BERKELEY LIBRARY UNIVERSITY Of CAIIPORNIA A SYSTEMATIC QUALITATIVE CHEMICAL ANALYSIS A Theoretical and Practical Study of Analytical Reactions of the More Common Ions of Inorganic Substances BY GEO. W. SEARS, Ph.D. ASSOCIATE PROFESSOR OF CHEMISTRY IN THE UNIVERSITY OF NEVADA NEW YORK JOHN WILEY & SONS, Inc. LONDON: CHAPMAN & HALL, LIMITED 1922 Copyright, 1922 BY GEORGE W. SEARS LOAN STAO5 PRESS OF BRAUNWORTH a CO. BOOK MANUFACTURERS BROOKLYN, N. Y. QP 8 1 PREFACE THE author believes that a course in Qualitative Analysis should not only train the student in accurate and careful manipulation and so prepare him for the more careful work necessary in Quantitative Analysis, but should also serve to extend and supplement his knowledge of General and Theoretical Chemistry. To this end an attempt has been made to present the theoretical and practical parts in such a way that the student will understand the signifi- cance and see the practical applications of the theoretical part. The book has been divided into four parts, including the Appendix. Part I consists of the Introduction and Laboratory Suggestions. In the Introduction is given a brief discussion of the Ionic Theory and Law of Mass Action, as applied to the principles involved in Qualitative Analysis, and reference to its various sections is made throughout the text. Part II consists of Preliminary Experiments, Method of Analysis, Discussion and Review Questions on the metal ions. In -order that the student may get a comparative idea of the actions of the different metal ions in a group toward a given reagent, the Pre- liminary Experiments are arranged in a manner which differs somewhat from the usual one and in order that he may not be confused by too many reactions, only those reagents used in the Method of Analysis are employed. Preliminary Experiments whose reactions may not be clear are followed by Notes. In the Method of Analysis an attempt has been made to give clear and concise directions for procedure only, all explanations and conditional pro- iii [ 2C1 iv PREFACE cedures being placed together under Discussion, following each group or sub-group. It is believed that by frequent reference to this the student's attention will be brought more effectively to the theoretical explanations and to the reasons for careful work. In so far as practicable, only Quantitative reactions have been given and only those methods have been used which have been thoroughly tested. Part III consists of Method of Analysis, Dis- cussion and Review Questions on the acids. A new method of anion analysis is given which follows in general the method of procedure in metal analysis, in that a single sample is taken for analysis, and largely by means of precipitation methods, the separadon and detec- tion of anions are made. Many of the same reactions used in the course of metal analysis are also employed. For these reasons no preliminary experiments are given, the author having found that students are able to under- stand and follow the directions in the analysis of " unknowns." In the Appendix are given directions for the preparation of reagents and test solutions, also tables of solubilities and atomic weights of the more common elements. In the preparation of this text the author has made use of information from all convenient sources, including such textbooks on Qualitative Analysis as those of Tread- well-Hall, A. A. Noyes, Julius Stieglitz, and W. A. Noyes, to whom he wishes to acknowledge his indebtedness. Acknowledgment is also due to Dr. J. F. G. Hicks, Stan- ford University, and to Miss M. Dewar, University of Nevada, for helpful criticism and assistance. CONTENTS PART I PAGE Introduction I Laboratory Suggestions 19 PART II The Systematic Analysis (Cations) 22 Preparation of Solution 22 Discussion (l-io) 26 Group I. Preliminary Experiments 29 Outline of Analysis (Table I) 31 Analysis 31 Discussion (11-15) 3 2 Group II. Preliminary Experiments (Cu division) 34 Outline of Analysis (Table II) 38 Analysis (Separation into Cu and Sn divisions) 38 Discussion (16-20) 39 Outline of Analysis (Table III) 42 Analysis (Cu division) 43 Discussion (21-27) 44 Preliminary Experiments (Sn division) 46 Outline of Analysis (Table IV) 50 Analysis (Sn division) 50 Discussion (28-31) 52 Group III. Preliminary Experiments (Al division) 53 Outline of Analysis (Table V) 56 Analysis (Separation into Al and Fe divisions) 56 Discussion (32-37) 57 Outline of Analysis (Table VI) 60 Analysis (Al division) 60 Discussion (38-41) 62 Preliminary Experiments (Fe division) 63 Outline of Analysis (Table VII) 66 Outline of Analysis (Table VIII) 66 v vi CONTENTS PAGE Analysis (Fe division) 67 Discussion (42-48) 70 Group IV. Preliminary Experiments 73 Outline of Analysis (Table IX) 75 Analysis 75 Discussion (49-53) 77 Group V. Preliminary Experiments 79 Outline of Analysis (Table X) 81 Analysis 81 Discussion (54-58) 82 Questions for Review 84 PART III Acids (General statement) 88 The Systematic Analysis (Anions) 89 Preparation of Solution 89 Discussion (59-60) 90 Group I. Outline of Analysis, Division A (Table XI) 91 Analysis (Division A) 91 Discussion (61-63) 92 Outline of Analysis, Division B (Table XII) 94 Analysis (Division B) 94 Discussion (64-66) 96 Group II. Outline of Analysis (Table XIII) 98 Analysis 98 Discussion (67-69) 99 Group III. Outline of Analysis (Table XIV) 102 Analysis 103 Discussion (70-75) 105 Group IV. Outline of Analysis (Table XV) 107 Analysis , 107 Discussion (76-80) 109 Questions for Review 1 1 1 APPENDIX I. Preparation of Reagents 113 II. Preparation of Test Solutions 1 16 III. Table of Solubilities 118 IV. Table of Atomic Weights 119 A SYSTEMATIC QUALITATIVE CHEMICAL ANALYSIS PART I INTRODUCTION 1. In his study of General Chemistry the student has become somewhat familiar with the more important ele- ments, their preparation, physical properties and some of their principal compounds. In his laboratory work he has become acquainted with the different types of chemical reactions; viz., combination, decomposition, displace- ment, double decomposition, oxidation-reduction, etc. He has learned that certain reactions are reversible, that others are non-reversible, and that the point of equilibrium in reversible reactions is influenced by the relative amounts of the substances entering into the reaction. 