Main
Calculations of Analytical Chemistry
Calculations of Analytical Chemistry
Leicester F. Hamilton, Stephen G. Simpson
0 /
0
How much do you like this book?
What’s the quality of the file?
Download the book for quality assessment
What’s the quality of the downloaded files?
Categories:
Year:
1947
Edition:
4th
Publisher:
McGrawHill Book Company
Language:
english
Pages:
411
File:
PDF, 22.62 MB
Download (pdf, 22.62 MB)
 Open in Browser
 Checking other formats...
 Please login to your account first

Need help? Please read our short guide how to send a book to Kindle
The file will be sent to your email address. It may take up to 15 minutes before you receive it.
The file will be sent to your Kindle account. It may takes up to 15 minutes before you received it.
Please note: you need to verify every book you want to send to your Kindle. Check your mailbox for the verification email from Amazon Kindle.
Please note: you need to verify every book you want to send to your Kindle. Check your mailbox for the verification email from Amazon Kindle.
You may be interested in Powered by Rec2Me
Most frequently terms
gram^{1054}
per cent^{820}
grams^{781}
acid^{774}
sample^{773}
ans^{565}
percentage^{470}
calculate^{320}
molar^{295}
titrated^{290}
calculations^{288}
ions^{282}
precipitate^{274}
titration^{272}
excess^{263}
hydrogen^{245}
naoh^{235}
hc1^{222}
liter^{213}
normality^{212}
analytical chemistry^{210}
dissolved^{200}
milliliters^{194}
oxidation^{192}
ion^{187}
ignited^{178}
sodium^{177}
chloride^{175}
weighing^{170}
precipitated^{169}
weights^{167}
weighed^{158}
sulfate^{152}
indicator^{151}
formula^{151}
kmno^{148}
substance^{130}
potassium^{128}
mole^{122}
iodine^{121}
copper^{116}
molecular^{107}
ferrous^{105}
so4^{105}
equilibrium^{104}
carbon^{103}
equations^{102}
milliequivalents^{100}
gram sample^{97}
ammonium^{91}
na2^{90}
electrode^{89}
zinc^{89}
oxidized^{88}
calcium^{88}
milliliter^{86}
cao^{86}
solubility^{85}
neutralization^{84}
0 comments
You can write a book review and share your experiences. Other readers will always be interested in your opinion of the books you've read. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them.
1

2

> W s^ u< ^ ^D CD DO OU_1 58434 >m XW > OSMANIA UNIVERSITY LIBRARY, Accession No. Call Author ^fitoJk*. JT .*:> S % 'S'^^jfttf&jLJ stufned last 2<? 8 7 marked below. on cy before the date % INTERNATIONAL CHEMICAL SERIES Louis P. HAMMETT, PH.D., * Consulting Editor * * CALCULATIONS OF ANALYTICAL CHEMISTRY The quality of the materials used in the manufacture of this book is governed by continued postwar shortages. A SELECTION OF TITLES FROM THE INTERNATIONAL CHEMICAL SERIES Louis P. HAMMETT, PH.D., A msden Mahin Prcmedical for Chemistry Students Physical Arthur Lecture Demonstrations Chemistry Aithur and Smith in General Introduction to Quantitative Analysis Quantitative Analysis Mittard Physical Chemistry for Colleges Moore Semixuicro Qualitative Analysis Booth and Damerell Quantitative Analysis. Briscoe Structure and Pioperties of Matter Coghill and Stvrtevant An Introduction to the Preparation and Identification of Organic Compounds Crist A Consulting Editor Laboratory Course in General Chem History of Chemistry Morton in Organic Technique Laboratory Chemistry The Chemistry of Ileterocyclic Compounds Norris Experimental Organic Chemistry The Principles of Organic Chemistry Parr Analysis of Gas, Fuel, Water, and Lubricants istry Daniel* Mathematical Preparation Chemistry for Physical Daniels, Mathews, and Williams Experimental Physical Chemistry Desha Experimental and Theoretical Electrochemistry Eastman and Rollefson Physical Chemistry Optical Methods of Chemical Analysis of Rieman, Neuss, and Naiman Quantitative Analysis The Elements Dole The Theoiy Elementary Qualitative Analysis Theoretical Qualitative Analysis Robinson and Gilliland Organic Chemistry Glasatone, Laidler, Reedy and Eyring Rate Processes of Fractional Distillation Schoch, Felting, and Watt General Chemistry and Biffen Commercial Methods Snell of Analysis Steiner Introduction to Chemical Thermo dynamics Still well Crystal Che; mistry Griffin Technical Methods of Analysis Dunn, and McCullough Experiments in General Chemistry Stone, Hamilton and Simpson Calculations of Quantitative Chemical Analysis Physical Organic Chemistry Solutions of Electrolytes Henderson and Fernelius Inorganic Preparations Huntress Problems Thomas Colloid Chemistry Hammett in Organic Chemistry Leighou Chemistry of Engineering Materials Long and Anderson Chemical Calculations Timm General Chemistry An Introduction to Chemistry Watt Laboratory Experiments in General Chemistry and Qualitative Analysis William* and Homerberg Principles of Metallography Woodman Food Analysis CALCULATIONS OF ANALYTICAL CHEMISTRY FORMERLY PUBLISHED UNDER THE TITLE Calculations of Quantitative Chemical Analysis by LEICESTER F. HAMILTON, S.B. Professor of Analytical Chemistry Massachusetts Institute of Technology and STEPHEN G. SIMPSON, PH.D. Associate Professor of Analytical Chemistry Massachusetts Institute of Technology Fourth Edition Third Impression McGRAWHILL BOOK COMPANY, NEW YORK AND LONDON 1947 ING. CALCULATIONS OF ANALYTICAL CHEMISTRY Formerly published under the title Calculations of Quantitative Chemical Analysis COPYRIGHT, 1922, 1927, 1939, 1947, BY THE McGRAwHiLL BOOK COMPANY, INC. PRINTED IN THE UNITED STATES OF AMERICA All rights reserved. This book, or parts thereof, may not be reproduced w, any form without permission of the publishers. PREFACE The book has been clfanged from Calculations of Chemical Quantitative Analysis to Calculations of Analytical Chemistry because the subject matter has been expanded to cover the stoichiometry of both qualitative and quantitative analysis. In order to include calculations usually covered in courses in title of this qualitative analysis, some rearrangements of material have been made, new sections have been added, and chapters dealing with equilibrium constants and with the more elementary aspects of Alanalytical .calculations have been considerably expanded. together, the number of sections has been increased from 78 to 114 and the number of problems from 766 to 1,032. The greater part of the book of quantitative analysis. amperometric titrations is still devoted to the calculations Short chapters on conductometric and and a section on calibration of weights have been added, and many other changes and additions have been made at various points in the text. A section reviewing the use of logarithms has been inserted, and a table of molecular weights covering most of the problems in the book is included in the Appendix. It is felt that every phase of general analytical chemistry is adequately covered by problems, both with and without answers, and that most of the problems require reasoning on the part of the student and are not solved by simple substitution in a formula. LEICESTER F. HAMILTON STEPHEN G. SIMPSON CAMBRIDGE, MASS., February, 1947. CONTENTS v PREFACE PART CHAPTER GENERAL ANALYSIS I. MATHEMATICAL, OPERATIONS I. 1. Factors Influencing the Reliability of Analytical Results 2. Deviation Measures as a Means of Expressing Reliability 3. Significant Figures as a 4. Rules Governing the Use of Significant Figures in Chemical Means 1 ... . Com3 putations 5. Conventions Regarding the Solution of Numerical Problems Problems 118 7. Rules Governing the Use of Logarithms Method of Using Logarithm Tables 8. Use 6. . of the Slide .... .... 13 . Rule 14 15 CHAPTER 10. II. 11. lonization of Acids, Bases, 13. CHEMICAL, EQUATIONS Purpose of Chemical Equations Types of Chemical Equations 12. Ionic 14. Ionic 16 16 and Salts 17 Equations Not Involving Oxidation Oxidation 18 Number 20 Oxidation and Reduction Equations 21 24 Problems 2543 CHAPTER 15. 16. 17. III. 6 7 9 Problems 1924 9. 2 3 of Expressing Reliability CALCULATIONS BASED ON FORMULAS AND EQUATIONS Mathematical Significance Formula Weights of a Chemical P^ormula . 28 28 Mathematical Significance of a Chemical Equation Problems 4470 29 32 CHAPTER IV. CONCENTRATION OF SOLUTIONS ^ 18. Methods 19. Grains per Unit Volume of Expressing Concentration 36 3f> vii CONTENTS 20. Percentage Composition. . . . 36 . 21. Specific Gravity 36 22. Volume Ratios 37 23. Molar and Formal Solutions 37 Weight and Normal Solution 24. Equivalent 25. Simple 38 Calculations Involving Equivalents, Milliequivalents, and Normality Problems 7186 39 43 CHAPTER 26. Law of V. P]quiLiBRiUM CONSTANTS Mass Action 46 27. Ion Product Constant of Water 47 28. pll Value 48 Problems 8794 29. lonization 30. Common 49 Constant 50 Ion Effect. Buffered Solution 52 31. lonization of Polybasic Acids 32. Activity and Activity 53 Coefficients 33. Dissociation Constants of 55 Complex Ions 56 Problems 95128 34. Solubility 57 Product 60' 35. Fractional Precipitation 62 Problems 129159 64 36. Application of Buffered Solutions in Analytical 37. Control of Acidity in 38. Separations Hydrogen by Means of 67 Chemistry 68 Sulficie Precipitations Complexion Formation 69 Problems 160178 71 Ratio 73 39. Distribution Problems 179185 74 CHAPTER VI. OXIDATION POTENTIALS 40. Relation of the Electric Current to Oxidationreduction ("Redox") Reactions 76 41. Specific Electrode Potentials 42. Rules for Writing Equations 76 for Halfcell Reactions 43. Oxidationreduction Equations in 44. Relation Terms of Halfcell Reactions. 79 . . between Electrode Potential and Concentration 45. Calculation of the Takes Place 80 83 Extent to Which an Oxidationreduction Reaction 86 CONTENTS 46. Calculation of Equilibrium ix Constant from Electrode Potentials . . 88 Problems 186218 PART 47. Sensitiveness of the GRAVIMETRIC ANALYSIS II. CHAPTER VII. THE CHEMICAL, BALANCE 93 Chemical Balance 94 48. Method 49. Conversion of Weight in Air to Weight in Vacuo of 87 Swings 95 97 50. Calibration of Weights Problems 219240 CHAPTER 51 . 52. Law VIII. 99 CALCULATIONS OF GRAVIMETRIC ANALYSIS of Definite Proportions Applied to Calculations of Gravimetric Analysis 102 Chemical Factors 104 105 53. Calculation of Percentages 106 Problems 241266 54. Calculation of 109 Atomic Weights 109 Problems 267275 55. Calculations Involving 110 a Factor Weight Sample 112 Problems 276284 56. Calculation of the Volume of a Reagent Required for a Given Reaction 1 13 116 Problems 285304 57. Indirect Analyses 118 Problems 305331 121 CHAPTER IX. 58. ELECTROLYTIC METHODS Decomposition Potential 125 by Electrolysis Problems 332359 126 59. Analysis 130 CHAPTER X. CALCULATIONS FROM REPORTED PERCENTAGES 60. Calculations Involving the Elimination or Introduction of a Con135 stituent 61. Cases Where Simultaneous Volatilization and Oxidation or Reduction Occur 136 Problems 360380 138 62. Calculation of Molecular Formulas from Chemical Analyses 63. Calculation of Empirical Formula of a Mineral .... 141 144 CONTENTS x 64. Calculation of Formulas of Minerals Exhibiting Isomorphic Re placement 145 Problems 381424 147 PART III. VOLUMETRIC ANALYSIS CHAPTER XI. CALIBRATION OF MEASURING INSTRUMENTS 65. 66. Measuring Instruments in Volumetric Analysis Calculation of True Volume 153 Problems 425435 155 153 CHAPTER XII. NEUTRALIZATION METHODS (AOIDIMETRY AND ALKALIMETRY) 67. Divisions of Volumetric Analysis 68. 69. 158 Equivalent Weights Applied to Neutralization Methods 158 Problems 436456 160 Normality of a Solution Problems 457466 Made by Mixing Similar Components . . 161 162 70. Volumenormalitymilliequivalent Relationship 163 71. Adjusting Solution to a Desired Normality 163 Problems 467480 72. 73. 164 Volume and Normality Relationships between Reacting Solutions . Problems 481492 166 Determination of the Normality of a Solution 168 74. Conversion of Data to Milliequivalents 169 Problems 493514 170 75. Calculation of Percentage Purity 76. Volumetric Indirect from Titratioii Values Methods 173 176 Problems 515532 77. 166 177 Problems in Which the Volume of Titrating Solution Bears a Given Relation to the Percentage 180 Problems 533545 181 78. Determination of the Proportion in in a Which Components Are Present Pure Mixture 79. Analysis of 183 Fuming Sulfuric Acid Problems 546564 184 186 80. Indicators 188 81. Equivalence Point 189 82. Determination of 83. Calculation of the Value at the Equivalence Point Degree of Hydrolysis of a Salt pH 191 196 CONTENTS xi Problems 565591 84. Titration of 196 Sodium Carbonate 85. Analyses Involving the 86. Relation of Titration Use of Two 199 Indicators Volumes to Composition 199 of Sample Problems 592606 205 Phosphate Mixtures 87. Analysis of 204 Problems 607612 207 209 CHAPTER XIII. OXIDATION AND REDUCTION ("REDOX") METHODS (OXIDIMETRY AND REDUCTIMETRY) 88. Fundamental 89. Equivalent 211 Principles Weights 91. 