicators used in neutralization reactions are dyestuffs or related organic com- pounds. Starch is an ideal indicator for reactions involving iodine; the pink color of permanganate solutions serves admirably to show an excess of permanganate ions. Potassium ferricyanide, used as indicator in the dichi’omate method for iron, is an example of an external or “out- side’^ indicator. Recently, indicators which change color at the equivar lence points of oxidation-reduction reactions have come into use. End Points. When just enough standard solution has been added, as shown by the indicator, to complete the main reaction, the practical end point is reached. This is the point where the indicator changes color. Strictly speaking, it is slightly in excess of the amount required for the main reaction, but the excess should not represent more than a drop or two of dilute Standard solution. The practical end point must not be confused with the equilibrium point nor with the stoichiometrical point based on equivalent volumes. We must clearly distinguish between these three points, since each has a theoretical basis but in practice all three must closely coincide. The relation between these three points can be illustrated by the titration of HCl with NaOH. The stoichiometrical (or equivalence) point in this case is that corresponding to an amount of NaOH equiva- lent to the HCl present, and assumes complete reaction. The equi- librium point is reached when the solution is exactly neutral and this occurs when the concentrations of H"*" and OH“ are both 1 X 10“^, The practical end point for the indicators usually employed in this titrar tion is found at hydrogen-ion concentrations which are either slightly greater or less than the exact neutrality amount. It must be empha- sized that the indicator must give an end point which is as near bs possible to the equilibrium point and this in turn must agree closely with the stoichiometrical point. Accurate Measuring Vessels. Since the accuracy of volumetric methods is so largely dependent upon the correct measuring of volumes of solution, accurately calibrated burets, pipets and flasks must be pro- vided. The calibration should conform to the unit of volume set down by an official agency such as the Bureau of Standards. Although the unit of volume is the liter, measurements in an actual titration are made in milliliters. The procedure for calibrating volumetric apparatus ,an- stance (standard) and divide the product into the weight of pure sub- stance taken. A 0.5 N solution has one-half the reacting strengtii as a 1 solution since it contains, per liter, one-half of a gram-equivalent weight. There- fore twice the volume of a 0.5 N solution would be requiretl for reaction as for a normal solution. A very important and often used relationship follows from this. This rule can be staled thus: Normalities are inversely proportional to volumes. Expressed in symbols this inverse volume rule is Ni:N2^ V2 : Id To illustrate, if 25 ml. of a 1 solution are required for a certain reaction, then 50 ml. of a 0.5 N will be needed. For the purposes of calculation, the proportionality is best expressed as ViNi = V2N2 As a simple application of tliis relationship consider the following case. Wiat volume of a 0.1500 N solution will be equivalent to 30.00 ml. of a 0.1234 N solution? We have here 30.00 X 0.1234 0.1500 24.68 ml. This relationship is by no means confined to solutions of the same solute. In fact, it is employed throughout all volumetric procedures in which two solutions are employed and particularly it is the basis of standardizing a solution and computing its normality by titration against another solution whose normality is knowm. In most of the volumetric procedures described in the follo\ving chapters, two solutions of opposite character are prepared and standardized, the one by a primary standardization method and the second by comparison (sec- ondary standardization) against the first. This application of the equivalent volume relationship by the sec- ondary standardization of a base by means of a standard acid is shown in the following illustration. Suppose 27.25 ml. of a 0.5123 N solution of HCl are found to react with 28.50 ml. of a NaOH solution. What is the normality of the NaOH solution? From the equation we have EhcI X NhCI = I^NaOH X A^NaOH •^NaOH = 27.25 X 0.5123 28.50 = 0.3144 N 42 VOLUMETRIC ANALYSIS The ultimate objective of every quantitative determination is to find the amount of the constituent in question in the sample undergoing analysis. The calculation reduces itself to, usually, the ratio of the weight of constituent to the weight of sample, and when the result is to be expressed in terms of percentage, the computation takes the form Weight of pure constituent Weight of sample X 100 = Percentage In the volumetric method, the weight of constituent is obtained as the product of the volume of standard solution used and the value of 1 ml. in terms of the constituent being determined. That is Volume used X [Value of 1 ml.] = Weight of constituent When the strength of the solution is expressed directly in terms of the proper titer, i.e., grams of constituent brought into reaction by 1 ml. of the titrating solution the calculation of the weight of pure constituent is then simply made on the basis of the relation Volume used X [Appropriate titer] As an example, if 31.75 ml. of an HCl solution having a Na2C03 titer of 0.005467 are required to neutralize the Na2C03 contained in a sample of impure sodium carbonate the weight of pure Na2C03 present is 31.75 X 0.005467 = 0.1735 gram On the other hand, if the strength of the titrating solution is known in terms of its normality, the weight of pure constituent is obtained by the relation Volume used X [Milliequivalent weight X Normality] To illustrate, suppose that 31.75 ml. of an HCl solution having a normality of 0.1032 are required to neutralize the Na2C03 contained in a sample of impure sodium carbonate, the weight of pure Na2C03 present is 31.75 X (0.053 X 0.1032) = 0.1735 gram the factor 0.053 being the gram-milliequivalent weight of Na2C03. The results of these two illustrations are the same, because the HCl solution having a Na2C03 titer of 0.005467 is a 0.1032 N solution. If in the above illustrations, the weight of sample was 0.2000 gram, the proportion of pure Na2C03 in the sample is 0.1735 0.2000 0.8675 and the percentage is 86.75. PROBLEMS 43 PROBLEM SET 2 1. Calculate the gram-equivalent weighU for complete neutralization of the following acids and bases; (а) HCl Ans. ( б ) H2SO4 (c) NH 4 OH (rf) Il3r04 (e) Ba(OH )2 2. Calculate the gram-equivalent weights of the following substances: (а) NaCl ( б ) AgNOa (c) Na^COs (d) BaCIa (e) ICHC8H4O4 3 . Wliat is the normality of each of the following concentrated laboratory re- agents: (o) HCl (sp. gr. 1.190, containing 37.23 per cent HCl by weight); ( 6 ) NILOH (sp. gr. 0.900, containing 28.33 per cent NH 3 by weight)? 12.15 , , (?>) M.07.V 4 . Find the normality of each of the follow'ing concentrated laboratory reagent.s: (a) H 2 SO 4 (sp. gr. 1.84, containing 95.60 per cent H 2 SO .1 by weight); (b) HNO 3 (sp. gr. 1.420, containing 69.77 per cent HNO 3 by weight). 6. Wliat volume of concentrated hydrochloric acid (sp. gr. 1.190 containing 37.23 per cent HCl by weight) should be taken to prepare a liter of 0.1256 solution? An.s. 10.34 ml. 6. A certain KCNS solution has a AgNOa titer of 0.1699. How much Ag could be precipitated by 35.00 ml. of this solution? What is the normality of the .solution? 7. How many milliliters of 0.5 N NaOH are required to neutralize (o) 20.00 ml. of 0.6 N HCl; ( 6 ) 40.00 ml. of 0.5 N H 2 SO 4 ; (c) 60.00 ml. of 0.5 N H 3 P 04 ? Ans. (a) 20.00 ml. (6) 40.00 ml. (c) 60.00 ml. 8. How much NajCOs can be neutralized by 36.25 ml. of a 0.1000 N HCl solution? How much Ag could be precipitated by this volume of acid? How many grams of KCNS would be required to precipitate the same amount of Ag as could be pre- cipitated by 36.25 ml. of the above acid? 9. What volume of 0.1682 N NaOH solution will be neutralized by 25.00 ml. of a 0.1000 W solution of HCl? By a 0.1000 W solution of H2SO4? Why are the volumes the same? Ans. 14.8GmI. 10. Whai volume of a 0.1342 N HCl solution will be neutralized by 0.12(X) gram of NaOH? By 0.0567 gram of NaaCOj? 11. Calculate the normalities of each of the following solutions having the expressed titers: (a) 36.47 (5) 49.04 (c) 35.05 (d) 32.07 (c) 85.69 (o) HCl (Na^COa titer = 0.0523) (6) NajCOa (NaOH titer = 0.00393) (c) H2SO4 (H2SO4 titer = 0.00486) (d) KOH (H 2 SO 4 titer = 0.0985) Arw. (a) 0.987 W (5) 0.0981 N (c) 0.0991 N (d) 2.008 W < VOLUMETRIC ANALYSIS 12. If 27.52 ml. of an HCl solution will neutralize 28.50 ml. of an NaOH solution, what volume of HCl wiU be neutraUzed by 1 ml. of the NaOH? If the acid solution has a normaUty of 0.6000, what is the normality of the base? 13. What volume of a 0.500 N solution of Na 2 C 03 is equivalent to 36.0 ml. of an acid solution whose normality is 0.100 iV? 7.TO ml. 14. In titrating 20.50 ml. of KOH solution, 22.46 ml. of a 0.6000 N solution of H 2 SO 4 were used. Calculate the weight in grama of KOH which reacted. 16. If a certain acid solution had a normality of 0.1432 what is its (o) NaOH titer; (5) Na 2 C 03 titer; (c) HCl titer? («) 0.005728 (b) 0.007690 (c) 0.006222 16. Calculate the weight of (o) NaOH neutralized if a titration required 30.00 ml. of a 0.1 iV solution of an acid; ( 6 ) if a sample of NaHCOs required the same volume of the acid. 17. Calculate the value of 1 ml. of each of the following: (a) 0.2650 N NajCOs (b) 0.5000 iV H 2 SO 4 (c) 0.0100 iV NaOH id) 0.0500 N Ba(OH )2 Arts, (a) 0.01405 (b) 0.02452 (c) 0.004000 (d) 0.004285 18. If 15.0 ml. of concentrated HCl (sp. gr. 1.19, containing 37.23 per cent pure HCl) is diluted to a liter what is the (a) normality, (b) the HCl titer of the solution? 19. If 25.00 ml. of an HCl solution having a NajCOs titer of 0.005165 reacts with 30.00 ml. of an NaOH solution what b the normality of the base? Ans, 0.08121 N 20. Calculate the percentage of pure Na 2 C 03 in a sample of an alkaline substance if a 1.5000-gram sample required 35.00 ml. of a 0.1 N solution of HCl for titration. CHAPTER III NEUTRALIZATION METHODS GENERAL LABORATORY INSTRUCTIONS Preliminary Preparations. Upon being assigned a desk, check tlie equipment against the list furnished. Examine each piece of glassware for chipped edges, scratched bottoms and other defects. Make sure all items are present and in good condition; report shortages and defects to instructor or supply room. Each item should have its proper place in the desk and be returned at the close of laboratory periods to its appointed location. Glassware should be kept separate from ironware. Never return dirty apparatus to the desk. Before leav- ing the laboratory, wipe off the desk top with sponge and towel. Make sure your desk is locked before leaving. Cleaning reagent is to be used for burets, pipets and flasks; all other apparatus is cleaned with soap and brush. Before attending to other prepa- rations listed below, make 500 ml. of dichromate cleaning solution according to the following directions : Dissolve 50 grams of commercial Na2Cr207 in 150 ml. of water, warming the solution to effect Fig. 2 . Wash bottle, more rapid dissolving. Cool, pour into a large evaporating dish, and add slowly, while stirring constantly, 230 ml. of concentrated H 2 SO 4 ; use commercial sulfuric acid if available. W'lien the mixture has cooled, at the end of the laboratory period transfer to a 500-ml. wide-mouthed bottle. Use this cleaning reagent only for flasks burets and pipets. While the cleaning solution is cooling, procure some glass tubing and make a wash bottle, according to the design and type recommended by the instructor. A simple type is shown in Fig. 2. Cut three 8-inch lengths of glass rod and fire-polish the ends. These will be used as stirring rods. 45 46 NEUTRALIZATION METHODS Place some calcium chloride in the bottom compartment of the desicca- toi* . Place a thin coating of vasehne on the ground joint between vessel and cover. Desiccators are used to store dried samples^ crucibles, etc. Exercises with the Analytical Balance, Reread those portions of Chapter I which deal with the use and care of the balance. Go to the balance assigned; study its construction and operation. Familiarize yourself with the weights in the weight box. Determine the rest point of the empty balance. If not in satisfactory adjustment, call the instructor; do not attempt the adjustment yourself. Of the several different methods of canning out a weighing operation, two are here described. The first and simplest of all is by equal swings; the other is known as the sensitivity method. The instructor will specify the method of weighing which is to be used in the course. 1. Method of Equal Swings. Place an empty, clean and dry weigh- ing bottle or crucible on the left-hand pan, after the point of rest has been determined. Then place the 5-gram weight on the right-hand pan; release beam and pan support; the pointer will probably move more rapidly to the right, showing that the object weighs more than 5 grains. Add the 2-gram weight; if too heavy, replace by the 1-gram weight. A single deflection of the pointer to right or left is sufficient to decide whether the object is fighter or heavier than 5 grams. Let us assume that the weight of the object is between 6 and 7 grams, as determined by the brass weights. Now begin with the fractional weights. First try the 0.5-gram (500 milligrams), then the 200-milligram and so on down to the lO-milfigram weight. Suppose the weight now is established as 6.73. The next two units, namely, the milligrams and tenths of milligrams, must now be determined by the rider on the beam scale. Close the balance case and place the lider on the 5-milfigram division of the beam. If too heavy, move it to 2.5, otherwise to 7.5, Again subdivide the beam scale by a third setting of the rider. Finally, a fourth setting will prob- ably be necessary to equalize defimitely the loads between weight and object. It is only duifing the last two or three settings of the rider that the pointer swings need be observed carefully and the swings of the pointer in both directions definitely noted. If final equilibrium is deter- mined with rider at 4.5, the correct weight of the object is 6.7345. Although this method is more easily understood and performed by the beginning student, the sensitivity method furnislies the better training. 2. Sensitivity Method of Weighing. As already defined, by sensi- tivity is meant the number of pointer scale divisions through which the point of rest is moved by an excess load of 1 milligram. First determine GENERAL LABORATORY INSTRUCTIONS 47 the rest point of the balance; suppose it is 0.3 scale divisioas to the left of center. Then proceed to counterbalance a clean, dry weighing bottle or cru- cible until the fractional weights are exhausted. Then determine the two positions, with the rider, between wliicli the true weight lies. Sup- pose that the weight load is 6.73 and that the rider at the 5-niiIIigram position is too great and at the 4-inilligram too little, the tr.ue weight being between 6.7340 and 6.7350. Determine the sensitivity for this load by finding the points of rest for both loads, by averaging swings to left and right, in both cases. Take the difference between these two new positions of the pointer. Suppose that, with the rider on 4 milligrams, the new point of rest is at 1.5 on right and with the rider at 5 milligrams the new rest point is at 2.1 on the left. This gives a sensitivity of 3.6 scale divisions. The position where the rider would have been placed, between 4 and 5 milligrams, in order to equalize the balance and bring the point of rest back to 0,3 is now calculated. Since the pointer rest position is moved through 3.6 scale divisions for an extra load of 1 milligram, in order to bring the pointer to the original rest point of 0.3 it must be moved through 1.5 4- 0.3 or 1.8 scale divisions. To move the pointer rest position 1.8 scale divisions to the left, there must be added 1.8/3. 6 or 0.5 milligram (0.0005 gram). The weight of the object is therefore 6.7345 grams. Methods of Weighing Samples. There are in general two methods of weighing samples. In the one method, the sample, dried to remove moisture unless otherwise directed, is placed in a weighing bottle; the bottle with its contents is accurately weighed; portions of approxi- mately the desired weight are transferred to beakers or flasks for dis- solving; and finally the weighing bottle with remainder of sample is reweighed. The difference in weight is the portion of sample taken for analysis. The most convenient and certain means of transferring the sample from the weighing bottle is to fashion a small spoon or ladle-like tool from thin sheet aluminum. The transfer spoon should be short enough to fit into the weighing bottle when the cover is in place; it is kept in the bottle and weighed with the sample. Another method of weighing samples, though less desirable, is to weigh out onto a weighed watch glass or weighing scoop the desired portion of sample. For exact work tared watch glasses are used to correct for buoyancy. Care must of course be taken not to spill sample on the balance pan. The weighed sample is then brushed with a camel’s hair brush into the dissolving beaker or flask. Instructions will be given in the procedures or by the instructor as to the specific method to be employed. 48 NEUTRALIZATION METHODS Determinatioii of Moisture in a Sample. Clean two weighing bottles and their stoppers. Attach to each a small sticker with your initials, i ‘‘nmber the bottles “1^' and “2.” Dry them with stoppers placed in a slanting position for 1 hour in the drying oven at 105°C.; remove to desiccator, aUow to cool to room temperature, place stoppers in the bottles and weigh accurately. Repeat the drying of the empty bottles until the successive weights do not differ by more than 0.2 milligram. Transfer into the weighing bottles 1 to 2 grams of the sample issued for this determination, stopper the bottles tightly and weigh accurately. The increase in the weight is the weight of sample. Fig. 4. Weighing bottles. Return the weighing bottles to the drying oven, place stoppers in a slanting position, allow samples to remain there for at least 1 hour (2 hours or overnight is better). Remove from oven, stopper the bottles, cool in desiccator and weigh. Reheat and repeat the other operations. The loss in weight of the sample is the moisture removed by drying. This divided by the weight of sample and multiplied by 100 is the percentage of moisture. GENERAL CONSIDERATIONS OF ACIDIMETRY AND ALKALIMETRY The measurement of the acid strength of a substance, by titration with a standard solution of a base, is termed acidimetry. Conversely, alkalimetry is the determination of the basic strength of a material by titration with a standard acid solution. Both processes employ neutrali- zation reactions. By these methods practically all acids and bases and many salt-s of weak acids and bases are analyzed. For the preparation of standard acids, we are practically limited to hydrochloric, sulfuric and nitric acid, of which the first named is most frequently used. For bases, sodium hydroxide is most generally selected, determinations of alkalinity and acidity 49 although potassium hydroxide and barium hydroxide are occasionally employed. In strength they vai-y between 0.5 N and 0.05 TV, an ap- proximately 0.1 N solution being, in most cases, the most desirable strength. Since these reagents cannot be obtained in a pure fonn, solutions made from them must be standardized against pure primary standards. As standards, for hydrochloric acid, pure NaaCO^ is generally employed and, in case an independent primary standardization of sodium hy- droxide is desired, pure potassium acid ])hthalate, KHCsH.iO.i, is an excellent standard. It is highly important that the correct indicator be used for acid-ba.sc titrations. There are many organic dyestuffs which can be employed, because they have the property of changing color over definite narrow ranges of hydrogen-ion concentration; the indicator chosen is the one wliich changes color at the hydrogen-ion concentration at which the reaction comes to equilibrium. The two indicators which will be em- ployed in the following experiments are (modified) methyl orange and phenolphthalein. The latter is colorless in solutions having a hydrogen- ion concentration greater than about 10“® and pink in solutions of 10“'*^ or less. Ordinary methyl orange is pink or red in rather strongly acid solutions (hydrogen-ion concentration of 10“^) and orange in somewhat less acidic solutions. Because the color change for ordinary methyl orange is hard for beginners to distinguish in 0.1 TV solutions, this indi- cator is modified by the addition of another dye, xylene cyanole FF, which shows a reddish color at a hydrogen-ion concentration of about 10“®, passes through a gray transition range and is green at 10“^. The preparation of these indicator solutions is given in the list of indicators in the Appendix. For the titration of strong acids with strong bases either indicator may be employed. Phenolphthalein is ased for the titration of weak acids; it cannot well be used in precise work for titrations with bases which contain carbonates. Methyl orange or, better still, methyl red is used for ammonia titrations. For precise work, an indicator blank should be run. This is the volume of standard reagent necessary to affect the indicator in those titrations in which the equivalence point does not coincide with the hydrogen-ion concentration at which the indicator changes color. DETERMINATIONS OF ALKALINITY AND ACIDITY Practical application of neutralization reactions is made in the follow- ing experimental work in the determination of the alkaline strength of gQ neutralization methods samples of impure sodium carbonate, and of the acid strength of po- tassium acid phthalate or, alternatively, oxalic acid. ^Q^^ed for ^ch volumetric determinations are standard solutions of HCl and NaOH ; the preparation and standardization of these reagents foUow. PREPARATION AND STANDARDIZATION OF APPROXIMATELY 0.1 IV SOLOTIONS OF HYDROCHLORIC ACID AND SODIUM HYDROXIDE Calculations and Preparation. A normal solution of an acid contains 1 008 grams of replaceable hydrogen per liter of solution. A liter of hydrochloric acid of normal strength contains 36.465 grams of pure hydrogen chloride since in this weight of solute there are 1.008 grams of replaceable hydrogen. Since a solution of approximately 0.1 N strength is best suited for the following procedures, 0.1 of a gram-mole of HCl must be used. This requires 0.1 X 36.465 or 3.65 grams. The reagent from which this is obtained is concentrated hydrochloric acid, of specific gravity of 1.19 and containing 37.3 per cent of pure HCl. Each miUiliter of this solution contains 1.19 X 0.3723 or 0.443 grams of pure HCl. To furnish 3.65 grams requires therefore 3.65/0.443 or 8.2 ml. For the Bolution measure out about 8.5 ml. of concentrated hydro- chloric acid, dilute with about 200 ml. of distilled water, transfer to the clean liter measuring flask and fill to the mark with distilled water. Keep the solution in a Uter bottle. Label the bottle “approximately 0.1 AT HCl” and place your name on the label. A normal solution of NaOH should contain 17.008 grams of replace- able hydroxyl per hter, and this amount is contained in a gram-molecular weight, 40.005 grams. For a 0.1 AT solution, one-tenth of a gram-mole is required, that is, 4.0005 grams per hter. Since sodium hydroxide is not pure, weigh out on the rough balance about 4.2 grams of NaOH and dissolve in about 200 ml. of distilled water. Pour this solution into a hter measuring flask. Add distiUed water until about half fuUj shake thoroughly and then fiU the fiask to the mark with distihed water. Transfer the NaOH solution to a bottle provided with a rubber stopper instead of a glass one because NaOH will cement a groimd-glass stopper to the bottle. Label the bottle properly. Comparison of Solutions. Place the burets in the buret clamp on the stand and fill both with dichromate cleaning reagent, allowing the solu- tion to remain in them for a little while. Remove the cleaning agent, rinse out with water and grease the stopcocks with a thin layer of stopcock grease. Fill with water and note whether they leak. Prepare two meniscus reading cards, one for each buret. This is done / V \ PREPARATION AND STANDARDIZATION OF 0.1 N SOLUTIONS 51 by making two horizontal cuts, about one-half inch apart and two inches long, in a small flexible cardboard. The area below the lower cut is then blackened by inking with a pen. The card is slid onto the buret, with the black area placed at a level so that the top of the inked portion is about 1 millimeter below tlie bottom of the mcnLscus of the liquid in the buret. Fig. 5 shows such a card. When all is in readiness, rinse one of the burets with a few milliliters of the HCl solution and the other with the NaOH solution. After rinsing, fiU the burets with HCI and NaOH solutions and bring the solutions within the graduations, being careful that air bubbles arc not trapped below the stopcock. Adjust the reading card ; record the read- ings of the two burets; estimate to the second decimal place. After taking the readings, nm out into a clean 250-nil. Erlenmeyer flask 35 ml. of the HCl solution, using the left hand to manipulate the stopcock. Add 50 ml. of distilled water and two drops of modified methyl orange indicator. Run in NaOH solution with constant swirling of the flask with the right hand until the Fm. 5. Buret reading card, pink color has given place to a green. Add HCl solution dropwise until the pink color is again restored; con- tinue these additions until one drop of NaOII wall change the color of the solution to a faint green and one drop of HCl udil color the solution a faint pink. In titrating always choose the faintest color as an end point; also treat each portion in the same manner; thus if you stop on a faint pink end point here do likewise in subsequent titrations. Record the final buret readings. If the volumes used differ by more than 5 ml. add water to the bottle containing the stronger solution and repeat the trial comparison. Refill the burets and make another comparison titration, using ap- proximately the same volumes as before. Make a third comparison and then calculate the equivalent volume ratios as illustrated below. Calculations, Suppose that on comparing these two solutions one of the titrations showed that 32.61 ml. of NaOH reacted with 33.75 ml. of HCl. Then, since 32.61 ml. of NaOH are equivalent to (o) 33 75 ml of HCl, 1 ml. of NaOH 33.75 32.61 and, conversely, 1 1 r TTr^i 32.61 1 ml. of HCl Cs 33.75 ^ 1.035 ml. of HCl 0.966 ml. of NaOH ^2 neutralization methods In this manner, compute the ratios for the several comparison titrar tions rejecting those that do not agree closely and averaging the re- mainder. Submit the calculations to the instructor for criticism. If acceptable results are not obtained, carry out more comparison titra- tions. These equivalent-volume ratios will be used not only in correcting total volumes used later in standardization of the HCl solution, but also for computing the strength of the NaOH solution. InAhis first series of titrations, the student will experience little diffi- cult/in securing checking results. Agreement of the ratios within 2 to 3 parts per thousand is satisfactory. Standardization of HCl Solution with Pure NaaCOg. Of the several possible ways of standardizing a solution of HCl, the direct titration against weighed portions of pure NaaCOs as outlined below is recom- mended. Aside from the high accuracy of the method, it has the ad- vantage here that in the subsequent analysis of soda ash (impure sodium carbonate), to which use this reagent is to be put, the same con- ditions and reactions are involved. Hydrochloric acid solutions may of course be standardized against a previously standardized solution of a base if one is available. The strength may also be determined pavi- metrically by precipitation as AgCl, and it is possible to obtain by distillation of HCl solutions a constant boiling mixture, in which the percentage of HCl by weight is 20.22 (at a barometric pressure of 760 millimeters Hg). Outline of Method of Standardization. Portions of pure Na 2 C 03 are weighed and dissolv'ed. Methyl orange indicator is added and the solutions are titrated with HCl. Adjustment of the end point is made with the NaOH solution which has previously been compared with the HCl. Pure Na 2 C 03 especially made for standardization purposes may readily be purchased, or it may be prepared in the laboratory by heating NaHCOa to a temperature of 270° to 300°C. for 1 hour. In case it be- comes necessary to prepare pure Na^COs from sodium bicarbonate, pro- ceed as follows. Place an iron crucible on a triangle resting on an iron ring. Place within this crucible another triangle, and insert into the tri- angle a porcelain crucible containing about 6 grams of NaHCOaj adjust the triangle so that the porcelain crucible is wholly within the iron cruci- ble and an air space exists between the walls of the two crucibles. Sus- pend a 350°C. thermometer in the NaliCOa so that the bulb is covered. Heat the iron crucible with a small flame at first and gradually increase the heat until 270°C. is reached. During this time occaaonally stir the NaHCOa. Heat very slowly from 270° to 300°C. so that one hour elapses during this interval. Do not allow the temperature to rise above 300°C. Transfer the crucible and Na 2 C 03 to a desiccator, allow to cool, then PREPARATION AND STANDARDIZATION OF 0.1 N SOLUTIONS 53 transfer to a weighing bottle, stopper well and keep in the desiccator until ready for use. Procedure. Weigh accurately into three numbered 2o0-ml. Erlen- meyer flasks three portions of NuoCOa, previously dried at 140°C., of about 0.2000 gram each. Dissolve each portion in 80 ml. of water. Fill clean burets with HCl and NaOH and take the initial readings, being sure that no air bubbles are in the tip below the stopcock. Add 2 drops of modified methyl orange indicator to the flask containing the smallest weight of Na2C03. Run in HCl, vith constant stirring, until a faint pink coloration is obtained, and then run in NaOH until a faint green or gray appears. Adjust the end point after this fashion until a point is reached wliere one drop of acid or one drop of base will change the color of the solution. Record the final buret readings and repeat the standardization with the other two carbonate standards. Calculations. (1) Strength of (he IICl Solution. Let us suppose that in neutralizing 0.1620 gram of pure Na2C03, 2.00 ml. of the NaOH and 32.07 ml. of the HCl solutions were required for complete neutrali- zation. Then, since, as shown by the example illustrating comparison, 1 ml. of NaOH is equivalent to 1.035 ml. of HCl, the volume of HCl used by the NaOH is 2.00 X 1.035 or 2.07 ml. The net volume of HCl, namely, that used by the NaoCOa alone, is 32.07 — 2.07 or exactly 30.00 ml. If the titration required 30.00 ml. of PICl to neutralize 0.1620 gram of pure NajCOs, 1 ml. of the HCl will have neutralized 0.1620 gram/30.00 ml. or 0.005400 gram of NaoCOg. This, by definition, is the Na2C03 titer of the HCl solution. Since the normality is, by definition, the ratio of the concentration compared tp an exactly normal solution, the normality is found by dividing the weight of solute or its equivalent per milliliter by the gram-milliequivalent w'cight. Each milliliter of the HCl solution is equivalent to 0.1620/30.00 or 0.005400 gram of Na2C03, as found by direct titration. The gram-equivalent weight of Na2C03 is Na2C03/2 or 53.00 grams; the gram-milliequivalent weight is therefore 53.00/1000 or 0.05300 gram. Then the normality is 0.005400 0.05300 0.1019 N These steps may be combined into the single equation 0.1620 30.00 X 0.05300 = 0.1019 AT 54 NEUTRALIZATION METHODS 1 4 Calculate, as shown in this example, the normality of the HCl separ rately for each standardization. The results should agree within two parts per thousand; the deviations are thus a measure of the pre^sion of the standardization. If satisfactory results are not obtained, the standardization must be repeated. Close agreement in the normality obtained by the several runs gen- erally indicates, as well, that the standardization has been correctly carried out. But this is not necessarily true because, although all like operations were performed in exactly the same manner, a serious con- stant error could have been repeatedly introduced. Good precision does not always guarantee high accuracy. Failure of agreement in the standardization values is generally due to carelessness and inexperience and the students^ lack of appreciation of the care that must be exer- cised. A repetition of the standardization generally results in a more satisfactory agreement of results. If results do not agree on the set of titrations, the standardization must, of course, be repeated. (2) Strength of the NaOH Solution. From the equivalent volume ratio obtained from comparison data, together with the normality of the HCl solution, the normality of the NaOH solution is readily computed. For example, if 32.61 ml. of NaOH were found to react with 33.75 ml. of HCl, the NaOH solution is 1.035 times stronger than the HCl solution. Since the strength of the HCl, in the above example, is 0.1019 iV, the normality of the NaOH is 0.1019 X 1.035 or 0.1054 N. Use might also be made of the relation FhCI X NhCI = UnbOH X NNaOH 33.75 X 0.1019 = 32.61 X x X = 0.1054 N Compute the normality of the NaOH solution by multiplying the normality of the HCl solution by the number of milliliters of HCl neutralized by 1 ml. of the NaOH. DETERMINATION OF THE ALKALINE STRENGTH OF SODA ASH Soda ash is crude sodium carbonate and contains small amounts of sodium bicarbonate together with impurities. The NaHCOa contrib- utes to the total basicity of the sample. In reporting results it is cus- tomary to report the total alkaline strength as percentage of Na 2 COs, Procedure. Spread the sample issued you on a clock glass, attach a label with your name and place in a drying oven, at 140®C., for one hour; this is done to remove moisture from the sample. Then transfer the dried sample to a weighing bottle and place in the dedccator. THE ALKALINE STRENGTH OF SODA ASH 55 The method of securing a uniform sample in this determination con- sists in taking aliquot portions of the dissolved, weighed sample as described below. Weigh out accurately a sample of the dried soda ash of about 1 gram into a 200-ml. beaker. Add 50 ml. of distilled water to dissolve the sample. Transfer the solution to a 250-mI. measuring flask, wash out the beaker and add distilled water from tlie wash bottle until the lowest part of the meniscus is level with the graduation. Rinse out a 50-ml. pipet with some of this solution, thoroughly shake the sample in the flask and then pipet 50 ml. into a 250-ml. Erlcnmeyer flask. Add 30 ml. of water and two drops of modi- fied methyl orange indicator. With HCl and NaOH in readiness in burets, add HCI, wliile swirling constantly, until a faint pink is reached; then alternately add NaOH and HCl to adjust the end point. Repeat the determination with two other 50-ml. portions of the sample. Calculations. When methyl orange is used as indicator, both of the alkaline constituents in the sample are shown to have been neu- tralized by the acid at the point of color change. The total basicity may therefore be computed from the net volume of HCl used and expressed as a percentage in terms of the kquid to equivalent of Na 2 C 03 . To illustrate, suppose that in the titration of an aliquot portion (one-fifth of a 1.0000-gram sample) there were employed a total volume of 28.00 ml. of a 0.1019 N solution of HCl and, for back titration, 2.60 ml. of NaOH, 1 ml. of which is equivalent to 1.035 ml. of the HCl. Each milliliter of 0.1019 N HCl reacts with 0.053 X 0.1019 or 0.005400 gram of NazCOg (or its equivalent of NaHCOs) hi the sample. The net volume of HCl used was 28.00 (2.60 X 1.035) or 25.32 ml. Therefore, the weight of basic constituents in the' sample, expressed as grams of Na 2 C 03 , is 25.32 X 0.005400 = 0.1367 gram The percentage of NagCOa in a 0.2000-gram portion of the sample is therefore 0.1367 Q2000 ^ ^ 56 NEUTRALIZATION METHODS These steps can be consolidated into the single set of mathematical operations 25.32 X 0.1019 X 0.05300 0.2000 X 100 = 68.35 per cent In the way illustrated, calculate for each of the titrations the net volume of HCl used. Multiply this by the normality of the HCl solu- tion and by the mi Uiequi valent weight of Na 2 C 03 . Divide the product by one-fifth of the weight of the sample and multiply by 100. The result is the total alkalinity of the sample, expressed as percentage of Na 2 C 03 . Results which check within two to three parts per thousand are very good. Submit calculations and results, as recorded in the record book, to the instructor. If results are not acceptable, the experiment must be repeated. Not every beginner is entirely successful with this, his first volumetric analysis. A large discrepancy may be due to faulty standardization (even though the normality checked), to careless weighing, to incorrect sampling, to errors in reading burets, etc. Sometimes, even, the error is discovered in the computations. Minor errors creep into the analysis through the use of uncalibrated burets, pipet and flask and weights, end-point errors, temperature variations, etc. If a re-run must be made, aU the sources of error should be guarded against and minimized wher- ever possible. DETERMINATION OF THE ACID STRENGTH OF POTASSIUM ACID PHTHALATE Potassium acid phthalate, KHC 8 H 4 O 4 , is phthalic acid, H 2 C 8 H 4 O 4 , in which one of the hydrogen atoms has been replaced by potassium. It behaves as a weakly ionized monobasic acid. It is an excellent pri- mary standard, widely used for the standardization of solutions of strong bases such as NaOH. Samples of the impure salt will be used as an exercise in the following determination. Standardization of NaOH Solution. Although the NaOH to be used has already been standardized by comparison with HCl of a known strength, it is here suggested that an independent standardization of the base be made with pure potassium acid phthalate. Weigh accurately three portions of about 0.8 gram each of the stand- ard into 250-ml. Erlenmeyer flasks. Add 80 ml. of distilled water to each flask and heat to boiling. Cool, add two drops of phenolphthalein indicator and titrate without delay with the NaOH solution to a pmk color end point. THE STRENGTH OF AN OXALIC ACID SOLUTION 57 Calculations. The molecular weight of potassium acid phthalate, KHCsH-iO-i, is 204.2, hence the millicquivalcnt is 0.2042 gram. Ascer- tain from the instnictor (or otherwise) the purity of the standard and correct the weight of the portions used. Divide tliis corrected weight by the net volume of NnOH used; this will give the amount of pure KHC8H.1O4 which reacted with 1 ml. of the base. To find the normality divide this titer value by 0.2042. Ilesult-s should not deviate by more than one or two parts per thousand. Determination. Dr^'^ the issued sample of impure potassium acid phthalate for one houi' in an oven maintained at 120°C. Transfer to a weigliing bottle and place in desiccator. Weigh into 300-ml. Erlen- meyer flasks three separate portions of approximately 1 gram each. Dissolve the smallest weighed sample with about 80 ml. of distilled water, heat to boiling, cool to room temperature under the water tap, add two drops of phenolphthalein indicator and add without delay from the buret the standard base until the solution changes from colorless to a light pink. Repeat with the other samples, being careful to go slowly when the volume of base reaches that used by the first sample. Calculations. Multiply the volume of NaOH used by its normality and in turn by the milliequivalcnt weight of KHC8H.1O4 (0.2042), divide this product by the weight of the sample and multiply by 100. This will give the percentage purity of the sample in terms of pure potassium acid phthalate. Percentages should agree wdthin several parts per thou- sand. Have the instructor approve data and calculations before making out the final report. DETERMINATION OF THE STRENGTH OF AN OXALIC ACID SOLUTION Outline of Method. The solution of oxalic acid submitted as an unknown is diluted, phenolphthalein is added and the solution is titrated with standard NaOH to a faint pink. The pink is just discharged with standard HCI, after wliich the solution is boiled. If no color appears the titration is completed with the standard NaOH in the hot solu- tion. Phenolphthalein indicator is sensitive to weak acids and is especially valuable in the titration of organic acids. At high dilutions it loses its sensitivity as an indicator. The reaction involved in the titration of oxalic acid is 2NaOH + H2C2O4 = Na2C204 + 2H2O After the discharge of the pink coloration with HCI the solution is 58 I NEUTRALIZATION METHODS boiled. NaOH generally contains some Na2C03, and this gives rise to the formation of NaHCOa Na2C03 ~f- 02^2^4 “ Na2C204 -|- H2CO3 ^ H2C^3 d" Na 2 C 03 “ 2 NaE[C 03 Since NaHCOs does not furnish enough OH“ ion to cause phenol- phthalein to remain pink, evidently all the NaOH solution added will not be effective in the titration. However, after boiling the solution, Na2C03 is formed; this reacts with the HCl, and if the Na2C03 is in excess it will hydrolyze sufficiently to give a return of the pink colora- tion. The reactions involved are 2NaHC03 = Na2C03 -i- CO2 H2O (1) NasCOa + 2HC1 = 2NaCl + H2O + CO2 (2) NaaCOa + HOH = 2NaOH + H2CO3 (3) The OH“ ions due to (3) will cause a return of the pink coloration. Procedure. Measure out 20 ml. of the oxalic acid solution and dilute to 50 ml. with distilled water. Add two drops of phenolphthalein indi- cator, stir and run in the standard NaOH solution until the colorless solution turns pink. Carefully discharge the color with standard HCl and add a few drops in excess. Boil the solution for about three minutes. Should the pink coloration return, add acid as before and boil again for three minutes. If no color reappears, bring it back with the NaOH, add acid, a drop or two in excess, and boil as before. Complete the titration in a hot solution. If no color reappears on boiling, choose as an end point the faintest pink color which lasts about one minute. Note that in this titration the change is from colorless to colored. This is the most accurate way of judging end points. Repeat the above titration with another 20-ml. portion of oxalic acid. From the volumes of standard solutions used and the volume of oxalic acid sample neutralized, compute the normality of the oxalic acid solution. questions 1. In general, what acids are used for the preparation of standard solutions? Can nitric acid be used under all circumstances? What substances might you use for preparation of a standard acid solution without the necessity of a standardization titration? How could you prepare a solution of HCl as a standard solution without a standardization titration? 2. What substances arc available for the preparation of standard base solutions? Could one use Ba(OH) 2 , NajCOa, KOH, NH4OH? If not, why? 3. What pure substances are available for the standardization of acid solutions? For basic solutions? Distinguish between primary and secondary standardisation. PROBLEMS 59 4. ^Vhy use methyl orange in the standardization of HCl? Could phcnolphthalein be used as well? When must phenolphthalein be used as indicator? What indicator would you employ for the titration of NH4OH with IICl? (See section on indicator theory.) 6. What advantages and disadvantages are there in iLsing aliquot portions of the sample of soda ash? 6. Summarize the errors which might account for poor results obtained in deter- mining the alkalinity of soda ash. 7. Do the same as in 6 for the potassium acid phthalatc or oxalic acid sample. 8. How might errors be avoided or corrected (1) in taking aliquot portions, (2) in reading the burets, (3) in the use of indicators? 9. Summarize, with examples, the various types of chemical compounds which may be analyzed by acid or base titrations. STOICHIOMETRIC CALCULATIONS OF NEUTRALIZATION Calculations involving the data of acidimetry and alkalimetry may conveniently be arranged, for purposes of assignment and study, under the following three heads : 1. Normalities, titers, standardization, etc. 2. Adjustment of solutions. 3. Mixed alkali calculations. Illustrations have already been given, in both Chapter II and the foregoing procedures, showing the preparation, comparison and stand- ardization of solutions and computation of percentage of constituents. Additional illustrations, in the form of problems relating to data of this kind as well as involving weight of primary standard and weight of sample and supplying further applications of acid-base determinations, are given below in Problem Set 3. Calculations involving the adjustment of the strength of solutions as well as rruxed alkali titrations are discussed on page 91 and followed by Problem Set 5. Calculations of pH values and titration curves are illustrated beginning with page 61 and followed by Problem Set 4. PROBLEM SET 3 ACID-BASE CALCULATIONS 1. What weight of pure solute is required for the preparation of liter quantities of 0.5000 N solutions of the following reagents? (а) HC2H3O2 (б) H2SO4 (c) KOH (d) Ba(OH )2 Ans. (a) 30.03 grams (5) 24.52 grams (c) 28.05 grams (d) 42.85 grams 60 NEUTRALIZATION METHODS 2. What volume of ‘^concentrated” hydrochloric acid solution (sp. gr. 1.190, containing 37.23 per cent of HCl by weight) must be taken for the preparation of a Uter of 0.5000 N HCl? 3. If 32.56 ml. of a 0.1382 N HCl solution were found, on comparison, to react with 31.85 ml. of an NaOH solution, what is the normality of the base? Ans. 0.1413 AT 4. A certain acid solution has a normality of 0.1652. How many milliliters of 0.1267 N NaOH are required for the neutralization of 35.00 ml. of the acid? 6. A titration of 30.00 ml. of a certain HCl solution required 35.80 ml. of a solu- tion of NaOH. What is (o) the value of 1 ml. of the HCl in terms of NaOH, and (6) the value of 1 ml. of the NaOH in teims of HCl? Ans. (a) 1.19 (5) 0.838 6. A solution was made by dissolving 6.0 grams of NaOH, adding 5 ml. of HCl (sp. gr. 1.19, 37.23 per cent HCl by weight) and diluting to a liter. What was the normality of this solution? 7. Calculate the Na 2 C 03 titer for each of the following solutions: (a) 0.1847 N HCl ^ris. (a) 0.00979 (5) 0.2000 N NaOH (6) 0.01060 (c) 0.1000 AT H 2 SO 4 (c) 0.005300 8 . A sodium hydroxide solution was standardized directly against pure potassium acid oxalate, ICHC 2 O 4 . If 0.5321 gram of the KHC 2 O 4 were used and required 30.00 ml. of the sodium hydroxide solution, what was the normality of the solution? 9. What weight of pure Na 2 COs would you use to standardize an approximately 0.2 N solution of HCl so that not more than 35 ml. of the acid is to be used? Ans. 0.37 gram 10. In standardizing a hydrochloric acid solution it was found that 27.50 ml. of the acid completely neutralized 0.1456 gram of pure Na 2 C 03 according to the reaction 2HC1 + Na 2 C 03 = 2NaCl + CO 2 + H 2 O What is (a) the Na 2 C 03 titer of the solution; (6) the HCl titer of the HCl solution; (c) the normality of the solution? 11. Exactly 30.00 ml. of an NaOH solution was found to require 15.5 ml. of 0.2 N HCl plus 29.0 ml. of 0.1 N H 2 SO 4 for neutralization. What was the normality of the NaOH solution? Ans. 0.2 AT 12. A KOH solution was standardized with pure potassium acid phthalate, KHC 8 H 4 O 4 , using 0.7530 gram of the standard and 42.67 ml. of the base. Calculate the normality of the KOH solution. 13. (a) Calculate the normality of a hydrochloric acid solution having a specific gravity of 1.120. (5) What is the normality of a sulfuric acid solution containing 48.00 per cent of H 2 SO 4 by weight? Ans. (o) 7.316 N (6) 13.51 Af 14. Calculate the percentage purity of sodium hydroxide pellets, a 1.0000-gram sample of which required for neutralization 48.10 ml. of 0.5 N H 2 SO 4 . 16. An analysis of soda ash furnished the following data: weight of sample, 6.0000 grams; using 3^ aliquots required 37.60 ml. of 0.1320 N HCl and 2.35 ml. of NaOH solution, 1 ml. of the latter being equivalent to 1.22 ml. of the acid. Calculate the percentage of the Na 2 COs in the sample. Ans. 24.23 per cent 16. Calculate the percentage of oxalic acid, H2C2O4, in a 1.0000-gram sample if 26.51 ml. of a 0A167 N sodium hydroxide solution was used in the detenninarion. THE THEORY OF NEUTRALIZATION 61 17. In determining combined nitrogen by the Kjeldahl method, the sample is digested with sulfuric acid which converts the nitrogen to NH3. The latter is dis- placed with strong NaOH and distilled into a measured excess of standard acid. The excess is then determined by titration with a standard base. If in such an analysis, 50.00 ml. of 0.500 N acid is employed and for back-titration 40.00 ml. of 0.550 N base is used, what weight of N was present in the sample? Atis. 0.0420 gram 18 . A 25-mI. sample of vinegar (assume density of 1) required 41.60 ml. of 0.60(X) N NaOH for titration. Calculate the j)crcentjige of nC 2 ll 302 in the sain])lc. 19. Phosphorus in steel is determined by oxidizing the ferric phosphide to H 3 PO 4 , precipitating this as (NH 4 ) 3 P 04 * I 2 M 0 O 3 and treating the ammonium phospho- molybdate with a measured excess of standard base, and finally back-titrating with standard acid. In the dissolving reaction 23NaOH are equivalent to IP, therefore the gram-milliequivalent weight of phosphorus is 30.98 23000 or 0.001347. If in such a determination, a 2-gram sample of steel, upon proper treatment accord- ing to the above steps, required 20.00 ml. of 0.1085 NaOlI and 17.50 ml. of 0.09802 N HNO 3 , what was the percentage of phosphorus in the steel? Arw. 0.030 per cent 20. In standardizing an HCl solution, it was found necessary, in order to neutralize 27.65 ml. of the HCl solution, to use 18.50 ml. of NaOH solution, the HCl titer of which was 0.03567, and G.50 ml. of another NaOH solution having an HCl titer of 0.03456. What was the HCl titer of the HCl solution? THE THEORY OF NEUTRALIZATION Equilibrium of the Ions of Water. The fundamental reaction of all neutralization reactions is the union of hydrogen ions and hydroxyl ions, leading to the formation of water, as represented by the ionic reaction H+ + OH“ ^ HOH WTien equilibrium is reached, the concentrations of and OH” in equilibrium with non-ionized HOH can be evaluated. Water, con- sidered as a very weak electrolyte, ionizes to a slight extent into H"*' ions and OH” ions. The ionization constant for water is expressed by the equation X Cqh" Choh Since, however, the concentration of the non-ionized HOH is prac- tically constant and is enormous compared to the concentration of its ions, the equilibrium expression can be simplified by rewriting it in the form Ch* X CoH" = X Choh = = 1-2 X 10“^^ which shows that the product of the concentrations of the ions is equal to a constant, which, at 25°C., has the value 1.2 X 10”^^. 