2. " Qualitative Analysis treats of the identification of matter," while Quantitative Analysis, as its name implies, deals with its quantitative determination. An efficient system of qualitative analysis should consist not only in finding out what substances are present and what are absent but also in obtaining an estimate of the relative quantity of each constituent present. In order to accom- plish this with accuracy and certainty the analyst should know the principle involved in the chemical actions con- cerned, the reason for each reagent used, the result it produces and how this result is brought about. For the purpose of Qualitative Analysis it is usually desirable 1 2 INTRODUCTION and often necessary to separate the substances from each other and then apply some characteristic chemical reaction which can be easily and accurately recognized. 3. In the course of this outline the following terms will be used frequently, and the student should become familiar with them and their meaning at the outset. (a) A Reagent is a substance which produces a known reaction and is used for the purpose of obtaining a desired result. (b) A solution is a homogeneous mixture, the prop- erties of which change gradually with change in composition. Aqueous solutions are of most importance and are used almost exclusively in this outline. The student should learn, early in his course, to distinguish between solutions and colloidal suspensions. (c) Precipitate is a term usually applied to a solid substance which separates from solution on the addition of a reagent. It may be finely divided so that it settles very slowly and is not held by a filter paper. Such a precipitate is termed "colloidal." Precipitates are distinguished as crystalline, flocculent, curdy, colloidal, etc. (d) Residue is a term usually applied to that portion of a solid left undissolved by a given reagent or solvent. (e) Filtration consists in the mechanical separation of a solid from a liquid, by means of a suitable screen, which allows only the liquid to pass through. The liquid passing through is known as the filtrate, while the solid remaining on the filter is called the precipitate or residue. (/) Decantation consists in carefully pouring off the liquid from the solid which has been allowed to settle. This operation is employed with crystal- line precipitates where a thorough washing is IONIC THEORY 3 essential. The washing liquid, usually pure water in small portions, is thoroughly mixed with the solid and set aside until the undissolved portion has settled. The clear supernatant liquid is then poured off. The operation is repeated until the solid is sufficiently washed. In most of the operations of Qualitative Analysis suf- ficient washing may be obtained by blowing a fine stream of water from a wash bottle on to the filter containing the precipitate or residue. (g) Digestion is the term applied to a reaction between two or more substances that are mixed, hot or cold, and allowed to stand for some time with occasional stirring. 4. Ionic Theory. In his study of the effect of the solute on the properties of solutions, Raoult showed that if a non-volatile substance is dissolved in pure water cer- tain properties, viz., boiling-point, freezing-point, osmotic pressure and vapor pressure, vary with the concentra- tion of the solute. Furthermore, he showed that equi- molecular proportions of certain substances produce an equal effect on these properties, e.g., one molecular weight of cane sugar (Ci 2 H 2 2Oii =342 gms.) or of glucose (C6Hi 2 O6 = i7i gms.) or of glycerine (C 3 H5(OH) 3 = 92 gms.), when dissolved in 10 liters of water, lowers the freezing-point 0.186 C. and raises the boiling-point 0.052 C. From Avogadro's Law we learn that equal volumes of gases under the same conditions of temperature and pressure contain an equal number of molecules. A careful study of solutions has shown that a dissolved substance possesses properties similar to those of a gas having the same molecular concentration. A consider- ation of these and other facts leads to the conclusion that equimolecular proportions of all substances, whether liquid, solid or gas, contain equal numbers of molecules. The change in freezing-point, boiling-point, etc., is therefore 4 INTRODUCTION proportional to the number of molecules or particles of solute dissolved in a given amount of the solvent. Examina- tion of a great variety of substances in different solvents has demonstrated the truth of this conclusion. 