92. and Reducing Agents and Reduction Processes of Oxidizing 90. Calculations of Oxidation 211 216 Problems 613630 217 Permanganate Process Bichromate Process 225 219 93. Ceric Sulfate or Cerate Process Problems 631682 226 227 94. lodimetric Process 234 Problems 683712 238 CHAPTER XIV. PRECIPITATION METHODS (PRECIPITIMETRY) Weights in Precipitation Methods Problems 713734 95. Equivalent 243 245 CHAPTER XV. COMPLEXION FORMATION METHODS (COMPLEXIMETRY) 96. Equivalent Weights Problems 735751 PART in IV. Complexion Methods 249 252 ELECTROMETRIC METHODS CHAPTER XVI. POTENTIOMETRIC TITRATIONS 97. Potentiometric Acidimetric Titrations 255 98. Simple Potentiometric Titration Apparatus 99. Quinhy drone Electrode 256 100. Glass Electrode 101. Potentiometric 257 258 "Redox" Titrations 102. Potentiometric Precipitation Titrations Problems 752775 259 261 261 CONTENTS xii CHAPTER XVII. 103. CONDUCTOMETRIC TITRATIONS 266 Conductance 266 104. Mobility of Ions 105. Conductometric Acidimetric Titrations 106. Conductometric Precipitation Titrations 107. 267 270 Conductometric Titration Apparatus 271 Problems 776789 272 CHAPTER XVIII. AMPEROMETRIC TITRATIONS an Amperometric Titration Problem 790 108. Principle of PART V. 275 278 GAS ANALYSIS CHAPTER XIX. CALCULATIONS OF GAS ANALYSES 109. Fundamental Laws 110. Gasvolumetric 111. Correction for 279 Methods Water Vapor 112. Calculations of Gas volumetric Analyses Problems 791813 113. Absorption 114. 281 282 283 284 Methods 286 Combustion Methods 287 Problems 814836 292 PART VI. PART COMMON ANALYTICAL DETERMINATIONS vii. "PROBLEMS ON SPECIFIC GROUPS AND DETERMINATIONS A. Qualitative Analysis 309 Silver Group Hydrogen Sulfide Group Ammonium Sulfide 310 311 Group Alkaline Earth and Alkali Groups 312 Anion Groups 313 B. Quantitative Analysis Water 313 Sodium, Potassium 313 Ammonium, Ammonia, Nitrogen 314 Silver, Mercury, Gold, Platinum Halogens, Cyanide, Thiocyanate, Halogen acids 315 315 CONTENTS xiii Barium, Strontium, Calciuirif Magnesium 163 Limestone, Lime, Cement 317 Iron, 317 Aluminum, Titanium Cerium, Thorium, Zirconium, Uranium, Beryllium, Bismuth, Boron Copper, Lead, Zinc, Cadmium, Brass Silicon, . 320 321 323 Tin, Antimony, Arsenic, Bronze Carbon, Carbon Dioxide, . Tungsten, Molybdenum 324 Chromium, Vanadium 325 Manganese 327 Cobalt, Nickel 329 Phosphorus Sulfur, Selenium 329 General and Miscellaneous Analyses 335 APPENDIX 365 INDEX 379 331 PART I GENERAL ANALYSIS CHAPTER I MATHEMATICAL OPERATIONS Factors Influencing the Reliability of Analytical Results. Analytical chemistry is ordinarily divided into qualitative analysis 1. and quantitative analysis. A compound or mixture is analyzed by qualitative analysis to determine what constituents or components are present; a compound or mixture is analyzed by quantitative analysis to determine the proportions in which the constituents or components are present. Calculations in qualitative analysis are limited mostly to those pertaining to equilibrium constants and simple weight and volume Calculations in quantitative analysis are more exand are based upon numerical data obtained by careful measurement of masses and volumes of chemical substances. From the numerical data obtained from these measurements the desired relationships. tensive proportions can be calculated. It is found; however, that duplicate analyses of the same substance, even when made by experienced analysts following identical methods, rarely give numerical values which are exactly the same. Furthermore, the discrepancy between results is found to depend upon the method used, and an analytical result obtained by one procedure may differ appreciably from a similar result obtained by an entirely different procedure. The most important factors which thus influence the precision of analytical results are the following: (1) the manipulative skill of the analyst; (2) the experimental errors of the procedure itself, such as the slight solubility of substances assumed to be insoluble or the contamination of precipitates assumed to be pure; (3) the accuracy of the measuring instruments used; and (4) fluctuations of temperature and barometric pressure. In order, therefore, that a numerical result obtained from chemical meas urements may be of scientific or technical value, the observer should have at least a general idea of its reliability. l 2 CALCULATIONS OF ANALYTICAL CHEMISTRY In this connection, there should be kept in mind a distinction between accuracy and reliability. The accuracy of a numerical result is the degree of agreement between it and the true value; the reliability or precision of agreement between it a numerical result is the degree of and other values obtained under substan the same conditions. Thus, suppose duplicate determinations of the percentage of copper in an ore gave 52.30 per cent and 52.16 per cent, and suppose the actual percentage was 52.32. It can be assumed that the analyst would report the mean or tially average of the two values obtained, namely 52.23 per cent. This differs from the true value by 0.09 per cent, which represents the absolute error of the analysis. Expressed in parts per thousand, the error would be 0.09/52.32 1,000 =1.7 parts per thousand. This is known as the relative error of the analysis. X it Since in most chemical analyses the true value is not known, follows that the accuracy of a given determination is seldom known. We can speak only of the precision or reliability of the numerical results obtained. 2. Deviation Measures as a Means of Expressing Reliability. The numerical measure a result is known as measure which is of in careful and chemical work is particular importance physical the deviation measure. Suppose, for example, repeated independent readings of a buret gave the following values: its precision measure. of the reliability of type of precision A (a) 43.74 (/) 43.75 (6) 43.76 (p) 43.75 (c) 43.76 (h) 43.76 (d) 43.75 (i) 43.73 (e) 43.77 The most probable value tor this reading obviously the mean, 43.753, which is obtained by dividing the sum of the readings by the number of readings taken. The deviation of each measurement from this mean, regardless of sign, is is shown in the following: (a) 0.013 (/) 0,003 (6) 0.007 (g) 0.003 (c) 0.007 (h) 0.007 (d) 0.003 (i) 0.023 (e) 0.017 MATHEMATICAL OPERATIONS The mean 3 deviation, or average of these nine values, is 0.0092 represents the and amount by which an average single independent from the most probable value; it is therefore a reading differs measure of the reliability of a single observation. It is more important, however, to know the reliability of the mean than that of a single observation. It can be shown that the reliability of a mean or average value is numerically equal to the average deviation of a single observation divided by the square In the above, the root of the number of observations taken. = 0.0031, and the value average deviation of the mean is 0.0092/V9 0.0031. for the reading may be expressed as 43.753 (It is customary to use only two measmeasurements are involved in a comures.) putation, it is possible to calculate from the deviation measure of each measurement the deviation measure or precision measure of the final result and thus obtain a numerical measure of the probable reliability of that result. For methods of such calculation the student is referred to Goodwin's Precision of Measure When significant figures in all deviation several such ments. Figures as a Means of Expressing Reliability. In most chemical analyses relatively few independent readings or 3. Significant determinations are made, so that numerical precision measures In such cases the reliability or precision of a are not often used. numerical value is best indicated by the number of significant figures used in expressing that value. It is true that this method of expression gives only a result, an approximate idea but the importance of the reliability of of the retention of the proper number of significant figures in analytical data cannot be overemphasized. numerical result expressed by fewer or more significant figures A than are warranted by the various factors involved may give to an observer an impression nearly as erroneous as would be given by a result which is inaccurate. 4. Rules Governing the Use of Significant Figures in Chemical Computations. The following definitions and rules are suggested by those given in Goodwin's Precision of Measurements: A number is an expression of quantity. A figure, is any one of the characters 0, 1,2, 3, 4, 5, 6, which, alone or in combination, serve to express numbers. significant figure is a digit which denotes the amount of the 7, 8, 9, A or digit, CALCULATIONS OF ANALYTICAL CHEMISTRY 4 quantity in the place in which it stands. In the case of the number 243, the figures signify that there are two hundreds, four and three units and are therefore all significant. The charIt may be used as a significant is used in two ways. acter tens, be used merely to locate the decimal point. It is a significant figure when it indicates that the quantity in the place in which it stands is known to be nearer zero than to any other value. Thus, the weight of a crucible may be found to be 10.603 grams, in which case all five figures, including the zeros, are significant. If the weight in grams of the crucible were found to be 10.610, meaning that the weight as measured was nearer 10.610 than 10.609 or 10.611, both zeros would be significant. By analysis, the weight of the ash of a quantitative filter paper Here the zeros are not significant is found to be 0.00003 gram. figure, or it may but merely serve to show that the figure 3 belongs in the fifth place to the right of the decimal point. Any other characters except digits would serve the purpose as well. The same is true of the value 356,000 inches, when signifying the distance between two given points as measured by instruments which are accurate The zeros are not significant. In order to avoid confusion, this value should be written 3.56 X 10 5 inches. If the distance has been measured to the nearest 100 inches, it to three figures only. should be written 3.560 X 10 6 inches. Rule I. Retain as many significant figures in a result and in data in general as will give only one uncertain figure. (For very accurate work involving lengthy computations, two uncertain figures may sometimes be retained.) Thus, the value 25.34, representing the reading of an ordinary buret, contains the proper number of significant figures, for the digit 4 is obtained by esti mating an ungraduated scale division and is doubtless uncertain. Another observer would perhaps give a slightly different value for the buret reading e.g., 25.33 or 25.35. All four figures should be retained. Rule II. In rejecting superfluous and inaccurate crease by 1 the last figure retained if figures, in the following rejected figure number Thus, 16.279, the new value becomes 16.28. Rule III. In adding or subtracting a number of quantities, extend the significant figures in each term and in the sum or dif is 5 or over. in rejecting the last figure of the MATHEMATICAL OPERATIONS 5 ference only to the point corresponding to that uncertain figure occurring farthest to the left relative to the decimal point. For example, the sum of the three terms 0.0121, 25.64, and 1.05782, on the assumption that the last figure in each is uncertain, is 0.01 25.64 1.06 26.71 Here seen that the second term has uncertain figure (the 4) in the hundredths place, the following figures being unknown. Hence, it is useless to extend the digits of the other terms it is its first beyond the hundredths place even though they are given to the tenthousandths place in the first term and to the hundredthousandths place in the third term. The third digit of the third term The fallacy is increased by 1 in conformity with Rule II above. of giving more than four significant figures in the sum may be shown by substituting x for each unknown figure. Thus, 0.0121* 1.05782 26.7 Ixxx Rule IV. In multiplication or division, the percentage precision of the product or quotient cannot be greater than the percentage precision of the least precise factor entering into the computation. Hence, in computations involving multiplication or division, or both, retain as many significant figures in each factor and in the numerical result