62 NEUTRALIZATION METHODS -This relationship is of fundamental importance in the study of neu- tralization reactions, for it shows that in any aqueous solution, no matter whether it is an acid, a base, a salt solution or water itself, the product o' the gram-ion concentrations of and OH“ is always constant. A liter of pure water, since it ionizes to the same extent into H**" ions and OH“ ions, contains at room temperature approximately 1 X 10“^ gram- ion of hydrogen and an equal concentration of OH“ ions. It is, by definition, neutral. By definition an acidic solution is one which contains an excess of hydrogen ions^ and a basic solution is one which contains an excess of hydroxyl ions. A neutral solution is one which contains eqiml concentra^, lions of these two ions. In fact, it is impossible to have any aqueous solution whatsoever in which there are not both hydrogen and hydroxyl ions. It follows, from the water equilibrium which demands concen- trations such that X CoH“ = 1*2 X 10 that if the concentra- tion of one set of ions — say that of hydrogen ions — is known, the con- centration of the other set becomes fixed. It is customary to designate solutions with respect to their acidity or basicity by their hydrogen-ion concentration. A neutral solution is one having a hydrogen-ion concentration of approximately 1 X 10“^, that is, one containing 0.0000001 gram-ion of hydrogen ion per liter of solution. This value is significant in that it constitutes the dividing line between acids and bases. The character of any solution may be designated by writing the hydrogen-ion concentration in decimal form as 0.00001, or in exponential form as 1 X 10“®, or in terms of what is known as the pH value. pH VALITeS Designating the acidity or basicity of a solution by means of its pH value arose from the following considerations. A very accurate means of determining the hydrogen-ion concentration of a solution is to measure - the electromotive force which is set up in an electrolytic cell between a hydrogen electrode (a platinum electrode coated with platinum black and saturated with hj'^drogen gas) dipping into a solution which is normal with respect to ions and a hydrogen electrode dipping into the solution whose hydrogen-ion concentration is desired. (See Fig. 10 on page 88.) The formula, a modification of the Nernst equation, which relates the electromotive force of the cell with the hydrogen-ion con- centration, is e.m.f. = 0.059 log Ch* THE THEORY OF NEUTRALIZATION 63 in which 0.059 is a constant, 1 tlie concentration of ions in the nonnal (standard or reference) electrode and the concentration of H"*" ions in the second electrode. By measuring the e.m.f., the factor, log 1 /Ch% can be calculated. The term pH is merely a substitution for the factor, log 1 /Ch*, and accordingly is defined as the logarithm of the reciprocal of the hydrogen-ion concentration. Thus, if the is 1 X 10“® the pH is 5, or if (7 h+ is 1 X the pH is 10.3G. The calculation of the pH value from a given value of Cu* is a simple matter. If, for example, a certain acidic solution is known to have a Cu* of 1 X lO”"*, the conversion to pH is based on the relationship (r^) = log 1 - log (10”^) = 0 - (-4) = 4 Note that the pH is merely the exponential value of the hydrogen-ion concentration with the negative sign removed. This, indeed, is a fortu- nate mathematical relationship, converting the hydrogen-ion concen- tration to the pH scale by simply expressing the exponent of the loga- rithmic relation as a positive number. Another example: What is the pH value of a solution whose hydrogen- ion concentration is 1 X 10“®-^^? pii == log Ixlr^ = 0 - (-8.75) = 8.75 The calculation is somewhat more complicated when the value of Ch* is expressed as a mixed number, as in the following example. Calculate the pH of a solution having a of 2.6 X 10“^. pH = log = log 1 6 X 10 = log 1 — log 2.6 X 10 ^ 0 - (log 2.6 X 10-*) 64 NEUTRALIZATION METHODS It now becomes necessaiy to convert the coefficient 2.6 into a logarithm, ir* )rder to incoiporate it as part of the exponent. The logarithm of 2.6 L 4-0.41. Therefore 2.6 X 10-^ = 1.0 X 10“^+'^'^^ = 1.0 X 10"^ ®® Then pH = 0 - Gog lO-^*®®) = 0 - (-3.59) - 3.59 In a manner analogous to the designation of Ch* values in terms of pH scale, we might express hydroxyl-ion concentrations on a similar logarithmic scale. That is pOH = log Loh“ It is to be noted in particular that, as the numerical value of pH in- creases, the H“^-ion concentration decreases. pH values up to 7 desig- TABLE II Solution Percentage Ionization pH CoH" loathci 79.6 1 X lO-O.i 0.1 1 X 10-13.9 0.1 athci 94.8 1 X 10-1.02 1.02 1 X 10-12.98 0.01 N HCl 99.8 1 X 10-2 2.0 1 X 10-12 0.0001 N HCl 100.0 1 X 10-4 4.0 1 X 10-10 0.000001 N HCl 100,0 1 X 10"® 6.0 1 X 10-8 HOH (pure water) 0.062 1 X 10-7 7.0 1 X 10-7 0.000001 N NaOH 100.0 1 X 10-8 8.0 1 X 10-® 0.0001 N NaOH 100.0 1 X 10-10 10.0 1 X 10-^ 0.01 N NaOH 99.5 1 X 10-12 12,0 1 X 10-2 0.1 N NaOH 92.9 1 X 10-12.97 12.97 1 X 10-1.03 1.0 AT NaOH 76.6 1 X 10-13.88 * 13.88 1 X 10-0.12 nate solutions which are acidic; a pH of 7 is an exactly neutral solution; and pH values greater than 7 refer to basic solutions, since in solutions which are basic the hydrogen-ion concentration must have some value less than 10”^. THE THEORY OF NEUTRALIZATION 65 Table IT gives the hydrogen- and hydroxyl-ion concentrations and the pH values for HCl and NaOli solutions of the specified concentrations, calculated from the relation X Cou- = 1 X 10-'" The existing (or actual) hydrogen-ion concentration of a solution, ex- pressed either in terms of Cn* or in tenns of the pll value, must not be TABLE III Ionization Constants and Pkrcentaqe Ionization Electrolyte Equilibrium Ratio ■^lon Percentage Ionization for 0.1 A'^ Solutions HCl Cii^ X Ccr Ciici ( 1 . 0 ) 94.8 HNO3 Cu* X Cnoj- -h Ciinos ( 1 . 0 ) 92.0 H2SO4 Ch* X Cnso 4 ~ Cn^soi ( 1 . 0 ) 90.0 Cii* X Csor CiisOi" (0 or>) GO 0 H2C2O4 Cr* X CiiCoOi" -5- ChsCjOi 0 0.18 40 0 Ch* X Cc^or ChcjO*" 0.00005 1.0 H3PO4 Ch* CH2PO4* ^ CH3PO4 0.01 27.0 Ch* X CHPO 4 " CitjP04" 0.062 0.1 Ch* X Cpo*" CHpOi" 0 . 0 i 2 ‘l 0.0001 HNO2 ' Cb* X CNO2" Chnoj 0.0005 ?.o HC2H3O2 Ch* X CcjUsOa" - 5 - Cnciinoa 0.000018 1.34 H2CO3 Ch* X Cucoa" ChjCO, O.OeS 0.12 Ch* X Ccoj” -j- Chcos" O.O 106 0.0017 HCN Ch* X Ccn" Chcn O.O 97 0.01 H2S Ch* X Chs" ^ Cejs O.OyO 0.05 Ch* X Cs" ^ Chs" O.O 14 I 0.0001 (Ch *)" X Cs- O.O 22 II NaOH Cns* X CoH" CNaOH ( 1 . 0 ) 92.9 Ca(OH)2 Ccii** X (CoH')* -i- Cca(OH)j 0.03 75.0 NH4OH Cnh 4 * X CoH" -T- CNH 4 OH + Cnh, 0.000018 1.31 NaCjHsOz 77.9 NH4CI 85.3 NH4C2H3O2 80.0 confused with the total available acid sterngth of the solution. In other words, the H'^-ion concentration which may be present at any moment depends upon the concentration of the solution and its degree of ioniza- tion, whereas the total amount, which becomes available during neu- tralization, is a measure of the total acidity of the solution. Thus, a liter 66 NEUTRALIZATION METHODS of normal HCI and a liter of normal HC2H3O2 both have the same avail- able amount of hydrogen ion (1.008 grams or 1 gram-ion), since both require an equivalent amount of a base for neutralization, but the actual hydrogen-ion concentration of a normal solution of HCI is nearly one huvidred times greater than the hydrogen-ion concentration of a normal HC2H3O2 solution, since the former is a strongly ionized acid and the latter a weak acid in which the H'^-ion concentration is low. In a titration for the determination of total acidity it is of little consequence whether the acid being titrated is a strong or weak acid, in so far as total hydrogen-ion concentration is concerned, since as hydrogen ions are removed more of the non-ionized molecules will dissociate. How such a titration is conducted is, however, of great importance. Table III gives the degree of ionization and the ionization constants of a number of acids, bases and salts at room temperature. NEUTRALIZATION AND HYDROLYSIS Since the fundamental reaction which takes place when acids and bases interact is the union of H"*“ ions and OH“ ions to form water, neutralization means that the water equilibrium is involved, the value of Kyf must be satisfied and equilibrium, as expressed by the relation Ch^ X CoH- = = 1.2 X 10-'^ must be maintained. When equivalent quantities of an acid and base are mixed the final equilibrium is not necessarily at neutrality. Indeed, it is only for those pairs of acids and bases for which the extent of ioniza- tion is the same that final equilibrium comes at neutrality (pH of 7); all other pairs, when reacting in equivalent amounts, reach final equi- librium in the mixed solutions in either the acidic or basic region, that is, at pH values smaller or greater than 7. The final equilibrium point reached in an acid-alkali titration depends upon the nature of the salt formed in the reaction, and this, in turn, depends upon the degree of ionization of the acid and base interacting. It is highly important to know at what pH value final equiUbrium is reached, when equivalent quantities of reacting acids and bases are brought together in a titration in order that the proper indicator be selected. In order to understand why equilibrium at the equivalence point in a titration does not always come at strict neutrality, the possibility and extent of hydrolysis of the salt resulting from the reaction must be con- sidered. Hydrolysis may be defined as the reaction of the ions of water with the ions of salts which are formed from weakly ionized acids or NEUTRALIZATION AND HYDROLYSIS 67 bases. A hydrolysis reaction is the reverse of a neutralization reaction; in fact, in the reversible reaction between an acid and a base it is the interaction of the products of tliis reaction. To illustrate, when HC2H3O2 reacts ^^^til NH4OH HC2II3O2 + NILiOH non + NII4C2H3O2 the ions of water react with the ions of NPI4C2H302 PIOH + NH4C2H3O2 ^ HC2II3O2 4 - NI-I4OII to refonn certain definite amounts of HC2II3O2 and NH4OH. If more non-ionized HC2H3O2 should foiTn than non-ionized NH4OPI, more ions are used up than Oil” ions, and the solution would then be basic. In this particular case, as shown later, the amounts of both happen to be almost the same, resulting in a solution w^hich is practically neutral. Hydrolysis is the reaction of PI"'' ions or OH” ions with the anions or cations of \veakly ionized acids or bases. It can be regarded as the reverse of a neutralization reaction, since the ions of the salt produced react with hydrogen ci hydroxyl ions to reform the original acid or base. The degree to wliicb *oe acid and base ionize, that is, whether they are weak or strong electrolytes, determines the extent to which hydrolysis will take place. The behavior, point of equilibrium and titration require- ments of four typical pairs of acids and bases are discussed below. Potentiometric and Indicator Titrations, There are, as pointed oizt on page 31, two methods by which a titration can be conducted with respect to the location of the end point. Both methods can be applied to acidimetric and alkalimetric, as well as oxidation and reduction and precipitation reactions. The older and common method is to use indi- cators which give a shaip color change when the equilibrium point of the reaction is reached, and the selection of the correct indicator depends upon know'ing this equilibrium point. In acid-base titrations, w^hen indicators are used, the nature of the acid or base being titrated fur- nishes the guide for the selection of the indicator. A potentiometric titration (see page 87) consists in measuring the changes in the electro- motive force of an electrolytic cell made up of the combination of a standard reference electrode and another electrode which is put into the solution being neutralized. As the reaction gradually runs to comple- tion by the addition of measured quantities of the standard reagent, there will be corresponding changes in the voltage of the cell. In the case of an acid-base titration, a potentiometric titration consists, in fact, of a series of hydrogen-ion concentration measurements, as described on page 88, so that, when titrating an acid by a standard solution of base, the hydrogen-ion concentration gradually and then rapidly decreases. 68 NEUTRALIZATION METHODS The change in the voltage of the cell is so rapid in the vicinity of the equilibrium point that an experienced analyst judges the end point by the abnormally large deflection of the voltmeter needle of the potenti- ometer. By actually recording the voltages, as the neutralization pro- gresses and from the Nemst equation for the hydrogen electrode, e.m.f. = 0.059 log — — Ch'*’ the hydrogen-ion concentrations or, better still, the pH values directly can be calculated from the relation e.m.f. 1 0.059 "" A series of pH values is thus obtained. When these values are plotted on coordinate paper against added volumes of standard reagent, a graph is obtained similar to the ones illustrated in a later section. Such graphs are known as titration curves. The equivalence point and therefore the end point can then be more accurately located from the curve. The vertical portion of such a curve gives the range over which the hydrogen- ion concentration or pH is undergoing the most rapid change, and the bisection point of the vertical portion (the inflection point) locates the definite equilibrium point and is taken as the end point. Curves which show graphically the changes taking place during the addition of successive volumes of reagent can also be plotted by calcu- lation from proper data. Curves of this kind are of considerable value in following the course of a reaction and in finding the end point in an actual titration. Such calculated curves, wliich are followed more or less closely in a potentiometric titration and give a general idea of the changes in an indicator titration, are shown for typical combinations of acids and bases. Reaction between Strong Acids and Strong Bases. The first case to consider is the neutralization of a strong acid, such as HCl, by a strong base such as NaOH. Suppose that 1 gram-mole of HCl is allowed to react with 1 gram-mole of the NaOH. According to the reaction HCl + NaOH = NaCl + H 2 O 1 gram-mole of NaCl and 1 of HoO are produced. The equation makes it appear that the reaction has gone absolutely to completion and, if it were possible to ignore the effect of the solvent, such would be the case. But we know that the water used as the solvent causes greater or less NEUTRALIZATION AND HYDROLYSIS 69 ionization of all dissolved acids, bases and salts and results in a certain amount of hydrolysis. Then, since the neutralization is carried out in an aqueous solution, the firet matter to decide is to what extent the reaction will be completed in aqueous solution. If the HCl and NaOH solutions are mixed in such a way that the final volume is exactl 3 ' 1 liter the completeness of the reaction is deter- mined by the water equilibrium: Cu* X Coir = 1-2 X 10~^‘*. From this it is evident that there will remain in solution concentrations of hydrogen and lo'droxyl ions such that their product is equal to 1.2 X 10"*'. Since the HCl, NaOH and the NaCl produced arc all highly ionized and dissociate to about the same extent, tliere will be little or no tendency for the sodium ions to recombine with hydroxyl ions to form NaOH, nor for hj'^drogen ions to recombine with chloride ions to fonn HCl. The equilibrium concentrations, therefore, of II ions and OH“ ions remaining to satisfy the water constant arc in equal amount. Their concentrations are 1 X 10”^ for H"*" ion and 1 X for OH“ ion, and the resulting solution is neutral. The equilibrium point comes at the exact neutrality point, and the reaction stops when there still remain in solution 1 X 10“^ gram-ion of hydrogen and 1 X 10“^ gram- ion of hydroxjd. We have then, to sum up the conditions when equilibrium is estab- lished, for the neutralization of HCl b^' NaOH, the e(|uilibrium eejuations H+ -i- Cr H- Na+ H- OH" NaCl + HOH i It Na+ + cr -f- H+ + oir ( 10 - 7 ) ( 10 - 7 > In general, the equilibrium point is reached for the interaction of a strongly ionized acid and a strongly ionized base at the strict neutrality point where the concentration of both ions is 1 X 10“^. Hydrolysis is negligible with this pair. The Titration Curve of Strong Acid and Strong Base, (a) Ideal Case. We ^dll consider first an ideal case, the titration of 25 ml. of a 0.1 N solution of HCl by a 0.1 N solution of NaOH. At the outset, before any base is added, the normality of the solution is 0.1. Assuming complete ionization of this strong acid, the Ch is 1 X 10“* and therefore the pH is 1. The equivalent volume of 0.1 N NaOH is 25 ml. At the equivalence point strict neutrality will have been reached and the pH, as already shown, vn\\ be 7. The course of the titration can be followed b>’^ calculating the pH values of the resulting solution as successive increments of the base are 70 NEUTRALIZATION METHODS added. Thus, when 5 ml. of the base have been added, there remain 20 Lu. of unneutralized acid in a total volume of 30 ml. The nor- mality, originally 0.1 N, will, however, have changed in the total vol- ume of solution. The new normality is found from the inverse volume relationship ViNi = V2N2 20 ml. X 0.1 iV = 30 ml. X a: iV from which the normality is found to be 0.066. This corresponds to a Cn* of 6.6 X 10“^ and a pH of 1.18, In this manner, the pH values shown in Table IV have been calculated. TABLE IV The Neutralization op a Strong Acn> by a Strong Base MI. NaOH i Added i i Ml. HCI Remaining Total Volume of Solution Normality of 1 Solution Ch* « pH 0 25.0 25.0 0.100 1 X 10-^ 1. 6.0 20.0 30.0 0.066 6,6 X 10“2 1.18 12.5 12.5 37.5 0.033 3.3 X 10-2 1.48 20 . 5.0 45.0 0.011 1.1 X 10-2 1.96 22.5 2.5 47.5 0.0053 5.3 X 10-2 2.28 24.0 1.0 49.0 0.0020 2.0 X 10-2 1 2.70 24.5 0.5 49.5 0.0010 1.0X10“’ 3.00 24.9 0.1 49.9 0.0002 2.0 X 10-< 3.70 24.95 0.05 49.95 0.0001 1.0 X 10-< 4.00 25.0 0.0 50.0 0.0 1.0 X 10“’ 7.00 25.05 -0.05 60.05 0.0001 (NaOH) 1.2 X 10““ 9.92 26.1 -0.1 60.1 0.0002 6.0 X 10-11 10.22 26.0 -1.0 61.0 1 0.002 6.0 X 10-“ 11.22 Note that when one-half of the acid has been neutralized, in the resultr ing 37.5 ml. of solution, the pH is 1.48. When 25.95 ml. of base have been added and only a single drop of HCl remains, the pH is 4.00. At equivalence it is 7. Beyond this excess base is present, but a single drop, 0.05 ml., establishes the pH at 9.92, in the basic region. In the graph in Fig. 7 such pH values have been plotted. The most- significant thing about this curve is the wide range, from pH of 4 to pH of 10, over which the curve is practically coincident with the vertical axis representing the equivalent volume. A very small drop, about NEUTRALIZATION AND HYDROLYSIS 71 0.05 ml., causes an enormous decrease in the hydrogen-ion concentra- tion (a large increase in the pH) in the vicinity of the neutrality point. This has an important bearing on the selection of suitable indicators for such a titration. (6) PnACTiCAL Case. In a practical, rather than ideal, titration such as in the laboratory procedure for comparison of solutions, the two Fig. 7. Titration curve of 0.1 N HCl by 0.1 N NaOII. solutions are rarely of the same strength and adjustment must therefore be made in the calculation of pH values used in the plotting of the graph. This somewhat complicates the computations. As an illustration, sup- pose that in making the comparison titration of acid against base it was found that 25 ml. of HCl required 30 ml. of NaOH and that, upon standardization, the HCl solution was shown to be 0.1234 iV. At the point, for example, when 5 ml. of the acid will have been neutralized, the volume of base added is not 5 ml. as in the ideal case but 30/25 or 1.20 X 5 ml. = 6.0 ml. The total volume of solution is then 31.0 ml. of which 20.0 ml. are unneutralized acid of oiiginal nonnality 0.1234. The new normality is then found from the inverse volume relationship (PiiYi = V 2 N 2 ) and from this the Ch* and finally the pH value can be found.' 72 NEUTRALIZATION METHODS As a practical assignment, the student should calculate a series of pH values, for the titration of acid by base, on the basis of one of his com- parison runs, using the normalities of acid and base as established by standardization. The pH values should then be plotted and the graph submitted. Reactions between Weak Acids and Strong Bases. The extent to which the neutralization reaction for this pair takes place depends upon the extent to which the reverse reaction of hydrolysis takes place. Let us consider the neutralization of acetic acid, HC2H3O2, by NaOH. When an exact equivalent of NaOH has been added to the HC2H3O2 present, the reaction would be complete if we ignored the water equi- librium and the hydrolysis effect. The neutralization of HC2H3O2 by NaOH according to the equation HC2H3O2 + NaOH NaC2H302 + H2O leads to the foimation of NaC2H302, a salt which is highly ionized and which undergoes a certain amount of hydrolysis. The hj'^drolysis is dependent in this case upon the feeble ionization of acetic acid. Since acetic acid is a weakly ionized electrolyte, hydrogen ions will be removed from the solution, leaving an excess of hydroxyl ions and resulting in a solution which is basic and not neutral. Equi- librium is therefore reached after the true neutrality point is passed, when base is added to acid. Titration Curve of Weak Acid and Strong Base, (a) Ideal Case. The calculation of pH values, when, for example, 25 ml. of 0.1 AT HC2H3O2 are titrated with 0.1 N NaOH, is complicated by the fact that the presence of NaC2H302 formed in the reaction acts by common-ion effect and, as previously shown, the equivalence point is displaced from neutrality because of hydrol^^sis. The initial pH can be directly calculated from the normality of the solution and its degree of ionization: Ch* = 0.1 X 0.0134 = 1.34 X 10“^ from which pH = 2.9. Or, if the degree of ionization is not known, Cr* can be obtained from the Ostwald dilution formula^substituting N for M K- 1.8 X 10~^ (neglecting the small amoimt of HC2H3O2 which ionizes) MaX Ma _ A/(l - a) “ 0.1 = NEUTRALIZATION AND HYDROLYSIS 73 Whence a, the degree of ionization multiplied by M (or the noimal- ity), gives Ch* = 1.3 X 10~3 and pH = 2.9 When an equivalent volume of base has been added (25.00 ml. of 0.1 NaOH), equilibrium will be reached. The pH of the solution at the equivalence point is calculated in the following manner, llie hy- drolysis is dependent upon the amount of NaC2H302 formed; it is there- fore simpler to arrive at the value by considering the hydrolysis of a NaC2H302 solution. In an aqueous solution of NaC2H302 final equi- librium depends not only on satisfying the water equilibrium Cn* X Cou- = 1.2 X but also on the acetic acid equilibrium which is controlled by its ioniza- tion constant: Cn* X Cc,niOt~ ^hc2H|0} 1.8 X 10“® Tlie hydrolysis equilibrium is then jointly controlled by the two equi- libria and if we divide the former by the latter Cn* X CoH~ X CcsUiOi" C'hCjHjO, 1.2 X 10~^^ 1.8 X 10“® we get the hydrolysis expression g 0 H _ X Chchk). ^ g 7 ^ CcjHiOj' where 6.7 X 10“^® is the hydrolysis constant for NaC2H302 , Cqh" is the hydroxjd-ion concentration at equilibrium, ChcjH ,02 is the concentra- tion of acetic acid formed in the hydrolysis and CcjHjO," is the concen- tration of the ion of the salt formed. The value of Con* equals that of ChcjHjOj because for every ion of hydrogen combined to form HC2H3O2 a hydroxyl ion is left in excess. If in a titration we add 25 ml. of 0.1 NaOH to 25 ml. of 0.1 TV HC2H3O2 they vnll have reacted in equivalent proportions. The amount of NaC2H302 would be 0.1 normal, but since the volume of the mixture is now 50 ml. the concentration of the salt, and hence that of the C2H3O2- ion, is 0.05 normal (or molar). Letting x stand for the OH“ concen- NEUTRALIZATION METHODS 74 tration, as well as the acetic acid concentration formed by hydrolysis, we have xXx (0.05 - x) - = 6.7 X 10 —10 As an approximation, since the amount of C2H302" ion combined is negligibly small, we can write = 6.7 X 10-*° 0.05 X = 5.8 X 10 -« Since Cq* = K W CoH“ (from the water equilibrium) 1.2 X 10 —14 5.8 X 10 “6 = 2.07 X 10 —9 corresponding to a pH of 8.68, or approximately 8.7. The steps in this calculation can be shortened, for since Cqh~ X CHCaHjOi _ C'c*H|Oa“ -^ion and C'hcjHjO, is equal to Cqh" (Coh)" = 2 ^ X K. ton From the water equilibrium, CoH" = K w Cr and solving for Ch% Cs K^XK ion Cc,H,Oi" /l.2 X 10“^^ X 1.8 X ■ v 10 -8 2.07 X 10-® corresponding, as before, to a pH of 8.68. When 5 ml. of 0.1 N NaOH are added the total volume is 30 ml., of which 20 ml. are unneutralized 0.1 iV acetic acid. The new normality of the HC2H3O2 is, as in the case of HCl (see page 70), therefore 0.066. NEUTRALIZATION AND HYDROLYSIS 75 The C2H302“-ion concentration corresponds to 5 ml. of O.l N NaC2H302 in 30 ml. or, from the inverse relation KiA^i = F2A^2» the normality of the salt is 0.0167. Substituting these values in the equation C'HCtniO, X Kjon Cc,H,or 0.066 0.0166 X 1.8 X 10“* = 7.22 X 10'® It will be noted that the ratio 0.066/0.166, which represents the concen- tration of acetic acid and acetate ion, respectively, is the same as the Fia. 8 . Titration curve of 0.1 N HC 2 H 3 O 2 by 0.1 N NaOII. ratio of the volumes of 0.1 solutions, namely 20 ml. of acid and 5 ml. of NaC2H302. That is 0.066 20 ml. 0.0166 ^ ~5 ml. This simplifies the calculation, so that here, as well as at all other desired pH values, we can substitute volumes for normalities directly. $ 76 NEUTRALIZATION METHODS For all points up to and including the addition of 24.95 ml, of 0.1 iV NaOH, the following formula can be used: Volume of acid remaining Volume of NaOH added X 1.8 X lO-® Values of pH calculated in this way are tabulated in Table V and shown graphically in Fig. 8. TABLE V Neutralization op a Weak Acm by a Strong Base MI. NaOH Added Ml. HC 2 H 3 O 2 Remaining Total Volume of Solution Cu* 1 pH 0 25.0 25.0 1.3 X 10"® 2.9 6.0 20.0 30.0 7.2 X 10"® 4.1 12.5 12.5 37.5 1.8 X 10"® 4.7 20.0 5.0 45.0 4.6 X 10"® 5.4 22.5 2.5 47.5 2.0 X 10"® 5,7 24.0 1.0 49.0 7.5 X 10"^ 6.1 24.5 0.5 49.5 3.7 X 10"’ 6.4 24.9 0.1 49.9 7.4 X 10"* 7.1 24.95 0.05 49.95 8.6 X 10-« 7.4 25.0 0.0 50,0 2.07 X 10-® 8.7 25.05 -0.05 50.05 1.2 X lO-i" 9.9 25.1 -0.1 60.1 6.0 X 10-“ 26.0 -1.0 61.0 6.0 X 10-“ 11,2 (b) Practical Case. As an assignment it is suggested that the stu- dent either prepare a small amount of approximately 0.1 V acetic acid, titrating 25 ml. of it with standard NaOH or else that similar data be furnished him, from which he can calculate a series of pH values and construct the graph. Reactions between Strong Acids and Weak Bases. A third type of neutralization reaction is that between a strong acid and a weak base, such as, for example, that between HCl and NH4OH. Here the salt formed is NH4CI, according to the equation HCl + NH4OH NH4CI + H2O The extent to which the ions of this salt ^vill react with the ions of water depends upon the degree of ionization of NH4OH, the weakly ionized base concerned in the equilibrium. Since some hydro3^1 ions are used 4. » NEUTRALIZATION AND HYDROLYSIS 77 up in the formation of the salt of the weak base, an excess of hydrogen ions remains, giving a solution wliich at equilibrium is acidic. In fact, a calculation from the hydrolysis constant will show equilibrium to be reached for a 0.1 solution of ammonium chloride at a pll value of about 5.24. Titration Curve of Strong Acid and Weak Base. Tlie titration of a strong acid by a standard weak base Ls never carried out in practice, but the reverse, the titration of weak bases, such as NH4OII by standard HCl, is of frequent occurrence, llie course of the reaction is analogous to that just considered for HC 2 H 3 O 2 and NaOH, but the pH values are reversed. Let us consider the titration of 25 ml. of 0.1 NH.iOH with 0.1 AT HCl. The pH of the original solution may be found either from the degree of ionization or from the ionization con.stant. In the first method, the solution being 0.1 N and the degree of ionization being 1.31 per cent, the CoH- is 0.1 X 0.0131 = 0.00131 or 1.31 X 10“^ Therefore and 1.2 X 10“*** 1.31 X 10“^ = 1 X 10““°^ pH = 11.04 In the latter case, using the equilibrium ^ 1 ^ 4 * X CoH' C'nh40h = K, ion Therefore, CoH" = VA~XCNni^ = 1.32 X 10 ^ K W CoR- and pH = log 1 Cb pH = 11.04 For values of pH, up to the addition of about 24.95 ml. of 0.1 iV HCl, the normality of the remaining base and the normality of the NH4CI formed may be evaluated or, alternatively, as in the case of acetic acid, the ratio of volumes may be taken. By combining = Ch X CoH" K- = we have Cnh 4 ^ X CoH~ CNH4OH ion 78 NEUTRALIZATION METHODS For example, when 5 ml. of acid have been added there remain 20 ml. of unneutralized NH4OH and there were formed 5 ml. of NHiCl. r * = 1 V X 20 ^ 1.75 X 10-® = 1.7 X 10-‘® Therefore pH = 9.77 Calculation of other points is left for the student. Points beyond the equivalent volume, when an excess of acid is pres- ent, have the same pH values as already calculated, on page 70, for the HCI-NaOH titration. The pH value attained at equivalence is calculated, from the hydrol- ysis formula, in a manner analogous to that for NaC2H302. Here the water equilibrium and the ionization constant of NH4OH are the controlling factors. Whence Cjl* X CoH“ _ Kw X CoH~ -f^ion CNH40H Cu* X Cnh 40H C'nh** 1.2 X 10~^^ 1.75 X 10"® = 6.9 X As before, if 25 ml. of 0.1 HCI react with 25 ml. of 0.1 AT NH4OH, we have at equivalence a 0.05 N solution of NH4CI. Letting x eq\ial Ch* as well as CnhiOh, = 6.9 X lO-'o 0.05 Ch^ = 5.9 X 10"® Instead of using the value of the hydrolysis constant, equations can be combined to read Ch X Cnhi* Kion and we have Ch* 1.2 X 10~^^ X 5 X 10"^ 1.75 X 10“® - 5.9 X 10“® Thb corresponds to a pH of 5.24, NEUTRALIZATION AND HYDROLYSIS 79 The graph in Fig. 9 shows the course of the neutralization of NIT4OH by HCI. Reactions between Weak Acids and Weak Bases. The fourth case may be illustrated by the follo\ving reaction : HC2H3O2 + NH4OH ;:± NH4C2H3O2 + HOIi in which the hydrolysis reaction, that is, the reaction from right to left, is to be considered. In tliis case OH” ions react witli ions from Fig. 9. Titration curve of 0.1 N NH 4 OH by 0.1 N HCI. the NH4C2H3O2 to form the weakly ionized base and, at the same time, H+ ions react with C2H3O2 ions to produce the weakly ionized acid. For final equilibrium the ionization constants for both NH4OH and HC2H3O2 must be satisfied as well as the relation Ch* X CoH- = 1.2 X 10 “'^^ Since both acid and base are weakly ionized, hydrogen ions as well as hydroxyl ions are removed to form the acid and base. In this particular case it happens that HC2H3O2 and NH4OH are ionized to almost exactly the same extent. Consequently almost the same quantities of hydrogen 80 NEUTRALIZATION METHODS and hydroxyl ions are removed, resulting in a solution which is almost exactly neutral. The course of the reaction presents some interesting characteristics but, in practice, such a titration is not made, because no indicator can be employed for the titration, nor can the end point be found in a potentiometric titration. If we take, for example, the reaction between 0.1 iV* HC 2 H 3 O 2 and 0. 1 NH 4 OH the following characteristics are to be observed. 1. If the acid is being neutralized by the addition of base, the initial pH of the solution, as already computed on page 73, will be 2.9, and, conversely, if base is being neutralized by acid, the initial pH of a 0.1 iV NH 4 OH solution is 11.04. 2. pH values for points up to practically the equivalent point are identical with those already calculated or discussed for either the acetic acid or the ammonium hydroxide titrations as the case may be. 3. The equivalence point, since both acid and base are about equally ionized, will have a pH of practically 7. In cases where both reactants are weakly ionized, but one is somewhat stronger than the other, the Ch* can be calculated from the formula derived below. The ionization equilibrium for the weak acid is and for the base Cn* X Cc^HaOt" CoH' X CnH4^ C'nh40h — ^aoid = K base and for water Therefore Ch* X CoH” = ^ W Ch+ X CoH" K W C'h'*’ X ^^ 211802 ” ^ ^OH” X -^acid X -f^bEise X CaCiUiOi C'nh 40 H Theexpression just given is the hydrolysis equilibrium for NH4C2H3O2. In its simpler form it is K CuCiHgOi X CNH 4 OH _ X C^NH 4 * ^acid X iCbasa Since the salt is a strong electrolyte hyd ( 1 ) NEUTRALIZATION AND HYDROLYSIS 81 and assuming practically equal concentrations of acid and base formed in hydrolysis CmUOU = CuCaUjOa From equation 1 therefore and therefore or CnCaTliOa _ CxinOH C’cjUiOa" (CnCatliOa) (CcjIIaOa")' K w acid XK baso CnCatljOj = C'CjHjOa" ^j- K W ^ocid X K base The hydrolysis equation for the salt of a weak acid as given on page 74 is CnCsiijOi X Con- Kvr CcjIljOa' K * & A /«« acid from which CHCaHaOa = ^acid X Lou" ( 2 ) Equating 1 and 2 since both equal CncaHjOa^ Kxv X CcaHaOa" •f^acid X C^OH” = Cc. IlaOj •*^acid K W A' X K baso (i^w) Solving for Coh” (/Cacld)^ X (CoH-)" A’acid X Con Kvf '^^bas© •^ocid acid since Cqh” = K w Cn therefore Ch* =v^ X K acid K base Fi'om this equation the hydrogen-ion concentration and the pH value at equivalence can be calculated, provided that the ionization constants 82 NEUTRALIZATION METHODS for the acid and base do not differ too much. In the case of NH4C2H3O2, the acid and base constants are substantially 1.8 X 10~®, from which it follows that the pH at equivalence is practically 7. Titration of Sodium Carbonate Solutions. If a salt of a strong base and a weak acid or a salt of a weak base and a strong acid is present in a solution, it is possible to titrate such a solution with a standard acid or base. Examples of such salts are sodium carbonate, borax and aluminum sulfate. The case of sodium carbonate is important since this salt is used for the standardization of HCl and enters largely in deter- minations of alkalinity. Sodium carbonate is the salt of the strong base, NaOH, and the weak acid, H2CO3. It \vill therefore hydrolyze to a considerable extent, giving a solution which is basic: 2Na+ + COa^ + 2HOH 2Na+ + 20H- + H2CO3 due to the repression of H"*" ions in forming the slightly ionized carbonic acid. When acid such as HCl is added, the free OH” ions are neu- tralized. When one equivalent of HCl is added, the bicarbonate is formed; when two equivalents are added the reaction is practically complete : NasCOa + HCl = NaHCOa + NaCl (1) and NaHCOa + HCl = H2CO3 + NaCl (2) There are two equilibrium constants which must be satisfied, namely, Cu* X CoH" = and the ionization constant for H2CO3: (Ch-)" X Ccor OhjCOi = K; ion The weakly ionized acid, H2CO3, is also unstable and readily decom- poses into CO2 and H2O, most of the CO2 escaping as a gas and only enough remaining to maintain the equilibrium CcOt X ChjO tr — ^ = Ah, CO, UHjCO, By carrying out the titration in a hot solution the amount of CO2, and therefore H2CO3, remaining is reduced to an extremely small amount. In a cold solution the titration curve of sodium carbonate shows two vertical portions, corresponding to two equilibrium points, the first when one equivalent of acid is added, showuig the forpiation of the bicai^ INDICATOR THEORY 83 bonate, and the second, a better-defined and sharper portion of the curve, coining when two equivalents of standard acid are added and showing the complete neutralization of tlie carbonate. This second point occurs in the region sensitive to methyl orange indicator. INDICATOR THEORY An indicator is a substance which must give a sudden color change in the region of the equilibrium point. That is, it must change its color at a pH value which is as near as possible to the equilibrium point reached in the neutralization. With the facts of hydiolysLs in mind and having considered the electrometric method by which this point can be experimentally determined for typical pairs of acids and bases, the selection of the proper indicator now becomes "a simple matter. TABLE VI Indicator List Indicator Transition Range in pH Color in Color in Acid — ♦ Base Methyl violet 0.1- 3.2 Yellow — » Green — » Blue Tropeolin 00 1.3- 3.2 Red — » Yellow Methyl yellow 2.9- 4.0 Red — ► Yellow Methyl orange 3.1- 4.4 Pink — * Yellow Congo red 3.0- 5.2 Blue-violet ^ Red Methyl red 4.2- 6.3 Red — » Yellow Brom-thymol blue 6.7- 7.6 Yellow — » Blue Azolitmin (litmus) 5 . 0— 8 . 0 Red — » Blue Kosolic acid 6.9- 8.0 Brown — » Red Phenol sulfonephthalcin 6.8- 8.4 Yellow — ♦ Red Naphtholphtbalein 7.3- 8.7 Rose — * Blue Phenolphthalein 8.2-10.0 Colorless — * Pink Thymolphthalein 9.3-10.5 Colorless — ♦ Blue Alizarine yellow 10.0-12.0 Yellow — ♦ Lilac Tropeolin 0 11.0-13.0 Yellow — * Orange A large number of organic compounds, chiefly dyestulYs and similar substances, have been found to show a marked color change over small ranges of hydrogen-ion concentration. These are the substances lused as indicators. Table VI gives a selected list of indicators, showing the range in pH values over which the indicator changes color. 84 NEUTRALIZATION METHODS The color ranges of these indicators have been plotted in the accom- panying chart. This is a handy arrangement for comparing pH ranges over which the indicator is sensitive with the pH ranges over which the equilibrium points occur as obtained from titration curves. J4 13 12 tl 10 ^9 c Z 6 CD a 7 oc 6 5 A 3 2 I 3 ^ ^ ^ ^ CJ 'ft I t § Q rv* ^ A. ^ “ '• I V) s •5 I ■III ^ ^ c> ‘S o t v. ? 5 5* ^ -bi C5 ;§ I 41 1 1 1 9 C 1 i _Jj 1 U □ JX f * 3““c 1 3 • 2-5— fjM ] w J 0 i a 5- ! S) -5 < ^ CQ c ? jj a X p li p - IS Si H c 1 n i rv id c^_- s 4 4 : c a — s ^ J J c 1 p 1 % 5 1 — 1 1 lJ_ m ri ^ *5? 1 •/?ed _ 1 1 _ 1 _2_ •5i "I •I V> I Neufra/ i .r i tb J5 Indicator Charf*, The neutralization curve for the neutralization of a strong acid by a strong base shows a rapid change in,pH from about 4 to 9, as shown in Fig. 7, the equivalent point being theoretically at the neutrality point (pH = 7). From the indicator chart it appears that brom-thymol blue, azolitmin, rosolic acid or phenol sulfonephthalein, which change color near the neutrality point, are ideal indicators to use. However, since the rate at which the hydrogen-ion concentration is changing is rapid in the regions just below and above the neutrality point, an indicator sensitive in either of those regions may be employed with little error. Thus methyl orange, which changes color at about pH = 4, if used will INDICATOR THEORY 85 pve an end point slightly before the neutrality point is reached; and phenolphthalein, whicli changes color at about pll = 9, will give an end point after the neutrality point is passed. If 0.1 solutions of the acid and base are used and 25 ml. of the acid are being titrated by the base, the end point for methyl orange occui's when 24.9 ml. of the alkali arc added; and with phenolplithalein as indicator the end point comes at 25.1 ml. of added base. For this pair, therefore, cither of these indicators may be used. For the combination of weak acid and strong base, for which the equilibrium point lies in the region of excess hydroxyl-ion concentra- tion, the titration curve has its vertical portion in tlie range; of pll most sensitive to such indicators as phenolphthalein and naphtholplithalein. Phenolphthalein Is the indicator commonly used here. For the combination of strong acid and weak base, Hie equilibrium point is coincident with the range over which methyl orange is sensitive, that is, in a solution with high hydrogen-ion and correspondingly low Iiydroxjd-ion concentration. Methyl orange is the indicator commonly used for this pair. Finally, a titration cun'e for the neutralization of a weak acid by a weak base ttill show no abnipt change, the changes in y^II being gradual throughout the entire range. There Ls no indicator suited for this pair, and such a titration is impossible. Neutralization is, of course, possible, but there is no wa}'^ of determining when equi\'alent amounts of the standard solution have been added to the unknown. If attempted as an experiment it will be found that with either of the two common indicators a gradual, shifting color change takes place which is of no value in finding the end point. To sum up the conditions, the tabulation below will be useful. 1 Combination Nature of Hydrolysis Region of Equilibrium Point 1 Indicator Strong acid -h strong base None (neutral) 1 pH = 4-9 M.O. or P.P. Weak acid 4- strong base Somewhat (basic) 7jH = 7-9 P.P. Strong acid + weak base Somewhat (acidic) pH = 4-6 M.O. Weak acid -j- weak base ' Somewhat (neutral) 1 None None Several theories have been put forward to explain the cause of the color change of indicators. Twn of these will be considered here. They are the ionization theory and the chromophoric theory. 86 NEUTRALIZATION METHODS Ionization Theory of Indicators. According to this theory, first advanced by Ostwald, the substances used as indicators behave either as very weak acids or as very weak bases. That is, they ionize to a slight degree, and the color observed in a solution containing excess hydrogen or hydroxyl is due either to the color of the non-ionized mole- cule or to the color of the cation or anion of the indicator. In the case of methyl orange, this indicator is a very weak base and can be considered as ionizing into OH“ ions and a cation which may be designated by the symbol Mo"^. In water or in a basic solution the concentration of the cation Mo"^ is extremely slight, and the yellow color of the solution is due to the non-ionized molecule. On the addition of an acid, a salt will be formed between the cations of the indicator and the anions of the acid, and this salt, ionizing, will increase the concentration of the cations Mo"*", which are pink in color. We have then: MoOH ^ Mo+ -h OH- (Non-ionized (Cation : _ molecule; yellow pink in acid in basic solutions) solutions) In the case of phenolphthalein, the indicator behaves as a very weakly ionized acid, ionizing into ions and an anion which may be desig- nated as Pp“. This indicator is colorless in neutral and acidic solutions and pink in basic solutions. If a base is added, salt formation takes place betiveen the cation of the base and the Pp“ anion, and this, by ionizing, increases the concentration of the anion and causes the solution to turn pink. The ionization of phenolphthalein can be shown thus: HPp H+ (Non-ionized molecule; colorless in acid solutions) + Pp- (Anion; pink in basic solutions) That this theory does not suffice to explain all the facts will be appar- ent when the chromophoric theory is presented. Chromophoric Theory of Indicators. Investigation into the structure of organic dyestuffs, of which these indicators are representatives, has shown that the production of color or change of color is due to the presence of certain groups or radicals within the molecule, called chromophores, which possess a distinctive color. The quinoid group, = is such a chromophoric group. A change of color is < therefore due to an internal rearrangement of the atoms of the molecule, and tautometric forms exist, one in basic solution, the other in acid solution. POTENTIOMETRIC TITRATION 87 For methyl orange, the two structures are: N II N BcDzoid structure (yellow) Quinold structure (piuk) For phenolphthalein, the tautometric forms are : OH Benxoid structure (colorless) Quinoid structure (pink) In short, on the basts of this explanation, the change in structure takes place, in the case of methyl orange, at a pH value of about 4, and for phenolphthalein at a pH of about 8. Methyl orange is thus most sensitive at low concentrations of OH“ ion and is commonly used for the titration of weak bases such as NH4OH. Phenolphthalein, on the other hand, is sensitive to low H"^-ion concen- trations and is used for the titration of weak acids such as HC2H3O2. POTENTIOMETRIC TITRATION Reference has already been made to a mode of titration in which the end point is indicated by a rapid change in the potential of the system, 88 NEUTRALIZATION METHODS rather than by the use of a chemical indicator. Such methods can be applied to acid-base titrations as well as to oxidation-reduction and to precipitation reactions. The potentiometric method for neutralization reactions is here briefly described. A potentiometric acid-base titration consists, in brief, of making a series of pH determinations as standard base is being added to the acid or vice versa. As the equivalent volume is being approached there will be progressive rapid change in the pH value, and on passing through the equivalence point the change in pH will be most rapid. Fia. 10. Potentiometric titration assembly. In practice, one uses a standard reference calomel electrode, which consists of mercurous chloride (Hg 2 Cl 2 ) paste in contact with metallic mercury and KCl solution. One form of the calomel electrode is shown in the assembly of the electrolytic cell, Fig. 10. This electrode consti- tutes one of the half-cells of the system. The other half-cell consists of the beaker wliich contains the solution to be titrated and into which a second electrode is placed. This second electrode may be either (1) a standard hydrogen electrode, (2) a quinhydrone electrode or (3) a glass electrode. Since the hydrogen electrode is subject to variations and diflicult to maintain, the preference, in practice, is given to the quin- hydrone or glass electrode. A quinhydrone electrode is made by saturating the solution to be titrated with quinhydrone and dipping into it an unplatinized platinum electrode. Quinhydrone is a mixture of quinone and hydroquinone and the potential of the electrode depends upon tlie hydrogen-ion concen- tration. This electrode has certain limitations and is being replaced by PROBLE^iS 89 the glass electrode, which consists of a glass membrane in contact with a silver-silver chloride electrode. It has already been shouTi that the Ncnist equation, when applied to the measurement of liydrogen-ion concentration, takes the form or E = 0.0591 log Cn^ E 0.0591 In the cell consisting of the normal calomel electrode, as reference, and the hydrogen electrode in the solution whose pH is to be deter- mined, the e.m.f. is E = 0.280 - 0.0591 log Cu* or E - 0.280 0.0591 the value 0.280 being the difference between a normal hydrogen and normal calomel electrode. In an actual titration, the cell is connected with a potentiometer and standard solution is titrated into the half-cell containing the solution of ^he sample. T.he voltages, Ey are noted as the titration progresses, the corresponding pH value calculated and plotted on graph paper. These titration curves are similar in shape to those shown previously and calculated from volume and normality relationship. The equiva- lence point is obtained by dropping a line from the vertical portion of the graph to the volume (.r) axis. PROBLEM SET 4 EQUILIBRIUM CALCULATIONS IN NEUTRALIZATION REACTIONS 1. Calculate the H-ion concentration in solutions of the following monobasic acids, the molarity, or normality and degree of ionization being given. (o) HCI 0.1 AI or 0.1 AT 94.8% (6) HNO 2 0.1 M or 0.1 N 8.0% (c) HCHO 0.1 M or 0.1 N 4.5% (d) HC 2 H 3 O 2 0.1 M or 0.1 N 1.34% (e) HCN 0.1 iUor 0.1 AT 0.01% Arts, (a) 9.48 X 10~2 (6) 8.0 X 10-3 (c) 4.5 X 10-3 id) 1.34 X 10-3 (e) 1.0 X 10-5 2. Calculate the H-ion concentration of a 0.01 solution of acetic acid which is 4.17 per cent ionized. What is the H-ion concentration of 0.01 N solution of HCl which is 99.8 per cent ionized? 90 NEUTRALIZATION METHODS 3. Calculate the OH -ion and H'''-ion concentration of the following solutions of bases, the normality and degree of ionization being given. = 1 X OH- H+ (a) NaOH 0.1 N 90.5% Ans. 9.05 X 10"!^ 1.11 X 10-“ (b) KOH 0.01 iV 93.5% 9.36 X 10-® 1.07 X 10-“ (c) NH4OH 0.001 AT 11.7% 1.17 X 10“* 8.66 X 10-“ 4. Calculate the OH -ion and H'*‘-ion concentrations in solutions of NH4OH of the following normalities and degrees of ionization. (Kw = 1 X 10- (a) 1.0 AT 0.4% (b) 0.1 AT 1.31% (c) 0.01 AT 4.07% (d) 0.001 N 11.7% 6. For the following solutions, the H"** -ion concentrations of which are ra^en express the acidity in terms of pH values. Solution Ch* pH (a) HCl 9.48 X 10-2 Ans. (a) 1.02 (b) HNO2 8.0 X 10“2 (b) 2.1 (c) HCHO 4.5 X 10“® (c) 2.35 id) HC2H3O2 1.34 X 10“2 id) 2,87 ie) HCN 1 X 10“® (c) 6.00 6. Find the pH values for the following bases having the H-ion concentration corresponding to the normalities given. (a) NaOH 1.11 X io-*2 0.1 AT (b) KOH 1.07 X 10-‘2 0.01 N (c) NH4OH 8.55 X 10“'^ 0.001 N 7. Suppose that you mixed 20.00 ml. of 0.1 N NaOH with 30.00 ml. of 0.1 N HCl, What would be (a) the H-ion concentration, and (6) the pH value of the resulting mixture, assuming complete ionization? Ans. (a) 2 X 10“* (6) 1.70 8. If to 25.00 ml. of 0.1 N NaOH you added. 12.50 ml. of 0.1 N HCl, what would be (a) the OH-ion concentration, (6) the H-ion concentration and (c) the pH value of the resulting mixture, assuming complete ionization? 9. Calculate (a) the H-ion concentration and (5) the pH value of a 0.01 N solution of NaC2H302 . (Ch* X CcjH* 02 ")/C'hC|HjO| = l-S X 10“®; Ch* X Coh“ = 1-2 X 10“'\ Atw. (a) 4.76 X 10“® (b) 8.34 10. Calculate the pH value of a 0.01 N solution of NH4CI. (CNH4 * X Cqh")/ CNH 4 OH = 1.76 X 10“®. Ch* X CoH- = 1.2 X 10““ 11. Calculate the pH value of the mixture when 25.00 ml. of 0.1 N HCl are mixed with 76.00 ml. of O.l N NH4 OH, (Cnh^^ X Coh-)/Cnh 40H = 1.75 X lO"*, Ch* X Cbn- - 1.2 X 10“'^ Ans. 9.48 12. What is the pH value of the mixture obtained by adding 60.00 ml. of 0.Q N NaOH to 60.00 ml. of 0.6 N HCjHsOa? {Cn* X Cc 4 H,o.