5. When an acid, base or salt is dissolved in pure water, the change in freezing-point, boiling-point, etc., is greater than would be expected from the general rule. In their chemical relations these compounds show a marked dif- ference from those which follow the rule. Their reactions in solution are very rapid. In double decomposition reactions they seem to be composed of two or more radicals which act largely independently of each other. They are the only substances whose solutions conduct electricity. Furthermore, there are certain colored salts, e.g., copper chloride (CuCU, greenish-yellow), copper bromide (CuBr2, reddish-brown), and copper sulphate (CuSC^, blue), whose solutions on being diluted finally assume the same color (blue). In explanation of these facts Arrhenius, in 1885, proposed what is known as the "Ionic Theory.' 11 It assumes that acids, bases and salts, when dissolved in water, dissociate into two or more radicals or particles, that these particles carry an electric charge and that an equilibrium exists between the undissociated particles and their dissociation products. 6. When an electric current is passed through a solu- tion of an acid, e.g., HC1, H 2 SO4, etc., the hydrogen collects around the negative electrode and the remaining radical, (Cl), (SOO, etc., collects around the positive electrode. When the current is made to pass through a solution of a base, e.g., NaOH, KOH, etc., the hydroxide radical (OH) proceeds toward the positive electrode while the metal radical is carried toward the negative electrode. In the case of salt solutions, e.g., NaCl, Na 2 SO4, etc., the metal and acid radicals act in the same manner toward the electric current as if they were present as the base and acid respectively. A radical, therefore, proceeding under the influence of an electric current, always moves toward KINETIC THEORY AND IONIC EQUILIBRIUM 5 the same pole regardless of whether it is present as an acid, base or salt. From this it follows that the hydrogen radical must carry a positive electric charge and the hydroxide radical a negative charge. Similarly the metal radical will be charged positively and the acid radical negatively. A radical bearing an electric charge is called an ion and the process by which ions are formed from the undissociated molecules is called ionization. 7. Kinetic Theory and Ionic Equilibrium. When an acid, base or salt goes into solution the influence of the water causes it to dissociate, with the formation of posi- tively and negatively charged ions. On the basis of the kinetic theory we may assume that the continual move- ment and jostling about of the molecules causes them to split apart or dissociate. At the same time the dissociated particles or ions will collide with each other with the result that some of them will again unite to form undissociated molecules. At first the former reaction, dissociation, is more rapid, but as the number of ions increases their union becomes more frequent until after a time the rate of the two reactions will be equal, i.e., the number of molecules dissociating in a unit of time will be just equal to the number formed by the union of ions. When this condition is established the reaction is said to be in equi- librium. It is expressed by means of an equation as follows: A+B 8. Degree of Ionization. Experiment has shown that the proportion of a substance existing in the form of ions depends on its concentration in the solution, the more dilute solution having the higher per cent of ionization. This is explained on the basis of the kinetic theory, as follows: In concentrated solutions the ions are compar- atively close together and collisions will be relatively frequent, while in the more dilute solutions the ions are necessarily farther apart and the time between collisions will be greater, The result is that fewer of them will 6 INTRODUCTION unite, per unit of time, to form t non-ionized molecules. The per cent existing in the form of ions must therefore increase with increased dilution, a condition which agrees with experimental data. While it has been found that the degree of ionization in solutions made from salts is relatively high and approximately the same 1 for all salts of the same concentration, a wide variation exists in the cases of acids and bases. 9. The degree of ionization in normal and o.i normal solutions of the more common acids, bases and salts is given in the following table: Acids PER CENT IONIZED Bases PER CENT IONIZED N soln. o.i N oln. N soln. o.i N soln. H+cr H + NO 3 ~ H+HSO 4 - H+HC 2 O 4 ~ H+C 2 H 3 2 - H+HCO-T 78.4 82 5i 0.41 0.17 90 90 60 50(0.2 N) (soln.) 1-3 K+OH- Na+OH- Ba + + (OH) 2 - NH 4 +OH- 77 73 69 0.4 86 86 I-3I SALTS Approximate degree of ionization for salts in o.i N solution Type M+A-(e.g., KC1) 86 per cent Type M + +A 2 -(e.g., BaCl 2 ) 72 per cent Type M 2 + A (e.g., K 2 SO 4 ) 72 per cent Type M + + A (e.g., BaSO 4 ) 45 per cent 10. Law of Mass Action. From a careful study of chemical equilibria and rate of chemical action Gulberg and Waage showed that the speed of a reaction is directly proportional to the concentration of the reacting sub- stances. This is known as the Law of Mass Action. 