as are contained in the factor having the largest percentage deviation. In most cases, as many significant figures may be retained in each factor and in the result as are contained in the factor having the least member of significant figures. For example, the product of the three terms 0.0121, 25.64, and 1.05782, on the assumption that the last figure in each is uncertain, is 0.0121 for, if the first term in the last place, it X 25.6 X 1.06 = 0.328 assumed to have a possible variation of 1 has an actual deviation of 1 unit in every is CALCULATIONS OF ANALYTICAL CHEMISTRY 6 121 units, and its percentage deviation would be y^r X 100 Similarly, the possible percentage deviation of the second would be * QO X ,, X 100 = 100 = 0.04, 0.0009. and that The first lUOj/o^ of the third = 0.8. term term would be term, having the largest per centage deviation, therefore governs the number of significant figures which may be properly retained in the product, for the product cannot have a precision greater than 0.8 per cent. That is, the product may vary by 0.8 part in every hundred or by nearly 3 parts in every 328. The last figure in the product as expressed with three significant figures above is therefore doubtful, and the proper number of significant figures has been retained. Rule V. Computations involving a precision not greater than onefourth of l.per cent should be made with a 10inch slide rule. For greater precision, logarithm tables should be used. If the oldstyle method of multiplication or division must be resorted to, reject all superfluous figures at each stage of the operation. Rule VI. In carrying out the operations of multiplication or division by the use of logarithms, retain as many figures in the mantissa of the logarithm of each factor as are properly contained in the factors themselves under Rule IV. Thus, in the solution of the example given under Rule IV, the logarithms of the factors are expressed as follows: log 0.0121 log 25.64 log 1.05782 = = = 8.083  10  10 1.409 0.024 9.516 = log 0.328 Conventions Regarding the Solution of Numerical Problems. In the calculation of numerical results from chemical data which have been obtained under known conditions and by known methods, little difficulty should be experienced in forming an approxi5. mate estimate of the reliability of the various factors and of the In the case of numerical problems which are unaccompanied by any data to show the conditions under which the various measurements were made or the precision of the values given, the retention of the proper number of significant figures in results obtained. MATHEMATICAL OPERATIONS 7 the final computed results may be a matter of considerable judgIn such cases the rules listed above are subject to modi ment. but in any case the need for a certain amount of common sense and judgment in using them in no way detracts from their fication, value. In the solution of problems in this book, it may be assumed that the given data conform to Rule I, above. In problems containing such expressions as "a 2gram sample/ "a 25ml. pipetful," or "a tenthnormal solution/' it may be assumed that the weight of the sample, the volume of the pipet, and the normality of the . 7 solution are known to a precision at least as great as that of the other factors involved in the problem. It should also be remembered that the atomic weights of the elements are known only to a limited number of significant figures and, in the absence of further data, it may be assumed that the values ordinarily given in atomicweight tables conform to Rule I above, in that the last figure in each is doubtful. It follows, therefore, that the same attention should be paid to the precision of the atomic and molecular weights involved in computations as to that of any other data. happens that independent calculations from given data give results which disagree by only one qr two units in the last This is usually due to the fact that significant figure retained. have been at different stages of the operations rejected figures involved; but this is usually of no importance, since, when properly It often expressed, the last significant figure in the result is doubtful anyway. Analytical determinations are usually done in duplicate. In most of the problems in this book, however, data apparently covering only one determination are given. It may be assumed that such values represent determinations. mean values obtained from duplicate Problems 1. How many significant figures are implied in the value 2.20 the value 5,000.002? Ans. 2. as is Three. In the value 2.010 Seven. X X 10""9 ? In 10 5 ? Four. Calculate the molecular weight of OsCl 4 to as high a degree of precision warranted by the atomic weights involved. Ans. 332.0. CALCULATIONS OF ANALYTICAL CHEMISTRY 8 Express the velocity of light, 186,000 miles per second, in such a way as it has been measured to the nearest 100 miles per second. 3. to indicate that Ans. X 1.860 106 miles per second. Samples were sent to seven different chemists to be analyzed for percentage of protein. The values reported were 43.18, 42.96, 42.88, 43.21, 4. What is the mean value, the average deviation of a single value from the mean, and the deviation of the mean? If the correct percentage is 43.15, what is the relative error of the mean in parts per thousand? 43.01, 43.10, 43.08. Ans. 43.060,0.094,0.036. 2.1. An ore actually contains 33.79 per cent Fe 2 O 8 Duplicate determinations and 34.02 per cent, and the mean of these is reported. By how many 33.80 give do the duplicate results differ from each other? What is thousand parts per 6. the . mean value? What the absolute error? is What is the relative error in parts per thousand? Ans. 6.5. Two 6. 33.91 per cent. 0.12 per cent. 3.5. analysts, working independently, analyze a sample of steel and re port the following results: ANALYST B: ANALYST A: = Sulfur 0.042 per cent 0.041 per cent Sulfur = 0.04199 per cent 0.04101 per cent By how many parts per thousand do the check values agree Each man uses a 3.5gram sample weighed to the nearest tenth analyst B ability as Ans. justified in his report? Do in each case? of a gram. Is his figures necessarily indicate greater an analyst? No. 24 parts, 24 parts. No. It is necessary to solve the following: 7. 4  (0.0021764 X 0.0121) (1.7 X 10~ ) 0.00047) each term being uncertain in the last significant figure. Should you use arithmetic, logarithms, or a slide rule in the multiplications? What is the (1.276 final + X answer? Ans. 8. A Slide rule. 7.5 X 10~4 . value which has been found by duplicate analyses to be 0.1129 and is to be multiplied by 1.36 ml. as measured by an ordinary 0.1133, respectively, and the product is to be subtracted from the value 0.93742 which has been very accurately measured. Express the result by the proper number buret, of significant figures. Ans. 0.784. 9. If in the analysis of a given substance a variation of 0.30 per cent how many milligrams should a 10gram sample be weighed? is allowable, to 10. value How many significant figures are implied in the value 16.00 X 103 ? In the value 1.60 X 10~2 ? 16 X 103 ? In the MATHEMATICAL OPERATIONS 9 last figure in each of the three in the product as given should be many figures Express the product in such a way as to indicate the In the following multiplication the 11. factors is How uncertain. rejected as superfluous? correct number of significant figures. 2.0000 X 0.30 X 500 = 300.00 Hf (NO 8)4 to as high a degree of prethe atomic weights involved. 12. Calculate the molecular weight of warranted by on astronomy gives the polar diameter of the earth as 7,900.0 book 13. A miles. To what precision of measurement does this number imply? If the measurement had been made only to the nearest 10 miles, how should the value be expressed to indicate this fact? cision as is Assuming each term to be uncertain in the last figure given, solve the of significant figures: following and express the answer to the correct number 14. (1.586 16. ^ 29.10) + [162.22(3.221 X 1Q4)]  0.00018 2 containing 65.97 per cent Ba is given One analyst obtains 65.68, 65.79, and 65.99 for triplicate determiand reports the mean value. By how many parts per thousand does A sample of pure anhydrous BaCl for analysis. nations differ from the mean? What is the absolute error of the mean, and the relative error (parts per thousand) of the mean? 16. The percentage of carbon in a sample of steel is found to be 0.42 per The calculations involve only multiplication and division. To how cent. many decimal places would you weigh out a 1gram sample in order to duplicate each result what is the result? A sample of limonite was analyzed by 12 students at different times The values obtained for the percentage of iron during the college year. 17. were: 34.62, 34.42, 34.60, 34.48, 33.71, 34.50, 34.50, 34.22, 34.41, 35.00, 34.65, 34.44. What is the mean value, the mean deviation of a single result, and the deviation of the mean? If the correct percentage is 34.75 what is the absolute error of the 18. A mean and what is its relative error in parts per thousand? sample of material was sent to two chemists. method and reported the CHEMIST A CHEMIST 30.15 30.251 30.15 30.007 30.14 30.101 30.16 30.241 B mean value and its deviation measure. which mean value is the more reliable? Calculate in each case the conditions being equal, Each used the same results of four analyses, as follows: Other In calculations of and division where quantitative analysis involving multiplication foursignificantfigure accuracy is required, fourplace logarithms 6. Rules Governing the Use of Logarithms. CALCULATIONS OF ANALYTICAL CHEMISTRY 10 should be used; in calculations where two or threesignificantfigure accuracy is sufficient, a slide rule should be used. Grammarschool methods of multiplication and long division should not be employed. Although the theory and use of logarithms are ordinarily covered in preparatory and high schools, the following outline is given as a review of the essential points in this phase of mathematics. 1. The logarithm of a number is the exponent of the power to which some fixed number, called the base, must be raised to equal the given number. Thus, suppose ax then x is =n the logarithm of n to the base a and may be written x = loga n 2. The base common system in the and the commonly used to denote a log term log, without subscript, is arithm in this system. Hence, = = 10 10 1 10 2 103 100, log 10 log 100 1000, log 1000 10, 10~ 10~ 3 = = log 0.1 log 0.01 0.01, log 0.001 0.001, is 10, =0 log 1 1, lO^O.l, 2 of logarithms  1 =2 3 = 1 = 2 = 3 etc. It is evident that the logarithms of all and 10 and 100 and 1 and 0.1 and 1 10 will 100 will 1000 will 0.1 will 0.01 will be be be be be numbers between plus a fraction 1 plus a fraction 2 plus a fraction 1 plus a fraction 2 plus a fraction etc. 3. If a number is not an exact power of 10, its common logarithm can be expressed only approximately as a number with a continuing decimal fraction. Thus, 36= ioior log 36= 1.5503 MATHEMATICAL OPERATIONS 11 The integral part of the logarithm is called the characteristic; the decimal part is called the mantissa. In the case just cited, the characteristic is 1; the mantissa is .5563. Only the mantissa a logarithm is given in a table of logarithms (see next section) the characteristic is found by means of the next two rules. of 4. The 1 is 1 less ; characteristic of the logarithm of a number greater than digits to the left of the decimal point. than the number of For example, the characteristic of log 786.5 is 2; the characteristic of log 7.865 is 0. The tween characteristic of the logarithm of a decimal number beand 1 is negative and is equal in numerical value to the number of the place occupied 5. decimal. by the first significant figure of For example, the characteristic of log 0.007865 the is 3. 6. The mantissa teristic may a logarithm always positive; the characbe either positive or negative. For example, of log 36.55 log 0.08431 = +1 = 2 is + + .5629 = 1.5629 .9259 This last logarithm is more conventionally written as 2.9259 with the understanding that only the 2 is negative. Another common method of expressing this logarithm is 8.9259 10. 7. The mantissas of the common logarithms of numbers having the same sequence of figures are equal. For example, log 2.383 log 23.83 log 0.002383 The cologarithm of a rocal of that number. It 8. of the number from zero. = = 0.3772 1.3772 3.3772 (or 7.3772 number is  the logarithm of the recipfound by subtracting the logarithm is For example, log 7. 