“)/Cho»h»o» “ 1.8 X WT®, Ch* X CoH- « 1.2 X 10-1*. ADDITIONAL STOICHIOMETRIC CALCULATIONS 91 13. Suppose you were titrating 25.00 ml. of 0.1 N HC 2 H 3 O 2 with 0.1 N NaOH. (a) What w’ould be the pH value of the acid before titration was begun (degree of ionization 1.3 per cent)? ( 6 ) What would be the pH of the solution when one half of the equivalent of NaOH (12.50 ml.) had been added? (c) When 0.9 of the acid is neutralized? Atis. (o) 2.89 ( 6 ) 4.74 (c) 6.70 14. If you were titrating 60.00 ml. of 0.1 N HCl with 0.1 N NaOH, what would be the pH value (a) when 30.00 ml. of the NaOH solution were added, ( 6 ) when 47.50 ml. NaOH were added and (c) when 49.90 ml. NaOH were added? Assume complete ionization. 16. If, for convenience, we mLx solutions of 0.1 ^ NH 4 OH and 0.1 A HCl in such proportions that the total volume of solution is 100 ml., what will be the pH (a) when we mix 55.0 ml. acid and 45.0 ml. base; ( 6 ) when 50.0 ml. of each are mixed; and (c) when 40.0 ml. of acid are mixed with 60.0 of base? ifw = 1-2 X 10“*^; Aion NH 4 OH = 1.75 X 10-5 (qJ 4 q (5) 6.24 (c) 8.88 16. If we mix solutions of 0.1 N NaOH and O.l N H(^H 302 in the following pro- portions, what will be the pH values of the resulting mixtures? («) 65.0 ml. acid and 35.0 ml. base; (6) 50.0 ml. of each; and (c) 10.0 ml. of acid and 90.0 ml. of base. A'w = 1.2 X 10-^^ Aiod HC 2 H 3 O 2 = 1.8 X 10-5. 17. Calculate the pH of the mixture when 12.50 ml. of 0.1 N HCl have been added to 25.00 ml. of 0.1 N NH 4 OH. An.s-. 9.2 18. Imagine you are titrating 25.00 ml. of a 0.1 N NH 4 OH solution with 0.1 N HCl. What are the pH values when (a) 5 per cent, (5) 60 per cent, (c) 90 per cent and (d) 09.9 per cent of the base have been neutralized? 19. If you mix equivalent volumes of a base, whose Aiod is 1 X 10-5, with an acid, whose Aion is 1 X 10“^, what will be the H-ion concentration of the mixture? Ans. 1.1 X 10'® 20. Calculate the hydrolysis constant for NH4C2H3O2. ADDITIONAL STOICHIOMETRIC CALCULATIONS Adjustment of Solutions, (a) Adjustment to an Exact Normality. Solutions prepared and standardized for titration purposes are rarely of the exact strength desired. That is, they usually are of the approximate strength called for but, of course, their exact strength becomes known upon standardization and they are then used without further adjust>- ment of concentration. It sometimes becomes necessary or desirable to adjust the concentration to some definite desired value by addition of water or solute. The exceptions to this are those solutions which can be made by weighing out an exact amount of primary standard. For example, suppose that it is desired to change the strength of an HCl solution which is 0.1019 N to one which is exactly 0.1000 N. How much water must be added to 700 ml. of the 0.1019 N hydrochloric acid solution in order to have one of the desired normality? 92 NEUTRALIZATION METHODS The diluted solution must contain 0.003647 gram of HCl per^milliliter in order to be 0.1000 AT. The undduted solution contains 0.03647 X 0.1019 or 0.003716 gram per milliliter; in 700 ml. there are 0.003716 X 700 or 2.60 grams of HCl. This quantity of pure HCl must be present in the total volume of diluted solution such that each milliliter contains 0.003647 gram of HCl. If X represents the volume of diluted solution, then 0.003647 X X = 2.60 X = 713.3 ml. Then 713.3 — 700.0 = 13.3 ml., the amount of water to be added. Since the normality is inversely proportional to the volume, the volume of diluted solution, y, can also be directly calculated from the relation 700 ml. : y ml. = 0.1 : 0.1019 y = 713.3 ml. and the volume of water to add is 13.3 ml. Such calculations are made on the assumption that no shrinkage in the total volume takes place on mixing. (6) Adjustment of Solutions to Read Percentage Directly. In many routine and industrial analyses it is desirable to adjust either the weight of sample or the strength of the titrating solution so that the volume of standard solution will indicate directly the percentage of the desired constituent or bear a simple relation to the percentage. The problem resolves itself into either finding how much solute is needed in making a solution for titrating a definite weight of sample so that each milliliter of solution used in titrating the sample will represent 1 per cent of the desired constituent, or else fixing the weight of sample that must be taken so that vdih a specified strength of the solution the volume used will represent 1 per cent of the desired constituent. The following case will serve as an example. How much HCl must be contained in a milliliter of solution so that, with a 1-gram sample of impure Na 2 C 03 , each milliliter used in titi'ating the sample represents 1 per cent of Na 2 C 03 ? From the equation 1 ml. X Na 2 C 03 titer of the HCl solution we have 1 gram 1 X 0.01 X 100 = 1 per cent X 100 = 1 per cent 1 ADDITIONAL STOICHIOMETRIC CALCULATIONS 93 The value of 1 ml. of the solution in tcnns of Na2C03 is therefore 0.01. Then (2HCI/Na2C03) X 0.01 = value of 1 ml. in teiins of HCl and equals 0.00GS8 gram. A liter of this .solution must contain G.88 grams of HCl. If a 2-gram sample is used, and each milliliter is to indicate 1 per cent, the strength of the solution must be twice as great. Instead of adjusting the strength of the solution, which is a tedious matter experimentally, the weight of sample may be adjusted so that, with a solution of given strength, the volume of solution used represents directly the percentage of the desired con.stituent. What weight of sample of soda ash should be taken so that, when titrated with a hydrochloric acid solution having an HCl titer of 0.01820, each milliliter should represent exactly 1 per cent of Na2C03? Since 1 ml. X 0.01820 X NagCOa 2HC1 Weight of sample X 100 = 1 per cent the weight of sample is found to be 2.G15 grams. The same adjustments may be made on a normalit}'^ basis. What must be the normality of a hj'^drochloric acid solution so that, W’hen a 2.000-gram sample of soda ash is titrated, each milliliter of HCl represents 1 per cent of Na2C03? From the general relation: Volume used X Normality X Milliequivalent weight Weight of sample XlOO = Percentage we have 1 ml. X V X NazCOa 2 X 1000 2.000 X 100 = 1 per cent from which N = 0.3773. What weight of soda ash must be taken in order that the volume of a 0.500 iV HCl solution should represent the percentage of Na2C03? 1 ml. X 0.500 X 0.0530 Weight of sample X 100 = 1 per cent from which the weight of sample required is 2.G5 grams. Calculations of Mixed Alkalies. In the usual method of calculating the alkaline strength of a basic material, the total alkalinity is computed in terms of the percentage of the constituent which predominates. For 94 NEUTRALIZATION METHODS example, in the analysis of soda ash, which is largely Na^jCOs but besides inert material, usually contains small amounts of NaHCOa, the result is reported as percentage of Na^COa, whereas in reality both alkaline substances react with the acid used for neutralization. In cases of mixed alkalies, such as mixtures of the alkali hydroxides, carbonates and bicarbonates, the separate amounts of each constituent may be determined by titrations involving the use of two indicators. The following possibilities may occur with the above alkalies using the sodium compounds as typical in samples which may also contain inert impurities such as chlorides and sulfates: 1. NaOH as the only alkaline substance. 2. NaHCOa as the only alkaline substance. 3. Na2C03 as the only alkaline substance. 4. Mixtures of NaOH and Na2C03. 5. Mixtures of NaHCOa and Na2C03. Besides these, we may have dry mixtures of NaOH and NaHCOa or dry mixtures of NaOH, NaHCOa and Na2C03, but these mixtures can- not exdst in solutions because the hydroxide will react with the bicar- bonate to form the carbonate. Samples covering the above five cases can be identified and the percentages of the constituents determined by the behavior of methyl orange and phenolphthalein iijdicators toward these alkalies during titration. 1. Sodium Hydroxide. This base behaves alike toward both indi- cators so that, w'hen a titration is conducted with standard HCl, the end point comes at practically the same volume of added acid when methyl orange is used as w’hen phenolphthalein is used. Complete neutralization is indicated when methyl orange turns pink and phenol- phthalein turns colorless. 2. Sodium Bicarbonate. This base is practically neutral toward phenolphthalein. This is because the hydrogen-ion concentration de- rived from the iveakly ionized HC03~ ion (HCOs™ ^ H"^ + COs”) has about the concentration (1 X 10”^) necessary'' to change the color of this indicator, 'so that a minute amount of HCl is sufficient to indicate a neutral solution when phenolphthalein is present. Methyl orange, however, docs not change color from alkaline to acid until a concentration of hydrogen ion of about 1 X 10^ is reached. This ivill occur only when the bicarbonate is actually neutralized. A‘ titration of bicarbonate with standard acid can therefore be made only with methyd orange as indicator. 3. Sodium Carbonate. With phenolphthalein present, in a cold sofu- tion, a titration of sodium carbonate with standai'd HCl will sliow an ADDITIONAL STOICHIOMETRIC CALCULATIONS 95 end point when the solution is only one-half neutializcd, that is, when the Na2C03 has been transformed into NaHCOa, thus: NaaCOa + HCI = NaHCOg + NaCI The volume of acid used ^^^ll therefore be one-half of the total amount required for complete neutralization. With methyl orange the end point does not come until complete neutralization takes place. 4. A Mixture of NaOH and NA2CO3. Remembering from the fore- going cases that methyl orange changes color at complete neutralization for any alkali and that phenolphthalein shows a neutral reaction agaiast NaHCOs, the behavior of the above mixture can be easily deduced. When the mixture is titrated with acid, in the cold, in the presence of phenolphthalein as indicator, all the NaOH and one-half of the Na2C03 will liave been neutralized when the pink color Is discharged. A second titration, using methyl orange, will give the total volume of acid required for complete neutralization of both bases. The following problem will show the method of calculation. A 0.2-gram sample of a mixture of NaOH, Na^COa and impurities was dissolved in water and then titrated cold with 0.1 A HCI, using phenolphthalein as indicator. The volume of acid required w-as 30.00 ml. Another 0.2-gram sample of the mixture w-as taken, dissolved in water and titrated with 0. 1 N HCI, using methyl orange as indicator. This second titration required 40.00 ml. of the acid. Wliat were the per- centages of NaOH and Na2C03 in the sample? Since the total volume of acid required for complete neutralization of both compounds was 40.00 ml. and the volume required for the com- plete neutralization of NaOH plus the neutralization of one-half of the Na2C03 was 30.00 ml., the difference in volume between the first and the second titration was 10 ml., wiiich represents the volume required for one-half of the Na2C03. Therefore the volume required for the complete neutralization of the Na2C03 is 20 ml., and the volume re- quired for the complete neutralization of the NaOH is likewise 20 ml. The milliequivaJent weight of NaOH is 0.04000, and the milliequivalent weight of Na2C03 is 0.0530. Therefore and 20.00 ml. X 0.1 X 0.04000 0.2000 20.00 ml. X 0.1 X 0.0530 0.2000 X 100 = 40.00 per cent of NaOH X 100 = 53.00 per cent of Na2C03 96 NEUTRALIZATION METHODS 6. Mixture of NA2CO3 and NAHCO3. If a mixture of Na2C03 and NaHCOa is titrated with acid, using phenolphthalein indicator in a cold solution, the end point will come when one-half of the NagCOa is neutralized; this is equivalent to the transformation of all the carbonate into bicarbonate. The bicarbonate originally present, as well as that formed from the carbonate, will be neutral toward this indicator. A second titration, in the presence of methyl orange, will give an end point when both compounds are completely neutralized. As an example, suppose that a 0.2-gram sample of a mixture of Na2C03, NaHCOa and inert material was titrated with 0.1 N HCl using phenolphthalein as indicator. The volume of acid required was 10.00 ml. When the sample was titrated with methyl orange indicator the volume of 0.1 HCl required for complete neutralization was 30.00 ml. Since in the first titration the Na2C03 was half neutralized and none of the bicarbonate reacted with the acid, the volume of acid required for complete neutralization of the Na2C03 is 2 X 10.00 or 20.00 ml. In the second titration, since a total of 30.00 ml. of acid was required and 20.00 ml. of this was used by the Na2C03 the volume required for the NaHCOs was 10.00 ml. The milliequivalent weight of Na2C03 is 0.0530, and that of NaHCOs is 0.0840. Therefore and 20.00 ml. X 0.1 X 0.0530 0.2000 10.00 ml. X 0.1 X 0.0840 0.2000 X 100 = 53.00 per cent of Na2C03 X 100 = 42.00 per cent of NaHCOa Volume Relationships. The relationships in the volumes of acid used may be summarized as follows: Volume Rehii^ Same volume for both indicators No volume for P.P. Twice the volume for M.O. as for P.P. More than half the total volume for P.P, Less than half the total volume for P.P. Substance NaOH NaHCOs NaaCOs NaOH and NaaCOs NaHCOs and NaaCOs The important fact, which is the key to the solution of mixed alkali and double indicator problems, is that phenolphthalein is neutral to bi- carbonates when used in a cold solution. PROBLEMS 97 PROBLEM SET 6 ADJUSTMENT OF STRENGTH AND MIXED ALKALI CALCULATIONS (a) Adjuslmcnl 1. What volume of concentrated HCI (sp. gr. 1.19, containing 37.23 per cent of IICl by weight) must be added to 500 ml. of 0.10 A' IICI in order to make it 0.20 normal? Ans. 4.2 ml. 2. What volume of 0.5123 A'^ H 2 SO 4 muj-t be added to 500 ml. of 0.1901 N IICI in order to make a solution which is exactly 0.5000 A'? 3. How many grams of NaOH mast be added to 1 liter of IICI, 1 ml. of which will react with 0.001150 gram of NaOII in order to make it U.KXK) normal? An.s. 0.1180 gram 4. What weight of sample of pearl ash (impure K 2 CO 3 ) should be taken for analysis in order that the volume of IICI used (1 ml. = 0.01500 gram K 2 CO 3 ) will give the per cent of K 2 CO 3 ? 6 . What weight of sample of II 2 SO 4 should be used so that the per cent of SO 3 should equal one-half the volume of 0.1 A' NaOII o^ed? An.er- ccutage of Na 2 C 03 in the samjde? 9. In the determination of NaOH in commercial caastic soda by titration with 0.5000 N IICI, what should the weight of sample be so tliat the percentage of NaOII is tunce the number of milliliters of HCI used? An.?. 1.0000 gram 10. A sample of caustic soda was found to be 50 per cent pure. What is the maxi- mum weight of Kimple that can be titrated with 0.3750 N HCI without refilling a 50-ml. buret? (6) Mixed Alkalies 11. (a) What volume of 0.100 A HCI will be required to titrate a 0.2500-gram sample of pure Na 2 C 03 using phenolphthalein in a cold solution? ( 6 ) What volume is required to titrate the same weight of sample using methyl orange? Atw. (a) 23.6 ml. ( 6 ) 47.2 ml. 12. How many milliliters of 0.100 A HCI are required to titrate a 0.2500-gram mixture of equimolecular quantities of Na 2 C 03 and NaHCOa using (a) phenolphtha- lein and ( 6 ) methyl orange as indicator? 13. How many milliliters of 0.1000 A HCI are required to titrate a 0.2600-gram mixture of equal parts by weight of N&^COs and NaHCOa using (a) phenolphthalein and ( 6 ) methyl orange as indicator? Ans. (a) 11.78 ml. ( 6 ) 38.43 ml. 98 NEUTRALIZATION METHODS 14. An impure NaOH sample required 30.00 ml. of 0.200 N HCl for titration to the phenolphthalein end point and an additional 3.00 ml. of acid to cany the titration from the phenolphthalein to the methyl orange end point. How many grams of NaOH and Na2C03 were present in the sample? 16. Calculate the percentages of NasCOs and NaHCOs in a 1.4000-gram sample which, on titrating with 0.1234 N HCl using phenolphthalein, required 16.26 ml. and, on titrating the same weight with the standard acid using methyl orange required 37.28 ml. Ans, 14.25 per cent; 6.02 per cent 16. An alkali sample was found to contain 92 per cent NaOH and 6 per cent Na2C03. Find the normality (a) with respect to the phenolphthalein titration and (6) with respect to the methyl orange titration if 6.00 grams of the ftIWftit ^ere dis- solved and diluted to exactly 1 liter. 17. A sample is known to contain NaOH and NaaCOg. If a 0.3000-gram sample containing 35.00 per cent NaOH requires 33.25 ml. of 0.1000 N HCl for titration to the phenolphthalein end point, what volume of HCl will be required to complete the titration to the methyl orange end point? Am. 7.00 ml. 18. A sample of impure NaHCOs weighing 1.6000 grams required 9.93 ml. of 0.1622 HCl for a phenolphthalein titration. A 1.2420-gram sample required 43.72 ml. of 0.1522 N HCl for a methyl orange titration. Calculate the percentages of NaHCOa and Na^COa. 19. A mixture is known to consist of equivalent amounts of NaOH and Na^COa or of Na2C03 and NaHCOs. If the volumes of acid required for methyl orange and phenolphthalein titrations are approximately in the ratio of 4 : 3, which mixture is present? Ans. NaOH and Na2C03 20. A 1.00-gram sample of equal parts of NajCOa and NaHCOa was titrated using phenolphthalein as indicator. Another 1.00-gram sample was titrated using methyl orange. What volume of 0.556 N H2SO4 was required in each titration? CHAPTER IV OXIDATION AND REDUCTION METHODS GENERAL CONSIDERATIONS The reactions between oxidizing and reducing substances comprise the largest group of volumetric methods of analysis. Many excellent and verj^ accurate methods for the determination of a large number of substances are based on reactions between oxidizing and reducing agents. Before considering such methods in detail it will be well to review the general nature of the phenomena of oxidation and reduction and briefly consider the electrochemical aspects of tlie subject. Nature of Oxidation and Reduction. The earliest notions relative to the phenomena of oxidation and reduction regard(Kl such actions as direct additions of ox-ygen and of hydrogen. Later these idea.s were modified to include the addition of elements other than oxygen, such as sulfur or the halogens, as true oxidations. Still later we have come to regard an oxidation as an increase in positive charges or a decrease in negative charges associated with ions and atoms, and a reduction as a decrease in positive charges or an increase in negative charges. For example, when iron changes from the ferrous to the ferric condition, there is an increase in the number of positive charges from +2 to +3; this is an oxidation. When stannic tin, Sn++++ is changed to stannous tin, there is a decrease in the number of positive charges asso- ciated with the tin atom; this is a reduction. Oxidation and Reduction in Terms of Valence and Electron Trans- fers. An increase in the number of positive charges is the result of the loss of electrons by an atom, a radical or an ion. A gain of electrons results in a net decrease of positive charges. The excess or deficiency of electrons associated with an atom may be regarded as its valence. The ferrous ion, Fe"*"*", for example, has an excess of two positive charges; it has resulted from a neutral atom of iron losing two electrons, and its valence is said to be -f-2. When a substance loses electrons, its positive valence is increased or its negative valence is decreased. Conversely, when a substance gains electrons, its positive valence is decreased or its negative valence is increased. Since an increase in positive valence 99 100 OXIDATION AND REDUCTION METHODS « is an oxidation and a dec;*ease in positive valence is a reduction and the valence change results from a loss or a gain of electrons, the term oxida- tion means a decrease in the number of electrons associated with the ion. Reduction, likewise, in terms of electrons, means a gain in the number of electrons by an ion. In the broadest sense, then, oxidation means the loss of one or more electrons and reduction means the gain of one or more electrons. In every reaction in which there is a transfer of electrons from one kind of atom to another, the electrons lost by atoms of one element must be gained by atoms of another element. That is, when one element is oxidized, another element is reduced. Thus, in the reaction of metallic zinc with sulfuric acid, the change of valence is shown by writing the equation in ionic form, as: Zn° + 2H+ = Zn-»-+ + Hz® in which the neutral zinc atom with zero valence becomes oxidized to the divalent, positively charged zinc ion, while two hydrogen ions are reduced to zero valence, becoming a molecule of gaseous hydrogen. The same change expressed in terms of electrons transferred from the zinc to the hydrogen may be shown by the electron or “half-cell” equations Zn® Zn++ -h 2e 2H+ + 2e Hg® in which the symbol e is used to denote an electron. The equations show that the total number of electrons lost by the zinc are taken up by the hydrogen. This reaction is thus truly one of oxidation-reduction. For an element or ion to function as an oxidizing agent it must readily accept electrons from the reducing agent, which in turn must easily lose one or more of its electrons. The word redox has recently come into use to designate such chemical changes, the word itself being de- rived by contracting the expression reduction-oxidation. Since oxidation involves a loss of electrons and reduction a gain of electrons, the phenomena of oxidation and reduction are fundamen- tally electrical in nature. A reducing agent is a substance which is capable of losing electrons, being oxidized thereby; an oxidizing agent accepts electrons, becoming itself reduced. The transfer of elec- trons in oxidation-reduction reactions is therefore from the reducing agent to the oxidizing agent. When such reactions are made to take place by mixing a solution of an oxidizing agent with a solution of a r^ucing agent, as in titration procedures, the transfer of electrons is direct in the solution passing from the reducing ions to the oxidizing GENERAL CONSIDERATIONS 101 ions. If, moreover, the oxidizing solution is placed in one beaker and the reducing solution in another, and the two solutions arc joined by means of an inverted U-tubo filled with an electrically conducting solu- tion such as saturated KCl, and a wire conductor with metallic (usually platinum) electrodes used to complete the external circuit between the two beakers, electrons will flow through the wire from the reducing solution to the oxidizing solution and the same ultimate result is ob- tained as by mixing the two solutions. The flow of electrons constitutes an electric current, and the above- described arrangement is, in fact, an electrolytic cell. The production of an electric current, by chemical means, is essentially based on a r(*dox reaction. Conversely, an elcctnc current may be used, as in electro- analysis, to bring about reduction and deposition of metals and oxida- tion of anions at the two poles of an electrolyzing appaiatus. A strong oxidizing agent readily accepts electrons; a strong icducing agent easily gives up electrons. The relative ease with which a sub- stance accepts or loses electrons is a measure of its relative oxidizing or reducing power. It is important to know how a strong oxidizing or reducing agent can be distinguished from a weak one. The tendency to gain or lose electrons can be measured, relatively, by electrochemical means. The method used is somewhat too complicated to discuss fully here and is only briefly outlined on page 10-1 under “Electrochemical Theor>' of Oxidation. “ Oxidizing and Reducing Ions. Although these rclation.s between metals or ions are examples of true oxidations and reductions, the rela- tions between two species of ions of the same element in two states of oxidation arc of much greater importance in quantitative analysis, be- cause the oxidizing and reducing agents commonly used in titration procedures arc of the variable-valence type. For example, u'hen a ferrous salt solution is used as a reducing agent, the half-cell or electron reaction is Fe++ ^ Fe+++ + le and the equilibrium ratio of ferrous- to ferric-ion concentration is an important consideration. Or, again, when KMnO, is used as oxidizing agent in an acid medium, the relation between the oxidized and reduced forms is sho\\Ti by the reaction Mn 04 “ + 8H+ 5e~-> -|- 4 H 2 O Here the ratio of Mn 04 “ to Mn++ i.e., the amount of Mn 04 ~ remain- ing unreduced when equilibiium is reached is, in part, a measure of the oxidizing power of pennanganate. 102 OXIDATION AND REDUCTION METHODS The relation between the two forms of a few commonly used oxidizing agents is shown below, in the form of reactions between the two species of ions. Cr207= + 14H+ + 6e -> 2Cr+++ + 7H2O Mn04- + 8H+ + 5 e M11++ + 4H2O NO3- + 4 H+,+ 3 e NO + 2H2O Cu++ + le ^ Cu+ 2Hg++ + 2e Hg2 ++ SgOg^ + 2e 2SO4 lo*^ + 2e ^ 21" Likewise, the two forms of ions of a few typical reducing agents are shown below. Fe++ -> Fe+++ + le + 2e 21-->l2° + 2e 2S2O3"" — > 8400“ + 2e C 204 ^ ^ 2CO2 + 2e AsOa^ + H 2 O As 04^ H- 2H+ + 2e Further examples are given in the table on page 108. These half-cell reactions cannot proceed of their OAvn accord because some means must be provided for furnishing the necessary electrons for the oxidizmg agents and of getting rid of the electrons for the reducing agents. In order that reactions can take place, suitable oxidizing and reducing agents must be brought together. The final equilibrium . reached in the reaction as a whole ^vill then depend upon the equilibrium ratios of both pairs. Thus to promote the reaction Mn04" + 8H+ + 5e^ Mn++ + fflgO so that, when equilibrium is reached, the solution should contmn gible amounts of the Mn04“ ion, a suitable reducing agent, such as a ferrous salt, is used. When the reaction Fe++ ^ Fe++'*' + le GENERAL CONSIDERATIONS 103 is combined with that for permanganate, i.e., wiien Mn04 is allowed to oxidize fen*ous ion, the reaction Mn04“ + 8Pr+ + 5Fe+‘^ 5Fc+++ + Mn++ + IHaO nms practically to completion. Balancing Equations of Oxidation-Reduction by Valence Changes. The method of wTiting and balancing e(]iiation.s involving valence changes has been illustrated above for the simple case of the reaction of metallic zinc with sulfuric acid. A more complicated example ^vill now be considered, namely, the oxidation of ferrous chloride by potas- sium dichromate in an acid solution. The reaction involved, written in unbalanced molecular form, is FeCl2 + K2Cr207 + PICI — ► FeCla + CrCla + KCI -|- H2O sho\ving that the ferrous chloiide is oxidized to the ferric (trivalent) condition, whereas the chromium in the potassium dichromate is reduced to the chromic (trivalent) condition, the oxygen of the dichromate uniting with the hydrogen of the acid to form w'ater. In ionic form, tliis unbalanced equation becomes Fe^"^ 4- CvzOj^ 4- Fe+++ 4- Cr'^+‘^ 4- H2O The chromium atom in the dichromate ion has a valence of six; in the chromic state it has a valence of three; each chromium atom sulTeis a reduction of three positive valences. Since there are two atoms of chromium in each dichromate ion, each atom of which loses three posi- tive charges, the total loss will be six for each dichromate radical. The six charges lost by one Cr207“ ion must be accounted for by the gain of charges on the iron. Since each ferrous ion show's an increase of one positive valence on becoming a ferric ion, the six charges from one di- chromate ion are assumed by six ferrous ions. Finally the seven atoms of oxygen from the dichromate ion require fourteen atoms of iiydrogen from the acid to form seven molecules of w^ater. The balanced equation in ionic form can now be written, 6Fe‘^'^ 4- Cr207“ 4- 14H‘^ — > 6Fe'*'*^+ 4- 2Cr'*‘++ 4- 7H2O Balancing Equations of Oxidation-Reduction by Electron Transfers. Since change of valence means an exchange of electrons betw-een the reducing agent and the oxidizing agent, the WTiting and balancing of oxidation-reduction reactions by the electron transfer method involves merely an extension of the above method. Thus, to take the same 104 OXIDATION AND REDUCTION METHODS example, since each atom of chromium, in being reduced from the hex- avalent to the trivalent state, gains three electrons, and two atoms of chromium are involved in each dichromate ion, the total gain of elec- trons for one Cr 207 “ ion is six: ' CtzOj^ + 14H-*- + 6e 2Cr+++ + THaO The electrons gained by the chromium atoms are lost by ferrous ions, which in turn are oxidized to the ferric state. Since each ferrous ion gives up one electron, six ferrous ions are involved: eFe"^"*" -> 6Fe+++ + 6e The number of hydrogen ions furnished by the acid is determined, as before, from the number of oxygen atoms available; seven oxygen atoms require fourteen hydrogen atoms and yield THgO. The balanced equar tion can then be written by assembling the electron equations, and the final equation becomes 6Fe++ + Cr 207 ==' + 14H+ 6Fe+++ + 2Cr+++ + 7 H 2 O In general, when setting up and balancing equations by this method, first write the unbalanced equation in ionic form; second, determine the number of electrons gained by the one element, which must equal the number of electrons lost by the element undergoing oxidation; third, assemble the separate electron equations and prefix the proper coefficient to each set of ions, thus giving the balanced ionic equation. Equivalent Weights and Normal Solutions. A gram-equivalent weight is defined as that weight of substance which will involve, directly or indirectly, 1.008 grams of hydrogen. When an atom of hydrogen is oxidized to hydrogen ion, an electron is given up: H^H+d-e and this means that the gram-equivalent weight, 1.008 grams, of hydro- gen is associated with one electron transfer. Hence for oxidizing and reducing substances, to be equivalent to this, the gram-atomic or gram- molecular weight is divided by the number of electrons lost or gained. This corresponds to the change in valence. Electrochemical Theory of Oxidation-Reduction. A metal such as zinc placed in a solution of its ions shows a greater tendency to pass into solution than to deposit the zinc ions. Conversely, cupric ions show a greater tendency to deposit than for metallic copper to dissolve. The tendency to dissolve is known as solution pressure and the reverse GENERAL CONSIDERATIONS 105 tendency as osmotic pressure. Wliether a metal will dissolve in a dilute solution of its ions or come out of solution dcpcnd.s upon the relative magnitudes of its solution pressure and osmotic pressure. If the electro- lytic solution pressure is greater than the osmotic pressure, a certain amount of the metal will pass into solution and assume the ionic state. If the osmotic pressure is the greater, metal ions will be deposited; and if both pressures arc equal the metal ion.s will form at the same rate at wiiich metal atoms are dej>ositod. When equilibrium is reached obviously the tw'o opposing ctTccts arc the same. If positively charged ioiLs arc forming at a greater rate than the ions are being deposited, the stiip of metal will assume a negative charge and the solution in contact with it will be positively charged. There will thus be set up a potential dilTercnce between the metal and its ion.s. The greater the concentration of ions the greater this potential differ- ence. If ions are losing their positive charges faster than they are being formed, the metal strip will be positively charged and the solution nega- tively charged. At equilibrium, the rates in both directions are equal and the dilTerencc of potential becomes constant. The Nernst Equation. The potential difference set up at the contact bctw'ccn a metal and a solution of its ions is relatx'd to the osmotic pres- sure of the solution and the electrolytic .solution pressure of the metal by an equation first formulated by Nemst. The Nernst equation is where E = potential difference between the metal and its ions R = the gas constant = 8.32 joules per degree T = absolute temperature (273 + i°C.) n = valence change (number of electrons lost or gained per atom) F = one Faraday or 96,500 coulombs logs = the natural logarithm to the base e p = osmotic pressure P 5= electrolytic solution pressure. This equation can be transformed into a more usable one. By insert- ing the values for R and F and converting natural logarithms into common logarithms to the base 10 by multiplying by 2.303, the equation becomes „ 0.000198T , p login p n 106 OXIDATION AND REDUCTION METHODS for any absolute temperature T. For the temperature of 25®C, or 298® absolute, we have 0.059 n If the equilibrium concentration of ions, k, is substituted for p, and p is expressed in terms of the ionic concentration, C, since osmotic pressure is proportional to the concentration, the Nemst equation becomes „ 0.059 , G E = log 7 n k Measurement of Electrode Potentials. There is no direct way by which the potential difference at the contact of a metal and a solution of its ions can be determined, but relative values can be obtained by setting up a voltaic cell in which one electrode is arbitrarily taken as the standard and the other electrode as the one whose potential is desired. Two such standard electrodes are in use. One is the nonnal hydrogen electrode, in which a platinum foil coated with platinum black is im- mersed in a solution of acid, 1 M with respect to hydrogen ions, the solution and electrode being kept saturated with hydrogen gas. The potential of the hydrogen electrode is arbitrarily taken as zero. The other standard is the calomel electrode. This consists of mercury covered with a layer of calomel paste (Hg 2 Cl 2 ) which in turn is covered with a solution of KCl saturated with Hg 2 Cl 2 . If then we wish to measure the potential of a metal in contact with a certain concentration of its ions, the metal and a solution of its ions are made a half-cell of a voltaic cell, the other half-cell being a standard hydrogen or calomel electrode. The two half-cells are joined by a U-tube filled with a conducting solution such as KCl, and the electrical circuit is completed by joining the electrodes to a sensitive voltmeter or poten- tiometer. The voltage measured is that of the cell as a whole. The potential of the electrode is the algebraic difference of the observed voltage and that of the standard electrode. Molar Electrode Potentials, The potential of any metal in contact with its ions varies with the concentration of the ions. In order to arrange the metals in a series with respect to their relative potentials, the potentials are compared for unit gram-ion concentration of their ions, that is, for solutions containing 1 gram-ion per liter of the metal ion. Metal electrodes dipping into 1 M solutions of their salts would give this concentration of metallic ions, if such molar solutions were completely ionized. In tlie actual determination of molar potentials, the metal is GENERAL CONSIDERATIONS 107 made the electrode in a solution much more dilute, 0.01 or 0.001 A/, and the potential for a 1 71/ solution is then calculated from the Nernst equation. The Nemst equation E = may be expanded into the form 0.059 n 0.059, 1 , 0.059, ^ If the concentration of the metal ion is 1 M, the molar potential, usually designated by the symbol Eq, is expressed by the equation 0.059 n By substituting for 0.059/n log l/k, the Nemst equation for the electrode potential, E, for any concentration, C, of metal ion becomes 0.059 n logC From this, by measuring the electromotive force of a cell, using the hydrogen electrode as reference to wliich the value of 0 is arbitrarily assigned, the molar potential, Eq, can be obtained by direct calculation from equation 3. In this way the values of Eq have been obtained and arranged in the familiar electromotive series of the metals. By methods somcwliat more complicated, the molar potentials of most oxidizing and reducing ions have been measured. These molar potentials are shown in Table VII. Calculation of Equilibrium Constants for Reactions in Which One Metal Replaces Another from Solution. A strip of zinc placed in a solution of a copper salt will cause the deposition of metallic copper, some of the zinc going into solution as zinc ions. The reaction taking place is Zn® -f Cu++ ^ Zn+-^ -f Cu® in which the metallic zinc, Zn°, is the reducing agent and the copper ions, the oxidizing agent. The Law of Chemical Equilibrium, when applied to this reaction, takes the form ^Zn** X CcuO X Czno 168 OXIDATION AND REDUCTION METHODS TABLE VII. STANDARD OXIDATION- Oxidizisg Potential Referred to Normal Hydrogen Electrode - 0 Reaction BiOa“ + 6 H+ + 2e =: Bi+++4-3H»0 lOr + 2 H+ + 2e = 10,- + HjO 1/2 SaO,— — "f“ 6 “ SO4— — 1.82 1.78 1.68 1.69 1.6 1.49 1.45 1.45 1.42 1.36 1.36 1.35 1.33 1.3 1.23 1.21 1.20 1.14 1.09 1.09 1.06 0.94 0.92 0.90 0.86 0.84 0.80 0.80 0.78 0.76 0.75 0.74 Co+++ + e « Co++ 1/2 H2O2 + H‘''+ e HaO (action toward reducing agents) PbOa + 4 H+ + SO4 — + 26 = PbSO, + 2 HaO MnO," + 4 H+ + 36 = MnOa + 2 HaO MnOr -f 8 H+ -I- 6e = Mn++ + 4 HaO BrOr + 6 H+ + 5e = 1/2 Bra + 3 HaO Ce++++ + 6 = Ce+++ ClOr + 6 H+ + 66 = Cl- H- 3 HaO BrO,“ + 6 H+ + 6e = Br- + 3 HaO Au+++ H- 36 = Au 1/2 Cla H-e = Cl- ClOr + 8 H^- + 8e = Cl- + 4 HaO MnOa -f- 4 H+ + 26 = Mn++ + 2 HaO 1/2 CraO,- - + 7 H+ + 36 = Cr'H-+ 4. 7/2 HaO 1/2 Oa + 2 H^ + 26 == HaO (in acid solution) TI+++ -h 2e - T1+ lO,- + 6 H+ -h 5e = 1/2 la + 3 HaO Ortho-phenanthroline-ferrous ion (J. Am. Chem. Soc., 63 3908 [1931J) (red in reduced, blue in oxidized form) 10,- + 6 H+ -1- 66 = I- + 3 HaO HSeO*- 4- 3 H+ + 2e = HaSeO, + HaO 1/2 Bra + 6 = Br- NO," + 4 H+ + 3e = NO + 2 HaO VO,- + HaSO, + 2 H+ + e - VOSO4 + 2 HaO Hg++ + e = 1/2 Hga-^^ Hg++ + 26 = Hg Diphenylamine sulphonic acid (green in reduced, reddish in oxidized form; satisfactory in presence of tungstates. J. Am. Chem. Soc., 63, 2902 [1931]) Unsatisfactory in presence of tungstates. See J. Am. Chem. Soc., 62, 4179 (1930), and 63, 2903 (1931) SbO, + 2H+ + 2e» SbO, + H»0 HaSeO, 4-4 H+ + 46 aSo + 3 HaO 1/2 Hga-*^ 4- fi = Hg Ag+ 4- 6 = Ag Fe+++ 4- 6 = Fe+^ Diphenylbenzidene; green in reduced, violet in oxi- dized form Diphenylamine; colorless in reduced, violet-blue in oxidized form • Reprinted from Lundcll and noflfman, Outfine* ofMethod$ of Chomicol Analysia, John WUey & Sons. Ino., 1038. GENERAL CONSIDERATIONS 109 REDUCTION POTENTIALS • OxidiziDg Potential Referred to Normal Hydrogen Electrode « 0 0.70 0.G8 0.66 0.59 0.67 0.53 0.53 0.5 0.49 0.47 •0.4 0.4 •0.4 0.40 0.36 0.34 0.333 •0.3 0.280 0.23 0.2 0.2 0.17 0.17 0.15 0.14 • 0.1 0.04 •0.0 0.000 •- 0.1 - 0.12 -0.14 •- 0.2 - 0.2 - 0.21 -0.23 -0.29 -0.34 -0.38 -0.40 -0.40 -0.44 Reaction C6H402 + 2 II’*' + 2c = CflHiOaHj (quinhydrone electrode) 1/2 Oj + e => 1/2 HaOj (reducing action toward stronger o.xidants) AlnOi” -f“ c = MnO* MnOr + 2 HaO + 3c = MnOj + 4 OH" As 04 +2H+ + 2e = AsOi + HjO Methylene blue at pll 2.86 (Hygienic Lab. Bui. 171, p. 191 (1928)1 I/2I3 + c = I- M0O3 + 4 n+ + c = M0O+++ + 2 HiO Fe(CN)« + c « Fe(CN)a HaSO, -f 4 H+ + 4c = S -f 3 HjO VOj' + 6 + 3c = V++ + 3 H,0 VO^ + 2 + c = V+++ + H 2 O PtCle- “ + 2c « PlClr “ -f- 2 Cl- 1/2 O 2 + HjO + 2c => 2 OH- UO2SO4 H- 4 H+ + SO4- - + 2c = U(S04)a + 2 H ,0 Cu++ + 2c = Cu Decinormal calomel electrode WO3 + 4 H+ + e = \VO+^-+ + 2 H,0 Normal calomel electrode AgCl + c - Ag + CI- PtCU” “ + 2c - Pt + 4 Cl- Bi+++ + 3c = Bi S + 2H+ + 2e =HaS Cu**^ + e = Cu+ Sn++++ + 2e = Sd++ SO4-- + 4 H+ + 2e - H,SO, + HaO WO+++ + 2 H+ + e = W++++ -f HjO TiO-H- + 2 H+ + e = Ti-H-+ + HaO M0O3 + 6 H+ + 3c = Mo+++ + 3 HjO H+ + e = 1/2 Ha 1/2 CbaOe + 5 H+ + 2c = Cb+++ + 5/2 H,0 Pb++ + 2c ^ Pb Sn++ 4- 2c = Sn COa + H+ -f e = 1/2 H1C1O4 V+++ + e = V++ SbO-*- + 2 H* + 3c = Sb + HaO Ni-^ + 2c * Ni Co++ + 2c = Co TI+ + c = T1 In+++ + 3c = In Cd++ + 2c = Cd Cr+++ + e = Cr++ Fe++ 4- 2c = Fe 110 OXIDATION AND REDUCTION METHODS TABLE VII — Continued Oxidizing Potential Referred to Normal Reaction Hydrogen Electrode « 0 “0.5 Ga+++ + 3c = Ga -0.56 Cr-*^' + 2e = Cr -0.76 Zn-*-*- + 2e = Zn -0.83 HaO + c « 1/2 Ha + OH- -1.1 Mn++ + 2c = Mn -1.4 U++-^ -f 4e = U -1.69 Be++ H- 2c = Be -1.7 A1+++ + 3c = A1 -2.40 Mg++ -f- 2c = Mg -2.71 Na+ -I- c « Na -2.76 Ca"*^ -1- 2c = Ca -2.90 + 2c — Ba -2.92 Sr-H- + 2e = Sr -2.92 K+ + c = K -2.93 Rb+ 4- c « Rb -2.96 r Li+ -f c = Li * Approximate value or poaitioo. Since the factors Ccu© and Cz^o represent concentrations of metals in the solid foiTQ, their values are constant, and we have simply Ccxi** = K (equilibrium constant) This means that, when the concentration of zinc ions bears a certain ratio to that of copper ions, the reaction will have come to equilibrium. The equilibrium value can be calculated from considerations of the electrode potentials by applying the Nemst equation. If a strip of zinc immersed in a solution of a zinc salt is made one electrode and a strip of copper in a cupric salt solution the other elec- trode, and the two vessels are joined by means of a salt bridge and the electrodes are connected externally through a voltmeter, electric current will flow until the sj'-stem has come to equilibrium. This will occur when the electrode potential of the zinc equals thfft of the copper. From the Nemst equation, for the zinc electrode, we have i;. E. . 0.059, ^ ^»inc — T ^ lOg bzn++ for the copper electrode - _ , 0 . 059 , ^ ^ooppor — lioCooppor) T ^ lOg Ccu++ GENERAL CONSIDERATIONS 111 At equilibrium, when the potentials are equal, 0.059 _ 0.059 , ■^ 0 (*ino) "r 2 10gCzn + + “ ^0(copp0 + 02 + K^SO. (h) H3ASO4 + H2S -* 112O + s + H3ASO3 ■ (c) Na^Oa + CrCIs + NaOH -* NaaCrO.* + NnCI + H2O 2. (a) KMn04 + H2SO4 + H2C2O4 -> MnSO., + HjO + CO2 + K^SO. (b) H2O2 + FcCla + HCl --> FeCU + H2O (c) KoCraO? + HCl + SnCIa CrCIj + H.O + SnCI. + KCl 3 . (a) KaCraO? + HCl + H2S — » CrCb + H2O + S + KCl (6) CI2 + KI I2 + KCl (c) KMn04 + H2SO4 + KI -* MnS04 + H2O + I2 + K->SO^ 4 . (a) I2 + NajSaOs — » Nal + Na2S40c (h) KaCroO; + HCl + KI — ► CrCls + HoO + I2 + KCl (c) NaOCl + HI -+12 + H2O + NaCi 5 . (a) KMn04 + H2O + MnS04 -» MnOa + H2SO., + K^SO^ (6) HCIO3 + II2S -» H2SO4 + HCl ‘ (c) K2Cr207 + H 2 SO 4 + H 2 C 2 O 4 Cr2(S04)3 + II 2 O + CO 2 + KoSO. 6. (a) (NH4)2S20 s + H2O + MnS04 ^ (NH4)2S04 + MnOa + IlTsO ib) HNO3 (dil) + BiaSa -> NO + II2O + S + Ri(N 03)3 (c) MnOo + H2SO4 + H2C2O4 — * MnSO, + H2O + CO2 7 . (a) KMnO, + HCl + H2S — ♦ MnCl2 + H2O + S + KCl (6) KIO3 + I2 + HCl -» ICl + H2O + KCl (c) HNO3 + H2S -» NO + H2O + S 8. (a) I2 + HASO2 + HoO -* HsAsO, + HI (b) HNO2 + HaS S + NO + H2O (c) KMnO, + H2O + NaaSnOa MnOa + KOH + NaeSnOs 9 . (a) KMnO, + H2SO4 + HNO2 MnSO, + H2O + HNO3 + K2SO1 (b) MnOa + H2SO4 + Nal -> MnSO, + I2 + NallSO, + II2O (c) HNO3 (cone) + AS2S3 -* NO2 + H2O + H3ASO4 + H2SO4 10 . (a) KaCraO? + H2SO4 + H2SO3 Cr2(S04)3 + H2O + K«.S 04 (6) Mn02 + HCl MnCIa + H2O + CI2 (c) SbaSs + HCl (cone) — » SbCl3 + S + H2S REPRESENTATIVE OXIDATION-REDUCTION PROCEDURES ^ oxidation-reduction reaction must run practically to completion if it is to be used as the basis of a quantitative volumetric method of analysis. Not all oxidation-reduction reactions fulfill tliis requirement and it is essential to select the proper combination of o.xidizing and' reducing agents, one of which constitutes the substance being analyzed the other being the standard solution with which the sample is titrated! In practice it is often desirable to have at hand two standard solutions the one an oxidizing agent and the other a reducing agent, so that if an excess of one reagent is deliberately or accidentally added’, this excess may be determined by means of the other reagent. 114 OXIDATION AND REDUCTION METHODS The selection of proper combinations is further restricted by the addi- tional requirements of the standard solution. A solution, it must be remembered, in order to be used as the titrating medium (1) must react practically quantitatively with the material being titrated, (2) must not change strength, (3) must be standardized by a reaction which runs practically to completion, and (4) must be used in connection with an indicator whose color change coincides very closely with the stoichio- metric point of the reaction. These requirements place restrictions on the possible number of standard solutions but, in spite of this, redox methods constitute by far the most important class of volumetric methods. Classification of Oxidation-Reduction Methods. It is convenient to classify volumetric oxidation-reduction methods on the basis of the standard solution used in the fundamental reaction. Among oxidizing agents, potassium dichromate, potassium permanganate and iodine are found to be the most satisfactory as standard solutions. In addition, potassium iodate and potassium bromate are frequently used. Ceric sulfate is now coming into use as a standard oxidizing solution. In connection with, these oxidizing agents, ferrous salts, oxalic acid, sodium thiosulfate, sodium arsenite and potassium iodide may be employed; these are strong reducing agents and, ^vith the exception of potassium iodide, are generally used in the form of standard solutions. Suitable pairs of these reagents are: Potassium dichromate with ferrous salts. Potassium permanganate with ferrous salts. Potassium permanganate with oxalic acid. Ceric sulfate with ferrous salts. Iodine with sodium arsenate. Iodine with sodium thiosulfate. Potassium iodide with oxidizing agents. For the purpose of study and laboratory experimentation, these com- binations are considered under the following types of methods; A. Dichromate methods. B. Permanganate methods. C. Ceric sulfate methods. D. Iodine methods. Use is made of standard solutions of potassium dichromate and ferroiis sulfate in the determination of iron in the laboratory procedure described first. The determination of iron, involving the use of standard potassium permanganate and ferrous sulfate solutions, and the determination p( DICHROMATE DETERMINATION OF IRON 115 calcium, involving standard permanganate and oxalic acid solutions, are described under permanganate methods. Tiie use of standard ceric sul- fate solution is outlined. Methods which involve reactions of iodine or its compounds arc discussed and procedures based on these reactions are tal^en up under that heading. A. DICHROMATE METHODS Volumetric oxidation-reduction methods which involve the dichromate ion are comparatively few in niunber. The reaction between ferrous ions and the dichromate ion is the most important one, and it serves as a basis for the determination of both iron and chromium, as well as, in- directly, for a number of other substances. The fundamental reaction is Cr207” + CFe"*"'*’ -f = 2 Cr'^'^+ -|- GFe"’"^"^ + 7H2O A standard solution of K2Cr207 may be used to determine the amount of iron in a given sample and, on the other hand, chromium in ores and salts may be converted into the dichromate ion and titrated with a standard solution of a ferrous salt. Dichromate Determination of Iron In the dichromate method for the determination of iron, the iron is first reduced to the ferrous state by stannous chloride. The excess of stannous chloride is then oxidized and removed by mercuric chloride An acid mixture, containing sulfuric and phosphoric acids, is added and the titration is carried out with standard potassium dichromate solu- tion, using sodium diphenylamine sulfonate as indicator. Standard ferrous salt solution is prepared in case it is needed for back-titration. The potassium dichromate solution may be made from primary stand- ard K2Cr207 or else standardized with weighed amounts of iron wire of knowm purity, but the strength of the ferrous sulfate solution is obtamed by comparison with the dichromate solution. The acid mixture reduces the pH of the iron solution to a definitely acid region. This is necessary in order to have the oxidation-reduction potential of the ferrous-ferric system near that of the potential of the diphenylamine sulfonate indicator. The phosphoric acid, by combining with ferric ions produced in the titration to form a colorless acid phos- phate complex, eliminates the disturbing color effect of the yellow^ ferric chloride. Furthermore, the removal of ferric ions keeps the ferrous- ferric concentration ratio high, and prevents the oxddation of the indi- cator until the ferrous-ion concentration becomes negligible. 116 OXIDATION AND REDUCTION METHODS Preparation and Standardization of Approximately 0.1 iV” Solviions of Potassium Dichromaie and Ferrous Sulfate In the reaction between the dichromate ion and ferrous ions in the presence of acid, the chromium in the Cr207“ ion becomes reduced to Cr"*"'*"'*" ions, and the ions are oxidized to the ferric, Fe‘^+''", state. Each chromium atom in the dichromate ion is reduced from a valence of +6 to a valence of +3. Hence, since there are two atoms of chro- mium involved in one Cr207^ ion, the total valence change, for this ion, is 6. This means an acceptance of three electrons for each chro- mium atom or a total of six for both atoms, as shown by the electron or half-cell equation Cr207^ + 14H+ -f 6e ^ 2Cr+++ -j- 7H2O The iron atom loses an electron in being raised from a valence of -|-2 to a valence of 3, in accordance wth the equation Fe++ ^ Fe+++ + le Hence, to balance the equation for the reaction between Cr207^, Fe^"^ and H"^ ions, v:e have 01-207= -h 14H+ + 2Cr+++ 7H2O 6Fe++ ^ j6Fe++'^ -h 0^ 01*207= + 14H+ + 6Fe++ 20r+++ ^ 6Fe++‘^ -j- 7H2O This ionic reaction is the basis for the standardization of the dichromate solution as well as for the determination of iron. For the preparation of the dichromate solution, potassium dichromate is used. From the reaction in moleculai* form K20r207 + 6FeS04 + 7H2SO4 = 01*2(804)3 + 3Fe2 (804)3 + K2SO4 + 7H2O it is seen that one mole of K20r207 reacts with six moles of feiTous sul- fate. A normal solution of an oxidizing agent is one that contains a gram-equivalent weight per liter. The gram-equivalent weight for an oxidizing agent is defined as that weight which corresponds to one unit change in valence or the gain of one electron. For K2Cr207, since the total change of valence is 6, in order to make a normal solution there is required one-sixth of the molecular weight in granis per liter. Hie molecular weight of K2Cr207 is 294.21. A normal solution therefore PREPARATION AND STANDARDIZATION OF 0.1 N SOLUTIONS 117 calls for 49.03 grams, and an exactly 0.1 N solution for 4.903 grams of the pure salt per liter. Foi a 0.1 solution, of the ferrous salt, onc-tonth of the molecular weight should be taken, since each iron ion in being oxidized from the ferrous to the ferric condition undergoes a valence change of 1 . Ciystal- lized feirous sulfate, FeS 04 * 7 Il 20 , or ferrous ammonium sulfate, FcS 04 *(NPl 4 ) 2 S 04 * 6 H 20 , may be used, approximately 28 grams of the former or 40 grams of the latter for a liter of the solution of approximate 0.1 A'' strength. Preparation of Solutions. Procure 5 grams of KaCraOr and 28 grams of hydrated ferrous sulfate, FeS 04 - 7 H 20 or, if this salt is not to be had, 40 grams of ferrous ammonium sulfate, FeS 04 ■ (NIl 4 ) 2 S 04 -CHoO (Mohr’s Salt). 