1 Notable exceptions to this rule are HgCl? and Pb(C 2 HgO 2 )2, whose per cents of ionization are relatively very small. LAW OF MASS ACTION 7 When sodium chloride (NaCl) is dissolved in water the following equilibrium is established ; Since the number of NaCl molecules which dissociate in a unit of time is proportional to the concentration of the undissociated molecules, the speed of dissociation may therefore be stated mathematically as follows: Si =/lC N aCl where Si = speed of dissociation, C Na ci = molar concentra- tion of undissociated NaCl and f\ =the proportionality constant. In like manner the number of Na + and Cl~ which unite to form the undissociated NaCl per unit of time is proportional to the product of their concentrations and may be expressed mathematically as follows: S f c* + c* 2 J 2v^Na ' v_xci where S2 = speed of union, C Na + and C C i~ =the con- centrations of the Na + and Cl~ respectively and / 2 = the proportionality constant. When equilibrium is established the speeds of the opposing actions must be equal ; therefore /iCMaci = /2CNa + ' Ccf~ and by transposition C Na + -C C i- fi C NaC l /2 Since /i and / 2 are constants their ratio must be constant and C+ c* - Na * V^Ci -IT- P = K ^NaCl where K is a constant. This is the mathematical state- ment of the Law of Mass Action as applied to the ioniza- tion of sodium chloride, and shows that the product of the concentrations of the ions in the solution divided by the concentration of the non-ionized molecules is a constant 8 INTRODUCTION quantity which is independent of the source of the ions. K is known as the ionization constant. For highly ionized substances that are very soluble K varies consider- ably with the change in concentration, but for slightly ionized substances and those that are difficultly soluble it remains practically the same. The following examples will serve to illustrate. Experiment has shown that in a molar solution of acetic acid (HC2H3O2), 0.41 per cent of the acid is in the form of ions while 99.59 per cent of it remains in the non-ionized state. The ion concentration, therefore, is I X 0.004 1 =0.0041 and that of the non-ionized portion is 1X0.9959=0.9959, i.e., the condition of equilibrium in a liter of i molar acetic acid becomes HC 2 H 3 2 - H+ + C 2 H 3 2 - (0.9959 m l) (0.0041 mol) (0.0041 mol) Substituting these values in the Mass Law equation r* + r* ^-H ' V- K is found to have the value 0.0041 Xo.oo 4 i =OQ 0-9959 If the above molar solution is diluted to ten times its original volume the proportion of acid that exists in the form of ions will increase to 1.3 per cent. Using this value we obtain 0.1X0.013=0.0013 for the ion concen- tration and 0.1X0.987=0.0987 for the concentration of the non-ionized portion. Substituting these values in the Mass Law equation C H + C C2H302 - = 0.0013 xo.ooi 3 ooQQ CHC2H302 0.0987 a value is obtained which is in very good agreement with that obtained in the molar solution. In the case of ammonium hydroxide (NH 4 OH) experi- SOLUBILITY PRODUCT 9 ment shows that a o.i molar solution is 1.31 per cent ionized. The ion concentration, therefore, is o.i X 0.0131 = 0.00131 and the concentration of the non-ionized molecules is o.i Xo. 9869 =0.09869. Substituting these values in the mass-law equation we obtain for the ionization constant CW.CaH- = o.ooi 3 i XQ.OOI3I =0 . OOOOI73 C NH4 OH 0.09869 In a o.oi molar solution the per cent of ionization has been found to be 4.07, from which we obtain the following equation (o .01X0. 0407) X (o . 01 Xo . 0407) T ^ T r ^^ =K =0.00001 72 (o.oi Xo.9593) ii. Solubility Product. Since precipitation methods play an important part in Analytical Chemistry, a con- sideration of the mass law in its relation to saturated solutions is of considerable importance. When a sub- stance, such as sugar or salt, is placed in contact with a liquid, some of the molecules of the solid enter the liquid and in accordance with the kinetic theory move about in all directions within the liquid. After a time some of them will return to the solid, and as more of the solid dissolves the number of molecules returning to the solid will increase, until the number entering the liquid and the number leaving it in a unit of time are equal. When this condition prevails equilibrium is established and the number of dissolved molecules (molar concentration) is a constant. The solution is said to be saturated. It follows, therefore, that in a saturated solution of an acid, base or salt there must be an equilibrium between the undissolved solute, the undissociated molecules in solu- tion and its ions. In the case of the difficultly soluble salt, AgCl, this equilibrium may be expressed as follows: AgCl 1 -AgCl-Ag++Cl- 1 A line ( ) drawn beneath a symbol will be used to denote the undissolved solid, precipitate or residue. 10 INTRODUCTION and the equilibrium for the Law of Mass Action becomes C Ag + C C i~ = j^ C A gd Since the solution is saturated, C AgC i is a constant quantity and C A g+-Ccr =K-C AgC i=K Therefore, in a saturated solution of a given ionogen the product of the concentration of its ions 1 is a constant and is called the Solubility Product. In the following table are placed the solubility products at 1 8 C. of some of the more common substances met with in qualitative analysis. Substance K Substance K HgS 4.o-io- 53 Ag 2 Cr0 4 i.o-io- 12 CuS 8.5-IO- 45 Mg(OH) 2 3.4-IO- 11 CdS 3.6-IO- 29 BaCrO 4 I.6-IO- 10 PbS 4.2-IO- 28 BaSO 4 9.0- io~ 10 CoS 3.0-IQ- 26 AgCl 8.7-IO- 9 NiS I.4-IQ- 24 CaC 2 O 4 i.y-io- 9 ZnS I.2-IO- 23 CaCO 3 2.8-IO- 9 FeS i.5-io- 19 BaCO 3 I.9-IQ- 9 MnS I.4-IQ- 15 PbSO 4 i.o-io- 8 MgNH 4 PO 4 2.5-IO- 13 SrSO 4 2.8-IO- 7 The ion product, and therefore the solubility, of a substance may be altered in the following ways: (a) By the addition of a reagent containing a common ion. (b) By addition of a reagent which forms with one of the ions a slightly ionized compound. 1 This is true in the case of ionogens consisting of one cation and one anion. In other ases the solubility product should contain the ion concentration raised to a power equal to the number of ions that are alike in its formula, e.g., the solubility product for PbCl 2 should be written Cpb + + -C 2 ci~ =K. COMMON ION 11 (c) By addition of a reagent which unites with one of the ions to form a complex ion. (d) By addition of a reagent which alters the charge on one on the ions. (Oxidation and reduction.) (e) By addition of a strong acid or strong base to an amphoteric substance. 