130 colog 7.130 = = 0.8531 0.0000  0.8531 1.1469 or 10.0000  10  10 0.8531 colog 7.130 10) = 9.1469 CALCULATIONS OF ANALYTICAL CHEMISTRY 12 9. The antilogarithm of A number that has the is A for a logarithm. For example, =0.8531 log 7.130 = antilog 0.8531 10. The logarithm logarithms of its factors. log (7.180 11. of The logarithm X of 7.130 a product is equal to the sum For example, 586.3) = = of the + log 586.3 + 2.7681 log 7.180 0.8531 3.6212 a fraction equal to the 'logarithm of the numerator minus the logarithm of the denominator; it is also equal to the logarithm of the numerator plus the cologarithm of the denominator. For example, ~ 7 180 log = log 7.180  =  0.8531 = log 7.180 = = 0.8531  is log 586.3 2.7681 2.0850 (or 8.0850  10) or 7 ISO log The use + colog 586.3 + 3.2319 (or 7.2319 2.0850 (or 8.0850  of cologarithms is particularly  10) 10) advantageous when the multiplication and division of several factors are involved in the same mathematical process. This is shown in the example at the end of this section. The logarithm a quantity is equal to the logarithm of the quantity multiplied by the exponent of the power. For example, 3 = 3 X log 71.80 log 71. 80 = 3 X 1.8531 12. of any power 13. The logarithm logarithm of the of of 5.5593 any root quantity divided of a quantity is equal to the by the index of the root. For example, log ^002 = Y2 X log 5.002 = 2 X 0.6992 = 0.3496 y 7. Method of MATHEMATICAL OPERATIONS 13 The precision of or Using Logarithm Tables. dinary chemical analytical work is seldom great enough to permit the retention of more than four significant figures in the numerical data obtained and in the calculations made from such data. Hence a fourplace logarithm table such as is given in the back of this book is entirely adequate. use the logarithm table in finding a mantissa proceed as follows: First find the first two digits of the number in the column headed " natural numbers," then go to the right until the column To reached which has the third digit of the number as a heading. To the number thus found add the number which is in the same horizontal line at the righthand side of the table and in the column is of proportional parts headed by the fourth Thus the number the number. + significant figure of representing the mantissa of log = 3744, and the logarithm is 2.3744. Antilogarithms may be looked up in the antilogarithm table in the same way. Only the mantissa is used in looking up the num 236.8 is 3729 15 ber; the characteristic is used merely to locate the decimal point. Thus the sequence of digits in the number having a logarithm of 1.8815 is 7603 +9= number 7612, and the actual is 76.12 as de termined by the given characteristic of the logarithm. In actual calculations from analytical data the essential purpose of the characteristic in a logarithm is to locate the position of the decimal point in the final numerical value obtained. Since in most cases a very rough mental calculation is all that is needed to establish the position of the decimal point, the use of characteristics can be dispensed with. The retention of characteristics is, however, helpful in serving as a check on the other method. Calculations of quantitative chemical analysis in which logarithms are of value seldom involve operations other than those of multiplication and division. v ^ * LxAMPLE.Calculate i i u i 9.827 xi SOLUTION: Method A X 50.62 by logarithms: 0tQ05164 x 136>6Q (without using cologarithms) = = Sum = log 9.827 log 50.62 0.9924 1.7044 2.6968  CALCULATIONS OF ANALYTICAL CHEMISTRY 14 = = Sum = = log numerator = log denominator Difference = log 0.005164 log 136.59 antilog Method B = 3.7129 or 7.7129 2.1354 2.1354 T.8483 9.8483 2.6968 or 12.6968 1.8483 9.8483 2.8485 2.8485 705.5.  10  10  10 10 10 Ans. (using cologarithms) = = log 50.62 = colog 0.005164 colog 136.59 = Sum = = antilog log 9.827 0.9924 or 0.9924 1.7044 1.7044 2.2871 2.2871 3.8646 7.8646 2.8485 12.6485 705.5. 10 Ans. As previously mentioned, much time characteristics in the solution of the is saved by omitting all above problem and merely writing down the mantissas of each logarithm or cologarithm. The location of the decimal point is then determined by a simple mental calculation on the original expression. Thus, inspection shows that the two factors in the numerator of the above expression give a result approximating 500 and that the factors in the denominator give a result approximating 0.7. The answer must therefore be in the neighborhood of 700, which establishes the position of the decimal point. 8. Use of the Slide Rule. The slide rule is essentially a logarithm table, mechanically applied. On the scales used for multiplication and division the numbers are stamped on the rule in positions proportionate to their logarithms. Multiplication by means of the rule is merely a mechanical addition of two logarithms; divi a mechanical subtraction of two logarithms. Manuals covthe proper use of a slide rule are readily obtainable and are ering usually provided by the manufacturer of the rule. sion is The student of quantitative analysis should be proficient in the use of a slide rule, particularly in the processes of multiplication and division. The slide rule saves a great deal of time in making minor calculations and tions made by is logarithms. an excellent means of checking calculathe Although precision of the ordinary MATHEMATICAL OPERATIONS 10inch slide rule 15 limited to three significant figures, is it is sug gested that sliderule accuracy be permitted in solving quiz problems and home problems, even though the data given may theo The purpose of more to make sure that the methods of calculation retically require foursignificantfigure accuracy. the problems are understood than to give practice in fundamental mathematical is operations. Most laboratory figure accuracy, calculations, however, require foursignificant and fourplace logarithms are necessary. Problems 19. Using fourplace logarithms determine the following: log 0.0009289, (6) (c) colog 52.61, (d) colog 0.06003, log 4.1733, (g) aritilog 7.2068 Ans. (a) 2.5884, 20. 0.0001490, (g) + 0.00002591, (6) of 0.09508. Check these to three Ans. (a) 0.005864, (6) 64.53,  10, log 387.6, (/) anti (c) 2.2789 or 8.2789  10, 0.001610. calculate Using fourplace logarithms 0.05811 (a) antilog 2.4474, 10. 4.9679 or 6.9679 (b) (d) 1.2216, (e) 280.2, (/)  (e) the following: (a) 226.3 X fourth power of 0.3382, (d) cube root (c) significant figures with 0.0009005, (c) a slide rule. 0.01308, (d) 0.4564. Locate 21. Using fourplace logarithms find the value of the following. the position of the decimal point by mental arithmetic and also by the proper use of characteristics. Also check the answer to three significant figures with a slide rule. 0.0046191 X 51.42 Ans. 22. (6) 287.7 0.03061. Using fourplace logarithms determine the following: colog 0.9566, (d) colog 718.1,  10. antilog 6.0088 log 0.005903, log 2.0696, (g) 23. X 0.84428 (c) calculate Using fourplace logarithms (e) (a) log 67.84, (/) anti antilog 3.6482, the 'following: (a) 33.81 X 362.4, (c) cube of 0.09279, (d) square root of 0.0009915, (6) 0.5546. Check these to three significant figures with a slide rule. 0.1869 5 Using fourplace logarithms find the numerical value of the following expression. Locate the position of the decimal point by the proper use of Also check the answer to characteristics and check by mental arithmetic. 24. three significant figures by means of a slide rule. 5987.9 X 0.006602 1.864 X 0.4617 X 1053.3 CHAPTER II CHEMICAL EQUATIONS Purpose of Chemical Equations. When the nature and composition of the initial and final products of a chemical reaction are known, the facts can be symbolized in the form of a chemical 9. When properly written, the equation indicates (1) the nature of the atoms and the composition of the molecules taking part in the reaction, (2) the relative number of atoms and moleequation. cules of the substances taking part in the reaction, (3) the pro by weight of the interacting and resulting substances, the proportions by volume of all gases involved. These four principles applied to reactions which go to completion serve as the foundation of quantitative chemical analysis. Before the portions and (4) calculation of a chemical analysis can be made, it is important to understand the chemistry involved and to be able to express the reactions in the form of balanced equations. 10. Types of Chemical Equations. The determination of the nature of the products formed by a given reaction involves a knowledge of general chemistry which, it is assumed, has already been acquired from previous study, but the ability to write and balance equations correctly and quickly is acquired only by considerable practice. The following discussion is given to help the student attain this proficiency, especially in regard to equations involving oxidation and reduction, which usually give the most trouble to the beginner. With equations expressing the reactions of (1) combination, (2) decomposition, and (3) metathesis, it is seldom that much about equality between the atoms and molecules of the reacting substances and those of the products, for little more is involved than purely mechanical adjustment of the terms and an elementary knowledge of valence. As examples of the above types of chemical change in the order difficulty is experienced in bringing given, the following equations may be 1ft cited : CHEMICAL EQUATIONS 17 (1) (2) 2HgO > 2Hg + O2 (3) FeCl 3 + 3NH OH * Fe(OH) + 3NH C1 4 4 3 Equations expressing reactions of oxidation and reduction, although usually somewhat more complicated, offer little additional difficulty, provided that the principles underlying these types of chemical change are thoroughly understood. The above equations are molecular equations. For reactions taking place in aqueous solution (such as the third case above) equations are usually better written in the ionic form. To do so correctly requires a knowledge of the relative degrees of ionization of solutes and the correct application of a few simple rules. 11. Ionization of Acids, Bases, and Salts. Although the theory of ionization should be familiar to the student from his previous study of general chemistry, the following facts should be kept in mind because they are particularly important in connection with writing equations: " " Strong acids include such familiar acids as HC1, H S0 HN0 2 4, 3, HC103 HBr0 3 HIO 3 HC1O 4 and , , , , HBr, HI, HMnO 4. These acids in solution are 100 per cent ionized, although at ordinary concentrations interionic effects may give conductivities corre sponding to an apparent degree of ionization a little less than 100 per cent. In ionic equations (see below) strong acids are written in the form of ions. "Strong" bases include NaOH, KOH, Ba(OH) 2 Sr(OH) 2 and Ca(OH) 2 These bases in solution are 100 per cent ionized and , , . in ionic equations are written as ions. with very few exceptions, are completely dissociated into simple ions in solution, and in ionic equations are written as ions. Two common exceptions are lead acetate and mercuric chloride. Many acids and bases are ionized in solution to only a slight degree at ordinary concentrations. Table IX in the Appendix lists most of such acids and bases ordinarily encountered in anaSalts, lytical chemistry, and the student should familiarize himself with the names of these substances and have at least a general idea ol the magnitude of the degree of ionization in the case of the common mon ones. Certain acids contain more than one hydrogen replaceable bj CALCULATIONS OF ANALYTICAL CHEMISTRY 18 a metal (polybasic It will acids). and the degree in steps, be noted that these acids ionize of ionization of the first hydrogen is in variably greater than that of the others. Phosphoric acid, for example, is about 30 per cent ionized in tenthmolar solution to ~~ = + and 2 PO 4 ions give ions, but the concentration of HPO 4 S ions is is much less, and that of PQ 4 small. Sulfuric acid is very ~ 100 per cent ionized into H+ and HSO 4 ions, but the bisulfate + ions and ion is only moderately ionized further to H H give H SCV* ions. Equations Not Involving Oxidation. Most of the reactions of analytical chemistry are reactions between ions in solution. For this reason, although the molecular type of equation 12. Ionic serviceable as a basis for quantitative analytical calculations, the socalled ionic equation is usually easier to write and is gen is erally better. In writing ionic equations, the following basic rules should be observed : Include in the equation only those constituents actually taking part in the chemical reaction. EXAMPLE I. The addition of a solution of sodium hydroxide 1. to a solution of ferric nitrate results in a precipitation of ferric hydroxide. The ionic equation is as follows: Fe+++ + 30H > Fe(OH) * 3 The sodium ions from the sodium hydroxide and the nitrate ions from the ferric nitrate do not enter into the reaction and hence are not represented in the equation. In cases where a reac,tant or product exists in equilibrium with constituent ions, express in the equation that form present in 2. its greatest amount. weak acids, weak bases, and the slightly ionized should salts be written in the molecular form. Substances of this most often encountered in analytical chemistry are the type 2 O, 2 3 O2 4 OH, 2 S, 2 C0 3 2 following: HF, It follows that H Pb(C H HC H HgCl 2 The last Appendix). 