4 / 24 ^ Pulverize the 5 grams of potassium dichromate and dissolve in about 200 ml. of distilled water, warming if necessary. Transfer the solution to a liter measuring flask and dilute to the mark. This solution is approximately O.IN. It should be noted here that potassium dichro- mate can be procured of sufficiently high purity so that an accurately weighed amount (4.903 grams) will give a standard solution of the desired normality directly. Dissolve the ferrous salt in distilled water, add 5 ml. of concentrated H2SO4 and transfer to a liter flask, making up the volume to 1 liter Keep the solution in a liter bottle. This solution is likewise approxi- matcly 0.1 iV. Stannous chloride and mercuric chloride reagents are supplied in the laboratory. Tlie fonner is prepared by dissolving 50 grams of stannous chloride in 100 ml. of concentrated hydrochloric acid and diluting to 1 liter with distilled water. A few pieces of metallic tin added to the re- agent prevents oxidation to the stannic state. The mercuric chloride is a saturated solution. The acid mixture is made by slowly adding 120 ml. of concentrated sulfuric acid and 35 ml. of concentrated (85 per cent) phosphoric acid to 850 ml. of distilled w-ater. Be sure the solution is cool before using. Comparison of Solutions. Clean and rinse two burets; fill one with the ferrous salt solution and the other wdth the potassium dichromate solution. Into a 500-ml. Erlenmeyer flask run about 30 ml. of the ferrous sulfate solution, acidify it with 200 ml. of acid mixture. Add 6 to 8 drops of diphenylamine sulfonate indicator. Now, slowly and with constant swirling of the flask, mn the dichromate solution into the ferrous salt solution. When the color changes from green to bluish gray, add the dichromate solution drop by drop. The end point is a purple color which persists for at least one minute. If the end point should be passed, 118 OXIDATION AND REDUCTION METHODS add 1 or 2 ml. of the ferrous salt solution, and again approach the end point, by careful addition of dichromate solution. Refill both burets and make another comparison, using approximately the same volumes as before. From the volumes of the two solutions used to complete the reaction, calculate the number of railliliteis of ferrous salt solution required to react with 1 ml. of dichromate solution. Standardization of Dichromate by Iron Wire, Unless the potassium dichromate used is of sufficient purity so that the solution, as made up under the procedure described above, may be regarded as of the exact desired strength, it is necessaiy to standardize the dichromate solution. The best primary standard is electrolytic iron, but iron wire of high and known purity may seiwe the purpose here. The wire is dissolved in hydrocliloric acid and, since the oxygen of the air may oxidize some of the dissolved ii’on, it is necessaiy to make certain that all the iron is in the ferrous state. The solution is reduced with stannous chloride, ac- cording to the reaction 2Fe-^++ + Sn++ = 2Fe++ -f Sn++++ An excess of stannous chloride is unavoidable ; but this excess must be removed because, being a reducing reagent, it will react with the di- chromate solution and introduce a seiious error. The excess is removed by mercuric chloride, which reacts according to the equation SnCl 2 + 2HgCl2 = SnCU + Hg 2 Cl 2 Procure from the instructor about 1 gram of iron wire. Examine it careful^ for rust and, if rust is present, remove it by rubbing with sand- paper. Weigh out three separate portions of iron wh*e of about 0.2000 gram each. Ascertain the purity of the iron \vire. Into each of three 500-mI, Erlenmeyer flasks place 10 ml. of concen- trated HCl and 10 ml. of distilled water, cover ^vith watch glasses and bring just to boiling. Remove the flasks from the flame and into each flask drop the portion of the iron wire as weighed out. Cover with watch glasses and boil until the wire is dissolved. Wash down the sides of the flasks and the watch glasses vnth distilled water from the wash bottle. In continuing the procedure beyond this point, carry each inffividual solution of iron through to the end of the titration before starting to reduce the next one. Add stannous chloride (SnCb) solution dropwise to the first solution, wliich has been reheated to boiling, until it becomes colorless or slightly 119 DETERMINATION OF IRON IN IRON ORE green. Only a few drops should be necessary. Then add one drop in excess. Thoroughly cool under the water tap and add 10 ml of satu- rated HgCl 2 solution. If sufficient SnCla solution was u.sod in the reduc- tion of the standard iron, a silky white precipitate of mercurous chloride will be formed. If this precipitate is not obtaim-d, the reduction of the iron was not complete, and the solution must be discarded. Further- more, if too much stannous chloride was added, a black precipitate of free mercury will be formed; in this case the solution must likewise be discarded. Allow to stand for three minutes. Dilute with 200 ml. of acid mLxture. Add G to 8 drops of the indicator and titrate slowly with the K 2 Cr 207 solution until the green color changes through a gray to the purple end point. The purple color should jrersist for at leiist one minute. Reduce and titrate the second dissolved standaid as just described Do likewise with the third. From the net volume of K 2 Cr 207 used, calculate the iron titer of the dichromate solution and from this the normality of the solution. Also, from the comparison data calculate the normality of the ferrous solution. Repeat the entire standardization if the results do not check witliin 3 parts per 1000. Determination of Iron in Iron Ore Procedure. Weigh accurately three portions of the sample of iron ore of about 0.5 gram which has been dried and, if organic material was present, ignited, into three numbered 500-mI. Erlenmeyer flasks. Add 20 ml. of distilled water, cover the flasks with watch glasses and heat to boding. Add 25 ml. of hot concentrated HCl and continue to heat, '^is treatment should dissolve the ore sample in 15 to 20 minutes. From time to time replace the acid lost by evaporation. Wlien the sample is dissolved, there may be a residue of white siliceous material present, but no dark iron oxide particles. If the sample does not di.s- solve m half an hour, the addition of 1 ml. of stannous chloride solution may hasten the reaction. When complete solution is obtained wash do^vn the sides of the flasks and the watch glasses with distilled water. From this point on, as in the standardization, carry each individual solution of iron through to the end of the titration before starting the reduction of the next sample. To the first iron solution, reheated to boiling, add stannous chloride solution drop^vise until the solution becomes colorless or pale green. Considerably more stannous chloride ^\^I1 be used here than was used in the standardization. Add one drop in excess. Cool under the water tap. When thoroughly cold add 10 ml. of saturated HgCIa solution. 120 OXIDATION AND REDUCTION METHODS If sufficient stannous chloride solution was used to reduce the iron completely to the ferrous state, a silky white precipitate of mercurous chloride will be formed. If no precipitate forms, or if a dark precipitate of :ree mercury is obtained, the determination must be discarded. A^Jow the solution to stand for three minutes. Dilute with 200 ml. of acid mixture and add 6 to 8 drops of indicator. Titrate slowly with the K 2 Cr 207 solution. As the end point is approached the green color changes to gray. Now add the dichromate solution dropwise until a purple color lasting for at least one minute is obtained. Reduce and titrate the second sample of iron ore in a similar manner. Repeat with the third sample. _ From the data obtained calculate the percentage of iron in the sample. QUESTIONS 1. Define a normal oxidizing solution and a normal reducing solution. Show how the amount of KiCraOy and FeS04'7H20 required for 1 liter of solutions of 0.1 normality is determined. 2. Where, in qualitative analysis, was the stannous chloride-mercuric chloride reaction used? 3. If too much stannous chloride is used, and the subsequent addition of mercuric chloride results in a deposit of metallic mercury, why cannot the solution be titrated with potassium dichromate? 4. What is the qualitative test for ferrous ion? For ferric ion? What is Prussian blue? Turnbull’s blue? 5. Give the equation for the reaction between chromic ions and Na202. On the basis of this preliminary oxidation devise a method for the volumetric determination of a chromic salt. 6. Suppose you had on hand a standard solution of KMn04 and supplies of ferrous sulfate; how would you standardize a K2Cr207 solution? PROBLEM SET 6 DICHROMATE CALCULATIONS 1. Find (a) the Fe titer, (6) FeO titer, (c) Fe203 titer of a 0.1200 iV K2Cr207 solution. Ans. (a) 0.006702 (b) 0.008622 (c) 0.009582 2. What weights of K2Cr207 are required to prepare liter solutions of the following normalities: (a) 0.0500; (b) 0.1100; (c) 0.5000? Find the corresponding Fe titers. 3. A reducing solution was prepared by dissolving 45 grams of Mohr’s salt in sufficient water to make a liter of solution. How much KsCtjO? should be used to prepare 750 ml. of solution, equivalent to the reducing solution? Ans. 5.640 grams 4. A student standardized a K2Cr207 solution against iron wire which was 99.80 per cent pure and found that a 0.2250-gram sample of wire reqmred 38.46 ml, of K2Cr207 solution. Find the normality of the K2Cr207 solution. PROBLEMS 121 6. The following data wore obtained by a student during an iron determination. Comparison of solutions 33.89 ml. K2Cr207 o 30.65 ml. I'eS04 Standardization Weight of iron wire (99.8 per cent pure) = 0.2055 gram Vo ume of R2Cr207 used = 38 ,2 \'oIumc of FcS 04 used _ 2 G3 ml Analysis Weight of sample _ n rnoo Volume of K.Cr.O, used . 3, ,03 Volume of I- cb04 used = 150 ml Find the percentage of iron in the sample. 4 „ «q K2Cr207 solution will represent 1 per cent of Fe in the sainjilc? ’ ^ 7. An ignited residue of a mixture of FcsOi an Fe+++ + le Since each Fc'^'^ ion loses an electron, five Fe'^'*' ions react to furnish the fi\'e needed electrons to the Mn04~ ion. The balanced equation in ionic form, is then Mn04" + 5Fc++ + 8H+ = Mn++ + 5Fe+-^ + 4H2O This is the fundamental reaction used in the standardization of the required solutions ns w'cll as that involved in the determination of iron. PREPARATION AND STANDARDIZATION OF 0.1 N SOLUTIONS 127 For the preparation of the pennanganate solution tl>e potassium salt, KMn04, IS ahvays used to supply the Mn0.r ion; and to furnish the necessaiy Fe++ ions, either FeSO^-TH^O or FeSO^ ■ (NH.,)2S04 (irijO (Mohr’s salt) is employed. The reaction rewritten in molecular form is 2 KMn 04 + 10 FeSO 4 + SHjSO., = 2MnS04 + 5Fe2(S04)3 + 8II2O + K2SO4 doubling the relative proportions in order to balance the coefficients A normal solution of an oxidizing agent is, by definition, one that con- tarns one gram-equivalent weight per liter and corre.sponds to n gain of one electron (or one unit increase in positive valence). Hence for KMn04, since the change in valence is from +7 to -f 2 or a gain of five electrons, there is required for a liter of a normal solution, one-fifth molecular equation 2ICMnO4/10). This calls for 31.01 grams of KMn04 for a liter of .solu- tion of normal strength and 3.161 grams for a 0.1 A solution For a normal solution of a ferrous salt, since the valence change for the iron is from -|-2 to +3, or a lo.ss of one electron, there will be re- quired the moleculai’ weight of the salt in grams per liter The same conclusion can be reached by consideration of the molecular equation because, since two gram-moles of KMn04 react with 10 gram-moles of FeS04 and a normal solution of KMnO.! requires a weight corresponding to 2KMnO4/10, a weight equivalent to this must be taken for the fer rous salt, that is, 10FeSO4/10, or simply FeS04. This corresponds to 28 giams of crystallized ferrous sulfate, FeS04 • 7H2O, or to 40 grams of Mohr s salt, FeS04 • (NH4)2S04 • 6H2O, for a liter of approximately 0. 1 iV solution of the reducing agent. Preparation of Solutions. Weigh out about 3.25 grams of KAInOj dissolve it in about 900 ml. of hot water, dilute to about 1 liter, bring to a boil and boil for several ^utes. Allow the solution to cool, pour it into a liter bottle and set it aside for several days. Potassium permanganate usually contains some hydiated Mn02 which, if allowed to remain or to accumulate in the solution, causes decomposition and weakening of the solution. This impurity must be removed by filtering the solution through asbestos in the following man- ner. After the solution has stood for several days set up a suction flask and filter tube, place a layer of asbestos in the bottom of a large Gooch crucible and wash the asbestos with about 10 ml. of hot KUnOi solu- tion acidified with H2SO4. See page 192 for Gooch crucible assembly. This treatment oxidizes organic matter which may be in the asbestos and which would reduce the permanganate later on. Wash the filter 128 OXIDATION AND REDUCTION METHODS free from acid, then heat the permanganate solution almost to boiling and filter hot through the asbestos. Preserve the solution in a glass- stoppered bottle, preferably one of dark glass. Guard the solution against the action of strong light and the inclusion of dust. If at any time a deposit of brown manganese dioxide is noticed the solution must be re- filtered through asbestos or else siphoned off. To prepare the ferrous sulfate solution, weigh out on the rough bal- ance about 40 grams of Mohr’s salt (ferrous ammonium sulfate) or about 28 grams of FeS 04 -71120, dissolve in distilled water, add about 5 ml. of concentrated sulfuric acid, make up to a liter and transfer to a liter bottle. The ferrous sulfate solution prepared for the dichromate procedure may, of course, be used here. Comparison of Solutions. Glass-stoppered burets must be used for permanganate solutions since the rubber-tipped Mohr burets are apt to cause some reduction of the solution with consequent inaccurate results. Wash out one buret with KMn 04 solution, allow to drain and then fill with the peimanganate solution. Wash out and fill another buret with the ferrous salt solution. Measure out about 30 ml. of the ferrous solu- tion, dilute to 100 ml., add 25 ml. of “preventive solution (see page 129) and titrate vAth. the permanganate solution, establishing the end point when the solution is faintly pink. Read the upper meniscus of the per- manganate solution. Repeat the comparison. Calculate for each comparison the number of milliliters of ferrous so- lution oxidized by 1 ml. of KMn 04 and also the number of milliliters of KMn 04 reduced by 1 ml. of ferrous solution. If checks are not satisfactory, repeat the entire titration. Record the data and results in the notebook. Selection of Method of Standardization and Determination. Among the several methods available for the standardization of a permanga- nate solution the following are to be considered: (1) with iron wire, (2) with pure sodium oxalate, (3) by secondary standardization by means of an already standardized solution of a ferrous salt. If such a standard solution of ferrous sulfate is at hand (prepared in the dichro- mate procedure), this might of course be used here to standardize the permanganate although it is not recommended. In this case the com- parison of solutions serves as a standardization. For general purposes, standardization with pure sodium oxalate is usually recommended. It is particularly recommended here for student training, especially if time in the course does not permit making a volu- metric calcium determination, since it gives practice with tliis impor- tant reaction. The procedure for standardization .with sodium oxalate is given on page 137. ZIMiMERMANN-REINHARDT METHOD 129 There are two alternative methods described in the following pro- cedures for the standardization of the pennanganatc solution by iron wire and the application of these metliods to the determination of iron in iron-containing samples. The first method described is the Zimmer- maim-Reinhardt method and the second the Jones reductor method. The instructor will specify which of these methods is to be employed in the standardization and in the iron ore anal3^sis. There are certain limitations in the application of the pennanganate method to the determination of iron. If the method is to be apiihcd to the determination of iron in iron ores, compounds of titanium and vana- dium must be absent, because the.se elements are reduced in the Jones reductor and oxidized by pennanganatc; in this contingency some other reducing agent must be used. In the second place, in all titrations with permanganate, chlorides must either be absent or a modification must be introduced to overcome the effect of chlorides. Since non ores arc most readily put into solution with hydrochloric acid, the solution must either be fumed down with concentrated sulfuric acid to remove the cliloride or a special reagent must be added whereby it is possible to titrate the iron in the presence of chlorides. Chlorides, introduced into the solution through the use of hydro- chloric acid as solvent, or the use of stannous chloride as reducing agent (as in the reduction of the iron in the dichromate method), interfere because some chloride ions will be oxidized, thas rendeiing the results high for iron. The modification followed when chlorides are present makes use of a special reagent consisting of a mixture of idiosphoric acid, manganous sulfate and sulfuric acid. Zitnmermann-Rcinhardt Method This method for the standardization of a potassium permanganate solution with iron we as well as for the deteimination of iron in iron samples utilizes stannous chloride for the reduction of the iron, as in the dichromate process, and embodies further the use of a special re- agent mixture, loiown as "preventive solution,” to overcome the harmful effects of chlorides. This preventive solution is made b^^ dissohing 200 grams of manganous sulfate, MnS04-4H20, in a liter of water and to it adding a cold mixture of 400 ml. of concentrated H2SO4, 400 ml. of 85 per cent H3PO4 and 1200 ml. of water; the reagent will be supplied in the laboratory. Titration of iron solutions bj' potassium pennanga- nate in the presence of chlorides gives results for iron which are too high, becai^e some permanganate is reduced by the chloride ions. In cold solutions where the chloride concentration is low, this error may not be 130 OXIDATION AND REDUCTION METHODS very large. However, the addition of a manganous salt, in this case manganous sulfate, eliminates this diflaculty by lowering the oxidation potential of the permanganate. This decreases the oxidizing action of potassium permanganate on chlorides to a negligible extent without appreciably affecting its reaction with ferrous iron. There will also be supplied stannous chloride (50 grams dissolved in 100 ml. of concentrated HCl and diluted to a liter) and saturated mercuric chloride, to reduce excess SnCl 2 . Standardization. Ascertain the purity of the iron wire to be used for the standardization. Weigh out three portions of about 0.2 gram each of the cleaned wire into 500-ml. Erlenmeyer flasks, add 10 ml. of hot con- centrated HCl and 10 ml. of distilled water. Cover the flasks with watch glasses and gently boil until the wire is dissolved. Fill a buret with the permanganate solution. Wash down the sides of the flasks and watch glasses with distilled water from the wash bottle. From this point in the procedure, carry each individual solution of the standard through to the final titration before starting the reduction of the next portion. Bring the first solution to a boil again and add stannous chloride re- agent drop by drop until the solution becomes colorless or faintly green. Onl 3 '^ a few drops may be needed. Add no more than one drop in excess. Thoroughly cool under the cold water tap and add all at once 10 ml. of saturated mercuric chloride reagent. Allow the solution to stand for three minutes. A silky, white precipitate of mercurous chloride form- ing at this point indicates that the procedure has been correctly fol- lowed. If no precipitate foiTns, too little stannous chloride was added and the iron was not completely reduced; if a black precipitate of free mercuiy is obtained, too great an excess of SnCU solution was present. In both these cases the solution must be discai'ded. Add enough water to bring the volume to about 300 ml. Then add 25 ml. of the preventive reagent and titrate with permanganate without delay. The end point is reached when a pink color persists in the solu- tion for at least half a minute. Cany out these steps for the second portion of the iron standard in exactly the same manner. Repeat ^\'ith the third portion. To calculate the normality of the peimanganate solution, divide tlie weight of iron standard by the volume of permanganate used and divide this in turn by 0.05584, the gram-milliequivalent weight of iron. Determination of Iron, If the sample is a readily soluble salt, such as ferrous ammonium sulfate, weigh out three 0.5-gram poiiions into 500-ml. Erlenmeyer flasks, add 10 ml. of water and 10 ml. of concen- trated HCl, heat to boiling. While hot reduce the first one with a few ZIMMERMANN-REINHARDT METHOD 131 drops of stannous chlonde reagent dropwnse and proceed with tlie treat- ment of the reduced solution, according to the instructions given below for the treatment of the solutions obtained from iron ore samples. If the sample is an iron ore, dry it at 110°C. for one hour. Ask the instnictor whether the samples contain organic material and, if so, weigh into crucibles three 0.5-gram portions; place these in a bright red muflie furnace for half an hour or heat them strongly with a Moker burner* transfer the ignited samples, without loss, to SOO-ml. Erlenmcj'ci- flasks.' Should the sample not require ignition to destroy organic matter, weigh directly into the flasks. To each flask add 20 ml. of distilled water and cover the flasks with watch glasses. Heat to boiling and then add 25 ml. of hot concentrated hydrochloric acid. Continue to heat until the ore is in solution, loavino- only a small light^colored sediment of silica. A little stannous chloride aids in dissolving the ore in ca‘^e the acid treatment is slow. From time to time during the dissolving of the ore, replace the acid lost by evapora- tion. Wash off the watch glasses and rinse down the sides of the fhisks. From this point on, each individual solution must be carried through the titration before starting to reduce the next one. To the first hot solution add stannous chloride dropwise until the red or yellow color is discharged and the solution becomes a pale green. More of the reagent ^\^ll be needed here than in the reduction of the iron \Wre standards. An excess of more than two drops must be avoided. Now cool the reduced solution, whether it is from the ore or the soluble salt sample, under the water tap. Add 10 ml. of saturated mercuric chloride solution in order to oxidize the excess of stannous cliloridc. A silky white precipitate of mercurous chloride should fonn. If no pre- cipitate is obtained, too little stannous chloride was used in the reduc- tion and the solution must be discarded. If, on the other hand, the solu- tion turns gray or black, showing reduced, free mercury, because too large an excess of stannous chloride was added, it must be discarded because permanganate will oxidize the mercury and results for iron will be too high. Allow the solution to stand for three minutes. Dilute the solution to 300 ml. with cold water, add 25 ml. of the pre- ventive reagent and titrate without delay with the permanganate. At the end point the pink color should persist for one-half minute. Reduce the second sample of iron, whether from soluble salt or ore, and carry the procedure through exactly as given. Repeat with the third unknown. Calculate the percentage of iron in the sample, using both the titer and the normality methods of computatiori, 132 OXIDATION AND REDUCTION METHODS Jones Reductor Method _ X Standardization of Permanganate by Iron Wire. For the direct st»*ndardization of potassium permanganate, there are in use a number of primary standards, among which pure ii’on and sodium oxalate are the most satisfactory. The sodium oxalate method is described on page 137. The standardization with iron wire is here described. In getting the wire into solution some iron will become oxidized and must be re- duced before the standardization titration is undertaken. A number of reducing agents are available for this preliminary reduction, such as certain metals, stannous chloride, hydrogen sulfide and sulfurous acid. In the pi'esent method, zinc in conjunction with suKuric acid is used, because it does not introduce chlorides into the solution and does not necessitate the removal of excess reducing agent. The apparatus used is knomi as the Jones reductor. A device less cumbersome than the Jones reductor has been suggested. This consists of a long spiral of cadmium, zinc or aluminum wire. In this method about 0.2500-gram portions of iron wire (electrolytic iron is more accurate) are dissolved in sulfuric acid, placed in an Erlenmeyer flask and the spiral of metal added. WiUard and Fuiman recommend cadmium wdre or amalgamated zinc AN-ire, 2.5 to 3 millimeters in diameter and about 1 meter long, turned into a spiral with a hook at one end so that the spiral can be removed when the reduction is complete. The solution is boiled during the reduction, and for fifteen minutes after the solu- tion has become colorless. The spiral is then lifted out and AA'ashed off, the inside of the flask washed down and the reduced solution titrated with the 4 permanganate solution without delay. The procedure for the determination of iron is similar to that for the standardization. The Jones Reductor, This device, essentially a column of amalgamated zinc held in a vertical tube con- nected with a suction flask, is used here for the reduc- tion of the iron to the ferrous state. It is illustrated in Fig. 11. The reductor tube should have an internal diameter of about inch and be about 18 inches long. It is enlarged at the top and has a glass stopcock sealed to the bottom. Tlie extension beyond the stop- cock should be about 6 inches long, and should be supplied with a stopper which fits snugly into the neck of the suction flask. A second suction flask or a safety bottle should be inserted in the line as a trap to Fig. 11. Jones reductor. JONES REDUCTOR METHOD 133 prevent water from the suction pump from backing up into the first receiver. The amalgamated zinc is prepai'cd as follows. Dissolve about 5 grams of mercury in about 25 ml. of dilute nitric acid, warming if ncccssaiy to start the reaction. Then dilute to 500 ml. Add to the solution 250 grams of granulated zinc of 20 to 30 mesh. Pour off the solution and wash the zinc repeatedly with water until washings are no longer acid. Preseiwe the zinc in a wide-mouthed bottle, under water, and use as required. The reductor is prepared for use by placing a loose wad of glass wool in the constriction of the tube, on top of this some broken porcelain or glass and then a layer of asbestos, somewhat thicker than that used in Gooch cmcibles. Then the tube is filled to a depth of 12 to 15 inches with the amalgamated zinc, and a plug of glas.s wool is placed on top of the zinc column and the tube is filled at once with distilled water. Place a mark on the tube about 1 inch above the upper level of the zinc column, below which the liquid level must never be allowed to fall. The blank and the standardization can now be begun. Weigh accu- rately three portions of clean iron wire of about 0.2 gram each. Since the permanganate is approximately 0.1 each milliliter will oxidize about 0.005584 gram of iron, and for 10 ml. of solution there will be re- quired about 0.2 gram. Prepare at least 1 liter of dilute sulfuric acid of about 1.1 specific gravity. Dissolve each portion of wire in 100 ml. of this dilute sulfuric acid and boil the solutions for several minutes. The Blank. In order to be certain that the reductor is functioning properly and that no reducing impurities are present, a blank must be run. This consists of duplicating the exact conditioiLS of operation, ex- cept that the ii’on solution is omitted. It is carried out as follows. With the reductor properly prepared and connected with the suction flask, start the suction, partly open the stopcock at the bottom of the zinc column and slowly draw 200 ml. of distilled watci' through the column, being very careful not to have the water level fall below the mark above the top layer of zinc at any time. Discard the wash water. Follow the water treatment with a total of 200 ml. of warm, dilute sulfuric acid (sp. gr. 1.1), drawing the solution through at the rate of about 25 ml. per minute. Close the stopcock when the solution level is at the mark above the top of the zinc. This precaution is very impor- tant because if air enters the column it will form hydrogen j^eroxide which, in turn, will act on the peimanganate. Then draw 100 ml. of distilled water through the column, closing the stopcock when the water level is at the mark above the zinc. Detach the suction flask, cool the contents and quickly add a drop or two of the peimanganate solution 134 OXIDATION AND REDUCTION METHODS from the buret. If the solution remains pink, the absence of foreign reducing substances is assured and the apparatus is ready for the reduc- tion of the iron solutions; if not, repeat the acid treatment. The Standardization. When a satisfactory blank is obtained, and the iron standards are dissolved and heated almost to boiling, the stand- ardization can be completed. Pass the iron solution through the tube at the rate of about 25 ml. per minute. When the solution level has dropped to the mark, wash out the beaker which held the solution with 100 ml. of the dilute sulfuric acid and pass this through the tube. Finally follow the acid solution with 100 ml, of distilled water. Guard against allowing any portion of the zinc to be exposed to the air. Re- move the flask, cool the solution and titrate without delay, until a per- manent pink color is obtained. If the end point is passed, back-titrate with the ferrous sulfate solution. Repeat this procedure with the other iron standards. Calculate the net volume of permanganate required for each run. From this calculate the Fe titer of the permanganate solution. Calculate the normality by dividing the Fe titer by the gram-milliequivalent weight of iron (0.05584). Procedure for Soluble Iron Salts. Take samples of about 1 gram each, dissolve each portion in 5 ml. of concentrated H2SO4 and 50 ml. of water, heating the solution nearly to boiling. Add 50 ml. of water. If a day or more inteivenes between this analysis and the previous use of the reductor, run another blank. The procedure for the determination is the same as the standardization. Pass the solution through the apparatus at a rate of about 25 ml. per minute. Tlien pass 100 ml. of dilute sulfuiic acid (sp. gr. 1.1) through the column to wash out the iion solution, and follow the acid washing with 100 ml. of water, taking care that at no time the liquid level falls below the mark. Detach the suction flask, cool the reduced solution and titrate at once with standard potassium permanganate solution, using the standard ferrous sulfate solution in case the end point -is passed. From the weight of sample used and the values of the standard solu- tions, calculate the percentage of iron in the sample. Procedure for Iron Ores. Weigh out into porcelain crucibles three portions of ore, of about 0,5 gram each. Roast the ore if necessary at a dull redness for about a half-hour, in order to oxidize carbonaceous and nitrogenous materials which may be present. Place the cooled crucibles in casseroles or evaporating dishes, add 25 ml. of dilute hydrochloric acid (sp. gr. 1.12) and heat the samples until solution is complete. To remove the hydrochloric acid, add 5 ml. of concentrated sulfuric acid and start evaporation on the steam bath. Wlien the solution is PREPARATION AND STANDARDIZATION OF 0.1 N SOLUTIONS 135 clear, heat the dishes on the hot plate until white fumes of SO 3 are evolved. Add not over 100 ml. of water and heat the solutions. The reduction of the iron in tlie Jones reductor and the subsequent titration are carried out exactly as described above for soluble samples. Calculate the percentage of iron in the iron ore. QUESTIONS 1. What reducing agents are frequently used as standard solutions in permanganate methods? In the determination of iron, explain ^vhy standard oxalic acid is not recom- mended. 2. What primary standards are available for the standardization of KMnO^ solu- tion? Why use iron wire in the above method? Outline a method hy which one might standardize KMUO 4 indirectly. 3. What reducing agents might be used to reduce iron to the ferrous condition? Write and balance the equations for the reduction of ferric ion by (a) nascent hydrogen * ( 6 ) SnCb: (c) SO 2 : (d) H 2 S. ' 4. Write the essential reaction that takes place in the Jones reductor during the reduction of iron solutions. Why must a blank be run on the reductor? Why is it so important to prevent air from coming into contact with the zinc in the reductor? Write the reaction between KMn 04 and II 2 O 2 in H 2 SO 4 solution. 5. How docs KMn 04 behave in neutral solution? If a certain KMnOi solution is of normal strength in an acid solution what is its normality when u.sed in a neutral solution? 6 . Outline three methods for the determination of manganese in alloys such as steel. VoLUMETiuc Determination of Calcium The volumetric determination of calcium is based on the precipita- tion of calcium oxalate, its subsequent solution in sulfuric acid and the titration of the liberated oxalate ion with a standard solution of potas- sium permanganate. The method is an example of those permanganate processes in which a reducing substance liberated in a suitable reaction and estimated by titration may be made the basis of a quantitative method of analysis (see page 125). The present procedure involves the preparation of standard solutions of potassium permanganate and oxalic acid and the standardization of KlMn 04 by Na 2 C 204 , Avhich as a primary standard for KMn 04 is considered more accurate than an iron standard. Preparation and Standardization of Approximately 0.1 N Solutions of Pota^siu7n Permanganate and Oxalic Acid In the reaction between potassium permanganate and oxalic acid in the presence of sulfuric acid, the manganese is reduced from a valence of +7 to a valence of +2. The oxalic acid is oxidized to carbon dioxide 136 OXIDATION AND REDUCTION METHODS and water (when acting as a reducing agent and not as an acid). In the molecule of H2C2O4 the two carbon atoms have different valence charges one atom having 4 positive charges and the other atom 3 positive and 1 negative one, as evident from the formula The average valence number is + 3 . • When a molecule of H2C2O4 is oxidized to CO2 and H2O the increase in valence is 2. Since there are two carbon atoms, the average change per carbon atom is therefore an increase of 1, and for the molecule it is 2. From an electron transfer standpoint we have and C204=-^2C02 + 2 e Mn 04 ” + 8 H+ + 5 e Mn 04 “ + 4H2O This leads to the balanced ionic equation 2Mn04~ + 50204“ + 16 H+ = 2 Mn++ + IOCO2 + 8H2O and molecularly 2 KMn 04 + 5H2C2O4 + 3H2SO4 = 2MnS04 + IOCO2 + 8H2O + K2SO4 The reacting ratios are 2KMn04 : 5H2C2O4, A normal solution of KMn04 contains per liter of solution one-fifth of the molecular weight in grams (in the above molecular equation 2KMn04/10). The weight required for a liter of normal solution is therefore 31.61 grams, and of a 0.1 N solution 3.161 grams, of KMn04. The crystallized oxalic acid contains two molecules of water of crystal- lization. An oxalic acid solution, to be equivalent to this, must there- fore contain 5(1120204 •2H20y 10 or H2C204-2H20/2 gram-moles, that is, one-half the molecular weight in grams of the hydrated acid, for a normal solution. For a 0.1 iV solution tliis calls for 6.3 grams per liter. If the anhydrous acid, H2C2O4, is used, there vnll be required 4.5 grams. Preparation of Solutions. The potassium permanganate solution pre- pared for the determination of iron may be used here. If this solution is not at hand follow the directions given on page 127 for the prepara- tion and purification of an approximately 0.1 iV solution of KMn04. To make a liter of approximately 0.1 A' oxalic acid solution, weigh out about 4.6 grams of H2C2O4 or about 6.3 grains of H2C204*2H20. Dissolve in water, transfer to a liter flask and dilute to 1 liter. DETERMINATION OF CALCIUM 137 Comparison of Solutions. Clean and fill a glass-stoppered buret with the permanganate solution. Clean and fill another buret with the oxalic acid solution. Run out into a 500-ml. Erlenmeyer flask an accurately measured volume of about 35 ml. of oxalic acid solution. Dilute to about 200 ml. with distilled water, add 5 ml. of concentrated sulfuric acid and titrate rapidly with the pennanganate solution, "flien warm to about G0°C., and add KI\In04 -slowly. Near the end point add the solution drop by drop. Use the oxalic acid solution, if nccc.' Mn++, the equivalent weight is Mn02/2, from which the gram-milliequivalent weight and the value of 1 ml. of the KMn04 can be calculated in terms of Mn02. From this the purity of the ore, as percentage of Mn02, can be readily calculated. 140 OXIDATION AND BEDUCTION METHODS QUESTIONS 1. Define a normal solution of a reducing^ agent. Show how the amount of oxalic acid required for a liter of normal reducing solution is determined. 2. Supply the proper divisors to make the following ratios equivalent: KMnOi : H 2 C 2 O 4 : CaC 204 : CaO 3. If KMn 04 is to be employed for the analysis of impiue calcium carbonate* suggest a method of standardizing such a solution other than by use of iron or sodium oxalate. 4. Devise a method for the volumetric determination of calcium in limestone, which may contain silica, compounds of iron, aluminum and magnesium in addition to CaCOa. 5. Could you use a standard solution of ferrous salt if the end point were over- stepped in the determination of calcium? PROBLEM SET 7 PERMANGANATE CALCULATIONS 1. Find the normality of KMn 04 solutions containing the following weights of KMnO^ per liter: (a) 15.800 grams: ( 6 ) 2.655 grams; (c) 64.21 grams. What are the corresponding Fe titers? Ans. (a) 0.5000 iV; 0.02763 (5) 0.08401 W; 0.004691 (c) 1.716 AT; 0.0958 2. Find (a) the Fe titer, (5) FeO titer, (c) Fe 203 titer of a 0.1500 AT KMn 04 solution. 3. How many grams of FeS 04 - (NH 4 ) 2 S 04 ' 6 H 20 should be used to prepare a liter of a solution equivalent to a KMn 04 solution containing 3.644 grams per liter? A ns. 45.22 4. Find the normality of a KMn 04 solution of which 36.23 ml. is equivalent to 1.811 grams of FeSO.i(NH 4 ) 2 S 04 - 6 H 20 . 6 . A student standardized a KMnOi solution against iron wire which is 99.8 per cent pure and found that a 0.1906-gram sample of wire required 33.85 ml. of KMn 04 solution. What is the normality of the solution? What is the Fe titer of the solution? Ans. 0.1006 A^; 0.005618 6 . The following data were obtained by a student during an iron determination: 1.00 ml. of KMnO.j 1.12 ml. of FeS 04 ; 0.2152 gram of pure iron wire required; 35.70 ml. of the KMn 04 solution and 2.65 ml. of the FeS 04 solution for back-titration. A 0.3126-gram sample of iron ore required 29.15 ml. of KMn 04 solution and 3.60 ml. of the FeS 04 solution for back-titration. Wliat was the percentage of iron in the sample? 7. What weight of iion ore shotfid be tahen so that each milliliter of a 0.1250 iV KMn 04 solution will represent 1 per. cent of Fe 2 Cb in the sample? Ajis, 0.9982 gram 8 . What should be the normality of a permanganate solution in order that each milliliter used will represent 0.5 per cent of Fe in a 0.5000-gram sample? 9. A certain FeS 04 (NH 4 ) 2 S 04 ' 6 H 20 solution has an iron titer of 0.005720. To Nvhat volume must a liter of this solution be diluted in order to make it exactly O.IW? Ans. 1024 ml. CERIC SULFATE METHODS 14^ 10. A commercial hydrogen peroxide solution was analyzed hy titration with potassium permanganate. A 10.00-mI. sample (sp. gr. 1.01) required 32 75 ml rrf 0.6 iV IvMn04. Find the percentage of H2O2 in tlic sample 11 An ignited of a mixture of Fe^O, and Ab03, obtained in the analysis of limestone, weighed 0.3400 gram. Ihis was fused with KIISO^ and dissolved in sulfuric acid, reduced, and the iron titrated with 0.1250 N KMnO.i, requiring 27 75 ml Find the weight of FC2O3 and AhOs in the sample. An,-i 0 2770- 0 0(330 12. Calculate the normality of KMn04 solutions having the following titers- (a) H2C204 titer of 0.004672; {/)) CaO liter of 0.02608; (r) FCSO4 tiler of 0 01620 13. What arc the milicquivalent weights of the following as reducing agents; (а) H2C^04 -21120 (б) KHCiOi-H'-CoOi -21120 (c) FeSO-i - 71120 ((/) Na2C204 Ans. (a) 0.0G303 {!>) 0.06354 (r) 0.2780 (d) 0.06700 14. If a 0.1985-gram sample of pure calcitc. CaCO.i. is used in standardizing a KMn04 solution, and 33.78 ml. of the permanganate solution arc requiicd, what is (a) the normality, (5) the CaCOs titer, (c) the CaO titer, and (d) the Ca titer of the permanganate solution? 15. It was found that 25.00 ml. of a certain 1120204 solution was neutralized by 28.70 ml. of a 0.1000 N NaOH solution. Find the normality of a KMn0.i solution 33.75 ml. of which reacts with 30.35 ml. of the II2C2O4 solution. Ans^ 0 1032 V 16. The calcium in a 0.1981-gram sample of limestone was precipitated as CaC20 This was then dissolved in sulfuric acid and titrated with 29.(H ml. of KMnO, w^ich had a Na2C204 titer of 0.006480. What was the iiercentage of CaO in the limestone? 17. The CaO titer of a certain KMn04 solution is 0.003806. How many grams of K2Cr207 must be taken to make a liter of potassium dichromate of the .same strength? Aras. 6.653 grams 18. How much water should be added to a liter of KMn0.i solution Iiaving a CaO titer of 0.002912 wliich is to be used for oxidation in an acid medium in order to make the solution 0.1000 N? 19. The following substances were mixed: 35 ml. of 0.1000 KMn04 solution 3.50 gi'ams of Na2C204, and 4.30 grams of FeS04 -7H20. The mixture was diluted to 250 ml. and acidified wath H2SO4. Show by a calculation whether the resulting solution is oxidizing or reducing. Am. Reducing 20. Pyrolusite (impure Mn02) is analyzed by treating the mineral with a known excess of reducing agent in the presence of acid and then back-titrating the reducing agent with standard KMnOi solution. The first reaction is described by the equation Mn 02 + H2C2O4 + H2SO4 — ♦ MnS 04 + 2CO2 4 - 2H2O Calculate the percentage of Mn02 in a sample from the following data. A 0.2200- gram sample was treated with 50.00 ml. of 0.1024 A^ H2C2O4 solution in the presence of sulfuric acid. The excess oxalic acid was titrated with 36.75 ml. of 0.1096 N KMn04 solution. * C. CERIC SULFATE METHODS Ceric sulfate, Ce(S 04 ) 2 , a newcomer to the group of strong standard oxidizing reagents, is in many respects similar to potassium permanga- 142 OXIDATION AND REDUCTION METHODS nate. In oxidizing power it is nearly as strong as the latter and stronger than potassium dichromate. In the reaction with ferrous ion Fe++ + Ce++‘''+ = Fe'*'++ + Ce+++ and with oxalates €204“ + Ce++++ = 2CO2 + 2Ce+++ the change in valence is from +4 to +3 or a change of one unit. Qe++++ -f e Ce+++ The equivalent weight is therefore the atomic weight, and to make a 0.1 solution, one-tenth of the gram-molecular weight of the salt should be taken. Ce(S04)2-2(NH4)2S04*2H20 is usually used to pre- pare the solution; this has a molecular weight of 632.5; therefore about 63 grams are taken, dissolved in 25 ml, of concentrated H2SO4 and di- luted to 1 liter for a 0.1 A'' solution. This is best standardized against pure iron as primary standard or, indirectly, against a previously stand- ardized solution of FeS04. In the standardization and reactions with iron, orthophcnanthroline is used as indicator, changing color from red to blue or purple. The indicator, when in a ferrous solution, consists of the probable complex, Fe(Ci2H8N2)3''"’' and, when oxidized, of Fe(Ci2H8N2)3'^“*“^. The reactions, unlike those with permanganate, can be conducted in the presence of hydrochloric acid. In carrying out reactions of iron vdth ceric sulfate, the iron is reduced with SnCl2 as in the dichromate or permanganate method. The titration then proceeds according to the procedures already de- scribed. D. IODINE METHODS Volumetric methods of analysis based upon the reactions of iodine and its compounds are extensively employed in analytical chemistry. Iodine methods constitute by far the most important and largest class of volumetric procedures. Three forms of the element iodine — namely, the free element, I2, the iodide ion usually introduced as IQ and the iodate ion used in the form of its potassium salt, KIO3 — are employed. Iodine is a good oxidizing agent and can be used as a standard solution to oxidize directly a number of reducing substances. The iodide ion, I”, is an excellent reducing agent and reacts with a large, number of oxidized substances to liberate free iodine, which may then be measured by titration with a standard reducing agent. Potassium iodate is a good oxidizing agent and may be used directly as a standard solution or in a mixture with potassium iodide. In most of the iodine methods starch IODINE METHODS 143 is used as indicator; it gives an intense blue color in the presence of a slight excess of free iodine. Fundamental Reactions. Iodine reacts quantitatively with sodium thiosulfate, according to the equation I2 4- 2Na2S203 = NaaS^Ofl -f- 2NaI (j) This reaction when applied to titration procedures is one of the most accurate and best-suited reactions of quantitative analysis. The com- pletion of the reaction is admirably marked by the starcli indicator Another fundamental reaction is that between iodine and the arsenite ion, AsOa^, the reaction being quantitatively complete in the presence of sodium bicarbonate: I2 "b NasAsOs 2NaHC03*" — NagAsO.! + 2NaI -f 2CO2 + ll^O (2) Starch indicator is likewise used in this reaction. These two reactions are of considerable importance because they are used to measure the amount of iodine liberated by the action of an oxi- dizing agent on potassium iodide. Tlie action of an excess of (unstand- ardized) potassium iodide on a substance capable of being reduced can be shown schematically thus: Oxidizer + 2I~ I2® + reduced form of oxidizer the liberated iodine being then titrated by a standard solution of either sodium thiosulfate or sodium aisenite. The amount of standard reducer used becomes then a measure of the amount of oxidizer contained in the sample. Classification of lodimetric Methods. 1. Methods of analysis which depend upon the consumption of iodine are termed iodimetric methods. Such methods arc carried out cither directly by titrating the constituent being determined with a standard solution of iodine, or indirectly by adding a measured excess of standard iodine and back-titrating the excess with a standard solution of sodium ai-senite or sodium thiosulfate. 2. Reactions in which iodine is liberated from potassium iodide by the action of the constituent being determined and the iodine then titrated by a standard solution of sodium thiosulfate or sodium arsenite are called iodometric methods. 3. lodate methods employ a standard solution containing KIO3 either alone or in admixture with KI. Closely allied to the iodate methods are those that employ potassium bromate. Time and space do not permit describing these methods in detail nor offering laboratoi’y pro- cedures involving these interesting methods. They are briefly summa- rized on page 155. 