12. Common Ion. When a soluble chloride, such as HC1 or NH 4 C1, is added to an acid solution of a silver salt, silver chloride (AgCl) is precipitated. The solution will be saturated with respect to silver chloride when the ion product, C A g + -C C i~, reaches the solubility product value, K Ag ci, for silver chloride. Any further addition of chloride tends to increase the ion product above the solubility product value. This in turn disturbs the equilibrium AgCl-Ag++Cl- so as to decrease the concentration of silver ions. Since the concentration of the silver ion can be decreased only by uniting with chloride ions to form non-ionized AgCl the solution tends to become supersaturated with respect to AgCl molecules. Some of the silver chloride will there- fore be precipitated and the following equilibrium will result : At 25 the ion concentration of a saturated solution of silver chloride has been found to be 1.2 -io- 5 . From this is obtained for the solubility product, C Ag + Cci~ = K = i .44 io- 10 If to I liter of this solution o.oi mole (0.535 g m O of ammonium chloride (NH 4 C1) (86 per cent ionized) is added, the concentration of the chloride ions is increased by 0.01X0.86=0.0086, and C c r will be 0.0086+0.000012 = 0.008612 or about 700 times as large as in the original solution. Since CAg+'Ccr = 144- io~ 10 it will be seen 12 INTRODUCTION that the concentration of silver ions must decrease to about TU~O of its original value. The student should especially note that the completeness with which a given ion may be removed from solution in this way depends on the concentration of the non-ionized molecules in a saturated solution, since their concentration is not decreased by addition of a common ion. Complete precipitation, from the standpoint of Analytical Chemistry, is obtained only when this value is very small. 13. Slightly Ionized Compounds. The preceding para- graph has shown how the addition of a common ion may be used to remove a given ion from solution when the compound formed is difficultly soluble. In the same way a given ion may be reduced to almost nothing when the compound formed is soluble but very slightly ionized. If to a o.i molar solution of acetic acid (HC 2 H 3 O 2 ) (1.3 per cent ionized) a o.i mole of some soluble acetate, such as sodium acetate (NaC2H 3 O 2 ) (86 per cent ionized), is added, the large excess of acetate ions tends to increase the speed of union of H+ and C 2 H 3 O 2 - and so shifts the equilibrium HC 2 H 3 O 2 ^H+ +C 2 H 3 O 2 - to the left. From the equation o 2 - = o. 0013X0. 0013 = I 7 . IO - 5 C HC2 H 3 o 2 0.0987 it will be seen that C H C2H 3 02 cannot be appreciably increased owing to the small concentration of H + available. The product C H + 'C C2 H 3 o 2 ~ must therefore recover approxi- mately its original value. Since on the addition of sodium acetate, C C2H3 o 2 ~ becomes 0.086+0.0013 = 0.0873 or about 60 times its original value, C H + must be decreased to about ^o of its former magnitude. While the per cent of ionization of the salt must also decrease because of the presence of the acetate ions from the acid, the amount is negligible in proportion to its original value since the number of acetate ions is relatively so few (0.0013 : 0.086). Therefore when the compound formed is soluble a given HYDROLYSIS 13 ion can be reduced to almost nothing only when the com- pound is very slightly ionized. 14. Hydrolysis. Pure water ionizes to a slight extent into H f and OH~. Although the ions of water may be neglected when all the substances concerned in a given reaction are highly ionized, they become quite appreci- able and must be taken into consideration when the re- action involves substances that are very slightly ionized or difficultly soluble. Experiment has shown that the concentration of H+ and therefore of OH+ in pure water at 25 is io~ 7 . The concentration of the non-ionized molecules is therefore very large in comparison and may be considered constant. From the Law of Mass Action, then, the product of the concentrations of the ions becomes a constant. Since C H + =C OH ~ =IO" 7 we have for the ion product C H + -C H~ = io- 14 . In any solution, there- fore, the concentration of H + multiplied by the concen- tration of OH~ must equal the ion product constant, io~ 14 . An increase of H+ must result in a decrease of OH- and vice versa. When sodium acetate (NaC2H 3 O 2 ), a highly ionized salt, is dissolved in water the concentration of C2HaO2~ may become so large that its product with the H+ of the water will exceed the ionization value for the slightly ionized acetic acid (H^HsC^). Some of the C2HaO2~ and H + therefore unite to form the non-ionized acid which momentarily reduces the ion product for water below io~ 14 . The result is that more water ionizes until the product C H + -C OH ~ again reaches io~ 14 . The student should note that C H + is now less than C OH ~ and therefore the solution becomes basic. On the other hand, if ferric chloride (FeCls) is dissolved in water the concentration of Fe +++ from the highly ionized salt, multiplied by the concentration of OH~ already present in the water, may exceed the ionization value for the very slightly ionized ferric hydroxide and form non-ionized Fe(OH) 3 . The ion product for water is thus momentarily reduced below 14 INTRODUCTION io~ 14 . More water must therefore ionize until the prod- uct C H + -C H~ = io~ 14 is reached.' C H + is now greater than COH~ and the solution reacts acid. It will be seen therefore that the ions of water must be taken into con- sideration when a substance is involved either one of whose ions may unite with one of the ions of water to form a very slightly ionized compound. Should the compound be difficultly soluble the equilibrium may be shifted to completion and a given ion removed from solution. 