2 3 2) 2 , , NH H H H P0 H C O and H SO , , HNO , 3 4 2 2 4 2 3 (see Table IX, three of these are borderline cases since they , , * It is desirable to underline formulas of precipitates. The use of downwardpointing arrows is equally satisfactory. If desired, formulas of gases may be overlined or denoted by upwardpointing arrows. CHEMICAL EQUATIONS 19 are ionized to a moderate degree to give hydrogen ions and H PO4~, HC2 O4~, lead 2 acetate HSO ~ and ions, respectively. 3 and mercuric chloride may be dissociated The salts somewhat into " [e.gr., Pb^HaC^)" in the former case] but are relato give the metal ions. They are therefore ionized tively usually written in the molecular form. EXAMPLE II. The addition of an aqueous solution of ammo1 complex ions little nium hydroxide to a solution of ferric nitrate results in a precipitation of ferric hydroxide. The ionic equation is as follows: Fe+++ + 3NH OH > Fc(OH) + 3NH + 4 3 4 ammonium hydroxide is ionized into ammonium ions and hydroxyl ions, the ionization is comparatively slight and only the undissociated ammonium hydroxide molecules In this case, although are expressed in the equation. 1 EXAMPLE III. The addition of a solution of hydrogen sulfide to an acid solution of copper sulfate gives a precipitate of copper sulfide: Cu++ + H S > CuS + 2H ' 2 The fact that the original solution is acid docs not require that hydrogen ions be on the lefthand side of the equation. The equa tion merely indicates that the solution becomes more acid. EXAMPLE IV. When a solution containing lead nitrate is treated with sulfuric acid, a white precipitate of lead sulfate is obtained. This precipitate dissolves in a solution of ammonium acetate, and the addition of a solution of potassium chromate then causes a The yellow precipitate to appear. ionic equations for these re actions are + HS0 ~ > PbS0 + H+ PbSO + 2C H   Pb(C H O + S0 Pb(C H O + CrO * PbCr0 + 2C H O Pb++ 4 4 As a matter tration of tion of 3 2 3 2 1 ammonia 2 3 4 4 2) 2 4 2 2 3 not entirely certain that an appreciable concenThe equilibrium existing in an aqueous solumore generally and better expressed as follows: of fact, it is 2 2) 2 is NH OH exists at 4 4 all. NH + H O 3 2 <=> [NH OH 4 (?)] ^ NH + 4 Here again, the concentration of OH"~ is relatively low, and the equation for the above reaction can therefore be written 3NH + 3H O > Fe(OH) + 3NH + 3 2 8 4 CALCULATIONS OF ANALYTICAL CHEMISTRY 20 EXAMPLE V. and footnote above) silver is AgCl + 2NH  Ag(NH 4 3 ammino ion, its Example II : + 2NH OH > Ag(NH dissociated into an aqueous solution written as follows (see AgCl or The Silver chloride dissolves in The equation of ammonia. like 3) 2 + 3) 2 + + 01 + H + 01 most complex ions, constituents: 2 Ag(NH 3 ) + 2 is > only very slightly Ag+ + 2NH 3. EXAMPLE VT. A nitric acid solution of ammonium molybdate ((NH 4 ) 2 MoO 4 added to a solution of phosphoric acid results in ] the precipitation of ammonium phosphomolybdate. 12MoOr + H P04 + 3NH 3 4 + + 21H+ > (NH 4 ) 3 P0 4 .12Mo03 + 12H2O Note here that sponding 24 for every 12 molybdate ions only 3 of the corre ammonium The ions present enter into the reaction. nitrate ions of course take no part in the reaction. Number. Although the term "valence" usually refers to the degree of combining power of an atom or radical, it is likely to be applied somewhat differently in the various branches of chemistry. For this reason, in inorganic chemistry the term "oxidation number" is to be preferred in expressing state of 13. Oxidation oxidation. assumed that the student already familiar with the general aspects of the periodic table and with the combining power of the elements he has thus far studied. It will be recalled that (1) the oxidation number of all free elements is zero, (2) the oxidaIt is number is +1 (except in the case of the relatively rare metallic hydrides), (3) the oxidation tion number of of hydrogen in its compounds is sodium and potassium in their compounds is +1, and number of oxygen in its compounds is 2 (with (4) the oxidation few exceptions). sum of the oxidation numbers of the elements of a given compound is zero, the oxidation number of any element in a compound can usually be readily calculated from those of the other elements making up the compound. Thus, the oxidation Since the algebraic number of Cl in HC1O 3 is +5, since +1 +5+ [3 X (2)]  0. In number of the C10 3 radical is 1, since it is combined with the +1 hydrogen. The oxidation number of S 2 in this case the oxidation CHEMICAL EQUATIONS is sulfur +12 since Na2 = +2 and atom The 7 oxygen atoms = therefore has an oxidation number of +6. oxidation number of an ion 1, that of the sulfate ion (SC^) S phate ion (P0 4 ) is 3. A few the sum Each 14. the same as the charge of the nitrate ion (N0 3 ~) is Thus, the oxidation number bears. 21 is 2, and that is of the phos HCNO may give trouble. Thus, in the compound of the oxidation numbers of the carbon and nitrogen cases it atom is obviously +1, but this would be true if C = +4 and N = 3,. or C = +3 and N = 2, or C = +2 and N = 1, etc. However, since the oxidation number of carbon is so often +4 (e.g., GO 2 > and that of nitrogen is so often 3 (e.g., NH 3 ), these would be the most likely oxidation numbers to take. A compound like Fe 3 O 4 shows an apparent fractional oxidation number for the metal constituent, in this case 2%. Actually two of the iron atoms have an oxidation number of +3, and one iron atom has an oxidation number of +2. This is called a mixed oxide (FeO.Fe2 3 ). A similar case is the salt oxidation number of each sulfur atom is 2^. Na S O6 2 4 ; the average, In socalled peroxy acids and salts of these acids, one (,or more) oxygen atoms has an oxidation number of zero. For example, in hydrogen peroxide, H 2 2 the oxidation number of one oxygen atom is 2; that of the other is 0. Sulfur forms analogous of the , persulfur acids. 14. Ionic Oxidation and Reduction Equations. In the case of oxidationreduction equations the two rules given in Sec. 12 should It will be found convenient in most cases to also be observed. write equations systematically according to the following steps: Write the formula of the oxidizing agent and of the reducing These should conform agent on the lefthand side of the equation. to Rules 1 and 2. a. Write the formulas of the resulting principal products on the These should likewise conform to righthand side of the equation. b. and 2. Under the formula of Rules c. 1 the oxidizing substance, write the expressing the total change in oxidation of all of its con Under the formula of the reducing substance, write number expressing the total change in oxidation number of its stituent elements. the number number constituent elements. CALCULATIONS OF ANALYTICAL CHEMISTRY 22 the number under the formula of the oxidizing agent in the as the coefficient for the reducing substance; use the number equation under the formula of the reducing agent in the equation as the cod. Use efficient for the oxidizing substance. e. Insert coefficients for the principal products to conform preceding f. step. common If possible, divide all the coefficients by the greatest divisor, to the or, if necessary, clear of fractions by multiplying all the by the necessary factor. g. If the reaction takes place in acid solution, introduce the formulas H^O and //+ in amounts necessary to balance the atoms of coefficients oxygen and hydrogen on the two sides of the chemical equation. If introduce the formulas in amounts necessary to balance the atoms of oxygen the reaction takes place in basic solution, HtO and OH" and hydrogen. charge on each of the two sides of the equation. They should be the same. EXAMPLE I. When a solution of chlorine water is added to h. Check the equation by determining the a sulfuric acid solution of ferrous The total net ionic sulfate, the iron is oxidized. stepbystep formulation of the equation for this reaction is as follows: + + RESULT C1 2 > Ke +++ C1 2 > Fe^ + + C1 2 > 2Fe 4++ STEP a, 6 c Fe ++ Fe++ 2FeV + 2C1~ None None / g 4+ = 4 + h EXAMPLE Cl~ Gl~ 2 1 d, e + + When a dilute nitric acid solution of stannous chloride is treated with a solution of potassium dichr ornate, tin is oxidized (from 2 to 4) and chromium is reduced (from 6 to 3). Neglecting the partial formation of complex ions (e.g. Slide") the II. development of the equation is as follows: RESULT STEP a, 6 c Sn ++ Sn+ + 2 rf, / g h e 6Sn++ 3Sn+ + 3Sn++ + + Cr2 Or > Sn++++ Cr2 O 7  > 811++++ + Cr 343 + 2Cr,O7 * 6811++++ + + Cr O ~ > 3Sn +++ + + + Cr O ~ + 1411+ > 3Sn++++ + 2 7 2 7 18+ = 18+ 2Cr+ ++ + 7H O 2 CHEMICAL EQUATIONS 23 in writing this equation in the molecular form one be at a loss to express the products correctly. The ques Note that would tion would whether to write stannic chloride and chromic nitrate or stannic nitrate and chromic chloride. As a matter of ionized fact, none of these is formed since the salts are completely arise in dilute solution. When hydrogen sulfide is bubbled into a dilute sulfuric acid solution of potassium permanganate, the latter is reduced (to manganous salt) and a white precipitate of free sulfur EXAMPLE is III. obtained. RESULT > 2 STEP a, b c MnOr + H S MnOr + H S 2 / g h EXAMPLE Mn++ + S Mn+ + + S 2 5 d, e > 2MnOr + 5HiS > 2Mn++ + 5S None 2MnOr + 5H S + 6H+  2Mn++ + 4+ = 4+ 58 2 TV. 8II 2 O an excess of oxidize a chromic salt to of sulfuric acid In the presence potassium permanganate solution + will clichromatc. RESULT Mn++ t 3Mn + 6Mn +  f + 6Mn ++ + EXAMPLE of V. When a nitrate in caustic gas is metallic alkali, aluminum the latter is 22TI added to a solution reduced and ammonia, is evolved. RESULT STEP a, 6 Al c Al 3 d, e 8A1 h 3 3 8 + 3NOr  8A1O2 ~ + 3NH 3 None f g + N0   AlOr + NH +N0  A10 +NH, 3 8A1 + 3NO  + 5OH + 2H O 2 3 8 = 8 1 SAlOr + 3NH 3 CALCULATIONS OF ANALYTICAL CHEMISTRY 24 EXAMPLE VI. Solid cuprous sulfide is oxidized by hot concentrated nitric acid forming a cupric salt, sulfate, and 2 gas* N0 RESULT STEP + NO CujS + NO CutS *. b c  Cu++  Cu++ 3 3 + SO  + NO + SO ~ + NO 1 1+1+8  * 10NO 2Cu ++ CuaS + 3 d, e + CuzS g h 10NO 3 4 2 4 2 + SOr + None 12H+ * 2Cu++ + 10NO2 + SO4 2+ = 2+ + 10NO + 6H O 2 2 Problems What is the oxidation number of each of the elements (other than = and hydrogen oxygen) in each of the following: (a) N2 O3 (6) SbS 3 (c) H 4 P2OS (d) K 2 Pt(N0 2 ) 4 (e) S 8 (/) Co(NH 3 ),+++; (g) Cu 3 [Fe(CN) 6 ] 2 (h) NaCHO2 ? 26. ; ; Ans. ; ; ; ; +3; (b) +3, 2; (c) +3; (d) +1, +2, +3; (e) 0; (/) +3, 3; (O) +2, +3, +2, 3; (A) +1, +2. 26. What is the oxidation number of each of the elements (other than hydrogen and oxygen) in each of the following: (a) MnO2 (6) A12 (SO4) 3 (c) NaCu(CN) 2 (d) (VO) 3 (PO 4 ) 2 (e) Fe(ClO 3 ) 3 (/) HAsOr; (?) CdS 2 O 6 .6H 2O; (h) (UO2 )(C1O 4 ) 2 .4H 2 O? (a) ; ; ; Ans. +5; (/) (6) +3, +6; (c) +1, +1, +2, 3; +2, +5; (h) +6, +7. +4; (a) +5; (g) 27. State the oxidation H Se; (6) N 0; <A) (NH 4) 2SO4 2 oxidation (a) K Mn0 (g) Na2S O.,.5H O; 4 7 (e) ; +4, +5; (e) +3, N 2 ; Na2Cr2O 7 ; of each element in the following: (a) HCN; K Fe(CN) 4 (/) 2 2 (h) number of each element (d) Pb 3 O 4 (c) Mg P 2 O 7 NH2OH; (t) IIN 3 IIC1O4 (6) ; 2 (d) 6 (g) ; . the 28. State number Mn O (d) 2 (c) 2 ; ; 2 ; ; ; (e) the in K II Sb O 2 2 2 7 following: ; (/) Na 2 S2 ; . number of each element in the following: (a) A1O2 ~; Cu(NII 3 ) 4 ++; (d) Ag(CN) 2; (e) SnS 3 ~; (/) MgNH 4 AsO4 29. State the oxidation (6) Fe(CN) (g) Na B 2 4 6 B (c) ; 7; (h) ; HC O ; 2 4 (0 WF . 