144 OXIDATION AND REDUCTION METHODS Methods Involving the Consxtmption op Iodine (loDiMETRic Methods) A already pointed out, certain materials in the lower state of oxidar ti*:.n can be quantitatively oxidized by iodine. The sample may either be titrated directly with a standard solution of iodine, or else an excess of standard iodine may be added and the excess back-titrated by means of standard arsenite or thiosulfate solution. In either case, the deter- mination rests upon the direct oxidation of the constituent by standard iodine. For example the reaction between iodine and sodium thiosulfate may become an accurate method for the determination of thiosulfates by titration of the sample with an iodine solution. (Conversely, free iodine in a sample of material may be determined by titration with standard thiosulfate.) In a similar manner, the reaction between iodine and the arsenite ion in a sodium bicarbonate solution is the basis for a direct iodimetric ' determination of trivalent arsenic. This method is described in detail in the followng procedure. Determinations of antimony and tin are based on similar reactions. Sulfites may be oxidized by iodine, according to the reaction H 2 SO 3 + I 2 4" H 2 O = H 2 SO 4 4- 2HI Likev4se, sulfides, by the reaction II 2 S + I 2 = S 4- 2HI The latter reaction is involved in the so-called evolution method for the determination of sulfides in steel. In actual practice, the steel is dissolved in hydrochloric acid and the evolved hydi'ogen sulfide is passed into cadmium chloride or zinc sulfate solution, by which the hydi'ogen sulfide is retained as CdS or ZnS. The precipitated sulfide is then dis- solved in an acidified solution containing an excess of standard iodine and the excess back-titrated 'with thiosulfate solution or, alternatively, titrated directly with standard iodine or iodate. The determination of arsenic in impure arsenious oxide samples is selected as a typical example of a direct iodimetric method and will give practice in the use of iodine and arsenite solutions. Volumetric Determination of Arsenic The method is based on the oxidation of arsenic from the arsenious to the'arsenic state, by direct titration with standard iodine. The deter- mination involves the preparation and standardization of solutions of PREPARATION AND STANDARDIZATION OF 0.1 N SOLUTIONS 145 iodine and sodium arsenite, which are then used in the determination of arsenic in samples of impure AS2O3. Preparation and Standardization of Approximately 0.1 N Solutions of Iodine and Sodium Arsenite In the reaction l2° + AsOa^ + H2O = 21“ + As 0 .r + 2H + the iodine is reduced to the iodide ion, each atom of iodine undergoing a valence change of 1. because in the reaction I2 + 2c ^ 21“ each atom of iodine gains one electron. A normal solution of iodine therefore contains the atomic weight of iodine in grams per liter. A liter of one-tenth normal solution of iodine will contain one-tenth the atomic weight of iodine or 12.693 grams. The solubility of iodine in water Is slight but it dissolves readily in a solution of potassium iodide. Wien iodine dissolves in potassium iodide, it probably does so according to the reaction I2 + KI = KI3 forming the double salt KI-I2 which ionizes into K'*’ ion and the tri- iodide ion, 13“ A solution of iodine in potassium iodide behaves like I2 and will be so regarded here. A solution of an arsenite, in order to be noimal as a reducing agent, must contain the gram-equivalent weight per liter of solution. When the AsOa” ion is oxidized to ASO4” the valence change is 2, since two electrons are lost, as can be seen from the half-cell reaction AsOa^ + H2 O -> As 04^ + 2II+ + 2e Tlierefore a normal solution contains one-half the molecular weight of the compound in grams per liter. The starting material for a sodium arsenite solution is usually arsenious oxide. This dissolves in sodium hydroxide, as shown in the equation AS2O3 H- 6NaOH = 2Na3As03 3H2O Since in a molecule of AS2O3 there are two arsenic atoms, each of which changes valence from H-3 to -f-5 on being oxidized, there is required one-fourth the gram-molecular weight of arsenious oxide per liter for a normal solution. For a liter of normal solution there are required there- fore AS2O3/4 = 197.82/4 = 49.45 grams and for a 0.1 A'' solution 4.95 grams. The reaction between iodine and sodium arsenite I2 H" Na3As03 -j- H2O ^ 2HI -f- Na3As04 146 OXIDATION AND REDUCTION METHODS is decidedly reversible and reaches equilibriiim while still large amounts' of arsenite remain in the solution, owing to the interaction of hydriodic acid with the arsenate ion. If means are taken to prevent the hydrogen- ion concentration from building up, the reaction can be made to go to practical completion in the desired direction. Such a means is avaB- abJe in a sodium bicarbonate buffer, the action of the bicarbonate being to neutralize the HI as fast as it is formed. In the presence, then, of sodium bicarbonate, the following reaction is satisfactorily quantitative: I 2 + NasAsOa + 2 NaHC 03 = 2NaI + Na 3 As 04 + 2 CO 2 + H 2 O Preparation of Solutions. It is possible to make a solution *of iodine of the exact desired strength, from carefully purified and resublimed iodine, with the exercise of extreme care to prevent loss. For the present purpose, how’ever, it is better to make the solution of an approximate strength and to standardize it against pure AS 2 O 3 . For a liter of an approximately 0.1 iV solution of iodine, weigh gut about 13 grams of iodine and 20 grams of KI. Rub these together in a mortar with small portions of water until dissolved. Transfer the solu- tion to a liter flask, make up to 1 liter. Transfer the solution to a liter bottle. This should be sufficient for this and the following copper deter- minations. Wrap dark paper around the bottle (an amber-colored bottle is better) to protect the solution from decomposition by direct sunlight. For 500 ml. of an approximately 0.1 fV solution of sodium arsenite, weigh out about 2.5 grams of arsenious oxide. Prepare a sodium hy- droxide solution by dissolving 20 grams of NaOH in 250 ml. of water. Add enough of the sodium hydroxide to the arsenious oxide to dissolve it (10 ml. should be sufficient). Dilute to 500 ml. and transfer to a liter bottle. If starch indicator solution is not supplied, prepare a quantity by rubbing about 1 gram of potato or soluble starch into a paste with cold w^ater and then pouring this into about 100 ml. of boiling waten Set the solution aside for some time and then decant the clear solution. Comparison of Solutions, Measure from a buret about 35 ml. of the sodium arsenite solution. Dilute to 100 ml. Add a drop of phenol- phthalein and then dilute hydrochloric acid (sp. gr. 1.12). The phenol- phthalein indicator may show an abnormal behavior in that, when first added to the above solution, it is colorless even in an alkaline solution, and addition of acid will show a color change first to pink and then to colorless. Add a few drops of acid in excess of the neutrality point. Dilute the solution to 150 ml., provide a cover for the flask and add 7 grams of sodium bicarbonate, covering the flask at once to prevent loss by effervescence. Wash the cover glass and sides of the flask. Run into QUESTIONS 147 the flask from a glass-stoppcred buret about 30 ml. of the iodine solution, add 2 ml. of the starch indicator and titrate to the appearance of a permanent blue color. Repeat the comparison with fresh portions of the solvitions. Calculate the number of milliliters of arsenite reduced by 1 nil. of the iodine solution. Standardization of the Iodine Solution by AS2O3. Arsonious oxide is a satisfactory primary standard for the direct standardization of iodine. The oxide can be procured in a high state of purity. A gram portion should be dried in a drying oven to be certain that it is free from moisture, unless that supplied has already been dried. Weigh out into 250-ml. Erlenmcyer flasks three jiortions of aliout 0.2 gram each. Dissolve each portion in 10 ml. of the dilute sodium hydroxide. Dilute to 100 ml., add a drop of phenolphthalein indicator, neutralize with dilute hydrochloric acid and add tlirce drops in exce.ss. Dilute the slightly acid solutions to 150 ml., add 7 grams of NallCO.-j, cover with watch glasses until action is over and then wash down the sides of the flasks and the cover gla-sses. Calculate the number of milli- liters of 0.1 N iodine required to oxidize 0.2 gram of AssOg. Add from a buret about 5 ml. less than the required amount, add 2 ml. of starch solution and titrate to the end point. If the end point is ])as.sed, back- titrate with the sodium arsenite solution. From the net volume of iodine required and the compaiison data calculate the normalities of both solutions. Determination of Arsenious Oxide. Weigh out portions of the impure AS2O3 samples of about 0.25 gram each. Dissolve each portion in 10 ml. of dilute sodium hydroxide solution. Dilute to 100 ml., add phcnol- phthalein indicator and neutralize with dilute hydrocliloric acid, adding three drops in excess. Make up the volume to 150 ml. ^\^th water, add 7 grams of sodium bicarbonate and titrate with the standard iodine solution, adding the starch indicator after 15 or 20 mi. of the iodine have been added. If the end point is passed, back-titrate with standard sodium arsenite solution. Calculate the percentage of AS2O3 in the sample. QbTESTIONS 1. Name four reducing substances ^^hich may be determined by direct titration with iodine. Supply the equations for the reactions involved. 2. What precautions must be observed in making up and preserving a solution of iodine? Why dissolve the iodine in potassium iodide? 3. Explain how NaHCOs acts as a buffer during the titration of arsenites by iodine. Why is the addition of NaHCOs required? WTiy not use NaOH? 148 OXIDATION AND REDUCTION METHODS 4. How might one standardize a solution of iodine other than by pure A 82 O 3 ? 6. Outline a method for the determination of hydrogen sulhde. 6. Explain the action of starch as an indicator and give reasons for the care required \u its preparation. PROBLEM SET 8 CALCULATIONS INVOLVING THE CONSUMPTION OF IODINE 1. Calculate the milliequivalent weights of the following: (o'! iodine; (b) NajAsOs; (c) AS2O3. Ans. (a) 0.1269 (b) 0.09695 (c) 0.04946 2. What weight of pure AS2O3 should be used when standardizing a 0.1000 N iodine solution so that not more than 35.00 ml. of iodine solution will be used during the titration? 3. If 27.25 tcA. of a 0.1356 iV iodine solution is found to react with 28.75 ml. of a Na2As03 solution, what is the normality of the arsenite solution? Ans. 0.1285 N 4. How much water must be added to 950 ml. of a 0.1026 N iodine solution in order to make a solution which is exactly 0.1000 N? 6. Calculate the percentage of AS2O3 in a sample of material if a 0.2200-gram sample requires 37.32 ml. of 0.1138 N iodine solution. Ans, 95.6 per cent 6. In what proportions should 0.1048 N and 0.00987 NasAsOa solutions be mixed to give a 0.1000 N solution? 7. It is found that 26.41 ml. of a NasAsOa solution reacts with 30.12 ml. of a 0.1236 N iodine solution. What is the sulfur titer of the NajAsOs solution? Ans. 0.002260 8. What is the percentage of AS2O3 in a sample of disodium arsenite, if 32.65 ml. of 0.1136 N I2 are used in titrating a 0.1975-gram sample? 9. A 1.500-grara sample of steel was treated with HCl, the evolved H2S precipi- tated as CdS and the CdS dissolved in an acidified solution containing 50.00 ml. of 0.1250 iV iodine solution. The excess iodine was titrated with 45.65 ml. of 0.1236 N NasAsOa solution. Calculate the percentage of sulfur in the steel. Atis. 0.650 per cent 10. Calculate the .4S2O3 titer of a 0.1682 W Na2As03 solution. Calculate the normality of an arsenile solution having an AssOs titer of 0.1865. 11. A student secures the following data during an arsenious oxide analysis: Comparison of solutions 34.69 ml. I2 — 36.81 ml. NaaAsOs Standardization Weight of AS2O3 = 0.2212 gram Volume of I2 *= 38.47 ml. Volume of NasAsOa = 3.82 ml. Analysis Weight of sample = 0.2882 gram Volume of I2 = 31.28 ml. Volume of NaaAsOs - 2.50 ml. Calculate the percentage of AS2O3 in the sample. Ans. 63.64 per cent 12. Calculate the percentage of antimony in a stibnite (815285) ore, 0.5623 gram of which requires 33.26 ml. of a 0.1324 N iodine solution. METHODS INVOLVING THE LIBERATION OF IODINE 149 13. What is the percentage of AS2O3 in a sample of paris green if a 0.4000-gram sample required 37.51 ml. of a 0.1100 AT solution of iodine for the oxidation of the arsenic? Ans. 51.12 per cent 14. Wliat is (a) the H2SO3 titer, (6) the Na^SOa titer, (c) the SO2 titer of a 0.2000 solution of iodine? 16. How much water must be added to 875 ml. of a 0.21G7 N iodine solution in order to make an exactly 0.2000 N solution? Aits. 73.1 ml. 16. Calculate the S titer of a Na^AsOa solution, 41.26 ml. of which reacted with 40.00 ml. of a 0.1362 N solution of iodine. 17. If a certain iodine solution is 0.2120 N, what weight of Na3As03 should he added to 500 ml. of this solution in order to make it exactly 0.2000 .V? An.'i. 0.6045 gram 18. Calculate the percentage of antimony in a sample of antimony ore, which on analysis required 32.65 ml. of an iodine solution, the latter having been standardized by causing 0.1478 gram of pure AS2O3 to react with 30.00 ml. of the iodine. The weight of antimony ore was 0.9637 gram. 19. What would be a suitable weight of pure rcsublimed iodine to u.=c for the stand- ardization of an approximately 0.01 N solution of NasAsOa, if not more than 40.00 ml. is to be used? An,*?, 0.0508 gram 20. A 5.000-gram sample of steel was dissolved in IICl and the II2S al)sorhed in an ammoniacal ZnS04 solution. The precipitated ZnS was dissolved in the prescnc>e of 25.00 ml. of 0.03568 N iodine. The excess iodine was then titrated with 0.03662 N Na2S203 requiring 18.55 ml. Calculate the percentage of sulfur in the steel. Ans. O.OG8 per cent Methods Involving the Liberation of Iodine (loDOMETRic Methods) In the iodometric methods considered in the present section, the amount of iodine liberated by the action of potassium iodide on the sub- stance being analyzed is made the basis of the method. In methods of this kind, the sample must be in the oxidized form and is reduced by KI. Tlie liberated iodine is then titrated with a standard solution of sodium thiosulfate or sodium arsenite, using starch as indicator. This type of method allows much variation in experimental arrangement so that a greater number of determinations can be carried out by this indirect iodiraetric method than by any other volumetric method. Two such variations in technique are discussed below'. In the first place, an excess of strong potassium iodide solution may be allow'ed to react directly with the material under analj'sis, the free iodine being titrated with standard reducer. As the first example under this head may be mentioned the determination of copper, which is based on the reaction 2Cu++ + 41- = I 2 + CU 2 I 2 and for which the procedure is given in detail below. 160 OXIDATION AND REDUCTION METHODS Free chlorine and free bromine may be determined by similar reac- tions : CI2 + 21- = I2 + 2cr and Br2 +21 = I2 + 2 Br Hydrogen peroxide reacts with KI to liberate iodine as shown by the equation H2O2 + 2 KI + H2SO4 = I2 + 2H2O + K2SO4 Hypochlorites in presence of HCl will liberate iodine from potassium iodide : HOCl + 2 KI + HCl = I2 + H2O + 2 KC 1 Bromates may be satisfactorily determined by a method based on the reaction HBrOg + 6 HI = Slg + HBr + SHgO in solutions not too strongly acid with HCL Permanganates, dichromates and chromates react in acid solutions with potassium iodide, and may be determined by such reactions. Use may be made of standard solutions of KMn04 or of K2Cr207 to stand- ardize a Na2S203 solution by the intermediate liberation of iodine. Certain indirect determinations may be carried out by allowing the material, in the oxidized form, to react with HCl or HBr, by which free clilorine or free bromine is evolved, which in turn is passed into strong potassium iodide solution, and the iodine thus liberated is titrated with standard reducer. The method is applicable to the determination of such substances as manganese dioxide, nitrates, nitrites, vanadates, etc. As an example, the analysis of pyrolusite, the ore of manganese di- oxide, may be cited. The sample is treated with HCl in a reaction flask, when the following reaction takes place: MnOa + 4 HC 1 = MnClz + 2H2O + CI2 The chlorine is passed into a receiver which contains a strong solution of KI: CI2 + 2 KI = 2 KC 1 + I2 and the iodine liberated is titrated with Na2S203 solution. Compare this method for the analysis of pyrolusite with the permanganate method previously described on page 139 . PKEPARATION and standardization of 0.1 N SOLUTIONS 151 Volumetric Determination of Copper This procedure is an example of an iodomctric determination in which iodine is liberated from pota.ssium iodide and titrated with standaid sodium tliiosulfatc solution. The copper is reduced from the cupric to the cuprous state, being precipitated as cuprous iodide in accordance with the reaction 2Cu++ + 41- = I2 + CU2I2 The method is applicable to the determination of copper in cupric compounds and, by modifications, can be applied to the determination of copper in ores, minerals and alloys. Provision is made below for both types of samples. The procedure calls for standard solutions of sodium thiosulfate and iodine. Preparaiion and Standardization of Approximately 0.1 N Solutions of Iodine and Sodium Thiosulfate In the reaction 2Na2S203 "1*12 = Na2S40o 2NaI the sodium thiosulfate is oxidized to sodium tetrathionate and the iodine is reduced to sodium iodide. The half-cell reactions are 2S2O3 — > 8406"^ "f" 2e and I2 -f 2e-> 21- giving the balanced ionic equation 28203"^ "I* I2 = 8403“ -j- 21— From these electron equations it is seen that an equivalent weight of iodine is one-half the molecular weight or, directly, the atomic weight, and a gram-equivalent weight therefore is 120.93 grams. A liter of 0. 1 V solution of I2 hence contains 12.093 grams of iodine. In the oxidation of 28203“ to 8403^ two electrons are involved, hence the gram-equivalent weight of Na28203 is 2Na2S203/2 or directly the gram-molecular weight (158.11 grams) of the anhydrous salt. The salt to be used is Na2S203-5H20; therefore, 248.2 grams of the crystallized pentahydrate in a liter of solution will give a normal solution of this reagent, and one-tenth of this amount (24.82 grams) should be used for &0.1 N solution. This fact can be independently deduced from valence considerations alone. In the molecule of Na2S203 one sulfur atom has a valence of -f 6, the other a valence of —2, the average valence being -f 2. In Na2S406, 152 OXIDATION AND REDUCTION METHODS two sulfur atoms each have an apparent valence of +6, the third 0 and the fourth —2, the average valence number being +2)^. There is therefore an average apparent valence increase of -}-34 for each atom of sulfur; and, since two atoms of sulfur are contained in one molecule of Na2S203, the gram-molecular weight per liter will give a normal solu- tion of this reagent. Preparation of Solutions. The solution of iodine used for the analysis of arsenious oxide should be used here if available. If not, make up the iodine solution as follows: Grind together in a mortar about 13 grams of pure iodine and 20 grams of KI with a little water. When dissolved, transfer to a liter flask and fill to the mark with distilled water. Store the iodine solution in a dark-colored or paper-wrapped bottle, since it is acted upon by strong sunlight. Make a liter of approximately O.l N sodium thiosulfate solution by dissolving 25 grams of Na2S203'5H20 in freshly boiled and cooled dis- tilled water. This boiling and cooling of the water is necessary to free it from carbon dioxide, the presence of which would gradually decom- pose sodium thiosulfate. Prepare starch indicator solution as directed on page 146 unless al- ready prepared and available in the laboratory. Prepare a solution of potassium iodide by dissolving 30 grams of KI in water and diluting to 100 ml. A solution of KCNS containing 20 grams per 100 ml. will be supplied. Comparison of Solutions. Place the iodine in a buret provided with a glass stopcock, since lubber is attacked by strong oxidizing agents. From the other buret run out about 35 ml. of the thiosulfate solution, dilute to about 150 ml., add 2 ml. of the starch indicator and then add iodine until a permanent blue color is produced. Add thiosulfate solu- tion if the end point has been passed. Calculate the number of milliliters of thiosulfate oxidized by l^ml. of iodine and likewise the volume of iodine used by 1 ml. of thiosulfate solution. Repeat the comparison until satisfactory agreement in the ratios is obtained. Obviously, if the iodine solutioif is the one already prepared and standardized for the ai-senic detci-mination and not separately prepareil for the copper deteimination, this comparison constitutes, at the same time, by proper calculation, a standardization of the thiosulfate solu- tion. It is nevertheless recommended that the thiosulfate solution be independently standardized according to the following procedure. Standardization of the Thiosulfate Solution by Copper. Thiosulfate solutions may be standardized in a number of ways as, for instance, directly against pure iodine or a standardized solution of iodine, against PREPARATION AND STANDARDIZATION OF 0.1 N SOLUTIONS 153 potassium bromate, permanganate, clichromate or against pure copper. The last-named standard is preferable in the present instance, since the solution is to be used for the determination of copper. The method depends upon the liberation of iodine from KI by a known weight of pure copper, and the subsequent titration of the iodine with the thio- sulfate solution. The reactions involved arc 2Cu++ + 41- = I2 + CU2I2 and I2 + 2Na2S203 = Na2S40G -h 2NaI From the relationships, 2Cu++ ^ ^ 2Na2S203, the gram-equiva- lent weight of copper is 03.57, and the gram-milliequivalent weight is 0.06357. Hence the copper titer of a tenth-nonnal solution is 0.00G357. If 35 ml. of the thiosulfate are anticipated for standardization, a suitable weight of pure copper to use would be 35 X 0.000357 or about 0.22 gram. Weigh out accurately and place in 250-mI. Erlenmeycr flasks three portions of pure copper in the form of foil or wiie, of 0.2 to 0.25 gram each. To each flask add 3 to 5 ml. of concentrated HNO3 and an equal volume of water, warming if necessary until the metal Ls dissolved. When it is dissolved add 15 ml. of water and 5 ml. of the urea solution supplied. Tills reagent was prepared by dissolving 50 grams of urea in a liter of water and is used to eliminate oxides of nitrogen. Boil for a minute and cool. Add dilute ammonia solution dropwise to the cooled solutions until a precipitate of Cu(OH)2 begins to form. If too much ammonium hy- droxide is added as shown by an intense deep blue color of the solution, carefully add a few drops of dilute HNO3 and again cautiously neutral- ize with ammonia solution. Add to the solution L gram of potassium acid phthalate and allow it to dissolve. This reagent will keep the solution at the proper acidity for titration with the tliiosulfate solution. Fill a buret with the thiosulfate solution, in readiness for the standard- ization. Compute the volume of thiosulfate solution which would be required for the titration of the smallest of the three standards on the assumption that the solution is exactly 0.1 N. Now to the flask containing the smallest amount of copper standard add 10 ml. of the potassium iodide solution and allow to stand for only a few minutes. Run in the thiosulfate solution rapidly to wuthin about 1 ml. of the calculated volum : This will also be indicated by the solu- tion assuming a pale yellow color. In order to secure the best possible end point add at this stage of the titration a drop of starch indicator and 10 ml, of potassium thiocyanate 154 OXIDATION AND REDUCTION METHODS reagent and shake the solution. This step is introduced to dislodge the adsorbed iodine from the cuprous iodide precipitate and results in a sharper end point. Finally add 5 ml. of the starch solution and titrate to the end point by adding the thiosulfate solution dropwise. The end point is some- what obscured by the presence of the precipitated CU 2 I 2 but should be quite sharp, as marked by the disappearance of the blue color. Treat another portion of standard copper solution with potassium acid phthalate and follow the same procedure as above. Do the same with the third portion. From the data thus obtained here and in the comparison calculate the normalities of the sodium thiosulfate and iodine solutions. Calcu- late also the Cu titers of the two solutions. Determination of Copper in Copper Salts. Weigh out three portions of the sample, as submitted, of about 0.5 gram each. If the samples are water soluble, dissolve each in about 50 ml. of water. If copper oxide samples are to be used, dissolve in a little nitric acid, heating if neces- sary. From this point the determination proceeds as with the standard- ization. Calculate the percentage of copper in the sample. Determination of Copper in Ores. If the determination of copper is to be made on ores, minerals or other mixtures which contain oxi^zing substances, the copper must first be freed from these interfering elements or modifications must be introduced to overcome these interferences. / In the modification recently proposed by Park these interferences are overcome by the use of a mm onium bifluoride and potassium acid phthalate. Dry the sample at 105° C. for one hour. Weigh out three 1-gram portions of the dried ore into 300 ml. beakers, add 10 ml. of concentrated nitric acid, cover with watch glasses, warm until the sample has practically gone into solution. Then rinse off the cover glasses and begin the evaporation on a water bath until fumes no longer ai'e evolved. Add a little more acid and digest until the sediment is white. Filter the solution into 250-ml. Erlenmcyer flasks and wash the residue and filter paper vith weakly acidulated water until all copper color has been removed. Do not allow the solutions to exceed about 75 ml. in volume. Treat the filtrate ^vith 10 ml. of bromine water and boil. Cool and add dilute ammonium hydroxide until a precipitate begins to form. Now add 2 grams of ammonium bifluoride. This reagent forms a complex ion, FcFe^, with the ferric ion. Shake the flask imtil the brown precipitate dissolves. Finally add 1 gram of potassium acid phthalate. lODATE AND BROMATE METHODS 155 From this point on the procedure is the same as that described on page 152 for the standardization. When each sample in turn has been treated Avitli tiie potassium iodide solution and titrated, calculate the percentage of copper in the sample. lODATE AND BrOMATE MeTIIODS Potassium iodate finds, as already indicated on page 143, extensive uses as an analytical reagent. Because it can be obtained in pure condition and consequently can be u.sed directly as a primary standard, some chemists prefer it to iodine or certain other oxidizing agents. In fact, when it is in the same solution with potassium iodide and the solution is weaklv acidified with HCl, iodine is liberated according to the equation KIO3 + 5KI 4- GIICl = 3I2 -f 3H2O + CKCl and the mixed reagent reacts like a solution of iodine itself. A conven- ient means of making a standard, nojTnal solution of iodine is to weigh out the equivalent weight of pure KIO3, add excess of KI and acidify when being used. The use of KIO3 alone as a standard oxidizer without admixture of KI finds many applications. The conditions under which the titrations are carried out are here briefly described. In a strongly acid solution the final reduction product of KIO3 is iodine monochloride, ICl. When KIO3 reacts with reducing agents the intermediate product is I 2 , which in turn reacts with the iodate to destroy the iodine. Titrations of this kind reach an end point when the color of the liberated iodine is dis- charged from a chloroform or carbon tetrachloride layer in the solu- tion being titrated. The reactions by which ICl fonns can be seen from the follo\ring, where KI is the reducing agent: KIO3 + SKI H- 6HC1 = 3I2 4- 3H2O 4- 6KC1 and KIO3 4- 2I2 4- 6HC1 = 5ICI + 3II2O 4- KCl If the titration is stopped when the iodate has been reduced to iodine, which can be accomplished in solutions of low acidity, the liberated iodine can be titrated with standard thiosulfate. Thus iodides can be oxidized to free iodine by the reaction KIO3 4- SKI + 6HC1 = 3I2 4- 3H2O 4- 6KCI and detemained in this w'ay instead of effecting the foimation 01 locune monochloride. 156 OXIDATION AND REDUCTION METHODS From the equation KIO3 + 2I2 + 6HC1 = 5IC1 + 3H2O + KCl it is seen that iodine can be determined by direct titration with a stand- ard solution of KIO3. Likewise, iodides, such as KI, can be determined by a similar titration: KIO3 + 2KI + 6HC1 = 3IC1 + 3H2O + 3KC1 Among the many other substances that can be determined by titrar tion with KIO3, with the formation of iodine monochloride, are arsenites, sulfites, thiosulfates, persulfates, tin, antimony and copper thiocyanate. In the reaction for the determination of arsenites according to the equation KIO3 + 2H3ASO3 + 2HC1 = ICl + H2O + 2H3ASO4 + KCl the change of IO3" to ICl is shown by the equation IO3" + 2HC1 + 4c -> ICl + H2O + 20 + Cl- From this it is seen that, under these conditions, the equivalent weight of KIO3 is KIO3/4 or one-fourth the gram-molecular weight. Certain substances, moreover, can be determined by use of a mixture of potassium iodate and potassium iodide, which react to liberate iodine. The iodine, in turn, can then be titrated with thiosulfate. The reaction just given for the determination of iodides 6HC1 + KIO3 + 5KI - 3I2 + 3H2O + 6KC1 may be used, as well, for the determination of strong acids like HCl and H2SO4 and, by special adaptation, for H3BO3 and certain hydrolyzable salts. Standardized solutions of potassium bromate, KBrOa, are frequently used in the determination of many substances. This oxidizing agent may be used either alone or along with KBr or KI in a variety of ways. The titration may be stopped either when the bromate has been reduced to bromide or reduced to free bromine. As example of the first mode of procedure, the determination of such substances as the elements tin, iron, arsenic, titanium and antimony may be cited. In the case of anti- mony, the reaction proceeds in the following way: 3SbCl3 + KBvOz + 9H2O = 3H3Sb04 + 9HC1 + KBr The reaction of ICBrOa with iodides in an acid solution is an impor- tant one: I^BrOg 4- 6KI + 6HC1 = 3I2 + 3H2O + KBr + 6KC1 and furnishes still another method for the determination of iodides. PROBLEMS 157 QUESTIONS 1. Show how the amount of iodine and sodium thiosulfate required for a liter of normal solution is calculated. 2. Write reactions which represent the standardization of sodium thiosulfate against (a) pure, resublimed iodine; (b) pure copper; (c) KBrOa. 3. Suppose you had on hand a standard solution of KMnO^ or of KoCViiOy. How could you use cither of these solutions for the standardization of a thiosulfate solu- tion? Balance the following equations: MnOi" + I" + H+ Mn++ + I 2 + H 2 O CraO;” + 1“ + -4 Cr+++ + I 2 + H 2 O 4. How is the following reaction: CI 2 + 2KI = I 2 + 2KCI applied (a) in qualitative analysis, (6) in quantitative analysis? 5. Outline an iodimetric and a permanganate method for the determination of hydrogen peroxide. 6. Outline an iodimetric and a permanganate method for the analysis of pyrolusite (Mn02). 7. How could you proceed with the determination of copper if ferric iron were also present in the sample? PROBLEM SET 9 CALCULATIONS INVOLVING THE LIBERATION OF IODINE 1. What are (a) the A 820 a, (6) the As, (c) the S and (d) the Cu titers of a 0.1 V iodine solution? Ans. (a) 0.001946 (6) 0.003746 (c) 0.003206 id) 0.006357 2. Find the normalities of solutions having the following liters: (а) Cu titer * 0.006475 (б) S titer = 0.003820 (c) Cr titer = 0.001800 (d) H 2 SO 3 titer = 0.004022 3 . data: A student standardized a sodium thiosulfate solution, obtaining the following 33.85 ml. I 2 35.48 ml. Na 2 S 203 After a proper treatment of the sample he found that 0.2202 gram of electrolytic copper required 38.12 ml. of the Na 2 S 203 solution and 2.24 ml. of the I 2 solution for back-titration. Find the normality of the Na 2 S 203 and I 2 solutions. Ans. 0.0968 A': 0.1014 V 4. A Na 2 S 203 solution was found to have a normality of 0.1024. If it is desired that each milliliter of this solution should be equivalent to 0.000357 gram of copper, to what volume should a liter of this solution be diluted? 158 OXIDATION AND REDUCTION METHODS 6. What normality should a Na 2 S 203 solution possess if each milliliter is to repre- sent 1 per cent of copper in 0.2000-gram samples. Ans. 0.0314^ N 6. A 0.2984-gram sample of copper ore required 33.79 ml. of 0.0909 N Na 2 S 203 to titrate the iodine liberated from KI by the copper. Find the percentage of copper in the ore. 7. A Na 2 S 203 solution was standardized with a KMn 04 solution by titration of the I 2 liberated from KI by the KMn04. To a solution containing KI, 32.26 ml. of 0.1021 N KMn04 were added. Find the normality of the Na 2 S 203 if 30.34 ml. were necessary for titration of the liberated iodine. Ans. 0.1085 N 8. A sample of lead paint was dissolved, the lead precipitated as PbCr 04 and this dissolved in acid. To the acid solution KI was added and the liberated iodine was titrated with 35.65 mL of 0.1010 N NaaSaOs. Find the weight of lead in the paint. 9. It is possible to determine the amount of free chlorine in swimming pool water by testing its action on KI : CI 2 4- 2KI = I 2 + 2KC1 and by titrating the liberated iodine with a standard thiosulfate solution. If a 100.00-ml. sample when treated with KI and titrated with 0.0102 N required 6.75 ml., what was the concentration of CI 2 in the water in terms of parts per million by weight? 244 parts per million 10. A student’s Na 2 S 203 solution had been standardized with a K 2 Cr 207 solution and the Na2S203 solution was then used in the analysis of a copper ore; 36.69 ml, of a 0.1 K 2 Cr 207 solution were allowed to react with an excess of KI and the liberated iodine was titrated with 34.92 ml. of Na 2 S 203 . The 0.2109-gram sample required 22.50 ml. of Na 2 S 203 to react with the iodine liberated from KI by the dissolved sample. Find the percentage of copper in the sample. 11. A portion of a mixture of KBrOs and KI was treated with HCl and the I 2 liberated in the reaction KBrOa -h 6HC1 + 6KI - KBr -h 3 I 2 + 3 H 2 O + 6KCI was titrated with 34.73 ml. of 0.1000 N Na 2 S 203 . Find the weight of KBrOs in the sample taken for analysis. Ans. 0.0967 gram 12. If 1.00 ml. of KMn 04 solution can liberate 0.1527 gram of iodine from KI, what weight of pure Mn 02 can be reduced by the same volume of oxalic acid as is oxidized by 32.20 ml. of the above permanganate solution? 13. A sample of Mn02 ore was treated with HCl, liberating chlorine: Mn02 + 4HC1 = CI 2 + MnCl 2 + 2 H 2 O and the chlorine passed into a solution containing excess KI: CI 2 + 2KI =12 + 2KC1 The iodine wtis titrated with a 0.1250 N Na 2 S 203 solution requiring 38.55 ml. for a 0.4850-gram sample of the ore. Calculate the percentage of Mn02 in the ore. Ans. 43.20 per cent 14. What weight of potassium iodate will liberate sufficient iodine to react with 35.25 ml. of Na 2 S 203 having an arsenious oxide titer of 0.04530? (Reaction similar to that in Problem 11.) 15. Hydrogen peroxide may be determined by an iodomotric method as shown by the following reaction: II 2 O 2 + 2KI + 2HC1 = 2 H 2 O + I 2 + 2KC1 PROBLEMS 159 If a 10-mI. sample of H2O2 (sp. rf. 1.01) requires 30.30 rnl. of 0.5000 N Na-S^O.^ for reaction with the liberated iodine, what is the percentage of H2O2 in the sample? /Irw. 2.55 per cent 16 . From the reaction KIO3 + SKI + 6 HC 1 = 3I2 + 6KCI + 3II2O calculate the gram-equivalent weight of KIO3. 17 . How many grams of (a) FeS04 -7H2O, (b) 21120 and (r) Na2S20,v5H2O must be used to make a liter of each of these reagents, each equivalent to a K2Cr207 solution of which 1 ml. can liberate 0 . 012(>8 gram of iodine from KI? ^ns. (a) 27.80 (6) 0.30 (c) 2-1.82 18 . A Na2S203 solution was standardized by means of a previou.sly standardized solution of KMn04, basing the method on the reactions; 2Mn04- + 101 - + 16 H+ = 2 Mn++ + 5I2 + 8H2O and I2 4 * 28203“ = 8406“ 4- 21- Using the following data, calculate the strength of the thiosulfate solution: KMn04 used, 35.00 ml.; Na28203 used, 38-72 ml.; strength of KMn04, 0.1260 N. 19 . How much iodine will be liberated from 35.07 ml. of a 0.1482 K2Cr207 solution when treated with KI and HCl? A71S. 0.6703 gram 20 . A sample of chromium ore was oxidized to Na2Cr04 by a sodium peroxide fusion, and this, after removal of iron and other impurities, was treated with HCl to convert it into Na2Cr207. An excess of KI was then added, liberating iodine in accordance with the reaction Na2Cr207 4 - 14 HCI 4 - OKI = 3I2 4 - 2 CrCl 3 4 * 2 NaCl + 6 KCI The liberated iodine was titrated with 38.65 ml. of a 0.1000 N solution of Na2S203. If the weight of sample was 1.2230 grams what was the i>ercentage of chromium in the sample? CHAPTER V VOLUMETRIC PRECIPITATION METHODS Precipitation methods depend upon the practically complete precipi- tation of a compound containing the element whose determination is undertaken. Volumetric precipitation methods differ from gravimetric precipitation methods in that, instead of filtering, purifying and weighing the precipitate, the amount of reagent required to complete the reaction is carefully measured, the completion of the reaction being indicated by a sharp change in color of an indicator or by use of electrical instruments as in potentiometric titration. The fundamental requirements for all volumetric precipitation processes are (1) a precipitation reaction which goes to practical completion, (2) a standardized solution of a precipi- tating reagent and (3) a means of indicating the end point. The third condition, namely, a means of indicating the end point, is the hardest requirement to fulfill, and it may be said that in general many precipi- tation reactions might be adapted to the volumetric technique if it were not for the lack of suitable indicators. Indicators, which must show a pronounced change in color immediately upon completion of the main reaction, are comparatively hard to find, and determinations for which such indicators are available are few in number. Some examples are considered below. Typical Precipitation Methods. Chlorides, bromides and iodides can be accurately determined by titration with a standard silver nitrate solution. For chlorides, the reaction is Cr + Ag+ = AgCl Reactions analogous to the above take place with the bromide and the iodide ion. The process may be reversed for the determination of silver by use of a standard solution of sodium chloride. In the method of Mohr, for the determination of chlorides, the indi- cator is potassium chromate, wliich reacts with silver nitrate to form a deep red precipitate of Ag 2 Cr 04 when a drop of the standard AgNOa beyond the equivalent volume has been added to the chloride solution. In the procedure given for chlorides in the following section, an adsorp- tion indicator, dichlorofluorescein, is utilized. 160 161 THEORY OF VOLUMETRIC PRECIPITATION A fairly satisfactory method for zinc is based on the reaction between zinc and pot^sium ferrocyanide, giving a precipitate of the approxi- mate composition K2Zn3(Fe(CN)o)2, iiranyl acetate being used as indi- cator. Lead may be dcteriniiuxl volumetrically by use of a standard potassium chromate solution, tannin being used as indicator. The determination of silver by titration with a standard solution of ix)tassium thiocyanate, according to the method of \'o]hard, i.s another typical example of a volumetric precipitation procedure, the procedure being given in detail later in this chapter. THEORY OF VOLUMETRIC PRECIPITATION Equilibria in Volumetric Precipitation Reactions. Tlie chief applica- tion of the Law of Chemical Equilibrium to precipitation reactions con- cerns the Solubility Product Principle. This principle states that, in a solution in which a precipitate has formed, the product of the gram-ion concentrations of the ions in the saturated solution sun ounding the solid is always equal to a constant, called the solubility product constant. In order for a solution to become saturated with respect to the ions which enter into the chemical composition of the precipitate, enough ions of the precipitating agent must be supplied so that their concentra- tion, multiplied by the concentration of the ion whose precipitation is sought, exactly equals the numerical value of the constant for the pre- cipitate in question. For the precipitate actually to form the constant must be exceeded. In simple cases such as the precipitation of AgCl by the ionic reaction Ag+ + Cr = AgCI the solubility product equation takes the form: in which and Cci- are the concentrations (gram-ion quantities re- ferred to a liter) of Ag*^ ion and Cl” ion in equilibrium with each other, and Ka,p. is the solubility product constant of AgCl. In more complicated cases, such as the formation of Ag 2 Cr 04 , in the reaction the equation is 2Ag+ -b Cr04“ = Ag2Cr04 X Ccr04 A'3.p.(AgjCr04) Solubility product constants are usually calculated from solubility data, and a table of such values is given on page 182. A liter of water, at room temperature, will dissolve approximately 0.00001 of a gram- 162 VOLUMETRIC PRECIPITATION METHODS molecular weight of AgCl; the saturated solution is 1 X 10“® molar with respect to the dissolved salt. In such a solution the concentration of Ag"^ ions is 1 X 10“® gram-ion, and the concentration of Cl~ ions is also 1 X 10"®. The solubility product equation is therefore X CqI- — A^a.p. 10"® X 10"® = 1 X 10"^° Although in this particular solution, which was formed by the dissolv- ing of AgCl, the concentrations of each set of ions are equal, such need not always be the case in order for equilibrium to be reached. Since the constant is a product of two factors, these may be varied at will. Thus in order to reach equilibrium and shortly thereafter to have a precipitate form, in a solution which contains a chloride-ion concentrar tion of, say, 10"^, a concentration of Ag"^ ions of 10"® would be required for saturation and equilibrium, and a slight quantity beyond this results in the formation of solid AgCl. Such equilibrium concentrations give the quantities required for the initial formation of the precipitate. But in order to have quantitative (practically complete) precipita- tion, an amount of Ag+ must be added which is equivalent to the Cl~ present, Tlxis is the condition in a quantitative titration precipitation process. The titration must stop whep an equivalent amount of pre- cipitating agent is added, this amount being signified by the color change of the indicator. No advantage can be taken in volumetric precipitation processes of an excess of precipitating reagent. When an equivalent amount has been added, the concentrations of Ag*^ and Cl“ remaining in solution in equilibrium with the precipitated AgCl are equal and, in fact, are identical to those obtained by dissolving AgCl in water. When the titration is completed there remain, in a liter of solution, 10"® gram- ion of Cl" and 10"® gram-ion of Ag"*". In 100 ml. the volume of the solution at the end of the titration, there w'ould remain 1 X 10"® gram- ion of unprecipitated chloride ion. This corresponds to 1 X 10"® X 35.46 or about 0.00005 of a gram of chloride. Although this represents a loss and therefore an inherent error in the determination, the amount is less than that which can be determined by the apparatus used. VOLUMETRIC DETERMINATION OF CHLORIDES There exist two general titration methods for the determination of chlorides, (1) the Volhard method, which employs standai'd solutions of silver nitrate and potassium thiocyanate, a modification of which is described later in the determination of silver, and (2) the Mohr method^ which employs a standard solution of silver nitrate with potassium PREPARATION OF A 0.1 N AoNOj SOLUTION 163 chromate as indicator. A modification of the Mohr mctliod introduced by Fajans utilizes an adsorption indicator and i.s here outlined. Tliis gives practice vfith an example of these newly devclo|)e(l intcrcstinir indicators. Outline of Method. The reaction which takes place is the simple ionic one leading to the formation of AgCI Ag+ + cr = AgCl A solution of AgNO.,, approximately 0.1 A, is prepared and .standard- ized with pure NaCl. The ad.sorption indicator is diclilorofiuore.scein, wiiich produces a pink color duo to the adsorption of Ag+ ions on the particles of AgCl. This color develops at the equivalence jjoint in the titration. Preparation and Standardization of an Approximately 0.1 A/^ AcNOa Solution A normal solution of AgNOg should contain the gram-molecular weight of AgNOa in a liter of solution, since the valence of Ag, the precipitation ion, is 1 and is directly equivalent to one hydrogen atom. To eoasen^e this expensive chemical, prepare a half-liter of the solution by weighing 8.5 grams of AgNOs, dissolving and making up to 500 ml. Weigh accurately three portions of pure NaCI (previously carefully prepared and dried) of about 0.2 gram each, into three 2o0-ml. Erlcn- meyer flasks. Dissolve each portion of the standard in 50 ml. of distilled water. To each add 8 to 10 drops of tlie dichlorofluorcscciii indicator and about 0.1 of a gram of dextrine, the latter for the purpose of dispersing the AgCl when precipitated. Titrate the solutions, with constant swirl- ing, and out of contact of strong sunlight. Tlie end point is reached when a pink color appeal's in the solution. Calculate the Cl titer and the noimality of the AgNOs solution. Determination of Chloride in a Sample of Soluble Chloride. Pro- cedure. Weigh out accurately about a 1-gram portion of the dried chloride sample. Dissolve in 50 ml. of distilled water and make up to 250 ml. in a volumetric flask. Pipet a 50-mL aliquot portion, add the indicator and dextrine as in the standardization and titrate with the standard silver nitrate. Calculate the percentage of chloride in the sample. Notei It is suggested that the same sample of soluble cliloride be used in the gravimetric determination of chlorides (see page 191) and that results be compared for the two methods. Obvioush'^, the instructor should withhold the true percentage of chloride in the sample until both reports have been submitted. 164 VOLUMETRIC PRECIPITATION METHODS VOLUMETRIC DETERMINATION OF SILVER The method here outlined is that first proposed by Volhard and con- sists of titration of the dissolved silver by standard potassium thiocya- nate, a ferric salt being used as the indicator. The fundamental reaction involved is Ag-^ + CNS" = AgCNS » _ wherein the silver thiocyanate forms as a white precipitate. After (prac- tically) all the silver has been precipitated the reaction between the ferric ions of the indicator solution and the thiocyanate ions sets in: Fe+++ + 6CNS- = Fe(CNS)6^ producing a complex ion which is red, thus furnishing the indication that the main reaction has run to completion. ^ The determination of silver by this method involves the preparation and standardization of a dilute solution of KCNS and the subsequent use of this standard solution in the titration of the silver samples. Pkbpabation and Standardization of an Approximately 0.05 N KCNS Solution The solution of KCNS should be dilute, about 0.05 N, A normal solution of a precipitating reagent should, by definition, contain an equivalent weight of the precipitating ion per liter of solution, namely, that amount of ion which will replace or be equivalent to 1.008 grams of hydrogen. Since the valence of the anion, CNS“, is 1, the molecular weight of the potassium salt, KCNS (97.17 grams), in a liter of solution constitutes a nonnal solution of tliis reagent. To make an exactly 0.05 N solution, 0.05 of the gram-molecular weight, that is, 97.17 X 0.05 or 4.8585 grams, of the pure salt are required for a liter of solution or 2.4293 grams for 500 ml. It is possible to procure or prepare KCNS of sufficient purity to serve as a primary standard, thus obviating the necessity of standardization by titration against pure AgNOs. If pure KCNS is available, weigh out exactly 2.4293 grams of the pure salt, dissolve in a little distilled water, transfer to a 500-ml. volumetric flask and fill to the mark with distilled water. The following procedure provides for standardization with pure AgNOs in case pure KCNS is not supplied. Procedure for Standardization. Procure from the supply room or from the instructor the following supplies: 1 gram of AgNOs, 2.5 grams of KCNS and 10 grams of ferric alum. Grind the AgNOs crystals in a mortar, place the powder on a watch cover it with a clean sheet of filter paper and place in a drying DETERMINATION OF SILVER IN SILVER ALLOYS 165 oven at 105-1 10°C. for about one-half hour. In the meantime, weigh out about 2.5 grams of KCNS and dissolve it in a small amount of distilled water. Transfer the solution to a 500-ml. measuring flask and dilute to the mark. Pour the solution into a clean liter bottle. This solution is approximately 0.05 normal. Prepare the ferric alum indicator solution by saturating 30 ml. of distilled water with ferric alum and adding enough HNOa (sp. gr. 1.20) to impart a faint yellow color to the solution. The weight of AgNOg to be taken for standardizing the KCNS solu- tion can be determined by a simple calculation. Assume that the KCNS solution is exactly 0.05 normal and that 35 ml. will be used for one standardization titration. A liter of noimal thiocyanate solution will react with 169.89 grams of silver nitrate; a liter of 0.05 normal solution will therefore react with 0.05 X 169.89 or 8.