15. Complex Ion. When ammonium hydroxide is added to a solution of a copper salt, copper hydroxide (Cu(OH)2) is at first precipitated, but on the addition of an excess of the reagent the Cu(OH) 2 precipitate is dis- solved and a deep blue solution is obtained. It would seem from the foregiong discussion and the principle of the Law of Mass Action that an excess of the reagent should produce a more complete precipitation as was found to be the case with silver chloride (see 12 above). It will be remembered from the study of general Chemistry that when ammonia (NH 3 ) is dissolved in water only a small portion of it reacts with the water to form ammonium hydroxide (NH 4 OH), the greater part of it remaining in the solution as ammonia (NHs). The following equilib- rium must therefore exist in the solution: H 2 O +NH 3 ^ NH 4 OH NH 4 + +OH-. An examination of the deep-blue copper solution .shows the presence of the complex ion Cu(NH 3 )4 ++ . From the principle of the solubility product it is evident that any increase of OH~ above that necessary to reach the solubility product for Cu(OH) 2 must result in a decrease in the Cu ++ concentration. The concen- tration of the Cu ++ may be decreased either by the form- ation of non-ionized Cu(OH)2 and consequent precipitation or by its union with free ammonia to form the complex AMPHOTERIC SUBSTANCES 15 ion Cu(NH3)4 ++ . The high proportion of free ammonia and the slight dissociation of the complex ion Cu(NH 3 )4 ++ both influence the equilibrium toward the formation of the complex ion, hence the net result is that the equilibrium Cu(OH) 2 - Cu(OH) 2 - Cu++ +20H- Cu(NH 3 ) 4 ++ will shift toward the formation of the complex ion and the ion product will be decreased below that of the solubility product value; more Cu(OH)2 will dissociate and the precipitate will pass into solution. The use, therefore, of a reagent which will react with a given ion to form a complex ion may be made in order to bring a substance into solution, to prevent precipitation or to remove an ion from the field of action. 16. Amphoteric Substances. An amphoteric element is one whose hydroxide in solution ionizes both as an acid and as a base, i.e., it produces both hydrogen and hydroxyl ions. When a strong acid, such as HC1, is added to a precipitate of aluminium hydroxide (A1O 3 H 3 ) the pre- cipitate is dissolved and experiment shows that the alu- minium is present in the solution as the positive aluminium ion (Al +++ ). On the other hand, when a strong base, such as NaOH, is added to the aluminium hydroxide precipitate, the precipitate is dissolved; but experiment shows that the aluminium is present in the solution as negative aluminate ions (A1O 2 ~). The following equi- librium is therefore assumed to exist in a neutral solution of aluminium hydroxide : A1+++ + 3 OH- A1O 3 H 3 ^ H+ +H 2 A1O 3 - ^ H+ +A1O 2 - +H 2 O When a strong acid, furnishing its high concentration of H + , is added the above equilibrium is disturbed owing 16 INTRODUCTION to the union of H + with the OH~ present to form the very slightly ionized water. Non-ionized A1O 3 H 3 then dissociates further to produce more OH~ with the final result that the aluminium hydroxide is dissolved and the aluminium remains in solution as A1+++; i.e., the equi- librium shifts to the left and A1O 3 H 3 acts as a base. When a strong base is added the high concentration of OH~ tends to use up the H + present in forming water as above. This causes a further dissociation of A1O 3 H 3 to produce more H + , with the final result that the aluminium hydrox- ide is dissolved and the aluminium remains in the solu- tion as A1O 2 ~; i.e., the equilibrium shifts to the right and A1O 3 H 3 acts as an acid. It should be noted that if a -weak base, such as NH4OH, is substituted for the strong base mentioned above the effect will be very much less noticeable, owing to the much smaller concentration of OH-. 17. Oxidation and Reduction. When iron is acted upon by hydrochloric acid, hydrogen is displaced and iron passes into solution according to the following equation: Fe+2HCl-FeCl 2 +H 2 Considered from the ionic standpoint this gives Fe +2H+ +2Cl--> Fe++ +2C1- +H 2 Now if a stream of chlorine gas is passed through the solu- tion a further change takes place as follows: C1 2 +2Fe ++ +4Cl-- 2Fe+++ +6CK In passing into solution the iron has become positively charged, while at the same time charged hydrogen has become neutral and neutral chlorine has become negatively charged. A free atom may be said to consist of a positively charged nucleus surrounded by a number of negatively OXIDATION AND REDUCTION 17 charged particles called electrons. These electrons are capable of existing independently of the atom and hence may leave one atom and attach themselves to another. The mechanism, therefore, of the above reactions may be stated briefly as follows: An atom of iron, capable of losing electrons, comes in contact with a hydrogen ion (hydrogen atom one electron). Two electrons leave the atom of iron and attach themselves to two hydrogen ions, which are in turn neutralized. The loss of these two electrons has, therefore, left the iron positively charged. In the second reaction a chlorine atom, capable of holding an additional electron, comes in contact with an iron ion and receives an electron from it. The chlorine, therefore, becomes