8 Give the oxidation number of each element in the following: 2 PtCl c (c) K 3 Co(NO 2 ); (d) SbOCl; (e) HC 2 H 3 O2 (b) (UO 2 ) 3 (PO 4 ) 2 LiH; (g) Bi(OH)CO3 (h) Hgl.HgNHJ; (0 Fe 4 [Fe(CN) 6] 3 30. (a) (/) K . ; 31. in (a) 32. How do you account for the unusual average oxidation number of sulfur (c) Na S O ? (6) FeS Na 2 S4O6 HO 2 ; 2 ; 2 8 following unbalanced equations do not involve oxidation and reConvert them into complete, balanced ionic equations. Introduce other constituents wherever necessary. Substances are in solution The duction. 2 ; ; ; and unless underlined. + NaOH > NaAlO + NH OH >Fe(OH) (a) A1C13 (6) Fe2 (SO4) 3 2 4 3 CHEMICAL EQUATIONS (c) (d) (e) (/) (g) (h) + NH OH  Cu(NH,) SO K Cd(CN) + H S Cd FeCl + K Fe(CN) > Fe [Fe(CN) H PO + (NH Mo0 + HNO > (NH ),PO .12MoO, Na AsS + H SO >As S + 1I S Na,Sb0 + II SO + H S > Sb S CuSO 4 4 4 2 4 3 4 3 4 33. The reduction. 2 4 2 2 2 5 2 U0 C1 > HC H O + PbO  2 2 Pb(C2 H 3O2 ) 2 2 3 t 4 4 3 4 2 4 2 6l3 4 4) 2 4 3 4 2 HC1 0) 26 following unbalanced equations do not involve oxidation and Introduce the necessary constituents and convert to complete, Substances are in solution unless underlined. balanced, ionic equations. (a) (&) (c) (d) (e) (/) to) (K) () 34. + KOH > KiPbOi NH OH  Ag(NII,) Cl + AgCl H H SnCle + S 8nS, Sb S + (NH S > (NH^jSbSi Hg(N0 + KI > K HgI Na SnO + HC1 + H S > SnS UO SO + KOH>K U O Pb(C H O + K CrO  PbCrO + KC H,O2 Pb(OH) 2 4 2 2 2 2 4) 2 6 3) 2 2 4 2 2 2 3 4 2 3 2 2) 2 2 2 7 4 2 2 4 Balance the following oxidation and reduction equations: ++ H +  Fe +++ + Cl" + H 2O ClOr (a) Fe ++ > Mn++ + CrOr MnO2 Cr H^ 2 + + + +H + ++ > aH+ Mn C1 H O MnOr + + + + MnO ~ + H S + H+ * Mn++ + S + H O I0  + I + H + > I + II MnO ~ + S O  + H+  Mn++ + S O  + H O ' (6) (c) (d) (e) (/) 36. 2 4 2 2 2 3 2 4 2 2 3 6 4 2 Complete and balance the following ionic oxidationreduction equations: (a) (6) (c) (d) (e) (/) to) (K) + H+ * Mn++ + I + H O Cr + H S + H+ * Cr+++ + S + H O Zn + OH  ZnO + H AsO  + Zn + H + > AsH + Zn++ + H O BrOr + I" + H+ * Br~ + I + H O NOr + Al + OH + H O * NH + AlOr Cr +++ + NaA + OH  CrOr + Na+ + II O Al+ ++ + S >A1(QH) + 8 + S0 MnOr + I 2 2  2 7 2 2 2 2 4 3 2 2 2 2 3 2 2 3 2 3 36. The following unbalanced oxidationreduction equations represent Convert them to balanced ionic reactions taking place in acid solution. equations, introducing (o) (6) (c) H + and H O wherever necessary. Cr O  + NO  Cr+++ + NO ~  Cr O + SO Cr +++ + S * Mn++ + O H O MnOr + 2 2 7 2 2 3 2 8 2 2 4 7 2 CALCULATIONS OF ANALYTICAL CHEMISTRY 26 (d) (e) + BiO > MnO ~ VO ++ + MnOr > VOr + Mn++ + Zn > U+ ++ + + Zn++ Mn++ 4 2 37. The following equations involve oxidation and reduction. They represent reactions taking place in the presence of acid. Convert them to balanced H O wherever necessary. Cr O + I > O+++ + I MnO  + H C O > CO Cr+++ + Bi0 > Cr O U++++ + MnOr  UO + + + Mn++ U0  + H+lKV+ + Ot  > I I + Fe(CN) + Fe(CN) S S 0r + I, + I ionic equations, introducing II + and 2  (a) (fe) (c) (d) (e) (/) (flf) 2 2 7 4 2 2 4 2 2 2 7 2 6 6 6 2 4 2 6 38. Write balanced ionic; equations for each of the following reactions taking place in acid solution unless otherwise specified. Introduce hydrogen ions and water or hydroxyl ions and water wherever necessary, (a) Bichromate reduced by sulfite giving chromic salt and sulfate; (6) chromic salt oxidized free chlorine to give chromate and chloride; (c) chromite oxidized in alkaline solution with sodium peroxide to chromate; (d) lead peroxide oxidized by by and free oxygen; (e) cupric salt and and metallic copper; (/) manganous salt and chlorate giving a precipitate of manganese dioxide and chlorine dioxide gas; (</) cobaltous chloride in alkaline solution with hydrogen peroxide permanganate giving mangaiious metallic salt aluminum giving aluminum salt to give a precipitate of cobaltic hydroxide. 39. Express the following reactions in ionic form, and in each case state fraction of the nitric acid employed serves for oxidation: what (a) (6) (c) (d) (e) (/) (g) 40. + HNO  Cu(NO + NO + H O Zn + IINO * Zn(NO + NO + H O Cu S + HN0 > Cu(NO + NO + II SO + H O Cu S + HNO * Cu(NO + NO + S + H O FeS + HNO * Fc(N0 + S + NO + H O FeS + HNO > Pe(NO,) + NO, + H SO + H O Sn + HNO + H O >H SnO + NO + H O Cu 3 2 3) 2 2 3) 2 3 2 3 2 3 2 3) 2 2 3) 2 2 2 2 3) 3 3 2 3 2 2 3 2 4 2 4 2 2 3 Write balanced ionic equations for the following reactions taking place in solution: (a) (6) (c) (d) (e) (/) + metallic aluminum + sodium hydroxide aluminate + hydrogen gas + ammonia. chromic salt + oxygen gas. Bichromate + hydrogen peroxide + acid chromic salt + iodine. Chromate + iodide + acid manganous salt + sulfate. Permanganate + sulfite + acid iodide sulfate + Cupric cuprous iodide precipitate + free iodine. cobaltous salt + oxygen gas. Cobaltic oxide + acid Nitrate > > > > > > CHEMICAL EQUATIONS Manganous (g) salt + permanganate dioxide precipitate. (h) Mercuric chloride chloride precipitate 27 neutral (in + stannous chloride + + H SnCle > solution) manganese mercurous hydrochloric acid 2 Write balanced ionic equations for the following reactions in which used as an oxidizing agent. Introduce H + and H2O where 41. nitric acid is necessary. (a) (6) (c) (d) (e) (/) (g) (fc) (0 + HN0  Cu++ + NO Zn + (very dilute) HNO  Zn++ + NH^ Fe++ + HNO  Fe +++ + NO Cu S + HNO > Cu++ + SO  + NO Cu S + HNO  Cu ++ + S + NO FeSj + HNO > Fe +++ + SOr + N0 Sn + HNO . H SnO + NO FesP + HNO  Fe+++ + HiPOr + NO +++ + HiSiO + NO FeiSi + HNO  Fe Cu 3 3 3 2 3 2 3 2 4 3 2 3 2 3 3 3 3 Write balanced ionic equations for the following reactions taking place Introduce H 2O and other simple constituents wherever necessary. 42. in solution. + H S0  CuSO + (NH SO + AgCN (precipitate) (6) NuAg(CN), + HNO > H Cl S + Ag S (precipitate) (f) Ag(NH KCN S0 + + NII OH > K Cu(CN) + KCNO (d) Cu(NH K Cd(CN) SO + KCN (e} Cd(NH Cl + HC1  CoCl (/) Co(NH 43. Balance the following molecular equations: (a) Se Cl + H O H SeO + IIC1 + Se; (b) RuQ + 1IC1  H RuCl + C1 + H O; (c) Ag AsO + Zn + II,SO  AsH t + Ag + ZnSO + II O; (d) Ce(IO + H (^ O Ce ((J O + I + CO (e) Fe(CrO + Na CQ + O  Fe O + Na CrO H(a) Cu(NH,)S0 4 4 2 4 4 4) 2 3 3) 2 2 3) 4 4 3) 4 4 3) 6 2 4 2 3 4 2 2 3 2 2 2 2 4) 3 (fusion). 2 2; 2) 2 2 6 4 3 4 CO 3 4 3 3) 4 2 2 3 2 2 2 3 > 2 3 2 2 2 2 4 4 CHAPTER III CALCULATIONS BASED ON FORMULAS AND EQUATIONS 15. law The Mathematical Significance of a Chemical Formula. of definite proportions states that in any pure compound the proportions by weight of the constituent elements are always the same. A chemical formula therefore is not only a shorthand method of naming a compound and of indicating the constituent elements of the compound, but it also shows the relative masses of the elements present. Thus the formula Na2 SO 4 (molecular weight = 142.06) indicates that for every 142.06 grams of pure anhydrous sodium sulfate there are 2 X 23.00 = 46.00 grams of sodium, 32.06 grams of sul= 64.00 grams of oxygen. The percentage of fur, and 4 X 16.00 2 X 23 00 X sodium in pure anhydrous sodium sulfate is therefore ^ 100 = 32.38 per cent. A grammolecular weight of a substance 16. Formula Weights. is its molecular weight expressed in grams. Thus, a grammolec' ular weight (or grammole, or simply mole) of Na 2 SO 4 is 142,06 mole of nitrogen gas (N 2 ) is 28.016 grams of the element. grams. A A formula weight (F.W.) is that weight in grams corresponding to the formula of the substance as ordinarily written. In most cases it is identical to the grammolecular weight, but occasionally the true molecular weight of a compound is a multiple of the weight expressed by the formula as ordinarily written in a chemical In practically all the reactions of analytical chemistry, however, it can be assumed that the value of the formula weight and that of the mole are the same. The gramatom or gramatomic weight is the atomic weight of equation. the element expressed in grams (e.g., 40.08 grams of calcium; 14.008 grams of nitrogen). A gramion is the atomic or formula weight of an ion expressed in grams (e.g., 40.08 grams of Ca++; 62.008 grams of NO 3~). 28 CALCULATIONS BASED ON FORMULAS AND EQUATIONS 29 A one thousandth of a mole; a milligramatom is one thousandth of a gramatom. A formula weight of hydrated ferric sulfate, Fe2 (SO4 )3.9H 2 0, millimole is for example, is 562.0 grams of the salt. It contains 2 gramatoms of iron (= 117.0 grams), 21 gramatoms of oxygen (= 33G grams), 9 formula weights (9 F.W.) of water, 3,000 milligramatoms of sulfur, 17. and in solution would give 3 gramions of sulfate. Mathematical Significance of a Chemical Equation. A chemical equation not only represents the chemical changes taking place in a given reaction but also expresses the relative quantities of the substances involved. Thus, the molecular equation H S0 + BaCl 4 2 2 * BaS04 + 2HC1 not only states that sulfuric acid reacts with barium chloride to give barium sulfate and hydrochloric acid, but it also expresses the fact that every 98.08 parts by weight of sulfuric acid react with 208.27 parts of barium chloride to give 233.42 parts of barium sulfate and 2 X 36.47 = 72.94 parts of hydrogen chloride, these numerical values being the molecular weights of the respective compounds. These are relative weights and are independent of the units If a weight of any one of the above four substances is the known, weight of any or all of the other three can bo calculated by simple proportion. This is the basis of analytical computations. chosen. EXAMPLE I. A sample of pure lead weighing 0.500 gram is dissolved in nitric acid according to the equation 3Pb + 8HN0  3Pb(N0 3 3) 2 + 2NO + 4H 2 HNO How many grams 3 are required? pure of Pb(NO 3 ) 2 could be obtained by evaporating the resulting solution to dryness? How many grams of NO gas are formed in the How many grams of above reaction? SOLUTION: Atomic weight of lead = 207 Molecular weight of HNO 3 = 63.0 Molecular weight of Pb(NO 3) 2 = 331 Molecular weight of NO = 30.0 (3 X 207) grams of Pb react with (8 X 63.0) grams of HNO 3 (3 X 207) grams of Pb would form (3 X 331) grams of Pb(NO3 ) 2 and (2 X 30.0) grams of NO , CALCULATIONS OF ANALYTICAL CHEMISTRY 30 Hence 0.500 gram of Pb would & v (\1 O = X tC o X 0.500 require  405 gfam of HNO  799 gram of Pb(N03 ) 2 ' and would form 0.500 X and X v O X 9 X 0.500 =  o nn *m = 0.0483 gram of NO grams of H 2 S would be required to precipitate the lead as lead sulfide from the above solution? How many milliliters of H2S under standard temperature and pressure would be required for the precipitation? (A grammolecular standard conditions 22.4 liters. under of a occupies gas weight EXAMPLE How many II. See Sec. 110.) SOLUTION: Pb++ +HS PbS 2 Atomic weight + 2H+ = 207 H 2 S = 34.1 of lead Molecular weight of 207 grams of Pb 4 4 require 34.1 grams of of Pb ++ Hence 0.500 " " gram 0.500 34.1 grams of Volume of EXAMPLE X ~ HS 2 HS2 III. HS 2 requires *? 0.0822 gram of H S. Ans. 2 occupy 22,400 ml. under standard conditions OR99  X 22,400 = 54.1 ml. In the reaction expressed by the equation: 2Ag2 C0 3 > 4Ag + 2 + 2C0 2 how many gramatoms of silver can be obtained from 1 F.W. of silver carbonate, (6) how many gramatoms of silver can be obtained from 1.00 gram of silver carbonate, (c) how many grams (a) of silver carbonate are required to give 3.00 grams of oxygen 2 ) are produced from gas, (d) how many moles of gas (CO 2 + 50.0 grams of silver carbonate, and of gas (C0 2 O2 ) are produced from + bonate? (e) 1 how many milliliters millimole of silver car CALCULATIONS BASED ON FORMULAS AND EQUATIONS 31 SOLUTION: (a) (6) Ag2 CO 3 F.W. Ag2 CO 3 2 F.W. 4 gramatoms Ag 1 2 gramatoms Ag. Molecular weight Ag2 CO 3 = 276 1 1 HA fin 1.00 > gram Ag2 CO3 = ^~ ^ = g 0.00363 (c) X 2 2 moles O2 = Ag2 CO 3 (e) 3.00 X = 552 276 KKO X ^d = > Ag2C08 Ans. 1 grams) give 51 .7 (= 552 grams) Ag2 C0 3 = 50.0 grams 0.00363 F.W. 0.00726 gramatom Ag. 2 moles Ag2 C03 (= 2 (= 32 grams) 3.00 grams (d) = = mole grams Ag2 C03 3 moles (O2 Ans. . + CO 50 ^ X 3 = 0.272 mole gas. 2 2) Ans. 1 mole Ag2 CO 3 > 1^ moles gas mole gas (standard temperature and pressure) = 22,400 1 mfflimole 1 EXAMPLE Ag2 CO3 = 1^ X = 33.8 ml. mL Ans. of gas. In the reaction expressed by the equation TV. MnO + 2NaCl + 3H SO 2 2 4 > MriSO 4 + 2NaHSO + 4 C\ 2 + 2H O 2 or MnO + 2C1 + 6H+ ~> Mn++ + C1 + 2H O (a) 2 2 2 how many gramions of Mn++ can be obtained from 1 milligrams of MnSO 4 can be obtained mole of MnO 2 (6) how many from 5.00 grams of MnO 2 (c) how many millimoles , , of MnO 2 are required to give 100 ml. C12 (standard conditions), and (d) if 1.00 gram of NaCl, and 5.00 grams of 1.00 gram of 2 SO 4 2 MnO are used, which is H , the limiting reagent, and how many milliliters of C1 2 (standard conditions) are evolved? SOLUTION (a) 1 mole : MnO 1 gramion Mn++ " > 0.001 gramion Mn 4 Ans. 1 millimole MnO 1 mole Mn0 (= 86.9 grams) > 1 mole MnS0 (= 151 151 5.00 grams MnO = 5.00 X SJTS oO.u 2 > 1 ". 