-19-15 grams, and 1 ml. of a solution of this normality w'ill bring into reaction 0.008-194 gram of silver nitrate. Since 35 ml. are assumed for the titration, tlie weight of AgNOg required is 35 X 0.008494 or 0.2973 gram. Weigh out accurately three portions of the dried AgNOg of about 0.3 gram each, and transfer them to 200-ml. beakers. Dissolve each in about 50 ml. of distilled water and add 10 ml. of dilute, freshly boiled HNOg (sp. gr. 1.20) and 5 ml. of the ferric alum indicator. Rinse out a buret w'ith a small amount of the KCNS solution, and fill the buret to the top of the graduations. Take the buret reading, place one of the silver nitrate solutions under the buret and nin in the thio- cyanate solution slowly with constant stirring. Let the precipitate settle at intervals, and watch for a change in the color of the supernatant liquid. When the first pink tinge appears which lasts for at least one minute take the final buret reading. Repeat with the other two solu- tions, but run these to 2 ml. of the end point, found b}^ calculating from the result of the first run the probable volume of solution required, and then slowly titrate to the end point. If the results as calculated below do not agree to within 2 parts per 1000, make another set of determinations. From the number of milliliters of KCNS solution and the number of grams of AgNOg used in each case calculate the Ag titer of the KCNS solution; also, calculate the normality of the KCNS solution. Record the titers and the normalities and average each set. Determination of Silver in Silver Alloys Samples of silver alloy, not exceeding 70 per cent of copper, are W'eighed out and dissolved in nitric acid. The solutions are boiled, cooled, diluted, and ferric alum indicator is added, after w’hich they are 166 VOLUMETRIC PRECIPITATION METHODS titrated with standard KCNS solution. Copper in the alloy, if present in too high a concentration, will obscure the end point. The purpose of boiling the solution is to expel lower oxides of nitrogen which tend to give a red color to the solution. and consequently a false end point. P ocedure. Weigh out three separate samples of the alloy of about 0.3 gram each, and place in beakers of 200-ml. capacity. Dissolve each sample in 15 ml. of dilute, freshly boiled HNO3 (sp. gr. 1.20), heating gently. If the samples do not dissolve readily, add more HNO3. Cool, dilute to 50 ml. with distilled water, add 5 ml. of the indi- cator solution and titrate with KCNS solution, following the same procedure used in standardization. If results do not check, repeat the determination. From the volume of KCNS solution used and the strength of the thio- cyanate solution calculate the w^eight of silver in the sample. Report the results as “fineness,” that is, parts of silver per thousand. Methods in Which Complex Ions Are Formed Closely allied with the simple precipitation methods and included in the same classification are those in which a complex salt is formed. Important among these is the determination of cyanides or, conversely, silver, as well as the use of a standard KCN solution in the determina- tion of nickel and copper. In a cyanide determination the reaction is AgNOa + 2NaCN = NaAg(CN)2 + NaNOa or ionically Ag+ + 2CN- = Ag(CN)2- and when an excess of standard silver nitrate is added, a precipitate of silver cyanide foi-rns. The end point is obtained by adding a mixture of NH4OH and KI. The NH4OH prevents AgCN from precipitating owing to the fomiation of Ag(NH3)2“^ ions but Agl is formed instead by the reaction Ag(NH3)2"^ + I” = Agl + 2NH3 When a yellowish opalescence of Agl appears, the end point is reached The determination of nickel by the cyanide titration is frequently used in alloy and steel analysis. The main reaction is the formation of complex nickel cyanide ions. Ni++ + 4CN- = Ni(CN)4" The indicator, as in the method for cyanides, is Agl, produced by adding PROBLEMS 167 KI and AgIsOa to the reaction mixture, but the end point is marked by the disappearance rather than the formation of an opalescence. A method for copper sometimes employed consists in titrating an ammoniacal copper solution with standard KCN, tlic reaction being expressed by the equation 2Cu(NI-l3)4(N03)2 + 7KCN + IIoO = K4[Cu(CN) 3]2 + NM4CNO -f ONIIa + KNO3 + NII4NO3 The end point is the disappearance of the blue color of the ammonia- copper complex ion. QUESTIONS 1. Give equations for the reactions involved in the most important volumetric precipitation procedures. Why are there so few such methods? 2. What indicator or indicators would you employ for the determination of (a) halides, (b) silver, (c) lead, (d) cyanide.s, (c) zinc? 3. Explain the action of an adsorption indicator. 4. Suggest a method for standardizing a hydrochloric acid solution using (a) Mohr's method, (6) Fajans’ method. 5 Outline a method for the determination of soluble thiocyanates. Cnu a standard solution of KCNS be made without standardization? C. For use in volumetric precipitation methods what pure substances could you use for standardizing (a) silver nitrate, (h) sodium cliloride, (c) pota.v.sium t hiocyanale? 7. Show how the equivalent weight of precipitating ions is related to the equivalent weight for hydrogen. 8. Write ionic reactions for the cyanide determination of (a) copper, (h) nickel, (c) silver. ' PROBLEM SET 10 VOLUMETRIC PRECIPITATION 1 . How many grams of sodium chloride must bo taken to prepare 1000 ml. of a solution which will be equivalent to 10 milligrams of Ag per milliliter? An-s. 5-1.1 grams 2. A solution of AgNOa is known to contain 21.857 grams of AgNOa per liter. Calculate the following titers. (а) Cl titer (б) Br titer (c) KCN titer (d) CNS- titer 3. What quantity of pure NaCl should be weighed out so that not more than 35 ml. of 0.100 N AgNOa will be used in a standardization? Ans. 0.205 gram 4. Calculate (o) the KCNS titer and (b) the Cl titer of a 0.1286 V solution of AgNOa- 6. Find the percentage of NaCl in a sample of rock salt, a 0.2600-gram sample of which required 42.25 ml. of 0.1000 N AgNOa for the precipitation of the chloride. A7ts. 95.00 per cent 168 VOLUMETRIC PRECIPITATION METHODS 6. What weight of pure AgNOa should be taken for the standardization of an approximately 0.08000 N solution of KCNS if not more than 40.00 ml. of the solution is to be used? 7. If 32.75 ml. of a KCNS solution, the normality of which is 0.04782, react with 34 62 ml of a AcNOs solution, what is the normality of the latter? Am. 0.04523 N 8. Calculate the percentage of silver in an aUoy, 0.2980 gram of which required 38.95 ml. of a 0.06125 N KCNS solution. 9. If a certain KCNS solution has a Ag titer of 0.01078, what is its normality? How much silver can be precipitated by 25.00 ml. of this solution? Am. 0.1000 0.2685 gram 10. How many grams of pure AgNOa are required for the precipitation of the CNS in 25.00 ml. of 0.05000 N KCNS solution? 11. If the lead in a sample of alloy is to be precipitated as PbCrO^, what volume of 0.1000 N K 2 Cr 207 will be required for a 0.5000-gram sample which contains 30 per cent Pb? 12 Calculate the error made in terms of unprecipitated Cl" in a volumetric deter- mination of chlorides by titration with standard AgNOs, if the total volume of solu- tion after titration is 100 ml. The Kb.v. of AgCl is 1.1 X 10 ^ . 13 What volume of 0.1020 N BaCla is required to precipitate the sulfate from a 0.2300-gram sample of reS 04 • 7 H 2 O which is 85.00 per cent pure? Am. 14.6 ml. 14. Zinc ores may be analyzed by dissolving the ore, separating the zinc and titrat- ing it with a standard potassium ferrocyanide solution according to the reaction SZnCL + 2 IC 4 Fe(CN )6 Zn 3 K 2 (Fe(CN) 6)2 + 6KCI Find the percentage of zinc in an ore, 0.5026 gram of which required 26.25 ml. of 0 1500 N ferrocyanide solution. 16 How many grains of KjCrOi must be taken to make a Uter of 0.1000 N solu- tion to be used for the precipitation of lead as PbCr04? What is the Pb titer of a 0.5213 N solution of KsCrO,? 16. To a 0.2652-gram sample of a mixture of pure NaCl and KCl were added 41 00 ml of 0 1 iV AgNOs. After precipitation of the chlorides as AgCl, the excess of AgNOa was titrated with 3.85 ml. of 0.1 N KCNS. What was the weight of KCl in . the mixture? 17. A 0.4S95-gram sample of a phosphate was precipitated as AgsPOi. The silver phosphate was dissolved in HNO 3 and the silver titrated with 38.60 ml. of a 0.05000 N KCNS solution. Calculate the percentage of PO4 in the sample. il « e 19. R(\ rkAY* 18. A 0.6324-gram sample of a material containing iodides was treated with 50 00 ml of a 0.1000 N solution of AgNOs, the precipitated Agl filtered off and the excess AgNOj titrated vdth 20.00 ml. of 0.05000 N KCNS. Calculate the percentage of iodide present. ^ . 19, Soluble cyanides may be determined by volumetric precipitation with standard silver nitrate. If a 0.200a-gram sample requires 25.26 ml. of 0.06000 N AgNOj, what is the percentage of KCN in the sample? Ans. 41.2 per c^t 20. It is desirable that the volumes of a 0.05 N solution of NaCl used for titrating samples of silver alloy should range between 25 and 40 ml. What weights of alloy should be taken for analysis if the samples are thought to contain approximately (o) 25 per cent; (5) 60 per cent;'(c) 76 per cent; (d) 96 per cent? PART III GRAVIMETRIC ANALYSIS CHAPTER VI THE THEORY, TECHNIQUE AND CALCULATIONS OF GRAVIMETRIC PRECIPITATION METHODS Gravimetric methods of analysis are those in which the desired con- stituent is separated and weighed in the fonn of a pure compound or a pure metal or in which a volatile product is evolved. These methods are therefore classified into (1) gravimetric precipitation methods; (2) evolution methods; (3) electrolytic precipitation methods. The tech- nique dilTers widely for these three types of methods. The gravimetric precipitation methods are by far the most important and numerous. All gravimetric precipitation methods are based on reactions in which a precipitate is produced through the action of a chemical precipitating agent. The precipitate thus obtained, which must contain all of the desired constituent, must be quantitatively separated from the solution, carefully purified, dried or converted into another compound whose chemical composition is definite and kno^vn and then finally weighed. Certain definite conditions must be fulfilled and certain requirements met in order that the selected reaction may become an accurate quan- titative method of determination. These fundamental conditions and the manner of satisfying them are discussed below. FUNDAMENTAL REQUIREMENTS OF GRAVIMETRIC PRECIPITATION METHODS Reactions which take place because a precipitate forms do not neces- sarily run far enouth to completion to make the reaction quantitative. Furthermore, the precipitated compound may not be in the condition, either chemically or physically, necessaiy for conversion and weighing in the form of a pure compound of definite, kno^m composition. Gravi- metric methods which give the accuracy demanded in exact chemical analysis must meet the following requirements. 169 170 GRAVIMETRIC PRECIPITATION METHODS 1. The reaction must ran practically to completion. 2. The precipitate formed must be practically insoluble in the medium in which it is precipitated and must contain (practically) all of the de- sired constituent. Furthermore, it must not contain any undesired constitutent. 3. The physical condition of the precipitate must be such that it can be readily filtered and washed free of impurities. 4. The precipitate must be capable of being dried or ignited to a pure form, of definite and known composition. It is evident that, if the nature of the reaction or of the precipitate is such that any one of the above conditions is not satisfactorily fulfilled, the method will be inaccurate and, if fulfilment is seriously lacking in any respect, the reaction cannot be used as a quantitative method. A consideration of these essential conditions will make it clear why reac- tions must be chosen vnth. care and why the operations attendant upon carrying out the process must be executed with the best possible tech- nique. It should be the aim of the student, in carrying out the follow- ing representative procedures, to subject each method to a thorough criticism with respect to these requirements. Completeness of Reaction. This first condition is self-evident, since the desired constituent is to be isolated and eventually weighed. No reaction runs absolutely to completion, since an equilibrium is reached jj 2 reaction. However, the important condition here is that the reaction should run as near to completion as possible. Many reactions in which relatively insoluble precipitates are formed are among those which do go, for all practical purposes, to completion. Insolubility of the Precipitate. This second condition can be regarded as a corollaiy of the first, because the mechanism by which these reac- tions run to practical completion is by the separation of an insoluble product. The equilibrium condition which determines the extent to which precipitation takes place is a direct application of the law of chemical equilibrium, and is embodied under the Solubility Product Principle. Its application to quantitative precipitation is of considerable importance and is discussed in more detail in a following section. The precipitate must contain (practically) all of the desired constitu- ent. Chemically it must be a compound which contains the ion whose determination Ls undertaken. It is not essential that the compound as precipitated be chemically pure or even have a definite chemical com- position while being precipitated. It is true that many precipitates do have a definite composition, but some do not and all are more or less contaminated with absorbed precipitating agent, with salts of other THE TECHNIQUE OF GRAVIMETRIC ANALYSIS 171 constituents which may be in the solution and with varying amounts of adsoibed water or water of hydration. Tliorough wasliing and drying or ignition are necessary to convert the precipitate into a compound of unquestioned definite composition. Physical Characteristics of the Precipitate. Tlie physical nature of the precipitated compound is as essential a condition os comjileteness of precipitation, for if the precipitate cannot be filtered easily and waslied free of impurities, it is useless for the purposes of gravimetric analysis. The ideal characteristic of a precipitate is a coarsely crystalline state, in which condition filtration and washing are most eiisily accomplished. Certain precipitates, however, are gelatinous and certain others finely divided, two conditions which are undesirable in handling gravimetric determinations. Finely divided and gelatinous substances tend to assume the colloidal state. A colloid is matter in a finely divided condition, the particles being so small that they cannot be separated from the solution by ordi- nary filtration methods. Colloids will not settle, will pass through the pores of filter paper or, if of a slimy nature, will clog the pores of the filter. These objectionable properties of colloidal precipitates, together with the fact that colloidal particles adsorb ions and salts on tlicir sur- faces, introduce difficulties in the filtration and washing of certain pre- cipitates. Adsorption of dissolved salts by finely divided and gelatinous pre- cipitates is especially objectionable, becau.se the adsorbed material is extremely hard to remove and thus contaminates the precipitate and renders impossible its conversion into a pure compound. In certain cases this contamination can be minimized by slow precipitation, by digestion and by repeated washing, but if these remedies fail the method will not give accurate results and must be discarded. Definiteness of Chemical Composition. The precipitate must be freed from impurities by washing and filtration, and then dried or ignited to a foim whose chemical composition is knowm, before it is finally weighed. This is evident from the fact that if the weight and foimula arc known, and the weighed compound carries all of the desired constituent and nothing else, the amount present in the sample can be calculated by directly applying the Law of Definite Composition. THE TECHNIQUE OF GRAVIMETRIC ANALYSIS The successful fulfilment of the fundamental requirements of gravi- metric precipitation processes depends very largely on the manner in which the experimental steps are carried out. The steps involved in 172 GRAVIMETRIC PRECIPITATION METHODS every determination of this kind are, in order : (1) securing a represent>- ative sample, (2) weighing a suitable amount of sample, (3) dissolvmg the sample, (4) precipitating the desired constituent, (5) filtering and washing the precipitate, (6) drying or igniting the compound and, finally, ^7) cooling and weighing the product. Sampli^ has already been considered in Chapter I. As in volumetric determinations, student samples are usually issued in uniform condition and require, in some instances, only drying. Weighing the Sample. The proper weight of sample to be taken for an analysis must be decided upon for each type of material. The rule which governs this is: Enough sample should be taken to give sufficient precipitate so that the dried or ignited product can be accurately weighed and the analysis conveniently carried out with a single set of apparatus but not so much that the operations of filtering, washing and drying consume too much time. The amount of sample to take is determined by the content of the constituent sought. If the constituent is suspected of being present in small amount a relatively larger sample must be taken; on the other hand, if the sample is thought to contain large amounts, a smaller portion of sample should be taken. The approxi- mate content in the sample, if not known, is usually detennined by a well-conducted qualitative analysis. In the case of students' samples the procedui*es provide for the proper weight to be taken. Dissolving the Sample. Water-soluble samples ordinarily will present no difficulty in dissolving. In some cases the water must be acidified. Dilute nitric acid is the usual solvent for water-insoluble samples, but in some cases dilute hydrochloric acid is preferable. In no case should the solvent exceed 100 ml., for it must be remembered that the larger the volume of resulting solution, together with the wash reagent which later must be used, the greater will be the loss of precipitate. The solu- tion should always be kept as small as possible. Materials which resist acid treatment must be fused to get them into solution. The commonest flux for materials of an acid nature, such as silicates, is Na 2 C 03 . For basic materials an acid flux such as KHSO4 is usually used. Precipitation. The chief concern in precipitation is so to arrange details, whenever possible, as to favor the production of large-grained crystals, since, as we have seen, precipitates that are finely divided or gelatinous are too hard to filter and wash. Slow, dropwise addition of precipitating agent, precipitation in warm solutions and digestion, i.e., heating the solution with its precipitate to nearly boiling for some time, all favor the formation of good precipitates. Stirring during precipitar tion is to be recommended. THE TECHNIQUE OF GRAVIMETRIC ANALYSIS 173 Tests for complete precipitation must always be made by noting whether or not a precipitate or turbid solution results on adding an additional portion of the precipitating re- agent to the clear, supernatant solution or to the filtrate. Wasliing and Filtering. Impurities ad- sorbed or mechanically entrained in the precipitate must be washed out as com- pletely as possible. Washing by decanta- tion, in which the precipitate is allowed to settle and tlie supernatant liquid poured , 2 . fi,ter paper, off through the filter, is always to be preferred. Several successive, small portions of the wash water should be used, stirring after each addition and, after the precipitate has set- tled, pouring off the solution through the filtering medium. The |>re- cipitate is then transferred to the filter paper or Gooch crucible by tip- ping the beaker, holding a stirring rod against the lip of the beaker, as showm in Fig. 13, and direct- ing a stream of water from the wash bottle into the bottom of the inverted beaker. Particles adhering to the sides of the beaker should be loosened and transferred to the filter paper by means of a mbber-tipped stirring rod known as a ‘‘policeman.” If washing by decantation is impracticable, the precipitate is transferred at once to the filter paper and there washed with successive, small (10-ml.) portions of the wash water until tests of the filtrate no longer show the presence of impurities. Silver nitrate is a good test reagent since in so many determinations chlorides are present and their complete removal is a good criterion of the Fig. 13. Transfer of removal of other impurities. precipitate to funnel. Specially prepared filter paper, treated with. hydrochloric and hydrofluoric acids to remove ash-forming constituents, must be used. This “ashless” filter paper leaves a negligible amount of ash on ignition and no correction for filter- paper ash need be made to the final weight of the ignited product. Drying and Igniting. Precipitates such as AgCl, which are decom- posed at the temperature required for burning off the filter paper, are separated from the solution by filtration through a Gooch filter. This consists of a crucible with a perforated bottom, on top of which is 174 GRAVIMETRIC PRECIPITATION METHODS formed a thin layer of asbestos which serves as filtering medium. The preparation of Gooch crucibles is described under the procedure for the determination of chlorides (see page 191). Porous porcelain or fritted glass crucibles may be used in place of Gooch crucibles. In analyses of this kind the crucible, properly prepared, is dried to constant weight in a drying oven at about 105°C. and, after receiving its charge of washed precipitate, is again dried to constant weight at the same temperature. Ashless filter papers are used whenever the precipitate is to be ignited, hleker burners or blast-lamp muffle furnaces are used for the ignition. If reducible compounds are thus ignited, the carbon resulting from the incomplete burning of the filter paper may reduce the compound; in this case it must be reoxidized and brought to constant composition. Specific directions are given under each procedure where special care in igniting precipitates must be exercised. Cooling and Weighing. The dried or ignited product, contained in a previously weighed crucible, should be allowed to cool somewhat in the air, after which it is placed in a desiccator to be brought to constant tem- perature and moisture content, A desiccator is a device that holds in its bottom compartment a drying agent, such as CaCb or other desic- cant, and that is provided with a shelf or tray upon which the crucibles rest. After remaining in the desiccator for a short interval the crucible may be weighed. Successive heatings and weighings are necessary until two successive weig hin gs do not differ by more than 0.0002 gram. It can then be safely assumed that the product has reached practically constant weight. Wlien the weight of the dried or ignited product and its chemical composition are kno^\’n, the amount of constituent which is contained in or is equivalent to it can readily be found by simple calculation. THE CALCULATIONS OF GRAVIMETRIC ANALYSIS In the general discussion of the calculations of quantitative analysis presented on page 13 it was pointed out that the analytical chemist has occasion to deal with calculations of different kinds, each type based upon some general law of chemistiy. APPLICATION OF THE LAW OF CONSTANT COMPOSITION In the present section attention is directed specifically to the funda- mental calculations based on the Law of Constant Composition and ap- plied to gravimetric procedures. This important law, the most charac- teristic feature of all chemical action, states that the weights of reacting substances are proportional to their atomic or molecular weights or APPLICATION OF THE I^W OF CONSTANT COMPOSITION 175 simple multiples of these weights. The law is use- sorbed, and from the increase in weight of the absorption tube its weight and percentage can be found. DESCRIPTIVE OUTLINE OF LIMESTONE ANALYSIS The rocks of the earth’s crust arc mainly of two types, namely, car- bonate and silicate rocks. Both types furnish vast quantities of raw materials for use in vaiious technical processes. The most important of the carbonate rocks are the limestones. A fairly pure limestone is essen- tially CaCOa; most limestones contain, besides other minerals, varying amounts of MgCOa, these being designated as dolomitic limestones. When the amount of MgCOa represents the composition CaIMg(C03)2 the rock is called a dolomite. Among the impurities arc silica and sili- cates or other minerals of iron, aluminum, titanium, manganese, etc., as well as small amounts of the alkalies. Generally, a proximate analysis of a limestone sample furnishes the best information as to tlie utilization of the material. Tlie proximate analysis consists of the following determinations: (1) loss on ignition, (2) silica, (3) combined oxides (reported as R2O3 and consisting princi- pally of Fe203 and AI2O3), (4) calcium, (5) magnesium. A separate determination of moisture may be desired as well as a determination of CO2- The results are generally calculated and reported in teims of the oxides; it should be borne in mind that these reported oxides actually do not exist as such in the original sample. Loss on Ignition. When a limestone is heated to about llOO^C., the principal reactions which occur are the decomposition of CaCOa and MgCOs with evolution of CO2: CaGOa = CaO + CO2 MgCOa = MgO -h CO2 The evolution of CO2, together with evolved moisture and combined water, accoimts for nearly all of the loss in weight. The CaO and 220 SYSTEMATIC ANALYSIS MgO further react with silica and alumina to render the sample more soluble in preparation of the silica determination. The procedure consists, in brief, of heating a weighed portion of the sample, gradually increasing the ignition temperature until about 1100°C. is reached. The ignited residue is cooled, weighed and re- ignited to constant weight. The result is calculated as percentage loss on ignition. Silica. The residue from the loss on ignition determination is trans- ferred to a casserole. The cake is broken up and the solution is evapo- rated to dryness. The resulting residue is treated with concentrated HCl, then with water, and digested. The precipitated silica is filtered off and washed with hot, dilute HCl and water. The filtrate and wash- ings are again evaporated and treated as before with the exception that the silica is washed wuth cold dilute HCl and water. The two filter papers are combined, charred and ignited to constant weight in clean, ignited porcelain crucibles and the silica is weighed. The result is the uncorrected silica. If a more exact determination is required, the impure silica is treated with, hydrofluoric acid, following the procedure pre- viously outlined. Combined Oxides. When the filtrate from the determination of silica is treated with an excess of ammonium hydroxide, a complex group precipitate is obtained. The precipitate may contain the hydrated oxides, Fe203’.TH20, .;Vl203*a:Il20, Ti02*xH20, Mn0i'*xH20 and such compounds as phosphates, arsenates, vanadates and silicates. These latter compounds exist mainly in combination with ferric u*on. The silicates contain the bulk of the Si02 not precipitated by previous acid dehydration. In refined analyses this group precipitate is analyzed for its constitu- ents. In a simplified procedure, the group as a whole is determined and recorded as combined oxides without reference to the amounts of indi- vidual oxides present. This determination is variously referred to as R2O3, ammonia precipitate, or combined oxides. The ignited product consists mainly of FC2O3, AI2O3, Ti02 and phosphates. The determination is carried out by treating the filtrate from the silica determination with HCl and bromine water and then boiling. Several drops of methyl red indicator are added and the solution is treated ^sdth ammonium hydroxide until the indicator turns ydlow. The precipitate is digested for se^^eral minutes, filtered, dissolved and reprecipitated in the same manner. Finally the solution is filtered, the precipitate is washed with a dikite NH4CI solution and ignited to con- stant weight. The filtrates are rcseiwcd for the subsequent determina- tions of calcium and magnesium. DESCRIPTIVE OUTLINE OF LIMESTONE ANALYSIS 221 Cdlcium. After the removal of the ammonium hydroxide precipitate, the solution will contain calcium, magiK'sium and alkali metals wnth possible traces of barium, strontium, silicon and manganese. The cal- cium is generally precipitated as the oxalate according to the following reaction: Ca+-^ + (NH^oCsO., + HoO = CaC204-Il20 + 2NII.|->- The filtrate from tlio combined oxide determination is treated with IICI and (NII.i)2C20.i solution. NILjOIl is next added to the warm solu- DIAGRAMM.\TIC SCHEME OF LIMESTONE ANALYSIS Sample CnCOs UgCOs Si02 Silicates of Fe, Al, Ti, etc. CO2 aiui H2O evolved — > loss on ignition Ignite at llOO^C. Residue CnO MgO Si02 Acid-solublc silicates, etc. Treat with IICI Dehy- drate, filter Residue = SiOexILO — » ignite and weigh as im pure Si02 Filtrate Ca++ Mg++ Fc+++ AI+++ etc. NII4OII l Precipitate Fe203-xH«0 Al203-xH20 etc. F iltraXe Ca++ 1 Mg++ etc. Igiute and weigh as R2O3 Precipitate CaC204 • H2O — + ignite and weigh as CaO (NH4)2C204 or dissolve and titrate ^^^th KMn04 Filtrate (NH4)2HP04 — * MgNH4P04 — ► ignite and weigh etc. I as Mg2P207 tion until the indicator, methyl red, just turns to yellow, thus precipi- tating calcium as CaC 204 • H 2 O. The precipitate is allowed to stand in the liquid for an hour and is then filtered and washed with a dilute oxalate solution. The precipitate is dissolved with hot HCl, and re- precipitated and filtered as before. Finally the precipitate is either ignited to CaO and weighed as such or dissolved in acid and titrated with KMn 04 . All filtrates and washings are resented for the subsequent determination of magnesium. Magnesium. After the interfering elements of the combined oxide group and of calcium have been removed, the solution is in the proper 222 SYSTEMATIC ANALYSIS condition for the precipitation of magnesium as MgNH4P04 • 6H2O, providing the magnesium is present in moderate amounts. If magne- sium is present in small amounts, ammonium salts must be destroyed before the precipitation is carried out. The precipitate is formed by adding an excess of (NH4)2HP04 first and then an excess of NH4OH. The following reaction occurs: Mg++ + NH4+ + P04^ -f 6H2O MgNH4P04 ■ 6H2O The filtrate from the calcium determination is treated with HCl and an excess of (NH4)2HP04 and NH4OH. The precipitate so formed is allowed to stand for at least four hours and is then filtered, washed, dissolved and reprecipitated, as before, under carefully controlled con- ditions. The precipitate is again allowed to stand and is then filtered and washed. It is finally ignited to constant weight at 1100 °C. and weighed as Mg2p207. DESCRIPTIVE OUTLINE OF THE ANALYSIS OF BRASS As typical of a complete, systematic analysis of an alloy, the analysis of brass is here described. Incorporated herein is also a brief discussion of the theoiy of electroanalysis, by which method copper is usually determined in non-ferrous alloys such as brass and bronze. Brass is an alloy consisting principally of copper and zinc and gener- ally containing small amounts of tin, lead and iron and possibly also aluminum and antimony. The usual analysis involves the quantitative separation and gravimetric determination of tin, lead, copper (some- times iron) and zinc. In the usual systematic separation and deter- mination of these metals, tin is first removed, as metastannic acid (11281103) and weighed as Sn02; any antimony present is usually deter- mined along with the tin. Lead is next removed as lead sulfate and weighed as such. Copper is usually determined by electrodeposition. Iron, if present, is precipitated (along with aluminum, if present), as hydrated ferric oxide and ignited to ferric oxide. Finally zinc is pre- cipitated as zinc ammonium phosphate, then ignited to and weighed as zinc pyrophosphate. THE THEORY OF ELECTROANALYSIS Electrolysis. Under certain specified conditions, when a direct cur- rent of electricity is passed through a solution of a metallic salt, the cations are reduced to the metallic fonii and are deposited on the cathode and the anions arc oxidized at the anode. Thus, for example, if a solu- tion of copper sulfate is electrolyzed with a suitable current strength, / THE THEORY OF ELECTROANALYSIS 223 using platinum electrodes, the reaction at the cathode is a reduction of the cupric ions to metallic copper: Cu++ -h 2e = Cu*" each ion gaining two electrons. At the anode, the SO.i“ ion is decom- posed, liberating free oxygen, which escai)cs as a gas, and reproducing an equivalent amount of sulfuric acid, the net effect being the decompo- sition of water at this electrode: II 2 O = 2H+ + -JO 2 + 2e Under suitable conditions of deposition tlie metal may be weighed, thus constituting a very exact and rapid method for the gravimetric determination of certain metals. The deposition of the metal and the nature of the deposit depend upon such factors as the voltage, the current density and the nature of the solution. Whether or not metallic ions are reduced and deposited on the cathode and free elements are liberated at the anode depends upon the voltage at the electrodes. There is a minimum potential required for dei)osition below which no deposition takes place. This minimum potential Is known as the decomposition potential of the dissolved salt in question. The decomposition potential is the algebraic sum of the deposition po- tential required to deposit the metal at the cathode and the potential required to produce an equivalent elTect at the anode. It varies with each metal and each anion. For a given anion such as in sulfate solu- tions, the lower the deposition potential for a metal, the more easily will it be reduced and deposited. The metals can thus be arranged in a series, in the order of increasing ease of their deposition. The order with certain omissions is as follows: alkali metals, alkaline-earth metals, zinc, cadmium, nickel, hydrogen, copper, silver, gold. For the anions the order is nitrate, sulfate, chloride, bromide, iodide, hj'droxyl. From this it can be seen that copper, for instance, is deposited at a lower potential than hj'drogen or zinc, and that the potential required to deposit copper from a sulfate solution is less than from a nitrate solu- tion. Numerical values, in volts, have been determined for the depo- sition-potentials for the series of metals and non-metallic elements for solutions containing equivalent (normal) concentrations of the ions. In practice, however, such data can be applied only approximately because, as electrolysis proceeds, the concentration of ions gradually diminishes and the voltage must be increased. A knowledge of the potential series (the electromotive series of metals) is nevertheless most useful in finding the order and relative ease ^\'ith which the metals can be separated and deposited. 224 SYSTEMATIC ANALYSIS In addition to this minimum potential required for deposition the additional voltage necessary to overcome the resistance of the solution must be taken into consideration in electrolysis. Overvoltage. The discharge of ions at the electrodes is further com- plicated by what is known as overvoltage. From the position of hy- drogen in the above series, it appears that hydrogen is more easily liber- ated than the metals above it (zinc, for example), and it should not ordinarily be possible to plate out zinc in preference to liberation of hydrogen. Hydrogen, however, is more easily liberated when its ions are in contact with certain metals than with others. The difference in voltage required to liberate hydrogen on a metal electrode during elec- trolysis and that required to liberate it on a platinized platinum elec- trode is called the hydrogen overvoltage for that metal. If the over- voltage is large and near that of the deposition voltage of the metal, the metal may deposit before hydrogen, or hydrogen itself may be evolved ahead of its turn when metals below it are being electrolyzed. On account of the high oven^oltage of hydrogen on zinc it is quite pos- sible, with a large current and too low an acidity, to plate out zinc during the electrolysis of copper solutions. Faraday’s Laws. The amount of metallic and non-metallic element liberated at the electrodes depends upon the quantity of electricity passed and can be calculated from the laws of Faraday. Faraday’s first law states that the weight of any element liberated at an electrode is proportional to the quantity of electricity passing through the solu- tion. His second law states that equivalent weights of different elements are liberated by the same quantity of electricity. A gram-equivalent weight of a substance is that weight in grams which wall react directly or indirectly with 1.008 grams of hydrogen. For metals, it is the gram- atomic w'cight divided by the valence change when the metallic ion is reduced. The unit of current strength is the ampere and is that amount of current which will deposit 0.001118 gram of silver per second from a solution of silver nitrate. The quantity of electricity is measured in coulombs. (A coulomb is one ampere flo^ng for one second.) Cou- lombs are related to amperes by the equation Coulombs = Amperes X time in seconds A current strength of 10 amperes, for example, flow'ing for 5 seconds will result in the passage of 50 coulombs. According to Farada 3 '’’s second law, in order to deposit a gram-equiva- lent or 107.88 grams of silver, there will be required 107.88/0.001118 or 96,500 coulombs. In general it has been found that 96,500 coulombs OUTLINE OF PROCEDURE 225 are required to liberate a gram-equivalent weight of any clement or compound. Tlie quantity, 96,500 coulombs, is called a faraday. A faraday will therefore liberate 107.88 grams of silver or 31.78 gram.s of copper, or 1.008 grams of hydrogen or 8.000 grams of oxygen from solutions of these ions. It should be noted that equivalent quantities of elements are liber- ated at each electrode. Thus in the electrolysis of a solution of copper sulfate, 90,500 coulombs (one faraday) will simultaneously deposit 31.78 grams of copper on the cathode and liberate 8.000 grams of oxy- gen at the anode. In electrochemical analysis, where the discharge of metallic ions must be complete, a direct calculation of the total current required cannot be made, because as electrolysis proceeds the concentration of ions diminishes and a somewhat larger current is required for the deposition of the metal than that calculated from Faraday’s law on the assump- tion that the ionic concentration remains constant. Current Density. Since the deposit is to be weighed, it must firmly adliere to the electrode. Tlie chief factor influencing the character of the deposit is the current density. Current density Ls the amperage divided by the area of the electrode. The unit chosen for the electrode area is one square decimeter (100 square centimeters). Current den- sities are expressed as amperes per square decimeter. For the decompo- sition of copper the current density must be low in order to give a good deposit, about 2 amperes per square decimeter of cathode area. For cylindrical gauze cathodes the area is computed as the total area, botli inside and outside, of a plane cylinder of like dimen.sions. In order to prevent impoverishment of ions in the immediate vicinity of the electrode the solution should be stirred, either by use of a separate stirrer or by rotating the anode by means of an electric motor. Most metals are deposited at the cathode in the metallic condition. Lead and manganese, however, are usually deposited as PbOa and Mn02 on the anode. Electrolysis is usually made from nitrate or sulfate solutions, because free chlorine liberated from chloride solutions attacks the platinum electrodes. OUTLINE OF PROCEDURE Determination of Tin. \Mien a brass is treated ^rith nitric acid, copper, zinc and lead pass into solution as the nitrates, but tin is con- verted to a colloidal hydrated stannic oxide (or metastannic acid, H 2 Sn 03 ) which is insoluble in nitric acid: 3Sn “b 4 HNO 3 = 3Sn02*H20 H- 4NO 2 H 2 O 226 SYSTEMATIC ANALYSIS As is common with colloidal precipitates, the hydrated stannic oxide easily adsorbs and coprecipitates other ions in the solution. This ad- sorption may be partially eleminated by digesting in concentrated nitric acid, in which case the precipitate is somewhat coagulated, thus cutting down the large adsorbing area. In the evaporation, the precipitate is not taken to diyness. If such happens, it is very difficult to wash impurities from the precipitate. The digested precipitate is filtered off, using paper pulp to prevent particles from running through the filter, and the oxide is dehydrated and ignited to Sn02- The percentage of tin in the sample is found by multiplying the weight of Sn02 by the factor for Sn in Sn02, 0.7877, dividing by the weight of sample and multiplying by 100. Determination of Lead. Two good methods are commonly used for the determination of lead. In the one method the lead is precipitated as the sulfate by evaporating the solution containing lead and H2SO4 to fumes of SO3. The other method involves the electrolytic precipi- tation as lead dioxide, Pb02, on the anode in a strong nitric acid solu- tion. Other methods such as precipitation as PbMo04 or PbCr04 are not so desirable because they introduce other elements into the filtrate and these may lead to complications in subsequent separations. The initial attack of the alloy by nitric acid converts the lead to lead nitrate: 5 Pb + 12HNO3 == 5Pb(N03)2 + 2 NO + 6H2O Upon addition of sulfuric acid and evaporation to fumes, the lead is precipitated as the sulfate and nitric acid is eliminated: Pb(N03)2 + H2SO4 = PbS04 + 2HNO3 The filtrate and w'asliings from the tin determination is treated with 5 ml. of concentrated H2SO4 and carefully evaporated over a low flame until dense w’liite fumes of SO 3 appear. The precipitated PbS04 is separated from the copper and zinc by use of a previously prepared Gooch crucible. The crucible is ignited at a relatively low temperature (500°-600°C.). The PbS04 is weighed as such; from such data the percentage of lead in the alloy can readily be calculated. Determination of Copper. The most convenient and one of the best methods for detennining copper in alloys and of separating it from zinc is by electrolytic precipitation. In this method, the cupric ion in solu- tion is electrolyzed, using platinum electrodes. The copper that is plated out is quantitatively determined by weighing the cathode before and after the reaction. OUTLINE OF PROCEDURE 227 The concentration of the filtrate from the lead doteiminatiori is ad- justed, the acidity is reduced by an ammonium hydroxide addition, and a few millilitei'S of nitric acid are added to insure a smooth, bright, copper deposit. The solution is then electrolyzed, using a circular jilatinum gauze cathode and a revolving anode. A few crystals of urea arc added to destroy nitrous acid. When all the copper lias been deijosited, the electrodes are slowly removed from the solution with the current still flowing, and waslied as the surface comes above the solution. The cathode is then washed with acetone and dried at 100° for several minutes, cooled and weighed. Determination of Zinc. One of the best methods for the determina- tion of zinc is precipitation with diammonium hydrogen phosphate in a solution of the proper acidity. The reaction is as follows: ZnS04 + 2(NH4)2 HPO.i + OII 2 O = ZnNH4P04 - 01120 + (NIl4)H2P04 + (NIl 4 ) 2 S 04 'Pliis precipitate is appreciably soluble in water and in acid or basic solutions. Therefore, the solution from which the precipitate is ob- tained must be very close to neutral (pH 5.5 to 7). The precipitate is washed with a 1 per cent (NH 4 ) 2 HP 04 solution and finally with a 1 : 1 alcohol-water solution to free the precipitate of soluble phosphates. The precipitate may be dried to 100°C. and weighed as ZuNIIjPO.i OII 2 O, but the better procedure is to ignite it to zinc pyrophosphate, Zn 2 P 207 , and weigh as such : 2ZnNH4P04*6H20-^ Zn2P207 + 2 NH 3 -f I 3 IT 2 O The solution from the copper determination is brought to exact neu- trality with NH 4 OH and heated to boiling. A solution of (NH 4 ) 2 HP 04 is added slowly, wnth stirring, and the precipitate is digested for 30 min- utes. The solution is then cooled and allowed to stand for an hour to increase the size of the precipitate particles. It is finally filtered through an ignited and weighed Gooch cniciblc, washed first with a phosphate wash solution and then with a 50 per cent alcohol solution. The pre- cipitate is finally ignited to ZU 2 P 2 O 7 and weighed as such. DIAGRAMMATIC SCHEME OF BRASS ANALYSIS Sample Sn 1 Pb Cu Zn Dissolve in HNO3. Filter. Residue: Sn02 -:rH20 —♦ ignite and weigh as SnOa Filtrate Pb++ C\i^ Zn^ H2SO4 to furaes. Filter through Gooch. Precipitate: PbS04 — » weigh Filtrate Cu* on cathode Cu+''‘ Zn++ [ Electrolyze * Zn"^"^ J • Z11 + + add (NH4)2HP04 -* ZaNH^POi-eHjO, ignite and weigh as ZniPjO?. 228 SYSTEMATIC ANALYSIS CALCULATIONS OF INDIRECT ANALYSIS Frequently, in chemical analysis, two closely related elements are present in the same sample. They will therefore usually react in the same v/ay toward reagents. Thus if, in the determination of iron, aluminum had been present in the iron sample, the precipitate would have consisted of a mixture of Fe(OH)3 and Al(OH)3, and on ignition a mixture of Fe203 and AJaOa would have been, obtained. In the chloride detennination, bromides and iodides, if present, would have pre- cipitated as AgBr and Agl along vidth AgCl. The alkali metals, sodium and potassium, are often associated with each other, giving on analysis a residue which consists of a mixture of the sulfates or the chlorides of these two elements. Similarly, mixtures of oxalates, carbonates, sulfates and sulfides are sometimes encountered in gravimetric analysis. Obviously, if a determination is to be made for the amount of each constituent present, a further experiment must be performed in order to find the amount of each present. There are various ways by which this ma3’' be done. In the first example cited, that of the mixture of iron and aluminum, experimentally the difficulty can be overcome by first precipitating the mixture with NH4OH, igniting the mixed Al(OH)3 and Fe(OH)3 to AI2O3 and Fe203 and weighing the mixture of oxides. The mixture of oxides may then be dissolved or another sample taken and, by a separ rate (volumetric) analysis, the amount of iron can be determined. Then, calculating the weight of Fe203 equivalent to the amount of iron found and subtracting this from the total weight of mixed oxides gives by difference the weight of AI2O3. In the case of materials containing potassium and sodium, these two elements will be left in solution after the removal of metals which pre- cede them in the qualitative grouping. The solution may then be fumed down with sulfuric acid, and the residue, consisting of a mixture of Na2S04 and K2SO4, weighed. The residue is then redissolved and the potassium precipitated as K2PtCl6 or as KCIO4. From these data the weight of potassium can be found, which, when recalculated as K2SO4 and subtracted from the weight of residue, gives, as the difference, the weight of Na2S04. Instead of determining one of the constituents by a separate pro- cedure and calculating the other by difference, as illustrated in the fore- going examples, it is possible with certain mixtures which have a common constituent to make a separate determination of this common constitu- ent. The calculations are then more complicated. A sample calcula- tion is made in the following problem. CALCULATIONS OF INDIRECT ANALYSIS 229 A sample of material contained, among other constituents, a chloride and a bromide. A weighed portion was dissolved and treated with an excess of AgNOa, precipitating a mixture of AgCl and AgBr weighing 0.8413 gram. By a separate analysis of tliis mixture the total amount of silver was found to be 0.5G78 gram. What are the weights of chlorine and bromine in the sample? We know the total weight of the mixture of AgCl and AgBr and the total weight of silver which is partly combined with the chlorine and partly W'ith the bromine. In order to find the amounts of chlorine and bromine, w^e must fimt calculate the separate weights of AgCl and of AgBr. Letting x stand for the weight of AgCl and y for the weight of AgBr, w^e have the equation X + 2/ = 0.8413 gram Tlie w^eight of silver combined as AgCl is found from the relation ( 1 ) Ag : AgCl = Weight of Ag in AgCl : x or Ag AgCl X = Weight of Ag in AgCl and the weight of silvet combined as AgBr is Ag : AgBr = Weight of Ag in AgBr : y Ag AgBr y = Weight of Ag in AgBr Tlie total weight of silver, wLich has been found by experiment to be 0.5678 gram, is therefore related to the w^eights of AgCl and AgBr by the equation Ag AgCl Ag AgBr y = 0.5678 gram Inserting the proper atomic and molecular w'eights, equation (2) becomes 107.88 107.88 143.34^ 187.80^ 0.5678 Using the above ratios as chemical factors of Ag in AgCl and of Ag in AgBr, we have the two equations 0.7526a: + 0.5744i/ == 0.5678 x-\- y = 0.8413 230 SYSTEMATIC ANALYSIS Sol ng these simultaneous equations, by the elimination of y we have 0.7526X + 0.57442/ = 0.5678 0.5744a: + 0.57441/ = 0.4832 0.1782X = 0.0846 X = 0.4747 Therefore x, the weight of AgCl, is equal to 0.4747 gram. The value for y may now be found from equation (1) : X + 2/ = 0.8413 y = 0.8413 “ 0.4747 y = 0.3666 Therefore i/, the weight of AgBr, is equal to 0-3666 gram. Finally the amount of chlorine is (Cl/AgCl) X 0.4747 = 0.1174 gram, and the amount of bromine is (Br/AgBr) X 0.3666 = 0.1560 gram. In general, data which involve a conunon constituent such as a mixture of silver salts, mixed sulfates of the alkalies or alkaline earths, mixed carbonates, etc., must be resolved into two equations, one of which usually embodies the chemical factors, and these equations must then be solved simultaneously. PROBLEM SET 14 INDIRECT ANALYSIS 1. A mixture of AgBr and AgCl weighing 1.2500 gram was treated with chlorine in order to convert AgBr into AgCl. After such treatment the mass weighed 1.1800 gram. What are the percentages of Cl and Br in the sample? Atw. 10.07 per cent Br; 18.88 per cent Cl 2. A mixture of FesOs and AI 2 O 3 weighing 0.3050 gram was obtained ^ igniting the mixture of hydroxides obtained from a 1.0000-gram sample of limestone. A separate analysis of the limestone for ii*on gave 6.58 per cent. What is the per- centage of aluminum? 3. A mixture of NaCl and NH 4 CI weighing 0.4500 gram was dissolved in water, treated with chloroplatinic acid, the (NH 4 ) 2 PtCl 6 filtered off and ignited, yielding 0.1063 gram of platinum. What are the percentages of NaCl and NH 4 CI represented in the mixture? Ans. 12.SG per cent NaCl; 87.13 per cent NH 4 CI 4. A certain mixture of AgCl and AgBr weighs 0.9000 gram. The silver in this mixture, determined independently, weighs 0.6500 gram. What are the weights of AgCl and AgBr in the mixture? 6 . A sample consisting of a mixture of K 2 SO 4 and NasS 04 weiglung 1.0537 grams, when dissolved and treated with BaCb, gave a precipitate of BaSOi weighing 1.6874 grams. What are the percentages of potassium and sodium? Ans. 17.80 per cent Na; 20.20 per cent E PROBLEMS 231 6. A 1.0000-gram sample of feldspar rock yielded a mixture of KCl and NaCl weighitig 0.2875 gram. After dissolving the mixed chlorides in water and adding AgNOa, the AgCl obtained weighed 0.0927 gram. Find the percentages of K and Na in the sample. 