negatively charged while the iron remains with a higher positive charge. The quantity of electricity equivalent to that carried by an electron is called a " unit charge " and may be either positive or negative in character. The number of excess " unit charges " carried by an atom or ion is numerically equal to its valence. Valence, therefore, may be either a positive or a negative number depending on whether the atom or ion holds less or more electrons than is sufficient to neutralize the positive nucleus. It follows, therefore, that the valence of an element in the free state is zero and that the algebraic sum of the positive and negative valences in any compound is zero. Oxida- tion consists in the loss of one or more electrons by an atom or ion, i.e., an algebraic increase in valence. Reduc- tion consists in the addition of electrons to an atom or ion, i.e., an algebraic decrease in valence. It will be seen, therefore, that oxidation and reduction must accom- pany each other and be equivalent in amount; i.e., in a given reaction if an element or ion loses one or more electrons those electrons must attach themselves to some other element or ion. In writing equations of oxidation and reduction the student should first write the skeleton equation; e.g., 18 INTRODUCTION when H 2 S is passed into a solution of HNOs free sulphur is produced and the HNOs is reduced to nitric oxide (NO). H 2 S+HN0 3 ->S+NO+H 2 He should then note any change in valence, i.e., what elements have lost or gained electrons during the reaction, and the number lost or gained by each atom. In the above equation sulphur has changed from a valence of negative 2 in H 2 S to zero in free sulphur; i.e., each atom of sulphur has lost 2 electrons. Nitrogen, on the other hand, has changed from a positive valence of 5 in HNOs to positive 2 in NO; i.e., 3 electrons have attached them- selves to each nitrogen atom. Since the total number of electrons lost by one element must be equal to the total number gained by the other, it is evident that 3 molecules of H 2 S will furnish just enough electrons to supply those necessary to change the nitrogen in 2 molecules of HNOs to NO. The balanced equation, therefore, becomes 3H 2 S +2HNO 3 -> 38 +2NO +4H 2 O For purposes of balancing equations of oxidation and reduction the valence of combined hydrogen should always be considered as positive i and that of combined oxygen as negative 2. A few of the more important oxidizing and reducing agents are given in the following table: Oxidizing Agents Reducing Agents 1. Halogens (Cl, Br, I) i. SnCl 2 2. HNO 3 2. H 2 S 3. Aqua regia 3. Nascent hydrogen 4. KC1O 4 4. SO 2 5- Na 2 O 2 5. H 2 C 2 O 4 6. K 2 Cr 2 O 7 6. Alcohol 7. KMnO 4 8. Pb0 2 LABORATORY SUGGESTIONS 19 LABORATORY SUGGESTIONS Qualitative Analysis has to do with both dry and wet reactions. The dry reactions are those used largely in blow-pipe analysis and will not be considered here. Since wet reactions will be used almost exclusively in this out- line a brief consideration of some of the most important processes is given. Filtration. Since a finely divided precipitate not only tends to pass through the filter and so make its separation difficult but also tends to clog the filter paper and thus render the process of filtration slow and tedious, it is necessary to have the particles as large as possible. This is accomplished most effectively by adding the precipitating agent slowly to a hot solution. Whenever permissible, time will usually be saved by filtering a solution while hot, since hot water passes through the filter paper more rapidly than cold water. The rate of filtration is also influenced very largely by the position of the filter paper in the funnel. The paper should fit closely against the funnel so that no air passages exist between. This may be accomplished by folding the circular filter paper in half and then folding again as shown in Fig I. The second fold should be pinched together at the point, opened between the longer fold so as to form an inverted cone and pressed gently into the funnel until it fits snugly against the glass all the way around. With- out removing it from the funnel the second fold may now be creased and the paper wet with a stream of water blown from the wash bottle to hold it in position (Fig. 2). Wash Bottle. After checking in his apparatus each student should make a wash bottle, using a 500 cc. or 750 cc. Florence flask. The wash bottle should be made sufficiently compact so that it can be easily held and the nozzle manipulated with one hand. A convenient form is shown in Fig. 3. Record of Results. The student should keep a careful and accurate record of all results obtained in his analysis INTRODUCTION of " unknowns." The keeping of this record not only enables the student to understand the principles involved and to follow the procedures more easily, but also it often enables the instructor to determine the causes of errors and so help the student to avoid repeating them. The FIG. i. FIG. 3. following form for " unknowns " has been found to be very easily kept and quite satisfactory: No. Substance Reagent Result Conclusion i Unknown NH 4 C1 White ppt. Group I present 2, Ppt. i 2N. HC1 White res. PbCl 2 , Hg 2 Cl 2 , AgCl 3 Fil. 2 H 2 S Soln. No Bi or Sb 4 Res. 2 Hot H 2 O White res. H g2 Cl 2 ,agCl 5 Fil. 4 H 2 SO 4 Soln. No Pb 6 Res. 4 NH 4 OH Black res. Hg present 7 Fil. 6 HNO 3 White ppt. Ag present LABORATORY SUGGESTIONS 21 The student will also find it of considerable benefit to keep a record of his results obtained in preliminary experiments by underlining and noting