2 (b) 4 2 2 = 8.69 grams MnSO 4. Ans. grams) CALCULATIONS OF ANALYTICAL CHEMISTRY 32 100 ml. C1 2 (c) = = ZZ.4 MnO 4.47 millimoles MnO = millimole Cl2 1 100 ml. C1 2 4.47 millimoles C12 = millimole 1 no ~~ = i i (d) MnO = 1.00 gram 1.00 gram NaCl  2 5.00 grams H SO 2 00 = 4 = ^, 2 (\r\ ~ 1 00 = = ^^ = 0.0115 mole 0.0171 mole oo.o = sir? yo. 1 X12OU4 Ans. 2. 0.0510 mole According to the equation these substances react in the molar The NaCl ratio of 1:2:3, or 0.0115:0.0230:0.0345. is therefore the limiting reagent and the other two are in excess. 2 moles NaCl > mole C1 2 = 22,400 ml. C12 1 ' NaCl > 0.0171 mole * 71 X 22,400 Zi = 192 ml. C1 2 Ans. . Problems How many grams of potassium and of carbon are contained in gram of K4Fe(CN) .3II O; (6) 1 F.W. of KHC H O ? 44. 6 Ans. 46. lead. (a) 2 4 4 0.0782 gram, 0.0360 gram; (b) 0.211 39.1 grams, 48.0 grams. A certain weight of lead phosphate, Pb (PO contains 0.100 gram of How many grams of phosphorus are present? What is the weight of the 3 What lead phosphate? Ans. is 0.00997 gram. 4) 2 , the percentage of oxygen present? 0.131 gram. 15.8 per cent. How many grams of oxygen are present in 1.00 gram following: (a) Fe O (b) BaSO (c) Fe(NO .6H O? 46. 2 Ans. (a) 4, 3, 0.300 gram, (b) 3) 3 0.275 gram, (c) (a) 18.7 per cent, (b) 25.8 per cent, each of the 2 What is the percentage by weight of sulfur in each BiA, (b) Na 2 S2 O3.5II2O, (c) K2 SO4 .A12 (SO4 ) 3 .24H2 O? Ans. of 0.686 gram. 47. (a) (a) 6 (c) of the following: 13.5 per cent. 48. Ignition of anhydrous magnesium ammonium phosphate forms magnesium pyrophosphate according to the equation: 2MgNH 4 PO 4 > Mg2 P2 O7 + 2NH 3 H2O. Calculate: (a) number of formula weights of Mg2 P2O7 produced from 1.00 F.W. of MgNH 4PO 4 (b) number of grams of NH 3 produced + , same time, (c) number accompanying the formation of at the Ans. (a) 0.500, (b) 17.0 of milliliters of 1 millimole of grams, (c) NH (standard conditions) 3 Mg P O 44.8 ml. 2 2 7. CALCULATIONS BASED ON FORMULAS AND EQUATIONS 33 What is the weight of the constituent elements in 0.717 gram of AgN03 ? Ans. Ag = 0.455 gram, N = 0.059 gram, O = 0.203 gram. 49. 60. Calculate the number of pounds of materials theoretically necessary for from CaO and 2CO3 (6) BaSO 4 the preparation of 1.00 pound of (a) from Na2S04.10II2 and BaCl2 .2H2 O. K KOH Ans. (a) (&) CaO = 0.500 pound, K CO3 = 1.23 pounds. Na2SO4 .10H2 O = 1.38 pounds, BaCl2 .2H2O = , 2 1.04 pounds. Balance the following equation and also write it as a balanced ionic BaSO 4 Calculate from it the followBaCl2  A1C13 equation: A1 2 (SO 4 )3 51. + + " . number of gramions of Al" "*" contained in 1 grammole of A1 2 (SO4)3, ++ (b) number of gramions of Ba reacting with 1 .00 gram of A1+ *~+, (c) number of grains of BaSO 4 obtainable from 2.00 grams of A1 2 (SO 4) 3 .18II 2 O, (d) number of grams of BaSO 4 produced by mixing solutions containing 3.00 grams of A12 (SO4 ) S and 4.00 grains of BaCU. 1 1 ing: (a) Ans. 52. (a) 2, (b) 0.0556, (c) From 2.10 grams, (d) 4.48 grams. the reaction: 4FeSa + 1 10 2 * 2Fe 2 O 3 + 8SO2 , calculate the fol lowing: (a) number of moles of FeS2 required to form 1 F.W. of Fe 2 O 3 (b) number of grams of oxygen required to react with 2.00 moles of FeS 2 (c) number , , of millimoles of SO2 equivalent to 0.320 gram of O2 , (d) volume of SO2 (standard Fe 2O3 conditions) accompanying the formation of 0.160 grain of Ans. (a) 2, (b) 176 grams, (c) 7.27, (d) . 89.6 ml. Complete and balance the following ionic equation for a reaction taking ~" ^Fe ++ ^ + Mn 4 +. Calculate Fe+ + + MnO4 from it the following: (a) number of gramions of Mn ++ produced from 1 gramion of Fe++, (6) number of millimoles of Fe 2 (SO 4 ) 3 .9H 2 O obtainable if 1 millimole of KMnO4 is reduced, (c) decrease in the number of gramions of H 4 53. place in the presence of acid: " 4"1 accompanying the formation of 1.00 gram of Fe "*, (d) number of grams of Fe 2 (SO 4) 3 obtainable by mixing solutions containing 1.00 gram of FeSO4 .7II 2O, 0.100 gram of KMnO4 and 1.00 gram of H 2 SO 4 . , Ans. 64. (a) y 5 , (b) What weight ing to the equation: Ans. 2%, (c) 0.0286, (d) 0.633 gram. NH is required to dissolve 0.120 gram of AgCl accord+ + Cl~? AgCl + 2NH > Ag(NH of 3 3 3) 2 0.0285 gram. H 66. How many grams of 2 S are required to precipitate the bismuth as Bi 2 S 3 from an acid solution containing 100 mg. of dissolved bismuth? Ans. 0.0244 gram. H 66. How many grams of 2 SO4 are required to dissolve 0.636 gram of 2H 2 SO4  CuSO4 SO2 metallic copper according to the equation: Cu 2H2 O? How many milliliters of gas are evolved (measured under standard + conditions)? Ans. 1.96 grams. 224 ml. + + CALCULATIONS OF ANALYTICAL CHEMISTRY 34 How many 67. grains of anhydrous chromic chloride (CrCl 3 ) could be ob2 S in the presence of HC1: K tained from 100 mg. of 2 Cr 2 O7 after reduction by Cr2O7 ~ 3H2 S 8H+ > 2Cr+++ 38 7II 2O? + + many milliliters Ans. (standard conditions) of 2 How many grams and how would be required? 0.0723 gram, 0.0347 gram, 22.8 ml. How many 68. What + HS + H is grams of chromium are present in 0.250 gram of the percentage of potassium in this compound? contains 0.200 gram of 2 SO 4 .A1 2 SO4.24II 2 O, weight of alum, is the percentage of oxygen in the compound? What weight of sulfur is present in an amount of 318 mg. of sodium, (fe) 1.00 gramatom of oxygen? How many 61. grams Na 2 S 2 O3 that gram FeSO4 .(NH4) 2 SO4 .6H2 O? 3 , , 62. contains of each of the of nitrogen are present in 1.00 NH 3 (b) Pb(NO ) 2 (c) What is the percentage of oxygen in FeSO4 .7H 2O, (c) K2 SO4 .Cr 2 (SO 4) .24H 2O? following: (a) (b) 2 What 60. (a) 2 K What 69. aluminum? K Cr C>7? each of the following: (a) H O, 2 3 63. Ignition of ing equation: bismuth basic carbonate takes place according to the follow 2BiOIICO 3 > BiA, + 2CO + H O. 2 2 Calculate the following: (a) number of formula weights of Bi 2 O 3 produced from 1 F.W. of the carbonate, (b) number of millimoles of OO 2 accompanying the formation of 1.00 gram of Bi 2 O?, (c) volume of CO 2 (standard conditions) formed from 0.0200 gram of BiOIICOs, (d) volume of gas (CO2 + water vapor) accompanying the forma tion of 1.00 millimole of Bi 2 O 3 . Convert the following to balanced molecular and ionic equations: Calculate from them the following: 3 >Fe(NO 3) 2 + AgCl. of of formula number AgCl obtainable from 1 F.W. of FeCl 3 (a) weights ++ of Fe of number '(b) produced per millimole of AgCl, (c) number gramions of grams of Fe(NO 3 )3.6H 2 O obtainable if 1.00 grammolecular weight of AgNO 3 is used up, (d) number of grams of AgCl obtained by mixing solutions containing 0.700 gram of FeCl 3 and 0.600 gram of AgNO 3 How many grams of which reactant are left over? 1 64. eC!3 + AgNO , *' . 66. Assuming that the reaction Na CO + NaNO takes 13Na CO + 14NaNO 2 2 5Fe 2 O ? 3 3 millimoles and how many (6) 1.00 gram of + 20Na 2 CrO4 milliliters are formed from that weight of Cr, for the fusion of the mineral place according to the equation: 3 3 + 7N + (measured at chromite with 10Fe(CrO 2 ) 2 + 13CO2 how many 760 mm. and 0C.) of gas Fe(CrO 2 ) 2 containing 2 , (a) 1.00 gramatom of Cr? NH How many milligrams of 3 are required to react with 27.2 mg. of 2NT H 3 >IIgNH 2 Cl the to Cl Hg Hg2 2 according equation: Hg 2 Cl 2 + Cl~? How many grams of free mercury would be formed? 4 66. + + + NH + How many grams and how many milliliters (standard conditions) of are 2 required to precipitate the arsenic from an acid solution containing a 6H + > As2 S 3 6H 2O? How 3H 2 S 0.100 gram of Na 3 AsO 3 2AsO 3 67. HS : many grams of the sulfide + would be formed? + + CALCULATIONS BASED ON FORMULAS AND EQUATIONS 35 68. Fe + ++ Balance the following equation: +HO 2 and calculate from it the MnOr + number Fe ++ + H+ > Mn++ + of grains of FeS04.7H 2 O re KMnO4 that contains 0.250 gram of Mn. ~ 69. Balance the following equation: Cr2 7 Fe++ H+ > Cr+++ +f+ of Fe If 1.00 grammolecular weight 2 O. 2 CrO4 is dissolved in water = 2H+ > Cr2 07 = and the solution acidified (2Cr04 2 0), how many grams quired to reduce that weight of +H + of FeSO 4 .(NH 4 ) 2 SO4.6H 2 O would be + + K +H + " required to reduce the chromium in the resulting solution? 70. When used for the oxidizing effect of its nitrate, which is the more economical reagent, potassium nitrate at 65 cents per pound or sodium nitrate at 50 cents per pound? How much is saved per pound of the more economical reagent? CHAPTER IV CONCENTRATION OF SOLUTIONS Methods Solution reagents of Expressing Concentration. used in analytical chemistry are usually either (1) laboratory 18. reagents the concentrations of which need be known only approximately, or (2) titration reagents the concentrations of which must be known to a high degree of precision. There are several ways of expressing concentration and it is important to have a In anaclear understanding of just what is meant in each case. of concentration are methods expressing lytical work the following most commonly used. 19. Grams per Unit Volume. By this method a concentration is expressed in terms of the number of grams (or milligrams) of solute in each liter (or milliliter) of solution. A 5gramper sodium chloride is prepared by dissolving 5 grams of the salt in water and diluting to one liter (not by adding one liter of water to the salt). This method is simple and direct but it is not a convenient method from a stoichiometric point of view, since solutions of the same concentration bear no simple relation to each other so far as volumes involved in chemical reactions are concerned. Chemical substances enter into reaction upon a moletomole basis and not upon a gramtogram basis. This method is on a percentage 20. Percentage Composition. in terms of grams of concentration and basis expresses byweight solute per 100 grams of solution. A 5 per cent of sodium chloride liter solution of made by dissolving 5 grams of the salt in 95 which of course gives 100 grams of solution. is grams of water, The specific gravity of the solution of a 21. Specific Gravity. of the concentration of the solute in the a measure is solute single Although occasionally used in analytical chemistry, it a cumbersome method, since it necessitates consulting a table solution. is in order to determine the percentagebyweight Tables of specific gravities of common 36 composition. reagents are found in the CONCENTRATION OF SOLUTIONS 37 handbooks and other reference books of chemistry. Tables covering the common acids and bases are also in the Appendix of this Here it will be found, for example, that hydrochloric acid text. of specific gravity 1.12 contains 23.8 in 100 22. grams grams of hydrogen chloride of solution. Volume Ratios. Occasionally in analytical work the con a mineral acid or of ammonium hydroxide is given in terms of the volume ratio of the common concentrated reagent and water. Thus HC1 (1:3) signifies a solution of hydrochloric acid made by mixing one volume of common, concentrated hydrochloric (sp. gr. about 1.20) with three volumes of water. centration of H SO a solution made by mixing one used concentrated sulfuric acid (sp. commonly This method of gr. 1.84) with three parts by volume of water. is cumbersome, particularly in work expressing concentrations where subsequent calculations involving the solutions are to be made. A molar solution is one con23. Molar and Formal Solutions. taining a grammole of substance dissolved in a liter of solution. This is usually identical to a formal solution which contains a formula weight of substance in a liter of solution (see Sec. 16). vSimilarly volume 2 4 (l"3) signifies of the A grammolecular weight of substance dissolved in a liter of water a molar solution, for the resulting solution constitute not does does not occupy a volume of exactly a liter. 1 A liter of molar (M) sulfuric acid solution contains 98.08 grams of 2SO4 a liter H ; (J^M, 0.5M, or M/2) sulfuric acid solution contains In this particular case 98.08 grams of 2S04 of 2 S04. of halfmolar H H 49.04 grams does not mean 98.08 grams of the ordinary concentrated sulfuric The concentrated acid contains acid, but of hydrogen sulfate. about 96 per cent of the latter. Since 1 mole of hydrochloric acid reacts with 1 mole of sodium hydroxide, a certain volume of sodium hydroxide solution will be exactly neutralized by an equal volume of hydrochloric acid of the same molar concentration, or twice the volume of hydrochloric acid of onehalf the molar concentration of the sodium hydroxide. 