7. An analysis of a limestone showed it to contain 98 per cent of a mixture of CaCOj and MgCOs. The CO 2 in the sample was 50.65 per cent. Calculate the percentages of MgCOs and CaCOa. An.s'. 90.80 per cent MgCOa; 7.14 per cent CaCOa 8. A 1.0000-gram sample of a certain material gives a mixture of SrC 204 and CaC 204 which weighs 0.5000 gram. By a volumetric method the oxalate content of the precipitate is found to be 0.2815 gram. What are the percentages of CaO and SrO in the mixture? 9. What is the percentage of K 2 O, NasO and SO3 in a sample made up by mixing equal gram-molecular parts of pure K 2 S 04 and Na 2 S 04 ? An.-;. 29.78 per cent KjO; 19.60 per cent Na20: 50.62 per cent SO 3 10. In an organic combustion analysis, oxygen was pa.ssod over an organic com- pound containing C, H and O. The carbon wjis oxidized to CO 2 and absorbed in a KOH absorption tube; the hydrogen was oxidized to H 2 O and was absorbed in a CaCb absorption tube. The sample weighed 0.7510 gram. The gain in weight of the CaCl 2 tube was 0.0987 gram and that of the KOII tul)c 0.7021 gram. Calculate the percentages of C, H and O in the sample. 11. If a 0.7500-gram sample of pure H 2 C 2 O 4 is placed in an organic combustion furnace and burned in a stream of oxygen, according to the reaction 2H2C2O4 + 02 = 2II2O + 4CO2 what W’eight of n20 and CO" should be collected in the absorption tubes? Ans. 0.1501 gram H 2 O; 0.7332 gram CO 2 12. In what proportions should CaCOs and MgCOs be mixed in order to make a sample containing 50.00 per cent CO 2 ? 13. A silicate rock sample weighing l.OOOO gram yielded 0.2500 gram of a mixture of Na:S 04 and K2SO4. After dissolving these sulfates and precipitating the potassium as K 2 PICI 6 , a W’eight of 0.4210 gram of K 2 PtCl 6 was obtained. Calculate the per- centages of K and Na in the sample. Ans. G.77 per cent K; 3.21 per cent Na 14. A mixture of MgCOa and SrCOs weighed 1.7500 grams. If these carbonates are converted to sulfates and the sulfate mixture weighs 2.2500 grams, what are the percentages of MgCOa and S 1 CO 3 in the sample? 16. A 0.2500-gram mixture of Na 2 S 04 and K 2 SO 4 was obtained from a 1.1250-gram sample of silicate rock. After dissolving these sulfates and precipitating with Ba++ as BaS 04 , a weight of 0.3870 gram of BiiSOa was obtained. What are the per- centages of K 2 O and Na 20 in the sample? Atvs. 7.47 per cent Na20; 2.76 per cent K 2 O 16. A mine owner ships a cargo of iron ore which analyzes 2.7 per cent moisture and 69.5 per cent Fe 203 on a wet basis at the mine. The chemist at the blast furnace finds that the moisture content has increased to 7.3 per cent dxiring transit. Wbat percentages of Fe 203 should be reported on (a) the wet and (6) the dry basis? 17. A sample of coal was dried, losing 8 per cent of its weight through evaporation of water. One gram of this dried sample gave 0.0173 gram of BaS 04 when the sulfur was converted to sulfate and precipitated as BaS 04 . What was the percentage of sulfur in the original coal? Ans. 0.22 per cent 232 SYSTEMATIC ANALYSIS 18. A sample of coal, as received, contained 8.2 per cent moisture; when the dried sample was analyzed for other constituents it was found to contain (a) 23.9 per cent volatile combustible matter, ( 6 ) 59.0 per cent fixed carbon and (c) 17.2 per cent ash. What are the percentages of these constituents in the sample as received? 19. A sample of iron ore was found to contain 6.5 per cent moisture and 57.83 per cent iron -What is the percentage of iron in the sample when calculated to the dry basis? Ans. 61.84 per cent 20. A certain soluble sulfate sample contained 2.13 pei/ cent moisture. On analysis 0.5682 gram of the original sample gave 0.7059 gram of BaS04. What was the per- centage of SOs in the moisture-free sample? SUMMARY OF ANALYTICAL METHODS In final summary, it will be evident to the student that for any ^ven element or ion, generally two or more methods of determination are possible and, sis a general rule, a choice is offered between a gravimetric and a volumetric method. Table IX, based on a similar one in Lundell and Hoffman, Outlines of MetJiods of Chemical Analysis^ summarizes the Table IX Methods Most Commonly Used in Determining the Elements Element Gravimetric Volumetric Aluminum AI2O3, AIPO4, Al-quinolate » lodometric, neutralization Antimony Sb2S3, Sb204 Oxidation, iodometric, iodimetric Arsenic AS 2 S 3 , AS 2 S 5 Ag on Ag 3 As 04 , iodometric, iodimetric Barium BaS 04 , BaCr 04 Iodometric Beryllium BeO Bismuth BioOs, BiOCl Colorimetric Boron Fixed B2O3 Neutralization Bromine AgBr Iodometric, precipitation Cadmium Cd, CdS 04 Calcium CaO, CnS 04 Oxidation of the oxalate Carbon Fixed CO2 Cesium CsCl, CSCIO4, Cs2PtCl6 Chlorine AgCl lodometric, precipitation Chromium CrjOa Reduction, iodometric, colorimetric Cobalt C0SO4, Co Columbium CbsOt Oxidation Copper Cu, CuCNS Iodometric, oxidation, colorimetric Fluorine PbClF, CaF2 Neutralization, colorimetric Gallium Ga203 Germanium Gc 02 , Mg2Gc04 Gold An - Hafnium HfOa Hydrogen H 2 O Indium 111203 Iodine Agl lodometric, precipitation 233 SUMMARY OF ANALYTICAL METHODS Table IX — Conlinucd Methods Most Commonly Used in Determining the Elements Element Gravimetric Volumetric Iridium Ir Iron FeaOs Oxidation, reduction, iodometric Lead PbS04, PbOa lodoinetric, precipitation Lithium LizSO^ Magnesium MgO, Mg 2 P 207 , MgS 04 , Mg-quinolate Iodometric, neutralization Manganese MnSOi, MnS, MU3O4 Oxidation, reduction, colorimetric Mercury HgS, Hg^CU, Hg Precipitation Molybdenum M0O3, PbMo04 Oxidation, colorimetric Nickel Ni-dimethylglyoxime, Ni Cyanide titration Nitrogen Colorimetric, neutralization (NA3) Osmium Os Oxygen H2O, AI2O3 Palladium Pd-dimethylglyoxime, Pd Phosphorus MgaPaO? Neutralization, colorimetric Platinum Pt Potassium KCl, KaPtCle, KCIO4, K2SO4 Radium Electroscopic Rare gases Gasometric Rare earths • Oxides Rhenium AgReO^, Re Neutralization Rhodium Rh Rubidium RbCl, RbsPtClc, RbCi 04 Ruthenium Ru Scandium SC2O3 Selenium Se Oxidation, iodometric Silicon Si 02 Silver AgCl. Ag Precipitation Sodium NaCl, Na 2 S 04 Strontium SrO, SrS04 Sulfur BaS04 Precipitation, iodometric, iodimetric Tantalum Ta 206 Tellurium Te Oxidation, iodometric Thallium Tl 2 Cr 04 , TI2O3, TLPtCU Oxidation, iodometric Thorium ThOa Tin Sn 02 Iodometric Titanium Ti 02 Oxidation, colorimetric Tungsten WO3 Neutralization Uranium U3O8 Oxidation Vanadium V2O6 Oxidation, reduction YfFrium Y2O3 Zinc ZnO, Zn2P207, ZnS Precipitation Zirconium Zr 02 , ZrP04 * Elementi) 57-71. All are trivalcnt in their principal valencoe. 234 SYSTEMATIC ANALYSIS methods most commonly used. Under the heading of gravimetric methods, the element can be weighed as metal, as oxide, or as a salt of definite composition. The volumetric methods involve neutralization, oxidation or reduction and precipitation or are some type of colorimetric determination. CALCULATION OF FORMULAS AND ATOMIC WEIGHTS Formulas. An important application of quantitative analysis is the use of analytical data to determine the empirical formulas of pure com- pounds and minerals. The analysis of a pure chemical compound will show the weight of each element in the weight of sample taken. The percentage composition can then be easily calculated. If the actual weight of each element is divided by the atomic weight of that element, the quotient wU be the number of gram-atoms of each element present. These gram-atomic weights bear a simple relation to one another in that there will be a common divisor; and, if each gram-atomic weight is divided by this common divisor, the ratio of the numbers of atoms of the elements present can be found. The percentages obtained from analysis may be used in the same way as the actual weights, since the percentage of an element is directly proportional to the weight of that element present. As a simple example, suppose that a 1.0000-gram sample of a pure salt is analyzed and found to be composed of 0.3934 gram of sodium and 0.6066 gram of chlorine. Dividing 0.3934 by the atomic weight of sodium, 22.997, gives 0.01711 gram-atom of sodium, and 0.6066 divided by 35.457 gives 0.01711 gram-atom of chlorine. There are just as many gram-atoms of sodium as there are of chlorine and, consequently, just as many atoms of one element as of the other. The ratio of the number of sodium atoms to the number of chlorine atoms in the compound, then, is 1 : 1; and the simplest formula is NaCl. The formula thus obtained is merely the empirical formula, that is, the simplest ratio of the number of atoms of each element which make up the compound. The actual molecular formula may be a simple mul- tiple of the empirical formula. A molecular-weight determination must be made in order to assign the proper formula to the compoxmd. Such determinations need not be too accurate, since it is only necessary to distinguish betw'een the empirical formula weight and its simple multi- ples. Various physical-chemical methods are available for determining molecular weights. In the Dumas method a known volume of a gaseous substance is weighed under controlled and readily measured conditions of temperar CALCULATION OF FORMULAS AND ATOMIC WEIGHTS 235 ture and pressure. Then, recalling that one gram-molecular weight, or mole, of a ga.s occupies 22.41 litei-s at O'^C. and 7C0 millimeters pre.ssure, the weight of gas occupying this gram-molecular volume under these standard conditions can be calculated. Tliis weight is the gram- molecular weight of the gas. For example, a Dumas bulb of 300.0 ml. capacity is filled at 25°C. and 740 millimeters pressure with a gas of unknowTi molecular weight. The weight of the gas in the bulb is found to be 0.526 gram. To calculate the molecular weight the volume of gas must fn-st be corrected to standard conditions. The student will recall that the volume of a gas varies directly as the absolute temperature and inversely as the pressure. The expression for the correction of volume is then 300.0 ml. X 273®C. X 740 mm. 273°C. + 25X\ 700 mm. which, when solved, gives 207.6 ml. as the volume corrected to 0°C. and 760 millimetei-s pressure. This volume of gas weighed 0.526 gram. To find the weight of 22.41 liters, or the gram-molecular weight of the gas, multiply 0.526 gram by 22,410 ml. and divide the product by 267.0 ml., the corrected volume. This gives 44.0 grams as the weight of the gas which occupies 22.41 liters, so 44.0 is the molecular weight of the gas. It is also possible, as in the Victor Meyer method, to vaporize a weighed amount of volatile liquid and measure under known conditions of temperature and pressure the volume of vapor produced. This volume is converted to standard conditions, 0°C. and 760 millimeters pressure. The weight of substance necessaiy to produce 22.41 liters of gas under these conditions may then be calculated. Attention should be called, however, to the fact that in laboratory determinations by the Victor Meyer method, tlie vapor from the volatile liquid displaces a volume of air which in turn is collected in a gas buret over water. In converting this volume of displaced air to standard conditions, the vapor pressure of the water in the buret must be subtracted from the obseiwed baro- metric pressure. With this exception, the method of calculation is the same as that for the Dumas molecular weight determination. A liquid to w’hich has been added a soluble, non-volatile non-electro- lyte has a higher boiling point than the pure liquid. This principle is the basis of still another method for the detennination of molecular weights. A 1 molal solution of such a substance in water has a boiling point of 100.52°C., the boiling point rise being directly proportional to the amount of solute present. Now, if a weighed amount of substance of unknown molecular weight is added to a kno\ra weight of boiling 236 SYSTEMATIC ANALYSIS solvent, and the increase in boiling point is measured, we have sufficient data to calculate the unknown molecular weight. To illustrate, 1.2 grams of solute was added to 17.5 grams of boiling water; the boiling pouit nse was 0.51°C. Since a 1 molal solution contains one mole of solute per 1000 grams of solvent, a conversion to a basis of 1000 grams of water is made first: 1.2 : 17.5 = x : 1000. The value of 68.6 grams is obtained for x. This amount of solute, however, produces only 0.51°C. boiling point rise, whereas the molal boiling point rise for water is 0.52°C. If 68.6 grams is multiplied by 0.52°C, and the product is divided by 0.51°C., 69.9 grams is found to be that weight of solute which must be added to 1000 grams of water to raise its boiling point 0.52°C.; this w'eight is the molecular weight of the unknown solute. The measured lowering of the freezing point caused by the addition of a soluble non-electrolyte to a pure solvent can be similarly used to calculate molecular weights. The molal freezing point lowering for water is 1.86°C. Tins method is probably the most accurate for the determination of molecular weights in solutions, and has helped estab- lish the true formulas of many organic substances from data of the analyses. Once we have found the empirical foimula of a compound from per- centage composition data, and the molecular weight by some physical- chemical means, it is a simple process to decide on the proper molecular formula. To illustrate, a calculation of the results of a combustion anal- ysis of a pure organic compound showed its empirical formula to be CH 2 O. When a freezing-point molecular-weight determination was made, the molecular w^eight of the compound was found to be 180. The sum of the atomic w^eights of the empirical formula is 30, whereas the true molecular weight is 180. By inspection, we can see therefore that the true molecular formula is six times as great, or C 6 H 12 O 6 . Atomic Weights. The entire structure of chemistry as an exact science rests upon the precision with which the atomic weights of the elements are knowm. The analytical procedures by w^hich the atomic weights are arrived at are unquestionably the most precise of all quanti- tative determinations. One method consists in making an exact quan- titative determination of the constituent in a veiy carefully purified sample of a compound containing the element the atomic weight of which is to be determined. Such a quantitative deteimination may not, how- ever, give the true atomic weight but rather a multiple or submultiple of it. What is really obtained is the combining weight and not necea- sarily the atomic weight. The combining weight is the proportion in which one element will combine with another; the atomic weight is the weight of an atom rela- / CALCULATION OF FORMUI^AS AND ATOMIC WEIGHTS 237 tive to that of oxygen taken as 10.0000. To illustrate, if wc analyze water we find that for every 1.0080 grams of hydrogen thcic an; 8.0000 grams of oxygen; tlie combining weight of oxygen is therefore 8.(K)00. If we determine the amount of copper and oxygen in black copper oxide we find that the copper and oxygen are united in the ratio of 31.78 : 8; whereas in the red oxide of copper the combining proportions of copper to oxygen as determined by experiment are 03.57 : 8. With tlio atomic weight of oxygen fixed at 10.0000, the combining weights of copper are 63.57 and 127.14. To arrive at Hie atomic weight of copper, we must decide which of these combining weiglits is the atomic weight. To decide whether the combining weight of an element is its atomic weight, or what multiple or submultiplc of the combining weight to use involves consideration of other criteria and the application of certain mles and laws. Among these arc: (1) the Law of Dulong and Petit, which states that the specific heat of a solid clement multijilicd by its atomic weight is, in most case.s, a constant having the value 0.4; (2) Avogadro’s HyiJothesis, that equal volumes of ga.ses (under the same conditions of pre.ssure and temperature) contain the same number of molecules; (3) the Periodic Law, in which the clement is placed in its proper family relationship; (4) most of the elements, in the gaseous state, are diatomic; (5) the least common multiple of several combining weights for the same clement will, in many cases, indicate the atomic weight. Applying the last rule to the case of copper previously referred to, where the combining weights are 63.57 and 127.14 respectively, this would indicate the atomic weight of copper to be 03.57. This can be checked by the rule of specific heats; for if wo multiply the specific heat of copper, 0.092, by 03.57 we get approximately 6, in sufficient agreement with the constant 0.4 to show that the selected combining weight is the tme atomic weight. During the first half of the nineteenth centiny, after Dalton’s atomic theory had been advanced and the combining weights of elements had been worked out with fair accuracy, especially by Berzelius, chemists were not in agreement as to the true atomic weights. The hypothesis of Avogadro had not been accepted because it seemed limited in its applicability, and Dulong and Petit’s Law had proved to be quite in- conclusive in practice. It was the contributions of the organic s^mthe- sists, Williamson, Hofmann, Wurtz and Kekul4, in fixing the formulas of simple substances like water, ammonia and methane, that paved the way for final agreement as to which multiple of the combining weight was the atomic w^eight. At a meeting in 1800 of the prominent chemists of that time, Cannizzaro summed up the existing knowledge, clarified 238* SYSTEMATIC ANALYSIS the interpretations of Avogadro’s Hypothesis, and made some sugges- tions as to how to develop a system of atomic weights. He proposed that the molecular weights of the volatile compounds of the elements should be determined. On analyzing these compounds, it could then be found what part of the molecular weight should be attrib- uted to each element. In selecting the atomic weight, the smallest weight of an element found in the molecular weights of its compounds should be chosen. Now that criteria had been established for determining which multiple or submultiple of the combining weight was the atomic weight, chemists entered again upon the problem of more accurate determinations. It had been more than twenty years since the work of Berzelius, and great improvements in apparatus and procedure had been made. Stas, a Belgian chemist, refined analytical techniques to a degree never before attained. His materials were carefully purified, his balances were of much greater accuracy, and his manipulations were conducted with utmost precautions. During the early part of this century, Richards, in the laboratories of Harvard University, further refined the methods of atomic weight determinations. The more general method of atomic weight determinations used today is the chemical convemion of one pure compound to another, Tliis is possible because now we can be certain of the chemical equivalency of the two compounds. The ratio of the equivalent weights of the two compounds is calculated. Then, by using those atomic weights pre- viously determined, the atomic weight of another element may be found. Let us take as an example Richards’ determination of the atomic weight of lithium. He weighed out many samples of carefully purified lithium cliloride on balances which were accurate to the hundred- thousandths of a gram. These samples were dissolved and caused to react with silver nitrate. From seven reported experiments he calcu- lated that 1.000000 gram of silver is equivalent to 0.392992 gram of lithium chloride. Then the following relationsliip was used: I Weight LiCl : Weight Ag = Molecular Weight LiClrMolecular Weight Ag The atomic weight of silver was accurately known to be 107.880, and that of chlorine had been determined as 35.457. The atomic weight of lithium is now the only unknown and can readily be calculated by sub- stituting (:r + 35.457) for the molecular weight of lithium chloride. On the basis of tins work Richards found the atomic weight of lithium to be 6.939. Most atomic weight determinations now consist in an elaboration of the care and precautions in carrying out the analytical procedures. PROBLEMS 239 In 1941 G. P. Baxter of riar\^ard Univereity was the Chairman of the Committee on Atomic Weights of the International Union of Chemistry. He and the other members, M. Guichard of the Sorbonne, Paris, O. Honigschmid of Munich, Germany, and R. \Miytlaw-Gray of the Uni- versity of Leeds, Leeds, England, each year consider the recent work on atomic weight determinations, and revise the table of atomic weights to conform to the most probable values. PROBLEM SET 16 FORMULAS AND ATOMIC WEIGHTS 1 . A sample of pure KMnO^ was found to contain 39.65 per cent of potassium. Taking K = 39.10 and O = 16.00, calculate the atomic weight of manganese. Atw. 54.89 2. A 0.7500-gram sample of a certain metal yielded 1.8430 grams of bromide. If the metal is monovalent, what is its atomic weight? If the metal is divalent, what is its atomic weight? 3. A 0.5000-gram sample of a certain pure compound yielded 0.1900 gram of chlorine; a 0.2500-gram portion of the salt upon the addition of (NH 4 ) 2 S 04 and the ignition of the precipitate gave 0.2800 gram of BaSOA- What was the compound? An^'. BaCb 4 . If, by an organic combustion, the percentage composition of a certain organic compound becomes established as 77.92 per cent carbon, 11.68 per cent hydrogen, and 10.40 per cent oxygen (by difference), wliat are the ratios, carbon: hydrogen: oxygen? What is the empirical formula? 6. The boiling pomt of a solution of 0.987 gram of a non-volatile non-electrolyte in 15.60 grams of water was 100.32®C. Calculate the molecular weight of the solute. Ans. 102.8 6. A compound of unknown molecular weight was dissolved in water and the freezing point of the solution was determined. The following data were obtained. Weight of sample 1.308 grams Weight of water 14.65 grams Freezing point of the solution 1.44®C. Calculate the molality of the solution and the molecular weight of the solute. 7 . If 3.2912 grams of pure silver yield 5.7295 grams of AgBr, what is the atomic weight of bromine? Ans. 79.92 8. In converting 8.5875 grams of AgBr to Agl, there was found to be an increase of 2.1494 grams. Calculate the atomic weight of iodine, assuming the atomic weights of Ag and Br to be 107.880 and 79.916 respectively. 9 . A sample of a pure compound gave on analysis 42.90 per cent zinc, 20.35 per cent phosphorus, and 36.76 per cent oxygen. What is the empirical formula of the compound? Atw. Zn^PgO? 10 . If 0.5000 gram of a certain mineral contains 0.2392 gram of copper, 0.0974 gram of arsenic, and 0.1635 gram of sulfur, what is the formula of the mineral? 11 . A compound when analyzed was found to have the following composition: carbon, 10.06 per cent; hydrogen, 0.84 per cent; chlorine, 89.10 per cent. A Victor 240 SYSTEMATIC ANALYSIS Meyer molecular weight determination was run. The vaporization of 0.1638 gram of the liquid displaced 35.62 ml. of air, which was collected over water and measured at 23.n‘’C. and 732.6 millimeters pressure. The vapor pressure of water at 23.0*C. is 21.1 millimeters. What is the formula of the compound? Ans. CHCI3 12. A chemist conducting an organic combustion analysis with an organic com- pound finds that with a 1.0000-gram sample, 2.5200 grams of CO2 and 0.0885 gram of water are obtained. Calculate the percentages of C, H, and O in the sample, and from this the atomic ratio of these elements. Give the empirical formula of the com- pound. 13. Given the formula of a pure compotmd as Mg2P207, calculate the percentage composition. Ans. 21.84 per cent Mg; 27.86 per cent P; 50.30 per cent 0 14. An evacuated Dumas bulb, when filled with a gas at 32.0®C. and 735 milli- meters pressure, gained 0.483 grams in weight. If the volume of the bulb had been previously found to be 283.5 ml., what is the molecular weight of the gas? 16. In making an atomic weight determination of sodium, five 1.0000-gram samples of pure NaCl yielded as an average of five closely agreeing results, 2.4521 grams of AgCl. Taking Cl = 35.457 and Ag = 107.880, what is the atomic weight of sodium? Ans. 22.997 16. Richards and Hoover foimd that the ratio of the weights of equivalent amounts of Na2S04 to Na2C03 was 1.340155 to 1.00000. Previously Richards had determined, the atomic weights of sodium and carbon as 22.995 and 12.005 respectively. On the basis of his data, calculate the atomic weight of sulfur. 17. In 1941 the atomic weight of holmium, one of the rare earths, was changed by the International Committee. Honigschmid and Hirshbold-Wittner compared silver with holmium chloride and found that the ratio HoCls/SAg was equal to 0.83829. Assuming the atomic weight ol silver to be 107.880 and that of chlorine to be 36.457, what is the revised atomic weight of holmium? Ans. 164.94 18. The atomic weights of silver and bromine are 107.880 and 79.916 respectively. From the ratios FeBr2/2Ag = 0.999646 and FeBr2/2AgBr = 0.574244, calculate the atomic weight of iron. 19. One of the most important ratios used in calculating atomic weights is the AgCl/Ag ratio. The follo^^*ing are the data which Richards and Anderegg obtained in their experimental determinations of the ratio. Ag AgCl 6 . 222 1 1 grams 8 . 26738 grams 5.16568 6.86361 5.18476 6.88876 6.92279 7.86933 5.00706 6.65256 Calculate each AgCl/Ag ratio, and the average ratio; and, assuming the atomic weight of silver to be 107.880, calculate the atomic weight of chlorine. Am. 1.32871; 1.32869; 1.32866; 1.32864; 1.32864; 1.32867 avg.; 35.457 at, wt. CL 20. If the atomic weight of sodium is 22.997 and Na2S04 contains 44.055 per cen* oxygen^ what is the atomic weight of sulfur? APPENDIX THE LITERATURE OF ANALYTICAL CHEMISTRY Modern chemistry owes its rapid advance, in part, to the exact measurements developed by the analytical cliomist. The pioneers in the field such as Lavoisier, Berzelius, Stas, Rose and I'resenius were ft-llowed. in later years, by a host of otiiers who brought to the present high state the art and tcrliniqtie of aimlyticid chemistry. The record of their accomplishments may be found in the chernice.l literature and if the student of chemistry is to keep abrea.st of current d(n-elopmenl.s in this field he must be on the alert for the published information. EspfciaDy if lie wants to look up a method of analysis, a new piece of analytical apparatus oi- some theoretical topic, he must know bow to u.sc the analytical literature. When an investigator perfects a new method of analysis, designs a new apparatus or contributes in some other way to the advancement of analytical chemistry, he publishes liis fmdings, as an original paper, in one of the chemical journals, preferably one on analytical chemistry, such as Zcitschrift fur analylical Chanic (the oldest, first published in 1841 by Frescriius), The Analyst (a British journal begun in 1877), Aniiales de chimie anahjtique cl de chimie cppliqntc vt Rcime de chiinie analylique rentes, the leading French journal in analytical chemistrj’, or. if he is an American, the Analytical Edition of Industrial and Engineering Chemistry. Soon after publica- tion, the article is abstracted and a brief abstract Is publi.shcd in an abstracting journal, of which Chemical Abstracts is the most important. British Abstracts and Chemische Zenlralblatt arc the other two abstracting journals. Later on, the article may be reviewed in a review journal and uUirnakly, if the original idea has merit, it finds its way into reference books, textbooks, and manuals of analysis. When the chemist, therefore, is searching for analytical information, he searches both the journal literature and the reference books arid similar i)ublications. It is best to start with the current biweekly Issues of Chemical Abstracts (section 7 and specialized fields), then refer to the annual indexes, finally working back to the decennial indexes. The other abstracting journals should also ho checked. Such a search wnll reveal information which may not have been incoriwrated into books on analytical chemistry. The books on analytical chemistry are very numerous and, if assembled, would fill many shelves in a library. They cover many phases of chemical analysis, both theo- retical and applied, in many languages. Below is given a selected list of analytical books, most of which should be available in college libraries; they will sui)ply the answer of many a student’s question: “What is a good book on such and such a topic in quantitative analysis?" I. M. Kolthoff and E. B. Sandell, Textbook of Quaniiiatu'e Analysis, New York, The Macmillan Co., 193G. W. F. Hillebrand and G. E. F. Lundell, Applied Inorganic Analysis, New York, John Wiley & Sons, Inc., 1929. 241 242 APPENDIX F. P. 'teadwell and W. T. Hall, Ancdylicdl Chemistryt Vol. II, New York, John Wiley & Sons, Inc., 1942. W. C. Pierce and E. L. Haenisch, Quantitative AnaXysisy New York, John Wiley & Sons, Inc., 2nd ed., 1940. H. H. Willard and N. H. Furman, Elementary QuaniiUUive AnalysiSf New York, D. Van Nostrand Co., 1940. N. H. Furman, ScotVs Standard Methods of Chemical Analysis, New York, D. Van Nostrand Co., 1939. G. E. F. Lundell, J. I. Hoffman and H. A. Bright, Chemical Analysis of Iron and Steel, New York, John Wiley & Sons, Inc., 1931. G. Lunge and C. A. Keane, Technical Methods of Chemical Analysis, New York, D. Van Nostrand Co., 1924r-l931. R. C. Griffen, Technical Methods of Analysis, New York, McGraw-Hill Book Co., 1921. A. H. White, Technical Gas and Fuel Analysis, New York, McGraw-Hill Book Co., 1920. I. M. Kolthoff and N. H. Furman, Potentiometric Titrations, New York, John Wiley & Sons, Inc., 2nd ed., 1931. I. M. Kolthoff and H. A. Laitenen, pH and Electro Titrations, New York, John Wiley & Sons, Inc., 2nd ed., 1941. L. M. Dennis and M. L. Nichols, Gas Analysis, New York, The Macmillan Co., 1929. C. J. Engelder, Gas, Fuel and Oil Analysis, New York, John Wiley & Sons, Inc., 1931. G. F. Smith, Special and Instrumental Methods of Analysis, Ann Arbor, Mich., Edwards Brothers, 1937. F. D. Snell and C. T. Snell, Colorimetric Methods of Analysis, New York, D. Van Nostrand Co., 1936. G. V. Hevesy, Chemical Analysis by X-Rays and Its Applications, New York, McGraw-Hill Book Co., 1932. C. A. Browne and F. W. Zerban, Physical and Chemical Methods of Sugar Analysis, New York, John Wiley & Sons, Inc., 3rd ed., 1941. I. M. Kolthoff, Add-Base Indicators, trans. by C. Rosenblum, New York, The Macmillan Co., 4th ed., 1937. M. B. Jacobs, Chemical Analysis of Foods and Food Products, New York, D. Van Nostrand Co., 1938. American Society for Testing Materials, Methods for the Chemical Analysis of Metals, frequently revised, Philadelphia. C. M. Johnson, Rapid Methods for the Chemical Arudysis of Special Steels, New York, John Wiley & Sons, Inc., 4th ed., 1930. U. S. Steel Corporation, Carnegie Steel Co., Methods of the Chemists of (he V. S, Steel Corporation: Methods for Sampling and Analysis of Alloy Steds, Pittsburgh, Pa., 1921. XHE PLAN OF TPii£ COURSE — SUGGESTIONS TO THE INSTRUCTOR In a onc-semestor course in quantitative analysis where, quite possibly the time aUottod might be reduced to one lecture, one recitation and six to eight hours of laboratory practice a week, the problem of fmditig sufficient time to cover the desired assignments is a perplexing one. Furthermore, students whose curriculum includes only a brief course presumably will terminate their analytical training with the com- pletion of such a course. It is therefore important that a judicious selection of sub- ject matter be made. Tlie author ventures here to make a few suggestions which might prove helpful in planning the content and distribution of assignments to fit the needs of various classes. The Lectures. These obviously will deal with the theoretical aspects, elaborate on the technique, give further explanatory details concerning the procedures and possibly offer opportunity for lecture demonstrations. The Recitations. If but one recitation period a week is available, drill on prolv lems and calculations seems the best utilization of the time. To provide for tliLs the text supplies fifteen weekly problem sets, designed to keep pace witli the labora- tory work. The author’s practice has been to make individual assignments for each student at the start of the semester, covering the entire semester's work, assigning for home study five odd-numbered problems (with answers) and five even-numbered problems (without answers) out of each set of twenty for each weekly assignment. The work is submitted regularly each week and the clas.sroom hour is partly devoted to solving problems, and partly to a writton quiz, using the unanswered problems for distribution among the class. If a second recitation period is scheduled, the time might be used in quizzes on the laboratory work. The Laboratory Schedule. Here there is considerable flexibility in the procedures presented in this book and the instructor can make suitable selection to meet his limitations of time. Here is the sequence the author follows: 1. Preliminary work on the balance. 2. Acid-base titrations (soda ash and potassium acid phthalate samples). 3. Dicliromate determination of iron; followed by 4. Permanganate determination of iron with same sample of iron ore. 5. Volumetric determination of calcium (to be later checked by the gravimetric method). 6. lodimetric arsenic determination. 7. lodometric copper determination. 8. Volumetric cliloride determination; followed by 9. Gravimetric chloride determination on same sample. 10. Gravimetric iron determination. 11. Gravimetric sulfate determination; and when time permits 12. Gravimetric calcium determination. 13. Gravimetric magnesium or phosphate determination. 243 244 APPENDIX The use of the same sample of iron ore, calcium and chloride reduces the number of separate samples to a minimum, lessens the administrative and expense burden and arouses the students* interest in comparative methods of analysis. This suggested program is geared to the pace of the average student. Any over- time required by the stow worker or careless student will have to be provided for separately. REAGENTS AND SUPPLIES (a) LIQUID REAGENTS FOR GENERAL USE Acetic Acid, HC 2 H 3 O 2 . Alcohol, Ethyl, CaHsOn, 95 per cent. Animonium Hydroxide, NH^OH, Concentrated. The concentrated “arnmnnin ■' sp. gr. 0.90, containing 28.33 per cent NHj by weight. ‘ ’ Ammonium Molybdate, (NHi) 2 Mo 04 . Add 100 grams of molybdic oxide to 400 ml. of cold water and then treat, with constant stirring, with 80 ml of eoncen trated NHrOH. Allow to stand for several hours and then filter into a solution of 400 ml. of concentrated HNOs in 600 ml. of water. After several days filler If « residue is present. (For PO 4 determination.) Ammonium Nitrate, NH 4 NO 3 , 10 grams per liter of solution. Ammonium Oxalate, (NH,);C 204 . 40 grams of (NiUljCjO, ■ II.O per liter of so.u- tion. (For Ca determination.) Barium Chloride, BaC^. 20 grams of BaClz ^IIsO per liter of solution. (For SO 4 determination.) Bromine Water, Bra. Saturated solution. (For Fe determination.) Diammonium Hydrogen Phosphate, (NH 4 ) 2 HP 04 - 12 H 2 O. 20 grams per 100 nj of solution. (For Mg determination.) Hydrochloric Acid, HCI, Concentrated. C. P. reagent. Sp. gr. 1.19, conlaininc 37.23 per cent of pure HCI by weight. ® Magnesia Mixture. 50 grams of MgCl 2 - 6 H 20 and 100 grams of NI^Cl dissolved in 500 ml. of water, to which is added 100 ml. of concentrated NH 40 H and the volume made up to 1 liter with distilled water. Filter before asing. (For POi determination.) Mercuric Chloride, HgCb. 60 grams of HgClo per liter of solution. Nitric Acid, HNO 3 . Concentrated. C. P. reagent. Sp. gr. 1.42, containing 69.77 per cent of HNO 3 by weight. ° Silver Nitrate, AgNOs- Precipitating agent for chloride determination. 20 grams of AgNOs per liter of solution. (For Cl determination.) Silver Nitrate, AgNOa. Test reagent. 10 grams of AgNOa and 10 ml. of concen- trated HNO 3 per liter of solution. Stannous Chloride, SnCla. 115 grams of SnCls ^HoO in 170 ml. of concentrated HCI and diluted to 1 liter. Add a few pieces of metallic tin to prevent oxidation of the reagent. (For Fe determination.) Sulfuric Acid, H 2 SO 4 , Concentrated. C. P. reagent. Sp. gr. 1.84, containing 95.6 per cent of H 2 SO 4 by weight. Sulfuric Acid, Concentrated. Commercial, for making cleaning mixture. The better grade (C. P.) may be used. \ 245 246 APPENDIX (b) SOLID REAGENTS FOR GENERAL USE To be supplied to reagent shelves, distributed by the instructor or dispensed from the supply room. Ammonium Chloride, NH 4 CI. Ammonium Nitrate, NH 4 NO 8 . Asbestos, ground, for Gooch crucibles. Calcium Chloride, crude, for desiccators. Litmus Paper, in vials, neutral, or red and blue. Silver Nitrate, AgNOg. Sodium Bicarbonate, NaHCOg. Sodium Dichromate, Na^CraO?, commercial, for cleaning mixture. Stopcock grease. Vaseline for desiccators. Zinc, pure, 20-30 mesh, for Jones reductor. (c) REAGENTS FOR PREPARATION OF STANDARD SOLUTIONS Sodium Hydroxide and Hydrochloric Acid. For standard base and acid. Potassium Dichromate. Can be obtained pure and use directly as a primary standard. Ferrous Sulfate or Ferrous Amm onium Sulfate. Potassium Permanganate. Oxalic Acid. (Ceric Sulfate). Arsenious Oxide or Sodium Arsenite. Iodine. Sodium Thiosulfate. Silver Nitrate. Potassium Thiocyanate. Can be purified and used as a primary standard. (d) REAGENTS FOR STANDARDIZATION Sodium Carbonate. For HCl standardization. Potassium Acid Phthalate. For NaOH standardization if an independent standard- ization is assigned. Iron Wire. For K 2 Cr 207 if the latter is to be independently standardized. Sodium Oxalate. For KMn 04 standardization. Arsenious Oxide. For iodine standardization. Copper Foil. For sodium thiosulfate standardization, if desired. Sodium Chloride. For AgNOg standardization. Silver Nitrate. . For KCNS standardization, if necessary. (e) INDICATORS Methyl Orange, ordinary. Dissolve 1 gram of the dye in hot water and dilute to 1 liter. The modified indicator is to be used in acid-base titrations. Methyl Orange, modified. In separate beakers dissolve 1 gram of methyl orange in 600 ml. of hot water and 1.4 grams of xylene cyanole FF in 500 eqI. of water. Mix APPENDIX 247 measur^ volumes of each solution (1 ml of each) and test with approximately 0 1 AT acid and base tor color change. Vary the proportions of the two snt,,.; ^ }■, mixture is obtained which gives the sharpest poasihlc color chance When iT “ ratio is established, mix the two solutions in proper proport ion.s correct Methyl Red. Dissolve 2 grams of the dye in 500 n.l. of Oo'i^r cent alcohol an,1 dilute to a liter. Pbenoli)hthalcin. Dissolve 10 grams of the dye in 500 ml. of 95 per cent nh-nhnl and dilute to a liter* Diphcnylamine Sulfonate. For 100 ml. of the indicator solution, wcich out 0 ao gram of barium diphcnylamine sulfonate, add 1 gram of Na,SO, and 100 ml f water, shake, allow the BaSOr to settle and use the aui>crnatant liquid " Ortho-Phenanthroline-Fcrrous Complex. Emidoyed only if the ceric sulfate method is used. The indicator can be purcha.sed from the G. Frederick Smith Cl ical Company, Columb\is, Ohio. Starch Solution. Stir 5 grams of soluble starch into 20 to 25 ml. of water and thor oughly mix until a uniform paste forms. Pour this into 500 ml. of boiling wat^/ When cool add 10 grams of KI. Dichlorofluorescein. Dissolve 0.1 gram of the reagent in 100 ml. of 70 per cent alcohol. Ferric Alum. To be prepared by student as needed for the Volhard method for silver. APPENDIX DENSITY OF STRONG ACIDS AT X5°C. IN VACUO (Accsording to Lunge, Isler, Naef, and Marchlewaky) Specific Gravity Per Cent by Weight 0.16 I. 15 2.14 3.12 4.13 5.15 6.15 7.15 8.16 9.16 10.17 II. 18 12.10 13.19 14.17 15.16 16.15 17.13 18.11 19.00 20.01 20.97 21.92 22,86 23.82 24.78 25.75 26.70 27.66 28.61 29.57 30.55 31.52 32.49 33.46 34.42 35.39 36.31 37.23 38.16 30.11 DNOs 0.10 1.00 I. 90 2.80 3.70 4.00 5.50 6. 38 7.26 8.13 8.99 9.84 10.67 II. 50 ■ 12.32 13.14 13.94 14.73 15.52 16.31 17.10 17.88 18.66 19.44 20.22 20.99 21.76 22.53 23.30 24.07 24.83 25.59 26.35 27.11 27.87 28.62 29.37 30.12 30.87 31.60 32.34 33.07 33.80 34.53 35.26 36.01 36.76 H2S04 0.09 0.95 I. 57 2.30 3.03 3.76 4.49 5.23 5.96 6.67 7.37 8.07 8.77 9.47 10.19 10.90 II. 60 12.30 12.99 13.67 14.35 15.03 15.71 16.36 17.01 17.66 18.31 1 18.96 19.61 20.26 20.91 21.55 22.19 22.83 23.47 24.12 24.76 25.40 26.04 26.68 27.32 27,95 28.68 29.21 29.84 30.48 31.11 Specific Gravity at -p- (Vacuo) Per Cent by Wei^t HNOi 37.51 38.27 39.03 39.80 40.56 41.32 42.08 42.85 43.62 44.39 45.16 45.93 46.70 47.47 48.24 49.05 49.88 50.69 51.51 52.34 53.17 54.04 64.90 55.76 56.63 1 67.54 58.45 59.36 60.27 61.24 62.21 63.20 64.22 65.27 66.37 67,47 68.60 69.77 70.95 72.14 73.35 74.64 75.94 U2SO4 1.7 2.2 2.8 33.43 34.00 7 4 1 19 38.61 9 7 15 13 6 0 4 44.28 44.82 45.35 45.83 46.41 46.94 47.47 48.00 48.53 49.06 49.59 50.11 50.63 5 61.66 52.15 52.63 53. 53.59 54.07 54.55 55 55 55 APPENDIX 249 Specific Gravity . 15® at-p (Vacuo) 1.170 1.475 1.4S0 1.4S5 1 . 490 1.195 1.500 1.505 1.510 1.515 1.520 1.525 1.530 1.535 1.540 1.545 1.550 1.555 1.5G0 1.5G5 1.570 1.575 1.550 1.555 1.590 1.595 1.600 1.605 Per Cent Weight ‘’y H.NO, 82 SI SG S7 89 91 94 96 95 99 99 8G 41 01 6G 80 56 04 34 05 02 62 56.90 57.37 67.83 68.28 58,74 59.22 59.70 60.18 60.65 61.12 61.59 62.06 62 . 53 63.00 63 43 03 85 64 . 20 64.67 65.20 65.05 06.00 6G 53 6G.95 07.40 67.83 63.26 68.70 GO. 13 Specific Gravity of 15" at- C^^ucuo) 1.610 1.GI5 1 020 1.625 1 630 1.635 1.040 1.645 1.650 1.655 1.060 1.G65 1.670 1.075 1.680 1.685 1.600 1.695 1.700 1.705 1.710 1.715 1.720 1.725 1.730 1.735 1.740 1.745 1 Per Cent by Weiglit HaSO, Specific Gravity . 15° (Vacuo) 69.56 1 . 7."^ 70.00 1.755 70.42 1.760 70.85 1.765 71 27 1.770 71.70 1.775 72.12 1.780 72.55 1.785 72.% 1.790 73 40 1.705 73.81 1.800 74 . 24 1.805 74 . CO 1.810 75.08 1.815 75.50 1.820 75 . 94 1.825 76.33 1. 830 76. 76 1 . 835 77.17 1.840 77.00 1 . 8 405 78.0-4 1.8410 78.48 1.8415 78.02 1.8410 70.36 1.8105 79.80 1.8400 80.24 1.8395 80.68 1.8390 81.12 1.8385 Per Cent l)y W’cight II 2 SO 4 81.56 82 00 82.44 83 01 83 51 84 . 02 84 50 85.10 85.70 86 30 86.92 87.60 88.30 89.16 90.05 91.00 92.10 93.56 95.60 95.95 96.33 97.35 98.20 98 52 98.72 98.77 99.12 99.31 Taken from TreadwcU-Hall, Analytical Chemittrp, Vol. 11, 0th od., published by John Wiley & Sons. Reprintod with pcrtnistiiou. APPENDIX DENSITY OF AMMONIA SOLUTIONS AT 16*0. (According to Lunge and Wiemik) SpeciEc Gravity [Per Cent NHs li Specific Gravity | Per Cent NHi 0.00 0.45 0.91 1.37 1.84 2.31 2.80 3.30 3.80 4.30 4.80 6.30 5.80 6.30 6.80 7.31 7.82 8.33 8.84 9,35 9.91 0.940 0.938 0.936 0.934 0.932 0.930 0.928 0.926 0.924 0.922 0.920 0.918 0.916 0.914 0.912 0.910 0.908 0.906 0.904 0.902 0.900 0.898 0.896 0.894 0.892 0.890 0.888 0.886 0.884 0.882 16.63 16.22 16.82 17.42 18.03 18.64 19.25 19.87 20.49 21.12 21.76 22.39 23.03 23.68 24.33 24.99 25.65 26.31 26.98 27.65 28.33 29.01 29.69 30.37 31.05 31.76 32.60 33.25 34.10 34.95 Taken from TreadwoU-HaU. Analytical CAamirfry. VoL II. 9thed..publiahedby JohnTOey A Sons. Reprinted with permission. APPENDIX 251 DENSITY OF WATER Temper- ature, Deg. C. Density Temper- ature, Deg. C. Density Temper- ature, Deg. C. Density 0 0.999S7 22 780 43 107 1 993 23 750 44 000 2 997 24 732 45 0.99025 3 999 25 0.99707 40 0.93982 4 1.00000 2G 681 47 940 6 0.99999 27 654 48 890 6 997 ’ 28 620 49 852 7 993 29 597 50 0.98807 8 988 30 0.995G7 61 762 9 981 81 537 52 715 10 0.99973 32 505 53 669 11 963 33 473 64 021 12 952 34 440 55 0 . 98573 13 940 35 0.99400 60 324 14 927 36 371 65 059 15 0.99913 37 330 70 0.97781 16 897 38 299 75 489 17 880 39 262 80 0.97183 18 862 40 0.99224 85 0 . 96865 19 843 41 186 90 534 20 21 0.99823 802 42 147 95 192 252 APPENDIX DATA AND REPORT FORMS’*^ VOLUMETRIC DETERMINATION OF Sample No Date DATA COMPARISON OF SOLUTIONS 12 3 4 burette reading at start burette reading at finish — ml. of used burette reading at start •- burette reading at finish ml. of used STANDARDIZATION 1 2 3 Initial weight of tube and standard or Weight of tare with standard Final weight of tube and standard or Weight of tare.., ““““ Weight of primary standard burette reading at start burette reading at finish — ml. of used burette reading at start burette reading at finish ; ml. of used ANALYSIS 1 2 3 Initial weight of tube and sample or Weight of tare with sample Final weight of tube and sample or Weight of tare — Weight of sample ^ burette reading at start burette reading at finish ‘ ml. of used burette reading at start burette reading at finish ml. of used • Sample pages from the author's Laboratory Record Book of Quonliiatwe Analysis. APPENDIX 253 report of volumetric analysis Determination of Student’s Sample Number Instructor’s Sample Number Results of Standardization titre of solution = Normality of solution = ^titre of solution = Normality of. solution = 8 high low Date Instructor’s Name and Lab. Section Grade • « • .Student’s Signature To be retained by li%structor> Run No. Per Cent Found Correct Per Cent DifTcrence Precision in parts per looo Results of Analysis I Remarks by Student high low high low To be returned to Student Determination of Run No. 1 a 3 Precision in parts per looo high high ^^^^^^high low low low Remarks by Instructor Date Instructor’s Signature Grade Student’s Signature. • . 254 APPENDIX GRAVIMETRIC DETERMINATION OF Sample No Date. DATA Initial weight of tube and sample or Weight of tare with sample Final weight of tube and sample or Weight of tare Weight of sample Weight of crucible Weight of crucible Weight of crucible. Weight of crucible with precipitate. Weight of crucible with precipitate. Weight of crucible with precipitate Weight of crucible with precipitate Weight of Precipitate RESULTS Per cent of — % %•••• % Average #•••% Remarks Approved by APPENDIX 255 REPORT OF GRAVIMETRIC ANALYSIS Detertaination of Student’s Sample Number Instructor’s Sample Number Run No. Per Cent Found. • , Correct Per Cent ■ — ■ ' ' Difference Precision in parts per looo I'isb High high low low low Remarks by Student Date Instructor’s Name and Lab. Section Grade Student’s Signature To be retained by Instructor To be returned to Student Determination of Run No. Precision in parts per xooo. . I high low high low high low Remarks by Instructor PERIODIC TABLE OF THE ELEMENTS 256 APPENDIX 2o7 CHEMICAL 1-ACTOllS* Sought Found Factor Log Fouglit Found Farlor Log Ag AgBr AgCl AgCN Agl AgNOj A UfO • 0.5711 0.7526 0 . 8055 0.4505 0.0350 0.8700 0.7731 0.7125 9.7593 9 . S7G6 9. 9000 9 . 0622 9 . 8028 9 . 9398 9 . 8882 9.8528 Ba 1 BaClj -211^0. . . li:4C03 BaCrO* BaiiO* Ihu^iFa 0 5021 (f.OOGl 0.5122 0 . 588.5 0. 11112 9.7493 9.8120 9.7341 9 7097 9.C913 Ag3r04 Ag4P:07 BiiO BisCOa BaCrO^ 0.7771 0 . 00.53 0,0570 9.8905 9.7820 9.8176 A1 \UOi 0.5294 0.2212 0.723S 9.3449 vUro4 Ca CaClj 1 0.3012 , 0.4005 0.5133 0.7146 0.2941 0 . 9003 9.5577 9 . 0025 9.7103 9.85-11 9.4089 1 9.9587 AljOi AIPO 4 0.4179 9 6211 CaF^ Pfi A Ab /\b203 . . • AS 2 O 3 A 0.7575 0.0521 0.0091 0.4832 0.4827 0.3939 9.8794 9 8143 9.7847 9 0811 9.6837 9.5953 V. CaSO* CO 2 As^Si MgjABjOj (NIl4MgAfl04)2. IIjO CaO CaCOj CaFj Ca(MC 03 ) 2 . . . . Ca(ILP04)2... . CallsI^O? Ca(ll.S 03)2 Cfxz{VOi )2 CaSOs CnS04 CaS04 - 2 ni 0 .. . CO} 0 . 5603 0.7182 0 3159 0 . 2391 0 . 2591 0.2773 0.5421 0.4007 0.4120 0.3257 1.274 9.748-1 9 . 8503 9.5359 9.3791 0.41-10 9.4430 9.7341 9.0091 9.6149 9.5128 9.1053 A 52 O 1 A520fi AsjSj AsjSft MgiAsjOT (NH4MgA604)3. Q 30 0.8609 0.8041 0.6378 0.0372 0.5109 9.9319 9.9053 9 . 8047 9.8013 9.7160 BqCOi BaCl 2 - 2 H 20 BaCr04 BaF} Ba(N08)a BaslPOOa | BaS BaSOi BoSOi BaS 04 0.ai58 1.017 1.085 0.0271 1.120 0 . 8599 0.7258 9.9272 0 0198 0 0356 9.7973 0 0491 9 9345 9.8608 Cd CdO CdS CdSOi 0 . 87.''/4 0 7780 0.6159 9,9122 9.8910 9.7895 oftoir 9 BRSO 4 BaS04 ; BaS04 Bi.O, 1 1 Bi BiOCl BiPO* 1 115 0,8946 0 7063 0.9003 0.6006 0.8118 : 0.4803 0.0472 9.9516 0 . 8844 0 . 9573 0 . 82.58 9.9095 9.6815 Bi Bi,0, BiOCl BiP04 BUSa BiAa04 0 . 8970 0.8024 0 687 1 0.8129 0.0000 9 . 9528 9.9014 9.8372 9.9101 9.7786 BiA-sOi BiONOs Di(N03)3-5Il20 Dr Ag ArHr 0.7108 0.4256 0.5576 0.9S74 1 9.8697 9 , 0290 9.7463 9.9945 AdsSj AS 2 S%, MgjAfljOT (NH 4MgAB04) H 20 0.9341 0.7410 0.7403 0.6040 9.9704 1 . 8698 1.8694 9.7810 • •K*'**'* •• •• •• • AgCl UBr C CO} BaCOs 0.2727 0.0608 9.4357 8.7839 B BsO, 1 KBF 4 HjBOa NaiB4O7l0H2O / 0.3123 0.CS651 0.1761 0.1142 9.4916 8.9371 9.2456 9.0577 Cl Ag AgCl AgNOa HCl 0.32S7 0.2474 0 2088 0 . 972-1 1 9.5168 9.3934 9.3198 9.9879 •From Trcadwell-HaH, Analytical Chetnisln/, Vo!. II, published by John Wiley & Sons. Beprinted with permisdoo. 258 APPENDIX CHEMICAL FACTORS— Conitniiei Sought Found Factor Log Cl I 0.2794 9.4462 KCI 0.4756 9.6772 MnOa 0.8158 9.9116 NaCl 0.6066 9.7829 NILCl 0.6628 9.8214 RbCl 0 . 2933 9 . 4673 Co Co(N 03 ) 2 - 6 H 20 0 . 2026 iin KaCoCNOa)*. . . 0.1304 9.1152 CoO 0.7866 9 . 8957 1 CO 0 O 4 0.7344 9 . 8659 C 0 SO 4 . • • 0.3804 9 . 5802 CoS04-7H20.. . 0.2097 9.3216 CoO Co 1.2713 9.1043 • Co(N 08 )a 6 H 20 . 0.2575 9.4109 K3Co(N02)6. . . 0 . 1657 9.2194 C 03 O 4 0 . 9336 9.9702 : C 0 SO 4 0.4S36 9.6844 CoS04-7H20.. . 0 . 2067 9.4259 Cr BaCr04 0 . 2052 9.3125 CraOs 0.GS42 9.8352 CdO CdS 0.8SS8 9.9488 Cd 1.142 0.0578 CdS04 0.6159 9.7895 Cr PbCrOi 0.1609 9.2067 K2Cr207 0.3535 9.5484 K2Cr04 0.1768 9.2474 CrjO^ BaCrOi 0.3000 9.4771 PbO04 0.2352 9.3715 CrOa 0 . 7602 9. 8809 CrO* BaCr 04 0.4578 9.6607 CraOa 1.5263 9.1837 PbCr 04 0.3589 0.6550 Cu Cu$0 0.8882 9 . 9485 CuO 0.7989 9.9025 CU 2 S 0.7986 9.9023 Cu3(CNS)a 0 . 5226 9.7181 CuS 04 0.3982 9.6001 CuSO^ 'SHiO. . . 0.2546 9 , 4058 CuO Cm 1.252 0.0975 CuaO 1.113 ' 0.0463 CuiS 0.9996 9 . 9998 Cu3(CNS)a 0.6541 9.8156 CUSO 4 0.4986 9.0976 CuS 04 * 6 H 20 * « • 0.3186 9.5032 Sought Fotmd Fe HF Fe FeO FetOi BaSiF( CaFj. . CaS 04 . HF... H2SiFe KiSiFu NaF. . SiF4... Factor Fe20a FeCl 2 . BaSiFe CaFa. . CaS 04 . F KaSiFe FeClsSHaO. . . Fe(HC03)2... FeO FgsOi FeP04 FeS FeS04 FeS04-7H20. FeS 04 (NH 4 )j SO4 'CHaO. , KaCraOr (titration) . , Fe FcCOa Fe(HCOj)2.... FoaOs FeP04 FeS FeS04 FeS04’7Ht0. . FeS04(NH4)2 S 04 * 6 Ha 0 .. . Fe FeClj FeCOa Fei04 FePOi FeS FeS* Fe3(S04)s (NH4)2804*Fet (S04}r24Hi0. 0.4076 0.4866 0.2791 0.6496 0.79X0 0.6170 0.4524 0.7302 Log 9.6102 9.6872 9.4467 9.9776 9.8982 9.7135 9.6655 9.8634 0.6994 9.8447 0.4405 9.6440 0.4292 0 . 6126 0.2939 1.053 0.5449 0.2066 0.3140 0.7773 0.7236 0.3701 0.6362 0.3676 0.2008 9.6327 9.7098 9.4682 0.0224 9.7363 9.3152 9.4969 9.8906 9.8595 9.5683 9.8029 0.5654 9.3027 0.1424 9.1536 1.139 0.0564 1.287 0.6202 0.4039 0.8998 0.4761 0.8172 0.4729 0.2584 0.1094 9.7925 9.6063 9.9542 9.6777 9.9123 9.6747 9.4123 0.1832 9.2630 1.430 0.4922 0.6892 1.035 0.5292 0.9082 0.6655 0.3993 0.1553 9.6922 9.8383 0.0149 9.7237 9.9582 9.8231 9.6013 0.1656 0.2190 APPENDIX 259 CHEMICAL FACTORS — Continued Sought Found Factor I.og HDr Ag 0 . 7505 9.8751 AfiBr 0.4311 9.6316 HCl Ag 0.3.3S0 9.5200 AgCl 0.2544 9. 10.56 CaCOa 0.72S8 9 . 8020 KCl 0.4801 9 . O^^l) 1 K ,0 0.7743 9.8889 NaCl 0 . 6239 9.7051 NajO 1.176 0 . 070.5 NllaCl 0.6817 9 . 8335 SnCli 0.5000 9.7480 NB2CO3 Methyl orange NajCjOa 0.6S81 9. 8377 standardization 0.5142 9.7358 HI Ae 1 1.186 0.0742 Agl 0.5448 9.7362 Pd 2.398 0.3709 Pdlj 0.7097 0.8510 HNOj KNOs 0 . 6233 0,7047 N 4.408 0.6.530 NaNOa 0.7113 9.8700 NH3 3.701 0 . 5683 NH4CI 1.178 0.0711 (NH 4 )iPtCle... 0.2S3S 9.4530 NO 2.100 0 3222 0.658 0.2196 1.370 0.1367 NaOa 1.107 0.0671 C20H16K4IINO5 0.1680 0 2252 Pt 0.6457 9.8100 1 H,P04 HPO3 1.225 0.0881 H«P :07 1.101 0.0457 MgjPsOr 0.8806 9.9148 P 3.159 0 . 4995 P2O6 1.380 0.1399 EsS BaSOi, 0.1460 9 1644 CdS 0.2360 0.3729 FeS 0 . 3877 0.5S85 S 1.063 0.0265 H 2 S 04 Al 2 (S 04)3 0.8594 9.9342 Ba(OH)a 0.5723 9.7576 BaSOi. • K 2 Al 2 (SO 04 * 0.4133 9.6163 24H2O 0.3101 9.4915 K2O 1.041 0.0174 Sought Found Factor KsO K2R04 0.5405 0.7328 KCIO 4 0 3399 9 6314 K 2 PtClc 0.1938 9 . 2873 Pt 0.4820 9.6836 KClOi AgCl 0 . 9007 9,9853 KCl 1.858 0.2691 ll:r04 KOII 0.8740 9.9415 KjSO* 0.5028 9 . 7503 NasCOj 0.9252 9 . 9662 ^04 0.7320 9.8045 (NI!4)-.iS04 0.7422 0,8705 lie HgiCli 0,8495 9.0292 iigcu 9.7388 9.8688 IIkO 0 9261 9.9606 Hg.S 0.8620 9.0355 I 1.177 0.0706 A«I 0 5406 9.7328 KI 0.7645 9 8834 Pci 2.379 0 3764 Pdl2 0 7041 0 8476 Na^S^Oj *01120 . 0 5113 9 . 7087 AgC! 0.8855 0.9472 K KCl 0.5244 0,7197 K 2 O 0.8300 9.9101 K 2 SO 4 0.4487 9 . 0520 KCIO 4 0 . 2822 0 4506 KjPtCU 0.1009 9.2064 Pt 0.4006 9.6027 KCl Ag 0,6911 9.8396 AgCl 0.5202 9.7162 K;S04 0 . 8557 0.9323 KC 104 0.6381 0.7309 KjPtae 0 . 3067 0.4868 Pt 0.7640 9.8831 K 2 O KCl 0.6317 9.8005 Mo M 0 O 3 0.6667 9 . 8230 MoS, 0.4995 9.6985 M 0 S 2 0.5996 9.7779 PbMo 04 0.2615 9.4174 N IINOa 0.2223 9.3470 NO 2 0.3045 9.4835 N 2 O 3 0,3686 9.5665 N:04 0.3045 9.4835 260 APPENDIX I CHEMICAL FACTORS— CorUinued Sought Found Factor Log Sought Found 1 Factor Log N N 2 O 6 NaNOa NHa Ft NH 4 CI 0 . 2594 0.164S 0.8225 0.1435 0.2619 9.4140 9.2169 9.9151 9.1570 0.4181 NHa (NH4)2HP04. , . (NH4)H2P04. . . (NH 4 ) 0 H (NH4)2Ptae... (NH4)aS04 0.2678 0.1480 0.4860 0.07670 0.2678 0.3153 0.1745 9.4113 9.1703 9.6866 8.8848 9.4113 9.4987 9.2418 K 2 SO 4 K KCA 2.229 1.1G9 1.850 0.3585 2.177 0.7465 0.3481 0.0679 0 . 2072 9.5545 0.3379 9.8731 • pt K 20 KaPtCJe SO 3 BaSOi Na 0.3034 0.3238 0.2235 0.4340 0.5476 0.2740 0.1534 0.7419 0.6760 0.3238 9 . 6940 9.5103 9.3493 9.6375 9.7386 9.4375 9 . 1858 9.8704 9.7697 9.5103 Mg MgO MgS 04 MgjPaO;. . . .fc. MgS04 0.6032 0 . 2020 0.21S4 0.2020 9 . 7804 9.3054 9.3392 9.3054 MgO \irrRni 0.3349 0.3021 0.4782 9 . 5249 9.5588 9.6796 # MgaPaOy MgCOa NaBr Ag AgBr EH 9.9806 9.7388 Mn MnCOa MnO MnaOj Mn304 MnS 04 Mn 2 P 207 MnS 0.4779 0.7744 0 . 6959 0.7203 0.3638 0.3869 0.6314 9 . 6793 9.8889 9.8426 9 . 8575 9 . 5608 9 . 5876 9.8003 NaCl Ag AgOl AgNOa NaaO 0.5418 0.4078 0.3441 1.886 9.7338 9.6105 9.5367 0.2756 Na2HAdO) MgaAsaOr I 1.095 ' 0.6697 MuO 0.6172 0.9301 0.4697 0.8153 0.4998 9.7904 9.9685 9.6719 9.9113 9.6988 NaaHAsOa MgaAsaO? I 1.198 0.7328 0.0786 9.8650 MnS NallCOa CO 2 B 0.2808 i>m2^ 2^7 NOj MT=r,n 1.159 0.2793 0.6354 0.1653 0.0641 9.4461 9.8030 9.2182 NaaO NaCl Na2S04 Na2COj 0.5303 0.4364 0.5S4D 0.3690 0.4365 0.2582 0.2792 0.2979 0.3047 0.7748 0.4364 9 . 7245 9.6399 9.7671 9 . 5670 9.6400 9.4119 9.4459 9.4741 9.5619 9.8891 9.0399 (NH4)2PtCl6, . . pt CaoHiaNjOa.. . . iNuQVV/2 Na2HP04 NaH2P04 NaaHaPjOT. . . . NaHSOa NaNOa NaOH N:Oa NO NHa NH 4 CI (NHOaPtCle. . . Pt CaoHiaNsOa — 1.800 3.171 1.010 0.2433 0.5534 0.1440 0.2552 0.5012 0.0041 9.3861 9.7430 9.1582 iia2^v4« • • « • • • • NO a Mn 1.533 0.1856 NH, 0.2704 0.3473 1.216 0.2408 0.2005 0.3184 9.4320 9.5407 0.0849 9.3923 9.3021 9.5023 •••••«*• H 2 SO 4 N NaNOa NaNOa NH 4 CI P MgaPaOj PaO824Mo0|. . . (NHOaPOi- I 2 M 0 O 1 PiOa 0.2787 0.0173 0.0165 0.4369 9.4453 8.2368 8.2184 0.6404 APPENDIX 261 CHEMICAL FACronS—ConUnued Sought r04 S04 Found I’aOi-21Mo03 (Nn4)jro4- I2MoOj Factor Log 0.S.535 9 9312 0.05283 8.7229 0 . 05063 8.7044 1 . 059 0 . 0250 0.3372 9 . 5279 0.o5125 8.9099 0.1848 9 . 2608 1 . i 0.7858 0 , 8953 0.2032 , 9.3079 : 0.3/92 0.2013 9 . •/7S8 0.3019 0 . 20S9 9.3200 ! 0.1371 9.1379 0.9408 9.9735 0,3270 9 5145 , 0.3048 9.5621 0.5346 9.7280 0.4108 0.6136 0.2745 9.4385 1 . 998 0.3007 0.3430 9 53.53 0.5881 9 . 7695 0.8163 9,0119 0.4595 9 . 6623 0 . 5636 9.7010 C.4U5 9.0144 0.3648 9 , 5620 0.5339 9.7275 0 8356 9.9220 0.752S 9 . 8767 0.7921 9.8988 0.7169 9.8555 O.C030 9 . 7803 0.4G76 9 . 6698 0.3093 9 . 5673 0.3051 9.4844 0.8482 9.9285 ; 0.S817 9.9453 0.6670 1 9.8242 Sought r:06 ZnO Found MriPjOj... PjOj-2-iMoOs (NU«)jrO«- 12MoO, . . . Factor Lor O.f.379 9.8048 0.03947 8.69R3 0.0378r) 8.5780 PbOj PbS in>S04 PbCrO* PbCI. PbCOa Pb(N'Os) 2 ... S S SnClj SnCl2-2rijO, K Sr(NO|)j SrCOj. . . SrS()4 . . . Sr(N03)j Cl 2 CO 3 . . I 0 92S3 0 . 8002 0 8059 0.6K31 0 0410 0 7149 0 :,C42 0.9331 0.932S 0.7359 0.6905 0 8025 0 83.53 0 6738 0 7881 0 8812 0.62C1 0 52C0 0.845G 0 5036 0.4770 0 4140 0.7020 0.5011 0.1896 WOj PbW04... ZnO ZnNH4p04 ZnaPjOr • • • ZnS ZnS ZnNH4p04 ZmPaO?. . . 0.8034 0 . 3663 0.4290 0.6709 0.0677 9.9376 9.9375 9.8315 9 . 806S 9.8721 9.7515 9 . 9699 9 . 9698 0 HOr.8 9.8302 9 . 901 1 9.9219 9.8285 9. 8006 9.9451 9.7907 9.7210 9.9272 9.7735 9.6785 9.6170 9.8463 9.7511 9 . 6S99 0.8519 9.9304 0.8718 9.9404 0.6165 9.7900 0.5605 9.7485 0.G144 9.7885 0.7931 9.8993 0.4435 9.6469 9.9049 9 . 5639 9.6324 9 . 8266 0.835X 9.9217 0.4561 9.0590 05340 9.7275 262 Five-Place Logarithms : 100 — 160 ^ ^ ^ Q 7 8 9 I Prop. Ports 100 00 000 043 087 130 173 217 260 303 346 389 432 00 860 01 284 01 703 02 119 531 02 938 03 342 03 743 475 618 561 903 945 988 326 368 410 745 787 ^8 160 202 243 572 612 653 870 284 694 689 732 775 817 *115 *157 *199 *242 536 578 620 662 912 953 995 *036 *078 325 366 407 449 490 735 776 816 857 898 979 *019 *060 *100 *141 *181 *222 *262 *302 383 423 463 503 543 583 623 663 703 782 822 862 902 941 981 *021 *060 *100 no I 04 139 1 79 218 ^58 297 ^36 T 76 415 '454 44 4.4 8.8 13.2 17.6 22.0 26.4 30.8 36.2 39.6 43 4.3 8.6 1 Z 9 17.2 21.6 26.8 30.1 34.4 38.7 42 4.2 8.4 12.6 16.8 21.0 25.2 29.4 33.6 37.8 532 04 922 05 308 05 690 06 070 446 06 819 07 188 555 571 961 346 729 108 483 856 225 591 610 650 689 727 766 805 844 883 999 *038 *077 *115 *154 *192 *231 *269 385 423 461 500 538 576 614 652 767 805 843 881 918 956 994 *032 145 183 221 258 296 333 371 408 521 558 595 633 670 707 744 781 893 262 628 930 298 664 967 *004 *041 *078 *115 *151 I 335 372 408 445 482 518 S 700 737 773 809 846 882 ® 120 I 07 918 954 990 *027 *063 *099 *135 *171 *207 *243 08 279 314 350 386 422 ~ 4 S 8 ^3 529 ’ sii ’ 636 672 707 743 778 814 849 884 920 08 991 *026 *061 *096 *132 *167 *202 *237 *272 09 342 09 691 10 037 380 10 721 11059 377 412 447 482 517 552 587 621 656 726 760 795 830 864 899 934 968 *003 072 106 140 175 209 243 278 312 346 415 449 483 755 789 823 093 126 160 517 551 585 619 653 687 857 890 924 958 992 *025 193 227 261 294 327 361 130 594 428 461 494 528 561 594 628 661 1 694 11 727 12 057 585 12 710 13 033 354 760 793 090 123 418 450 826 860 893 156 189 222 483 516 548 926 959 254 287 581 613 992 *024 320 352 646 678 743 775 808 840 066 098 130 162 586 418 450 481 872 905 937 969 *001 194 226 258 290 322 513 545 577 609 640 672 704 735 767 799 830 862 893 925 956 13 988 *019 *051 *082 *114 *145 *176 *208 *259 *270 14 301 333 364 395 426 457 489 520 551 582 140 1 613 644 675 706 737 768 799 829 860 "iiT 41 40 89 4.1 4 3.9 8.2 8 7.8 12.3 12 11.7 16.4 16 16.6 20.5 20 19.5 24.6 24 23.4 28.7 28 27.3 32.8 32 31.2 36.9 36 36.1 38 37 36 3.8 3.7 3.6 7.6 7.4 7.2 11.4 11.1 10.8 15.2 14.8 14.4 19.0 18.6 18.0 22.8 22.2 21.6 26.6 25.9 25.2 30.4 29.6 28.8 34.2 33.3 32.4 35 34 33 3.5 3.4 3.3 7.0 6.8 6.6 10.6 10.2 9.9 14.0 13.6 13.2 17.5 17.0 16.5 21.0 20.4 19.8 24.5 23.8 23.1 28.0 27.2 26.4 31.5 30.6 29.7 14 922 15 229 534 15 836 16 137 435 16 732 17 026 319 953 983 *014 *045 *076 *106 *137 *168 *198 259 290 320 351 381 412 442 473 503 564 594 625 655 685 715 746 776 806 866 897 927 957 167 197 227 256 465 495 524 654 987 *017 *047 *077 *107 286 316 346 376 406 684 613 643 673 702 761 791 820 850 879 056 085 114 143 173 348 377 406 435 464 909 938 967 997 202 231 260 289 493 522 551 638 667 696 725 754 782 811 84 33 3.2 6.4 9.6 12.8 16.0 19.2 , 22.4 8 125.6 28.8 31 80 3.1 3 6.2 6 9.3 9 12.4 12 16.6 16 1&6 18 21.7 21 24.8 24 27 ^ 27 Prop. Parts 203 Five-Place Logarithms : 150 — 200 li Prop . Parts W 0 1 2 1 3 ' mm ■a 8 150 17 609 667 696 725 754 782 811 840 j 51 17 898 926 955 984 *013 •041 *070 *009 BBSS BS 62 18 184 213 241 270 298 327 355 384 ■pB|9 1 2 2.9 2.8 6.8 6.6 53 469 498 526 554 683 611 639 667 T 1 * 724 3 A 8.7 8.4 54 18 752 780 808 837 865 893 921 949 977 4 tL 11.6 11.2 14 A 14 0 65 19 033 061 089 117 145 173 201 229 257 9 6 A*9.(J A*V>V 17.4 16.8 56 312 340 368 396 424 451 479 507 635 562 7 8 20.3 19.6 23.2 22.4 57 590 618 645 673 700 728 756 783 811 0 26.1 26.2 58 19 866 893 921 948 976 *003 •O.'O *058 *085 Vy xJ *1 12 59 20 140 167 194 222 249 276 303 330 358 385 ICO 412 439 : 4G6 1 493 548 575 602 629 656 A4 61 683 710 ' 737 763 817 844 871 898 <)'>r 4i 40 62 20 952 978 *005 *032 *085 *112 *1.39 *165 *192 1 2 2.7 2.6 6.4 6.2 63 21 219 245 272 299 325 352 378 405 431 A J 458 1 3 8.1 7.8 64 484 511 557 564 590 617 643 669 696 1 722 4 ! B 10.8 10.4 1 ‘I c 1 n 65 21 748 775 801 827 854 880 9U6 952 958 9R5 o 0 1^.0 iO.U 16.2 16.6 66 22 011 037 063 089 115 141 167 194 220 216 7 8 ! 18.9 18.2 21 6 20.8 67 272 298 324 350 376 401 427 453 479 50.5 V 1 9 1 24.3 23.4 68 531 557 583 608 634 660 686 712 737 763 69 22 789 814 840 866 891 917 ; 943 968 994 *019 170 23 015 070 096 121 147 172 ' 198 223 249 m 71 300 325 350 376 401 426 tm 477 602 25 72 553 578 603 629 664 679 704 729 754 77C) 1 2 2.6 6.0 73 23 805 830 855 880 905 930 955 1 930 *005 m 3 7.5 1 A /a 74 24 055 080 105 130 155 180 204 229 254 279 4 10.0 IOC 75 304 329 353 378 405 428 mm 477 502 527 o 6 1 4.0 16.0 76 551 576 601 625 650 674 699 724 718 773 7 8 17.5 20 0 77 24 797 822 846 871 895 920 944 969 993 *018 9 22.6 78 25 042 066 091 115 139 164 188 212 237 261 79 285 310 534 358 582 406 431 455 479 503 180 527 551 575 600 624 648 672 ' 696 720 744 81 25 768 792 816 840 864 888 912 935 959 983 a* vis 82 26 007 031 055 079 102 126 150 174 198 221 1 2 2.4 2.3 4.8 4.6 83 245 269 293 316 340 564 387 411 435 458 3 A 7.2 6.9 84 482 505 629 553 576 600 623 647 670 694 4 9.6 9.2 1 O i\ 11 C 85 717 741 764 788 811 834 858 881 905 928 6 G 12«0 11.8 14.4 13.8 86 26 951 975 998 *021 *045 1 *068 *091 ♦114 *138 *161 7 s 16.8 16.1 10 2 lA 4 87 27 184 207 251 254 300 323 346 1 370 393 o 9 2! 6 M 7 88 416 439 462 485 508! 531 554 577 600 623 W 1 W 89 646 669 692 715 738 761 784 807 830 852 190 27 875 , 898 921 944 967 9S9 *012 ’035 *058 *081 91 28 103 126 149 171 194 • 217 240 262 285 307 22 21 92 330 353 375 398 421 443 466 488 511 533 1 2.2 2.1 93 556 578 601 623 646 668 691 713 735 758 2 4.4 4.2 1 1 1 3 6.6 6.3 94 28 780 803 825 847 870 892 914 937 959 981 4 e» 8.8 8.4 95 29 003 026 048 070 092 115 137 159 181 203 O 6 11.0 10.6 13.2 12.6 96 226 248 270 292 314 336 358 380 403 425 7 a 15.4 14.7 1 7 f O 97 447 469 491 513 535 557 579 601 623' &15 O 9 I/.O 10.0 IQR lao 98 667 688 710 732 754 776 798 820 842 863 ^ 1 99 29 885 907 929 951 975 994 *016 *038 *060 *081 200 30 103 125 146 168 190 211 233 255 276 298 || Prop* Parts N mm 2 3 [5 6 8 9 264 Five-Place Logarithms : 200 — 250 0 wm 2 30 103 125 146 320 341 363 535 657 578 750 771 792 984 197 *006 218 387 408 429 597 618 639 31 806 827 848 32 015 035 056 222 243, 263 449 469 654 675 858 879 33 041 244 445 646 33 846 34 044 666 686 706 866 885 905 064 084 104 8 mm 276 298 492 614 707 728 920 942 *133 1 1*154 345 366 555 676 765 785 973 994 181 201 387 408 693 613 797 818 *001 *021 203 224 405 425 606 626 806 826 *005 *025 203 223 Prop. Parts 220 242 262 282 301 321 341 361 380 400 420 439 459 479 498 518 537 557 577 596 616 635 655 674 694 713 733 753 772 792 811 34 830 850 869 889 908 928 947 967 986 *005 35 025 044 064 083 102 122 141 160 180 199 218 238 257 276 295 315 334 353 372 392 411 430 449 468 488 507 526 545 564 583 603 622 641 660 679 698 717 736 755 774 793 813 832 851 870 889 908 927 946 965 35 984 *003 *021 *040 *059 *078 *097 *116 *135 *154 230 1 36 173 192 211 229 248 267 286 305 324 342 361 380 399 418 436 455 549 568 586 605 624 642 736 754 773 791 810 829 493 511 530 680 698 717 866 884 903 36 922 940 959 977 996 *014 *033 *051 *070 *088 37 107 125 144 162 181 199 218 236 254 273 291 310 328 346 365 383 401 420 438 457 475 493 511 530 548 566 585 603 621 639 658 676 694 712 731 749 767 785 803 822 37 840 858 876 894 912 931 949 967 985 *003 240 1 38 021 039 057 075 093 112 130 148 166 184 220 238 256 274 292 510 328 346 364 399 417 435 453 471 489 507 525 543 578 596 614 632 650 668 686 703 721 44 739 45 38 917 46 39 094 47 270 48 445 49 620 757 775 792 810 828 846 863 881 899 934 952 970 987 *005 *023 *041 *058 *076 111 129 146 164 182 199 217 235 252 287 305 322 340 358 375 393 410 428 463 480 498 515 533 550 568 585 602 637 655 672 690 707 724 742 759 777 250 39 794 811 829 846 863 881 898 915 933 950 ZZ 1 2.2 Z 4.4 3 6.6 4 8.8 6 11.0 6 13.2 7 15.4 8 17.6 9 19.8 21 2,1 4.2 6.3 8.4 10.6 12.6 14.7 16.8 18.9 6 10 6 12 7 14 8 16 • 18 19 1 1.9 2 3.8 3 5.7 4 7.6 6 9.5 6 11.4 7 13.3 8 15.2 • 17.1 18 1 1.8 2 3.6 3 5.4 4 7.2 6 9.0 6 10.8 7 12.6 8 14.4 9 16.2 17 1 1.7 2 3.4 3 6.1 4 6.8 6 8.5 6 10.2 7 11.9 8 13.6 • 15.3 Prop. Parts Five-Place Logarithms : 250 — 300 16 1.6 3.2 4.8 6.4 80 9.6 7 11.2 8 12.8 9 14.4 15 1.5 3.0 265 N 250 0 39 794 39 967 40 140 312 483 654 824 2 3 811 829 846 863 881 985 *002 *019 157 175 192 329 346 364 500 518 671 688 841 858 535 705 875 3 209 381 652 722 892 40 993 *010 *027 *044 *061 41 162 179 196 212 229 330 347 563 380 397 0 898 *071 243 415 586 756 926 *095 263 430 915 933 950 *088 *106 *123 261 278 295 432 449 466 603 620 637 773 790 807 943 960 976 *111 *128 *145 280 206 313 447 46-J 481' 497 514 531 547 664 581 597 614 | 631 I 647 664 681 697 714 731 747 830 847 863 880 896 913 41 996 *012 *029 *045 *062 *078 42 160 325 488 651 813 42 975 177 193 210 226 243 504 521 537 553 570 667 684 700 716 732 830 846 862 878 891 991 *008 *024 *040 *056 43 136 152 169 185 201 217 297 457 616 ' 775 43 933 . 44 091 77 248 78 404 79 560 313 473 632 791 949 107 329 345 361 377 489 505 521 537 648 664 680 696 807 823 838 854 965 981 996 * 122 138 154 264 279 295 311 420 436 451 467 576 592 607 623 280 716 731 814 979 *144 ' 308 472 619 635 781 797 94." 959 *104 *120 249 265 281 409 425 441 569 584 600 727 743 759 886 902 917 *044 *059 *075 185 201 217 232 749 911 *072 233 393 342 498 654 81 44 871 82 45 025 83 179 84 332 85 484 86 637 788 45 939 46 090 747 2 762 778 793 809 358 514 669 824 917 071 225 932 086 240 948 102 255 378 393 408 530 545 561 682 697 712 834 849 864 984 *000 *015 135 150 165 963 117 271 423 676 728 Prop. Parts 240 255 270 285 300 315 389 404 419 454 449 464 538 553 568 583 598 613 687 702 716 731 746 761 835 850 864 879 894 909 46 982 997 •012 *026 *041 *056 47 129 144 159 173 188 202 276 290 1 305 319 334 349 422 436 451 465 480 494 567 582 596 611 , 625 640 47 712 727 741 7o6 770 784 |[|Q^ w 2 3 4 5 373 389 529 .545 685 700 840 855 994 *010 148 163 301 317 4.54 469 606 621 758 773 909 924 *060 *075 210 225 330 345 359 374 479 494 509 523 627 642 657 672 776 790 805 820 923 938 953 967 *070 *085 *100 *114 217 232 246 261 363 378 392 509 524 538 654 669 683 799 813 828 407 553 698 842 9 266 Five-Place Logarithms: 300 — 350 320 840 2 3 4 6 6 7 8 741 756 770 784 799 813 828 842 01 47 857 02 48 001 03 144 885 900 914 929 943 029 044 058 073 087 173 187 202 216 230 958 972 986 101 116 130 244 2S9 273 04 287 302 316 330 344 359 373 387 401 416 05 430 444 458 473 487 501 515 630 544 658 06 572 586 601 615 629 643 657 671 686 700 07 714 728 742 756 770 786 799 813 827 841 08 855 869 883 897 911 926 940 954 968 982 09 48 996 *010 *024 *038 *052 *066 *080 *094 *108 *122 310 49 136 150 164 178 192 206 220 234 248 262 11 276 12 415 13 554 290 304 318 429 443 457 668 682 696 332 346 360 471 485 499 610 624 638 374 388 402 613 527 541 651 665 679 14 693 707 721 734 748 762 776 790 803 817 15 831 845 859 872 886 900 914 927 941 955 16 49 969 982 996 *010 *024 *037 *051 *065 *079 *092 50 106 120 133 147 161 174 243 256 270 284 297 311 379 393 406 420 433 447 188 202 215 229 325 338 352 365 461 474 488 601 515 629 642 556 569 583 596 610 623 637 21 651 22 786 23 50 920 24 51 055 25 188 26 322 664 678 691 705 718 732 745 759 772 799 813 826 840 853 866 880 893 907 934 947 961 974 987 *001 *014 *028 *041 068 081 095 108 121 135 148 162 176 202 215 228 242 255 268 282 295 308 335 348 362 375 388 402 415 428 441 27 455 468 481 495 508 521 534 548 661 574 28 587 601 614 627 640 654 667 680 693 706 29 720 733 746 769 772 786 799 812 825 838 330 31 51 983 32 52 114 33 244 851 865 878 891 904 917 930 943 957 970 996 *009 *022 *035 *048 *061 *075 *088 *101 127 140 153 166 179 192 205 218 231 257 270 284 297 310 323 336 349 362 34 375 388 401 414 427 440 453 466 479 492 35 504 517 530 543 556 569 582 595 608 621 36 634 647 660 673 686 699 711 724 737 750 37 763 776 789 802 815 827 840 38 52 892 905 917 930 943 956 969 39 53 020 033 046 058 071 084 097 853 866 879 982 994 *007 no 122 135 148 161 173 186 199 212 224 237 250 263 41 275 288 301 314 326 339 352 42 403 415 428 441 453 466 479 43 529 542 555 667 580 593 605 44 656 668 681 694 706 719 732 45 782 794 807 820 832 845 857 46 53 908 920 933 945 958 970 983 47 54 033 045 058 070 083 095 108 48 158 170 183 195 208 220 233 49 283 295 307 320 332 345 357 364 377 390 491 504 517 618 - 631 643 744 757 769 870 882 895 995 *008 *020 120 133 145 245 258 270 570 382 394 350 54 407 419 432 444 456 469 481 494 5061618 Prop. Parts U 1 1.6 2 3.0 8 4.6 4 6.0 6 7.6 e 9.0 7 10.6 8 12.0 13.5 U 1.4 2.8 4.2 6.6 6 7.0 6 8.4 9.8 11.2 12.6 13 1.3 2.6 3.9 6.2 6.6 7.8 . 9.1 8 i 10.4 11.7 13 1 1.2 3 2.4 8 3.6 4.8 6.0 7.2 8.4 8 9.6 10.8 Prop, Parts Five-Place Logarithms : 350 — 400 267 Prop. Parts 350 54 407 419 432 444 13 1 1.3 2 2.6 3 3.9 4 6.2 6 6.6 0 7.8 7 9.1 8 10.4 9 11.7 6 7 8.4 8 9.6 8 10.8 ii 1 1.1 2 2.2 3 3.3 4 4.4 5 6.5 6 6.6 7 7.7 8 8.8 8 9.9 10 1 1.0 2 2.0 8 3.0 4 4.0 6 6.0 6 6.0 7.0 8 8.0 9.0 3GO 370 551 654 777 54 900 55 023 145 267 388 509 630 751 871 55 991 56 110 229 348 467 585 703 643 555 568 667 679 691 790 802 814 913 925 937 0.35 047 060 157 169 182 279 291 400 413 522 534 478 597 714 490 608 726 502 620 738 820 832 844 855 56 937 57 054 171 74 287 75 403 76 619 77 634 78 749 79 864 380 67 978 81 58 092 82 206 83 320 949 066 183 299 415 530 961 078 194 310 426 642 972 089 206 550 660 770 879 59 988 60 097 60 206 wm 5 6 456 469 481 680 693 605 704 , 716 72,8 827 1 839 851 949 962 974 072 ! 084 096 , 194 206 218 315 1 328 340 437 449 461 658 670 682 678 691 703 799 811 823 919 931 943 •a>8 *050 *062 158 170 182 277 289 301 396 407 419 514 526 538 632 644 656 750 761 773 867 879 891 984 996 *003 101 113 124 217 229 241 334 345 357 449 461 473 565 576 1 588 680 692 1 703 795 807 818 910 921 933 *024 *035 *047 138 149 161 252 263 274 365 377 388 478 490 501 591 602 614 704 715 726 816 827 838 928 939 950 •040 *051 1 •062 151 162 173 262 273 284 373 384 395 483 494 506 594 605 616 704 715 726 813 824 835 923 934 945 *032 ♦043 *054 141 152 163 249 260 271 Prop. Parts mm 8 Ol 494 606 518 617 630 642 741 763 765 864 876 888 986 998 •oil 108 121 1.33 230 242 255 852 364 376 473 485 497 694 606 618 715 727 739 835 847 859 9.55 -A 967 979 *074 *086 *098 194 205 217 312 324 336 431 415 455 649 561 573 667 679 691 785 797 808 902 914 926 *019 ,*031 •043 136 148 i 159 252 264 1 276 368 380 ' 392 484 496 507 600 611 623 715 726 738 830 841 852 944 955 967 *058 *070 1 *081 172 1 184 195 286 297 309 399 410 422 512 524 535 625 636 647 737 749 760 850 861 872 961 973 984 *073 •084 *095 184 195 207 295 306 318 406 417 428 517 528 539 627 638 649 737 748 759 846 857 868 956 966 977 *065 *076 *086 173 i 184 196 282 1 293 304 mm 8 mm 268 Five-Place Logaritlims: 400 — 450 410 420 60 206 314 423 531 07 60 959 08 61 066 09 172 278 384 490 595 **700 805 61 909 17 62 014 18 118 19 221 325 448 548 649 440 ■■ 2 4 217 228 239 249 325 336 347 358 433 444 455 466 541 552 563 674 649 660 670 681 756 767 778 788 863 874 885 895 970 981 991 ♦002 077 087 098 109 183 194 204 215 289 300 310 321 395 405 416 426 500 511 521 632 606 616 627 637 711 721 731 742 815 826 836 847 920 930 941 951 024 034 045 055 128 138 149 159 232 242 252 263 335 346 356 366 439 449 459 469 542 552 562 572 644 655 665 675 747 757 767 778 849 859 870 880 951 961 1 972 982 053 063 073 083 155 165 175 185 256 j 266 276 286 357 367 377 387 458 468 478 488 558 568 579 589 659 669 679 689 759 769 779 789 859 869 879 889 959 969 979 988 058 06S 078 088 157 167 177 187 256 266 276 286 355 365 375 385 454 464 473 483 552 562 572 682 650 660 670 680 748 768 768 777 846 856 865 875! 943 953 963 972 040 050 060 070 137 147 167 167 234 244 254 263 331 341 350 360 Prop. PftrtB 8 818 829 1 1.1 2 2.2 8 3.3 4 4.4 5 3.6 C 6.6 7 7.7 8 8.8 9 9.9 2 2 . 3 3. 4 4.0 6 6.0 6 6.0 7. 6 I 8. 9 t 1 0.9 2 1.8 $ 2.7 3.6 4.6 6.4 6.3 7.2 6.1 Prop. Parts Prop. Parts 10 1 1.0 2 2.0 3 3.0 4 4.0 3 6.0 6 6.0 7.0 8.0 9.0 8 1 0.8 2 1.6 3 2.4 4 3.2 6 0 I 4.8 6.6 6.4 7.2 Prop. Parts Five-Place Logarithms: 450 — 500 260 65 321 418 514 610 706 801 896 2 .331 341 427 437 523 533 619 629 379 .389 ;9B 408 715 811 906 725 820 916 4 56 .552 648 466 562 658 475 571 667 400 470 460 *001 *011 096 106 191 200 744 753 763 8.30 839 849 8.58 925 935 944 954 *020 *0.30 *0.39 *049 115 124 154 1 210 219 229 238 276 285 295 304 314 323 332 370 464 558 652 745 839 66 932 67 025 117 380 389 474 483 567 577 661 671 755 764 848 857 94 1 950 034 043 127 136 398 492 586 417 511 605 427 521 614 680 689 699 708 77.3 78.3 792 801 867 876 885 89-1 960 959 978 987 052 062 071 080 145 154 164 173 210 219 228 237 247 256 265 311 321 330 405 413 422 495 504 514 348 3.57 440 449 532 541 4‘)5 504 .591 600 586 696 782 792 877 887 973 982 *068 *077 , 162 172 247 257 266 361 4.55 549 642 736 829 922 *015 108 201 284 293 376 385 468 477 560 569 .367 4.59 ,550 74 578 75 669 76 761 77 852 78 67 943 79 68 034 587 596 605 614 624 6.33 679 688 697 706 715 724 770 779 788 797 806 815 861 952 043 870 961 052 879 970 061 888 979 070 897 988 079 906 997 088 642 651 660 7.3.3 742 752 825 834 843 916 925 9.34 *006 *015 *024 097 106 115 124 133 142 151 160 169 178 1 187 196 205 215 224 233 242 251 260 269 278 305 314 323 332 341 350 359 368 395 404 413 422 431 440 449 458 287 296 377 386 467 476 84 485 494 502 511 520 529 538 547 5.56 565 85 674 583 592 601 610 619 628 637 646 655 86 664 673 681 690 699 70S 717 726 735 744 87 753 88 842 89 68 931 762 771 780 851 860 869 940 949 958 789 797 806 815 824 833 878 886 895 904 913 922 966 975 984 993 *002 *011 400 69 020 028 037 046 055 064 073 082 090 099 91 108 92 197 93 285 94 373 95 461 96 548 97 636 98 723 99 810 500 69 897 117 126 135 205 214 223 294 302 311 152 161 241 249 329 358 381 390 399 408 417 425 469 478 487 496 504 513 557 666 574 583 592 601 644 653 662 671 679 688 732 740 749 758 767 775 819 827 836 845 854 862 170 179 188 258 267 276 546 355 364 434 443 452 522 531 539 609 618 627 697 705 714 784 793 801 871 880 888 932 940 949 958 966 975 270 Five-Place Logarithms : 500 — 550 500 69 897 69 984 70 070 167 04 243 05 329 06 415 07 501 08 586 09 672 510 520 530 540 757 11 842 12 70 927 13 71 012 349 433 517 600 684 767 850 181 263 346 428 509 591 675 239 320 400 480 560 640 719 799 878 73 957 550 74 036 El 3 4 5 6 Di 8 mm 906 914 923 932 940 949 958 966 975 992 *001 *010 *018 *027 •036 *044 *053 *062 079 088 096 105 114 122 131 140 148 165 174 183 191 200 209 217 226 234 252 260 269 278 286 295 303 312 321 338 346 355 364 372 ! 381 389 398 406 424 432 441 449 458 467 475 484 492 509 518 526 535 544 552 561 569 578 595 603 612 621 629 638 646 655 663 680 689 697 706 714 723 731 740 749 766 774 783 791 800 808 817 825 834 851 859 868 876 885 893 902 910 919 935 944 952 961 969 978 986 995 *003 020 029 037 046 054 063 071 079 088 105 113 122 130 139 147 155 164 172 189 198 206 214 223 231 240 248 267 273 282 290 299 307 315 324 332 357 366 374 383 391 399 408 416 1 425 441 450 458 466 475 483 492 600 , 508 525 533 542 550 559 567 575 584 692 1 609 617 625 654 642 650 659 667 692 700 709 717 725 734 742 760 ! 759 775 784 792 800 809 817 825 834 : 842 858 867 875 885 892 900 908 917 925 941 950 958 966 975 983 991 999 *008 024 032 041 049 057 066 074 082 090 107 115 123 132 140 148 156 165 173 189 198 206 214 222 230 239 247 255 272 280 288 296 304 313 321 329 337 354 362 370 378 387 395 403 411 419 436 444 452 460 469 477 485 493 501 518 526 ' 534 542 550 558 567 675 583 699 607 616 624 632 640 648 656 665 681 689 697 705 713 722 730 758 746 762 770 779 787 795 803 811 819 827 843 852 860 868 876 884 892 900 908 925 933 941 949 957 965 973 1 981 989 *006 *014 *022 *030 *038 *046 *054 *062 *070 036 094 102 111 119 127 1 135 143 151 167 175 183 191 199 207 215 223 251 247 255 263 272 280 288 296 304 312 328 336 344 352 360 368 376 384 392 408 416 424 452 440 448 456 464 472 488 496 504 512 520 528 536 544 552 568 576 584 592 600 608 616 624 632 648 656 664 672 679 687 695 703 711 727 735 743 751 759 767 775 783 791 807 815 823 830 838 846 854 862 870 886 894 902 910 918 926 933 941 949 965 973 981 989 997 *005 *013 *020 ♦028 044 052 060 068 076 084 092 099 107 Prop. Parts 0 0.9 1.8 2.7 3 . 6 ’ 4.6 6.4 , 6.3 817.2 8 . 1 , 8 0.8 1.6 2.4 3.2 4.0 4.8 6.6 816.4 7.2 tl’ 21 1.4 3 I 2.1 2.8 3.6 6 4.2 7 4.9 8 6.6 6.5 Prop. Parts Five-Place Logarithms : 550 600 Prop. Parts 6 1 0.8 2 1.6 3 2.4 4 3.2 5 4.0 6 4.8 7 6.6 8 6.4 9 7.2 7 1 0.7 2 1.4 3 2.1 4 2.8 6 3.6 6 4.2 7 4.9 8 6.6 9 6.3 Prop. Parts 550 74 036 51 115 52 194 53 273 819 560 61 896 62 74 974 63 75 051 570 687 664 740 815 74 891 75 75 967 76 76 042 118 193 268 580 343 641 716 790 864 76 938 77 012 044 052 060 068 076 123 131 1.-9 147 155 202 210 218 255 280 288 296 304 312 359 367 374 382 390 437 445 455 461 463 515 523 531 539 547 593 601 609 617 624 671 679 687 695 702 749 757 764 772 780 827 834 842 850 858 904 912, 920 927 935 981 989 997 *005 *012 059 066 074 082 089 156 143 151 159 166 213 220 228 256 ' 245 289 1 297 305 312 520 366 374 381 389 397 442 450 458 465 475 619 526 534 542 549 595 603 610 G18 626 671 679 686 694 702 747 755 762 770 778 823 831 838 846 853 899 906 914 921 929 974 982 989 997 *005 050 057 065 072 080 125 133 140 148 155 200 208 215 223 250 275 283 290 298 305 350 358 365 373 580 425 433 440 448 455 500 507 515 522 530 574 582 589 597 604 649 656 664 671 678 723 730 738 745 753 797 805 812 819 827 871 879 886 893 901 945 953 960 967 975 019 026 034 041 048 271 9 099 107 170 240 327 •106 484 562 640 718 796 178 257 335 186 265 343 414 421 492 500 570 578 618 726 803 656 733 811 873 881 839 930 958 966 ‘023 *043 105 113 120 174 251 328 40-1 481 557 182 259 335 412 438 565 189 197 266 274 343 351 420 496 572 427 504 530 633 641 648 709 785 861 717 724 7.32 79.3 800 808 «68 876 884 163 238 313 170 245 320 178 253 328 185 2oO 335 388 395 403 410 477 485 552 559 626 634 701 708 775 782 849 856 908 9S2 056 916 989 063 923 930 997 *004 070 078 590 085 093 1 100 107 115 159 166 173 181 188 232 240 247 254 262 305 313 320 327 335 379 386 393 401 408 452 459 466 474 481 625 532 539 546 554 697 605 612 619 627 670 677 685 692 699 743 750 757 764 772 77 815 822 830 837 844 122 129 137 144 151 195 203 210 Yi7 ~2^ 269 276 283 291 298 342 349 357 364 371 422 430 437 444 495 503 510 617 568 576 583 590 600177 815 822 01 887 02 77 960 03 78 032 04 104 05 176 06 247 319 390 462 610 620 G40 895 967 039 111 183 254 326 398 469 2 830 902 974 046 118 190 262 333 405 476 8 837 909 981 053 125 197 269 340 412 483 4 844 916 988 061 132 204 276 347 419 490 8 9 851 859 866 873 880 931 938 945 952 *003 *010 *017 *025 075 082 089 097 140 147 154 161 168 211 219 226 233 240 283 290 297 305 312 355 362 369 376 383 426 433 440 447 455 497 504 512 519 526 Prop. Parts 533 540 547 554 561 569 576 583 590 597 604 611 618 625 633 640 647 654 661 668 675 682 689 696 704 711 718 725 732 739 746 753 760 767 774 781 789 796 803 810 14 817 824 831 838 845 852 859 866 873 880 15 888 895 902 909 916 923 930 937 944 951 16 78 958 965 972 979 986 993 *000 *007 *014 *021 79 029 036 043 050 057 064 071 099 106 113 120 127 134 141 169 176 183 190 197 204 211 078 085 092 148 155 162 218 225 232 239 246 253 260 267 274 281 288 295 302 309 379 449 518 588 657 727 796 865 316 323 330 337 344 386 393 400 407 414 456 463 470 477 484 525 532 539 546 553 595 602 609 616 623 664 671 678 685 692 351 358 365 372 421 428 435 442 491 498 505 511 560 567 574 581 630 637 644 650 699 706 713 720 734 803 872 741 810 879 748 817 886 754 824 893 761 831 900 768 837 906 775 844 913 782 851 920 789 858 927 G30 79 934 941 948 955 962 969 975 982 989 996 80 005 072 140 209 277 346 414 482 550 010 079 147 216 284 353 421 489 557 017 085 154 223 291 359 428 496 564 024 092 161 229 298 366 454 502 570 030 099 168 236 305 373 441 509 577 037 106 175 243 312 380 448 516 684 044 113 182 250 318 387 455 623 591 051 120 188 257 325 393 462 530 598 058 127 195 264 332 400 468 536 604 065 134 202 271 339 407 475 543 611 618 625 638 645 652 659 665 672 679 686 754 821 693 760 828 699 767 835 706 774 841 713 781 848 720 787 855 726 794 862 733 801 868 936 943 949 *003 *010 *017 070 077 084 137 144 151 204 211 218 271 278 286 338 346 361 KM 8 8 1 0.8 2 1.6 3 2.4 4 3.2 5 4.0 6 4.8 7 6.6 8 6.4 9 7.2 7 1 0.7 2 1.4 3 2.1 4 2.8 5 3.6 G 4.2 4.9 8 I 6.6 6.3 « 1 0.6 2 1.2 3 1.8 2.4 3.0 3.6 4.2 8 4.8 6.4 Prop. Parts Five-Place Logarithms: 650 700 273 Prop. Parts 7 1 0.7 2 1.4 3 2.1 4 2.8 6 3.5 C 4.2 7 4.9 8 5.6 9 6.3 Prop. Parts N 650 1 81 29 61 358 52 425 63 491 54 558 65 624 66 690 57 757 68 823 59 889 298 365 431 498 564 631 697 763 829 895 600 81 954 961 61 82 020 62 086 63 151 217 282 347 413 478 543 027 092 158 223 289 354 305 m 318 325 371 378 385 391 438 445 451 458 505 511 518 525 571 578 584 591 637 644 651 657 704 710 717 723 770, 776 783 790 836 842 849 856 902 908 915 921 968 974 981 987 033 040 046 053 099 105 112 119 164 171 178 184 230 236 243 249 295 302 308 315 360 367 373 380 426 432 439 1 445 491 497 504 510 556 562 569 575 398 465 631 698 664 730 796 862 928 405 471 538 604 671 737 411 478 544 351 418 485 611 617 677 684 743 750 803 809 8 1 6 869 875 882 935 941 948 994 *000 *007 *014 060 125 191 066 073 079 152 138 145 197 204 210 263 269 276 328 334 341 393 400 406 458 465 588 695 471 636 601 670 607 I 614 620 627 633 640 646 653 '659 1 666 71 72 73 74 75 76 77 78 79 672 737 802 866 930 82 995 83 059 123 187 679 743 808 872 937 •ool 065 129 193 685 750 814 879 943 *008 072 136 200 692 756 821 885 950 *014 078 142 206 698 763 827 892 956 *020 085 149 213 680 251 257 264 270 276 « 81 315 321 327 334 340 0.6 82 378 385 391 398 404 1.2 83 442 448 455 461 467 1.8 2.4 84 506 512 518 525 531 3.0 85 569 575 532 588 594 3.6 86 632 639 615 651 658 4.2 1 4.8 6.4 87 696 702 708 715 721 88 759 765 771 778 784 89 822 828 835 841 847 GOO 885 891 897 904 910 711 776 840 718 782 847 724 730 789 795 853 860 091 155 219 097 161 225 104 168 232 no 174 9 117 ISl 245 283 289 296 302 308 347 353 410 417 474 480 359 423 487 366 429 493 372 436 499 91 83 948 92 84 011 93 073 954 960 967 973 017 023 029 036 080 086 092 098 94 136 142 148 155 161 95 198 203 211 217 223 96 261 267 273 280 286 97 323 98 386 99 448 330 336 342 348 392 393 404 410 454 460 466 473 537 544 550 556 565 601 607 613 620 626 664 670 677 683 689 727 734 740 746 753 790 797 803 809 816 853 860 SG6 872 879 916 923 929 935 ^42 992 998 *004 055 061 067 117 123 130 173 180 186 192 236 242 248 255 305 311 317 367 373 379 429 435 442 491 497 504 700 84 510 516 522 528 535 541 547 5531559 566 274 Five-Place Logarithms: 700 — 750 If 700 84 510 710 720 730 740 07 84 942 08 85 003 09 065 126 187 248 309 552 612 673 733 153 213 273 332 392 451 510 37 747 38 806 39 864 923 41 86 982 42 87 040 43 099 wm wm MM 516 622 528 635 578 584 690 697 640 646 652 658 702 708 714 720 763 770 776 782 825 831 837 844 887 893 899 905 948 954 960 967 009 016 022 028 071 077 083 089 132 138 144 150 1 193 199 205 211 254 260 266 272 315 321 327 333 376 382 388 394 437 443 449 455 497 503 609 516 558 564 570 576 618 625 631 637 679 685 691 697 739 745 751 757 806 812 818 866 872 878 920 926 932 938 980 986 992 998 040 046 052 058 100 j 106 112 118 159' 165 171 177 219 225 251 237 279 285 291 297 338 344 350 356 398 404 410 415 457 463 469 475 516 522 528 534 576 581 687 593 635 641 646 652 694 700 705 711 753 759 764 770 812 817 823 829 870 876 882 888 ESi 955 941 947 988 994 999 *005 046 052 058 064 105 111 116 122 163 169 175 181 221 227 253 239 280 286 291 297 338 344 349 4 355 396 402 408 413 454 460 466 471 512 518 623 629 Prop. Parts 763 769 775 781 788 824 830 836 842 848 884 890 896 902 908 944 950 956 962 968 356 362 368 374 380 386 433 439 445 493 499 504 552 558 564 976 •035 093 151 7 1 0.7 2 1.4 8 2.1 4 2.8 6 3.6 6 4.2 7 4.9 8 6.6 0 6.3 • 1 0.6 2 1.2 3 1.8 4 2.4 5 3.0 e 3.6 7 4.2 8 4.8 9 6.4 $ 1 0.5 2 1.0 8 1.6 4 2.0 6 2.6 0 3.0 7 3.6 8 4.0 9 4.6 Prop. Parts Prop. Parte .6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 6.4 .5 1.0 1.5 2.0 2.5 3.0 7 3.6 8 4.0 9 4.6 Prop. Parts Five-Place Logarithms : 75o 800 N 750 87 506 51 564 52 622 53 679 737 , 795 852 910 87 967 88 024 275 512 518 523 529 535 1541 770 780 790 570 628 685 743 800 858 915 973 030 576 633 691 581 639 697 749 754 806 812 864 869 921 927 984 041 587 645 703 760 81H 875 933 990 047 593 651 708 766 823 881 772 829 887 777 835 892 616 668 674 726 731 783 789 84 1 846 898 904 081 087 093 098 104 9.38 944 930 955 o/:j 996 *001 *007 *013 *0i8 033 058 064 070 076 138 195 252 309 366 423 480 536 693 649 144 201 258 156 213 270 161 218 275 no IIG 121 127 167 v)15 o21 326 33 74 874 75 930 76 88 986 77 89 042 78 098 79 154 209 265 321 376 432 487 642 597 653 708 763 818 873 927 94 89 982 95 90 037 96 091 97 146 98 200 99 255 800 90 309 N 372 3S3 429 434 440 485 491 497 542 547 553 598 604 610 655 660 666 711 717 722 767 773 779 824 829 835 880 885 891 936 941 947 992 997 *003 048 053 059 104 109 115 159 165 170 215 221 226 271 276 282 326 532 337 582 387 393 437 443 448 492 498 504 548 553 559 603 609 614 658 664 669 713 719 724 763 774 779 823 829 834 878 883 889 933 938 944 988 993 998 012 048 053 097 102 108 151 157 162 206 211 217 260 266 271 332 3 SO 446 502 5:39 615 281 338 395 451 508 564 621 133 190 247 504 457 513 570 627 349 3.35 360 406 412 417 463 468 474 576 632 525 581 658 530 587 643 672 677 734 790 846 689 694 700 739 795 852 745 801 857 730 736 807 812 863 868 070 076 126 131 182 187 081 137 193 087 14.3 198 092 148 204 248 254 260 310 315 565 371 421 426 454 459 465 470 476 481 509 515 520 526 531 537 564 570 575 581 586 592 785 790 796 801 807 812 840 845 851 856 862 867 894 900 905 911 916 922 949 955 960 966 971 977 331 336 342 347 352 358 276 810 820 0 2 3 4 90 309 314 320 325 331 363 369 374 380 385 417 423 428 434 439 472 477 482 488 493 626 531 536 542 547 580 685 590 596 601 634 639 644 650 655 687 693 698 703 709 ^"41 747 752 757 763 795 800 806 811 816 849 854 859 865 870 902 907 913 918 924 90 956 961 966 972 977 91 009 014 020 025 030 062 068 073 078 084 116 121 126 132 137 169 174 180 185 190 222 228 233 238 243 275 281 286 291 297 328 334 339 344 360 381 387 392 397 403 830 8 342 347 352 396 401 407 450 455 461 504 509 515 . 558 563 569 612 617 623 666 671 677 720 726 730 773 779 784 827 832 838 Prop. Parts 840 249 502 355 886 891 897 940 945 950 993 998 *004 046 052 057 100 105 no 155 158 164 206 212 217 254 259 265 270 307 312 318 323 360 565 371 376 403 413 418 424 429 434 487 540 593 645 698 751 803 855 440 492 545 598 651 703 756 808 861 445 498 551 603 656 709 761 814 866 609 661 714 766 819 871 455 508 561 614 666 719 772 824 876 461 514 566 619 672 724 777 829 882 466 471 477 482 519 524 529 535 572 577 582 587 624 630 635 640 677 682 687 693 750 735 740 745 782 787 793 798 854 840 845 850 887 892 897 903 913 918 924 929 934 939 944 950 955 91 960 92 012 065 117 169 221 273 324 376 965 018 070 971 023 075 976 028 080 981 053 085 986 058 091 278 350 381 283 288 293 335 340 345 387 392 397 428 453 438 443 449 454 459 480 531 685 634 686 737 788 840 891 92 942 485 490 495 500 536 542 547 552 588 593 693 603 639 645 650 655 660 691 696 701 706 711 742 747 752 758 763 511 562 614 793 799 845 850 896 901 804 855 906 809 814 819 860 865 870 911 916 921 997 ’ ‘‘002 ’ *007 049 054 059 101 106 111 15.1 158 163 205 210 215 257 262 267 309 314 319 361 366 371 412 418 423 464 469 474 516 521 526 567 572 578 619 624 629 670 676 681 722 727 732 ! 773 778 783 824 829 834 875 881 886 927 932 937 978 983 988 6 0.6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 6.4 5 0.5 1.0 1.6 2.0 2.6 3.0 3.6 4.0 4.6 Prop. Ptits Five-Place Logarithms: 850^900 277 Prop. Parts e 0.6 1.2 1.8 2.4 30 3.6 4.2 4.8 A 0.4 0.8 1.2 1.6 2.0 2.4 2.B 3.2 3.6 800 880 800 Prop. Parts 92 942 947 952 m 92 993 998 * 00.‘5 *008 93 044 049 054 059 095 100 105 110 146 151 156 161 197 202 207 212 247 252 258 263 298 303 308 313 349 354 359 364 399 404 409 414 450 m 460 465 500 505 510 515 551 556 561 566 601 606 611 616 651 656 661 666 702 707 712 717 752 757 762 767 802 807 812 817 852 857 862 867 902 907 912 917 93 952 957 962 967 94 002 007 012 017 052 057 062 067 101 106 111 116 151 156 161 166 201 1 206 211 216 250 255 260 1 265 300 305 1 310 315 349 354 359 364 399 404 409 414 448 453 458 463 498 503 507 512 647 552 557 562 596 601 606 611 645 650 655 660 694 699 704 709 743 748 755 758 792 797 802 807 841 846 851 856 890 895 900 905 939 944 949 954 94 988 993 998 *002 95 036 041 046 051 085 090 095 100 134 139 143 148 182 187 192 197 231 236 240 245 279 284 289 294 328 332 337 342 376 381 386 390 95 424 429 434 439 0 EH 2 3 *013 064 115 166 217 268 518 369 420 470 520 571 621 972 022 072 121 320 369 419 463 517 667 616 967 973 *018 *024 069 075 120 125 171 176 222 227 273 278 323 328 mm is 379 450 475 480 526 551 576 581 626 631 676 682 727 732 777 782 827 832 877 882 927 932 977 932 027 032 077 082 126 131 176 181 226 2.51 275 280 325 3.30 374 379 424 429 473 478 1 522 527 571 621 626 670 675 719 724 768 773 817 822 866 871 915 919 978 985 988 *029 * 0 ^ 080 085 131 136 181 186 232 237 283 288 334 384 435 339 389 440 * 0.39 090 Ml 192 242 293 344 394 445 485 490 495 6.36 686 636 541 691 641 687 692 737 742 787 792 837 842 887 892 937 942 596 646 697 747 797 847 897 947 987 992 997 037 086 136 681 949 954 959 963 968 973 978 983 *007 *012 *017 056 061 066 105 109 114 153 158 163 202 207 211 250 255 260 299 303 308 347 352 357 395 400 405 444 1 448 453 4 5 ■a 191 196 240 245 290 295 340 345 389 394 438 443 488 493 537 542 586 591 635 640 685 689 734 738 783 787 832 836 880 885 929 934 978 985 *027 * 0.-^2 075 080 124 129 173 177 221 226 270 274 318 323 366 371 415 419 463 468 8 9 278 Five-Place Logarithms : 900 — 950 ■asm 8 9 448 453 458 463 468 497 501 506 511 516 545 650 654 559 564 593 598 602 607 612 641 646 650 655 660 689 694 698 703 708 737 742 746 751 756 785 789 794 799 804 832 837 842 847 852 880 i 885 1 890 895 899 Prop. Parts 910 472 477 482 487 621 525 530 635 569 674 678 683 617 622 626 665 670 674 713 718 722 761 766 770 776 809 813 818 823 856 861 866 871 ioT 909 914 918 923 928 933 1 938 1 942 1 947 920 952 957 961 966 95 999 *004 *009 *014 96 047 052 057 061 095 099 104 109 142 147 152 156 190 194 199 204 237 242 246 251 284 289 294 298 332 336 341 346 379 384 388 393 976 980 985 990 995 *023 *028 *033 *038 *042 071 076 080 086 090 114 118 123 128 133 137 161 166 171 175 180 185 209 213 218 223 227 232 256 261 265 270 275 280 303 308 313 317 322 327 350 355 360 365 369 374 426 473 520 567 614 661 708 755 802 431 455 440 478 483 487 525 530 534 572 577 581 619 624 628 666 670 675 713 759 806 717 722 764 769 811 816 930 848 853 412 417 459 464 506 511 553 558 600 605 647 652 694 699 741 745 788 792 834 839 881 886 895 942 96 988 97 035 081 128 174 220 267 900 946 993 039 086 132 179 225 271 313 317 322 327 359 405 451 364 368 373 410 414 419 456 460 465 635 6S1 727 97 772 918 923 928 932 937 965 970 974 979 984 *011 *016 *021 *025 *030 058 063 067 072 077 104 109 114 118 123 151 155 160 165 169 197 202 206 211 216 243 248 253 257 262 290 294 1 299 304 308 336 340 345 m 354 382 387 391 396 400 428 453 437 442 447 474 479 483 488 493 520 525 629 634 539 666 571 575 580 585 612 617 621 626 630 658 663 667 672 676 704 708 713 717 722 749 754 759 763 768 795 800 804 809 813 5 0.6 10 1.6 2.0 2.6 3.0 3.5 4.0 4.6 « 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 Prop. Parts Prop. Parts 4 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 Prop. Parts Five-Place Logarithms : 950 looo 279 950 97 772 61 818 823 52 864 868 63 909 914 54 97 955 959 65 98 000 005 66 046 050 970 980 091 137 182 096 141 186 2 782 827 873 918 964 009 055 100 146 191 227 232 236 272 277 318 322 327 363 367 372 408 412 417 453 457 462 498 502 507 643 647 652 688 592 597 632 637 641 677 682 686 722 767 811 726 771 816 731 776 820 856 860 865 900 905 909 945 949 954 77 98 989 994 78 99 034 038 79 078 083 123 127 131 167 171 176 211 216 220 255 260 264 300 304 308 344 348 352 388 392 396 432 476 520 436 480 524 441 484 528 564 568 572 607 612 616 651 656 660 695 699 704 739 743 747 782 787 791 826 830 835 870 913 99 957 00 000 874 917 961 878 922 965 \mwm 786 791 795 800 832 836 841 845 877 882 886 891 923 928 932 937 968 973 978 982 014 019 023 028 059 064 068 073 105 109 114 IIB 150 155 159 164 195 200 204 209 241 245 250 254 286 290 295 299 331 3.36 340 345 376 381 385 390 421 426 4.30 435 466 471 475 480 511 516 520 525 556 561 565 570 601 605 610 614 646 650 655 659 691 695 700. 704 735 740 744 749 780 784 789 793 825 1 829 834 838 1 869 874 878 883 914 918 927 958 963 967 972 *003 •007 •012 *016 047 052 056 061 092 096 ' 100 105 136 140 145 149 180 185 189 193 224 229 233 238 269 273 277 282 313 317 322 526 357 361 366 370 401 405 410 414 445 449 454 458 489 493 498 502 553 537 542 546 577 581 585 590 621 625 629 634 664 669 673 677 708 712 717 721 752 756 760 765 795 800 804 808 839 843 848 852 883 887 891 896 926 930 935 939 970 974 978 983 850 896 941 85.5 900 946 987 991 032 037 078 082 123 168 214 127 1 73 218 859 905 950 996 Oil 087 132 177 223 259 263 268 304 308 313 349 354 358 394 399 403 4.39 444 448 484 489 493 629 534 538 574 679 583 619 623 628 664 668 673 709 713 717 753 798 843 758 802 847 762 807 851 887 892 896 , 9.32 936 941 976 981 985 I *021 *025 *029 065 069 074 109 114 118 154 158 162 198 202 207 242 247 251 286 291 295 379 .383 467 511 555 471 515 559 599 603 647 691 734 909 952 996 030 035 Q 013 017 022 026 3 4 5 6 8 I 9 index A Accuracy, 12 Acidity, determination of, 49 Acids, 48 degree of ionization, 64, 05 standard solution of, 53 table of concentration, 248 Alkalinity, determination of, 49 Analysis, indirect, 228 proximate, 213 sj'stcmatic, 213 ultimate, 213 volumetric, 29 Arsenic, determination of, 144 Arsenite, standard solution of, 149 Atomic weights, calculation of, 236 table of, inside back cover B Balance, exercises with, 46 use of, 18 Balancing equations, 103 Base, standardization of, 56 Brass, analysis of, 222 Bromate methods, 155 Buffer solutions, 186 Buoyancy, correction for, 22 Buret, calibration of, 35 C Calcium, gravimetric determination of, 201 volumetric determination of, 135 Calculations of: atomic weights, 236 dichromate methods, 120 factor weight samples, 210 formulas, 234 gravimetric analysis, 174 indirect analysis, 228 iodometric methods, 157 ionic equilibria, 89 mixed alkalies, 93 Calculations of: {Continued) mixed halides, 200 neutralization methods, 59, 89, 97 pH^Jcrmanganate methods, 140 pH values, 62 reagents, 23 jjiuuuci constants, 181 volumetric analysis, 35 volumetric precipitation, 107 Calibration, of buret, 35 of flasks, 34 of weights, 21 Ceric sulfate methods, 141 Chemical Equilibrium, Law of 6 61 107, 101, 181 ’ * Chemical factors, 177 Chlorides, determination of, 102, 191 Cleaning solution, 45 Combining weights, 9 Common-ion effect, 184 Complex ions, 106 Computation rules, 13 Concentration, hydrogen-ion, 63 Copper, electrolytic determination of, 225 iodometric determination of, 151 Current density, 225 D Definite Composition, Law of, 8 Density tables, of acids, 248 of ammonia, 250 of water, 250 Desiccator, 48 Determination of : acidity of oxalic acid, 57 alkalinity of soda ash, 49 arsenic, 144 calcium, gravimetric, 201 volumetric, 135 carbon dioxide, 244 chlorides, 162, 191 copper, electrolytic, 226 iodometric, 151 281 INDEX 282 Determination of: {CoTUinued) hydrogen peroxide, 138 iron, by dichromate, 115 by permanganate, 126 gravimetrically, 195 lead, 226, 231 magnesium, 204 moisture, 217 normality, 38 oxalic acid, 57 phosphates, 206 potassium acid phthalate, 56 pyrolusite', 139 silica, 218 silver, 165 soda ash, 54 sulfates, 198 tin, 225 water of crystallization, 217 zinc, 227 Dichromate methods, 115 calculations of, 120 Dilution problems, 25 E Electrochemical analysis, 222 Electrode potentials, 106 Electron transfer, 103 End points, 32 Equilibrium, Law of Chemical, 6, 61 107, 161, 181 Equivalent weight, 38, 104 Errors of analysis, 1 1 Evolution methods, 216 Exponents, 14 F Factor weight samples, 210 Factors, chemical, 177 Faiaday’s laws, 224 Filtering, 173 Formulas, calculation of, 234 G Gooch crucible, preparation of, 192 Gravimetric precipitation, 169 Gravimetric procedures, 190 H Half-cell reactions, 102 Halides, calculations of mixed, 206 precipitation of, 190 Hydrogen electrode, 88 Hydrogen-ion concentration, 63 Hydrolysis, 66 Hydroxides, precipitation of, 194 I Ignition of precipitates, 173 Indicators, 31 chart, 84 list, 83 theory of, 83 lodate methods, 155 Iodine, standard solution of, 146, 152 table, 64, 65 lodometric calculations, 148, 157 Iron, dichromate determination of, 116 gravimetric determination of, 196 permanganate determination of, 126 J Jones reductor, preparation and use, 132 L X^ead, determination of, 226, 231 Limestone analysis, 219 Liter, 33 Literature, 241 Logarithms, table of, 262-279 use of, 14 M Magnesium, determination of, 204 Methyl orange, structure, 87 Milliequivalent, 38 Mixed alkalies, calculation of, 93 Moisture, determination of, 48, 217 Molar potentials, 106 N Nernst equation, 105 Neutralization methods, 45 ^ t QLLdHA IQBAL LIBRARY Normality, 38 calculation of, 39 Notebook, 2 O Overvoltage, 224 Oxalic acid, determination of, 57 Oxidation, electrochemical nature 104 Oxidizing agents, 102 table of, 108 P Permanganate methods, 122 calculation of, 140 uses of, 123 pH value, 62 of indicators, 83 table of values, 64, 70, 76 Phenolphthalein, 56, 87, 247 Phosphates, 203 determination of, 206 Plan of course, 243 Potentials, table of, 108 Potentiometric titration, 67, 87 Precipitates, ignition of, 173 washing of, 173 Precipitation, 169 electrolytic, 222 volumetric, 160 Precision, 1 1 Pyrophosphate, 205, 209 INDEX' 283 of. Silica, determination of, 218 Silver, determination of, 1G5 Solubility product principle, 8, 47 biblc of solubility products, 181 use of, 180 Solutions, adjustment of, 91 buffer, 186 standard, 30 standardization of, rrc Standardization Specftc gravity, ealcnlatioiui involving, tables, sec Density tables Standardization, 31 of solutions of: ferrous salt, 119, 130, 134 hydrochloric acid, 52 iodine, 145 potassium dichromato, 118 potassium perman-armte, 130, 134 potassium thiocyanate, IG4 silver nitrate, 103 sodium arsenite, 145 sodium hydroxide, 56 sodium thiosulfate, 153 Stoichiometric calculations, 59 Sulfates, determination of, 198 Tin, determination of, 225 Titration curves, 71, 75, 79 Titer, 37 R Reactions, between acids and bases 48 68,72 half-cell, 102 of oxidation, 113 Reagents, calculation of, 23 list of, 245 Redox indicators, 112 Reducing agents, 23 Report forms, 253, 255 V Valence, 99 balancing equations by, 103 Volumetric methods, calculations of, 35 W Water of crystallization, determination of, 217 Weights, calibration of, 21 S Sampling, 10 Significant figures, 13 Ziramermann-Reinhard method, 129 1 Zinc, deterihination of, 227 L. » INTERNATIONAL ATOMIC WEIGHTS 1043 Reprinted from the Journal of the American Chemical Society Sym- bol Atomic Number Atomic IVcioht • Aluminum . . . A1 13 26.97 Antimony. . . . Sb 51 121.76 Argon A 18 39 . 944 Arsonio As 33 74.91 Barium Ba 56 137.36 Beryllium Bo 4 9.02 Bismuth Bi 83 209 . 00 Boron B 5 10.82 Bromine Br 35 79.910 Cadmium. . . . Cd 48 112.41 CiUcium Ca 20 40.08 Carbon C 6 12.010 Cerium Co 1 58 140.13 Cesium Cs 55 132.91 Chlorine Cl 17 35.457 Chromium. . . Cr 24 •52.01 Cobalt Co 27 58.94 Columbium. . Cb 41 92.91 Copper Cu 29 63.57 Dysprosium. . Dy GO 162.46 Erbium Er 68 167.2 Europium. . . . Eu 63 152.0 Fluorine F 9 19.00 Gadolinium . . Gd 64 150.9 Gallium Ga 31 69.72 Germanium. . Ge 32 72.60 Gold Au 79 197.2 Hafnium Hf 72 178.6 Helium He , 1 , 2 4.003 Holmium. , . . Ho 67 164.94 Hydrogen H 1 1 . 0080 Indium In 49 114.76 Iodine I 53 126.92 Iridium Ir 77 1 193.1 Iron Fe 26 55.85 Krypton Kr 36 83.7 Lanthanum. . La 57 138.92 Lead Pb 82 207.21 Lithium Li 3 6.940 Lutecium Lu 71 174.99 Magnesium . . Mg 12 24.32 Manganese.. . Mn 25 54.9? Mercury Hg 80 200 Molybdenum. . Neodymium. Neon Nickel Nitrogen. . . . Osmium Oxygon Palladium . . . Phosphorus. . Platinum. . . . Pot.'issium. . . Praseodymium Protactinium. Radium Radon Rhenium. . . . Rhodium. . . . Rubidium. . . Ruthenium. . Samarium. . . Scandium. . . Selenium. . . . Silicon Silver Sodium Strontium. . . Sulfur Tantalum. . . Tellurium . . . Terbium. . . . Thallium. . . . Thorium. . . . Thulium. . . . Tin Titanium... ‘ Tungsten. Uraniur Vano'' Xe Sym- bol Mo Nd No Ni N Os O Pd P Pt K Pr Pa Ra Rn Ro Rh Rb Ru Sm Sc 50 51 Ag Na Sr S Ta To Tb Tl Th T' * Atomic Number 42 GO 10 28 7 70 8 40 15 78 19 59 01 88 86 75 45 37 44 02 21 34 14 47 11 38 16 73 52 6" V Atomic Wci{;hl 95 95 144.27 20.183 68.09 14.008 190.2 1 0 . 0000 100.7 30.98 195.23 39 . 096 140.92 231 226.05 222 186.31 102.91 85.48 101.7 150.43 45.10 78.96 28.00 107.880 22.99'’ 87.' ^4. o. %>. b- \ 3a V °4 %