1 Solutions containing a grammolecular weight of substance dissolved in 1,000 grams of water are useful in computations involving certain physicochemical phenomena. Such solutions are often referred to as molal solutions, but this standard is not used in general analytical work. CALCULATIONS OF ANALYTICAL CHEMISTRY 38 One molecule of hydrogen sodium hydroxide. sulfate will neutralize 2 molecules of H S0 + 2NaOH > Na S0 + 2H 4 2 To 2 neutralize a certain onehalf that tion volume volume of 4 2 sodium hydroxide solution, only same molar concentra of sulfuric acid of the would be required. Volumetric calculations are therefore when concentrations are expressed in terms of greatly simplified moles of substance per unit volume of solution; for, when so expressed, the volumes of reacting solutions of the same molar concentration, although not necessarily equal, bear simple numerical relationships to each other. What volume EXAMPLE. of 0.6380 M solution will neutralize 430.0 ml. of 0.4000 potassium hydroxide M sulfuric acid? SOLUTION: 1 mole H S0 ~ 2 4 2 moles KOH 430.0 ml. of 0.4000 molar solution contains x ml. H SO o = 0.1720 mole H SO 2 4 KOH KOH contains 0.0006380 mole KOH 0.1720 mole 1 0.4000 2 4 0.3440 mole Volume required = ^ r^r^ooT^ = U.UUUoooU 539.3 ml. Ans. 24. Equivalent Weight and Normal Solution. The equivalent weight of an element or compound is that weight equivalent in reactive power to one atomic weight of hydrogen. The milli one thousandth of the equivalent weight. The equivalent weight gramequivalent weight is the equivalent weight expressed in grams; is the grammilliequivalent weight is the milliequivalent weight ex The application of gramequivalent weights pressed in grams. to various types of chemical reactions will be taken up in detail in Part III, but simple cases, applying particularly to qualitative 1 analysis, will be considered briefly here. The equivalent weight of a substance, like the atomic or molecular weight, merely a number without a unit of weight; the gramequivalent weight is a definite number of grams. However, when the connotation is clear, the terms 1 is " equivalent weight" and "milliequivalent weight" are frequently used to signify gramequivalent weight and grammilliequivalent weight, respectively. CONCENTRATION OF SOLUTIONS The gramequivalent weight 39 an acid, base, or salt involved a simple metathesis such as a neutralization or precipitation is that weight in grams of the substance equivalent in neutralizing or precipitating power to 1 gramion of hydrogen (i.e., 1.008 grams of H+). A normal solution contains 1 gramequivalent weight of solute in 1 liter of solution, or 1 grammilliequivalent weight in 1 milliThe normality of a solution is its relation to a liter of solution. normal solution. A halfnormal solution therefore contains in a unit volume onehalf the weight of solute contained in its normal N, or solution, and this weight may be expressed as 0.5 N, N/2. The concentration of a normal solution is expressed simply of in ^ as N. Since the concentrations of solutions used in precise volumetric analysis are usually found experimentally, the concentrations can not often be expressed by whole numbers or by simple fractions. They are more likely to be expressed as decimal fractions, e.g., 0.1372 N. 26. Simple Calculations Involving Equivalents, Milliequivalents, and Normality. The use of equivalents, milliequivalents, and normality is and the terms them is essential so extensive in analytical chemistry are so fundamental that a clear understanding of at this time. More detailed discussions applying particularly to quantitative analysis will be given in Part III. Let us consider here only the simplest reactions between common acids, bases, and salts, and as an example let us take sulfuric A mole, or gramand a molar solution molecular weight, of 4 is 98.08 grams, 2 of the acid therefore contains this amount of pure hydrogen sulfate in a liter of solution. Since 98.08 grams of H 2 SO4 has a neutralizing power equivalent to 2 gramatoms (2.016 grams) of hydrogen as an ion, the gramequivalent of H 2 SO4 as an acid is 98.08/2 = 49.04 grams, which is equivalent in neutralizing power to 1 gramatom (1.008 grams) of hydrogen as an ion. The grammilliequivalent weight is 0.04904 gram. A normal solution of sulfuric acid therefore contains 49.04 grams of ILSO4 in a liter of solution, or 0.04904 gram of H 2 S04 in a millimeter of solution. A 1 molar solution of sulfuric acid is 2 normal; a 1 normal soluacid. The molecular weight of II 2 SO 4 is 98.08. H S0 tion of sulfuric acid is ]/2 molar. 40 CALCULATIONS OF ANALYTICAL CHEMISTRY Sodium hydroxide is a base with a molecular weight of 40.00. The gramequivalent weight of NaOH is 40.00 grams, since this amount is neutralized by 1.007 contains 40.00 grams grams of 11+ A normal solution of NaOH a liter of solution and is likewise 1 molar. . in ACIDS  =36.47 g These weights are all equivalent to ^2^=60.05 g. y2=40.03g. FIG. 1. Gramequivalent weights of some acids, bases, and salts. The gramequivalent weight of a simple salt is determined the same way as that of an acid or base, namely by reference in to 1.008 grams of 11+ as a standard. In the case of the salt of a metal, the equivalent weight is ordinarily the molecular weight of the salt divided by the total oxidation number represented by the atoms of metal in the formula. The equivalent weights shown in Fig. same standard, of a few acids, bases, and salts are Since each of these amounts is equivalent they are mutually equivalent to one another. 1. to the CONCENTRATION OF SOLUTIONS 41 In each case the specified amount when dissolved in one produce a 1 normal solution. liter of solution will N HC1 will Ba(OH) 2 or 1 It follows that 1 liter of 1 N N H2SO4 neutralize 1 liter of 1 N any onenormal any onenormal base. More generally, a certain volume of any acid will neutralize the same volume of any base of the same normality. NaOH, base. or 1 liter of 1 One liter of 1 N AgNO Similarly, 1 liter of 1 from 1 liter of 1 NaCl or N of i N Fe 2 (SO 4 ) 3 from the chloride will precipitate 3 liter of 1 will just precipitate the sulfate liter of , will also neutralize 1 liter of 1 N BaCl 2 and the , of 1 liter 1 latter N Na SO4 or 1 liter 2 . We found that when two solutions of equal molarity react, the volumes are in simple ratio to each other. But when two solutions of equal normality react, the volumes of the solution are equal. Since volumes of reagents in analytical chemistry are usually measured in milliliters rather than in liters, it is more convenient to consider a normal solution as containing Hence the number milliliter. I grammillicquivalent of grammilliequivalent weight per weights present in a solution can be found from the simple relationship : Number X of milliliters normality number = of grammilliequivalent weights or X ml. N = number of me. vvts. (See footnote, p. 38) N HOI contain Thus, 2.00 ml. of 0.00 X I1C1 12.0 millicquivalent weights, = This will 0.438 gram of hydrogen chloride. L,UUU exactly neutralize 12.0 milliequivalents of any base, for example, 4.00 ml. of 3.00 NaOH, or 4.00 ml. of 3.00 Na2 CX) 3 or 80.0 ml. or 12.0 TTWWX N N of0.150NBa(OH) 2 It follows that , , etc. when solutions A and B mutually interact to a complete reaction, ml.,i EXAMPLE What X N,i = m\. B X N* the approximate molarity and normality To what volume should of a 13.0 per cent solution of 2 SO 4 ? solution? 100 ml. of the acid be diluted in order to prepare a 1.50 I. is H N CALCULATIONS OF ANALYTICAL CHEMISTRY 42 From SOLUTION: Appendix, the specific gravity table in the specific gravity of the acid is 1.090. weighs 1,090 grams 1 liter = contains 1,090 X 0.130 mole II 2 SO 4 = 98.08 grams 1 liter 1 142 grams H S0 2 4 = 142/98.08 = 1.45 M. Ans. 1 gramequivalent H 2 SO 4 = H 2 SO 4 /2 = 49.04 grams Normality of solution = 142/49.04 = 2.90 N. Ans. 100 ml. contain 290 milliequivalents H 2 SO 4 Molarity of solution After dilution x ml. of 1.50 x EXAMPLE N contain 290 milliequivalents = 290 = x 193 X 1.50 ml. Ans. A solution contains 3.30 grams of Na CO II. 2 What in each 15.0 ml. With how many is its What normality? milliliters of 3.10 N is its acetic acid, 3 .10H 2O molarity? HC H O 2 3 2, will 2H+ + N H 2S04 25.0 ml. of the carbonate react according to the equation: C0 8  2 O C0 2 ? With how many milliliters of 3.10 H + carbonate react? will 25.0 ml. of the SOLUTION : Molecular wt. Na CO 3 .10H 2 O = 286 2 = Na2CO 8 .10H2 O = = 0.143 Milliequivalent weight Equivalent wt. 143 Zt 3 30 Solution contains y^ = 0.220 gram per ml. JLo.U 1 normal solution would contain 0.143 gram per ml. Normality = Molarity EXAMPLE , What . Q = = x A12 (S0 4 ) 3 ' III. X = 1.54 0.77 = x = = 3.10 (a) would be A of N. Ans. M. 25.0 Ans. X HC H O Ans. H SO Ans. 12.4 ml. 2 12.4 ml. 2 3 2. 4. M 0.100 solution of aluminum sulfate, what normality as an aluminum salt? normality as a sulfate? of the salt are contained in each (6) 1.54 (c) How many milliliter? (d) milliequivalents What volume of CONCENTRATION OF SOLUTIONS 43 N NH OH would be required to react with the aluminum in 35.0 ml. of the salt solution according to the equation: A1+++ + 3NH4 OH > A1(OH) 8 + 3NH4+ ? (e) What volume of 6.00 N BaCl2.2H 2 O solution would be required to precipitate the sulfate from 35.0 ml. of the solution? (/) How many grams of BaCl 2 .2H 2 O are contained in each milliliter of the above solution? SOLUTION 6.00 4 : mole A1 2 (SO4 ) 3 = 6 equivalents (2 A1+++ o 6H+) 0.100 molar = 0.600 normal as Al salt. Ans. = 0.600 normal as sulfate. Ans. 1 (a) (6) Ans. 0.600 milliequivalent per milliliter. = ml.* X N* (d) ml.A X N^ = 35.0 X 0.600 x X 6.00 (c) (e) = 3.50 ml. Ans. ml.A X N^ = ml* X N* x X 6.00 = 35.0 X 0.600 x = 3.50 ml. Ans. (/) 0.00 x = X Ans. 0.732 gram. Problems What 71. fraction of the molecular weight represents the milliequivalent weight in the case of each of the following acids, bases, (fc) H As0 3 Ans. (/) 4, (c) (a) HP 4 2 7, 1/2,000, (d) (b) Th02 (NII 4 ) 2 SO 4 (e) , 1/3,000, (c) , 1/4,000, (/) and (d) H SiF salts: (a) Zn3 (AsO 2 6, 3) 2? 1/4,000, (e) 1/2,000, 1/6,000. 72. tion? How many How many Ans. 73. 2 solution of 4 50.0 ml. of 0.200 N solu 4 2 5.00 millimoles. 0.872 gram. A K S0 are contained in K SO are present? grams of millimoles of H SO 2 4 has a specific gravity of 1.150. What the is normality of the solution? Ans. 4.90 N. 74. What is the normality of a solution of 15.0 ml. 4 2 make a 15.0 N. 4 2 4 be diluted to Aiis. NH OH having a specific gravity How many milliliters of 13.0 N H SO would be neutralized by of the NH OH? To what volume should 250 ml. of the 13.0 N H SO of 0.900? solution that 17.3ml. is 4 5.00 molar? 325ml. 30 per cent solution of H 3 PO 4 has a specific gravity of 1.180. What normality as an acid assuming partial neutralization to form HPO 4"? What is its molar concentration? 76. A is its Ans. 7.22 N. 3.61 M. CALCULATIONS OF ANALYTICAL CHEMISTRY 44 76. How many of grams SrCk.GHgO are required to prepare 500 ml. of N solution? What is the molarity of the solution? How many milliliters 1.00 N AgNOs would be required to precipitate the chloride from 20.0 ml. 0.550 of of the strontium chloride solution? Ans. 77. M. 11.0 ml. N How much Cr2 (SO4 )3.18H 2 O 0.200 0.275 36.6 grams. N NH OH 4 from 20.0 ml. Ans. water must be added to 50.0 ml. of a 0.400 solution of make it 0.0500 molar? How many milliliters of would be required to precipitate all the chromium as Cr(OH) 3 in order to of the original undiluted solution? 16.7ml. 40.0ml. A piece of aluminum weighing 2.70 grams is treated with 75.0 ml. of 1.18 containing 24.7 per cent After the metal 2 SO 4 by weight). 6H+ > 2A1+++ 3H 2 ) the solution is diluted completely dissolved (2A1 H (sp. gr. is + to 400 ml. + Calculate (a) normality of the resulting solution in free sulfuric of the solution with respect to the aluminum salt it contains, acid, (b) normality total volume of 6.00 4 (c) required to neutralize the acid and precipi N NH OH all the Ans. (a) tate 79. and aluminum 0.365 N, What is the salts: (a) as A1(OH)3 from 50.0 ml. of the solution. (6) 0.750 N, (c) 9.30 ml. gramequivalent weight of each of the following acids, bases II 2 C2O 4 .2H 2O, (6) H PO 3 3, (c) CaO, (d) Fe2 3, () SnCU, (/) (Ca 3 (P0 4