the color of all precipitates in outline form similar to that of Table I and the following tables. PART II THE SYSTEMATIC ANALYSIS Cations (metal ions) PREPARATION OF SOLUTION If the unknown substance is a liquid or solution treat by (i); if a solid non-alloy treat by (2) and if an alloy treat by (3). For the purpose of a complete qualitative analysis the unknown should be divided into four parts as follows: First part for organic matter and general infor- mation. Second part for analysis of cations (metal ions). Third part for analysis of anions (acid ions). Fourth part for special tests and in case of accident. (1) Unknown Liquid. Test with litmus for acidity. Evaporate a known volume to dryness in a porcelain dish and note the amount of residue. If organic matter may be present test the residue by (4). In case organic matter is known to be absent treat an amount of the solution which contains about i gram of solid by (10). (2) Unknown Solid (non-alloy). Treat a small portion (about o.i gm.) of the finely powdered substance for organic matter by (4). If organic matter is abse.nt add to another small portion in a test-tube 10-15 cc. of water, and shake the mixture thoroughly. If it fails to dissolve, heat to boiling. In case the substance is insoluble in water try to dissolve another small portion in 5 cc. of 22 UNKNOWN SOLID (NON-ALLOY) 23 6N. HNOs. From the knowledge gained by the above tests treat about i gram of the solid according to (a), (b) or (c). (a) If the substance is soluble in water or dilute HNOz, dissolve about i gram of it using as little of the 6N. acid as possible (see Discus- sion 3) and treat by (10). (b) If the substance is soluble in dilute HCl, dissolve about i gram. If more than 5 cc. of the 6N. HC1 is used evaporate the solution to 5 cc. (see Discussion 3) dilute with water to a volume of 100 cc. and treat by (20). (c) If the substance is insoluble in dilute acid, add to about i gram of the finely powdered sub- stance in a porcelain dish 6 cc. of I2N. HC1, cover the dish with a watch glass and heat gently. (See Discussion 4.) If a residue remains, cool, add 2 cc. of i6N. HNOs and heat the mixture. In either case finally evapo- rate just to dryness, moisten the residue with I2N. HC1 and again evaporate to dryness. Heat the residue to 120130 till it is thoroughly dry, keeping the dish in motion over a small flame. Loosen the residue with the end of a glass rod, add just 5 cc. of 6N. HC1 (see Dis- cussion 3) and pulverize with the rod any large particles. Cover the dish and warm the mix- ture, taking care that none of the acid evapo- rates. Add 10 cc. of water and heat to boiling. Filter while hot, treat the filtrate by (20) and the residue by (5), (6) or (7). (See Discus- sion 5.) (3) Unknown Solid (Alloy). To about 0.5 gram of the finely divided material in a porcelain dish add 10 cc. of 6N. HNO 3 , cover with a watch glass and warm gently as long as the action continues, adding small portions of 24 CATIONS (METAL IONS) l6N. HNOs from time to time if the action is renewed thereby. Finally evaporate just to dryness, add just 5 cc. of 6N. HNO 3 and 15 cc. of water. Heat to boiling, and if a residue remains, filter, treat the filtrate by (10) and the residue by (2, c). (See Discussion 6.) (4) Organic Matter. To determine whether organic matter is present in an unknown a small portion of the solid is placed in a hard glass test-tube or in a glass tube closed at one end and heated to dull redness. If the sub- stance chars (a black color may be due to certain metallic oxides) and emits a burnt odor, organic matter is present and should be removed as follows (see Discussion 7). Place about I gram of the solid substance (more if the amount of organic matter is large) in a porcelain dish and heat gently with 5 cc. of cone. H 2 SO4 until it is well charred. Cool, add slowly and with constant stirring i6N. HNOs until violent action ceases. Warm gently for a few minutes and then heat more strongly, keeping the contents well stirred, until the substance is thoroughly charred. Repeat the process until the mixture becomes light straw colored and remains so when strongly heated. Treat by (a) or (b). (a) If the substance has dissolved completely, evapo- rate under a hood to 1.5 cc., cool and pour the contents into 15 cc. of water. If there is a residue, heat to boiling and boil as long as it seems to be dissolving. Filter off any remaining residue, wash thoroughly and treat by (6). Treat the filtrate by (10). (b) If the substance has not dissolved completely transfer to a platinum crucible and treat by (5), or if the platinum crucible is not available, evaporate to 1.5 cc. as in the preceding para- graph, pour the contents into 15 cc. of water, heat to boiling, filter, wash thoroughly and treat the filtrate by (10). Treat the residue by (7). TREATMENT WITH Na 2 CO 3 SOLUTION 25 (5) Treatment with H 2 F 2 . Transfer the residue from (2, c) or the mixture from (4, b) to a platinum crucible, add enough cone. H 2 SO4 to make a total volume of 3 cc. (see Discussion 3) ; heat the mixture with a moving flame until the thick white fumes of H 2 SO4 appear. To test for silicate or silica add carefully from a loop of a platinum wire 56 drops of pure cone. H 2 F 2 and warm the mixture over a steam bath. The formation of gas bubbles shows the presence of silica or silicate. (See Discussion 5.) Now add 2-5 cc. more of the pure cone. H 2 F 2 , cover the crucible and digest the mixture on the steam bath for about fifteen minutes unless solution takes place more quickly. Remove the cover and evaporate carefully until the white fumes of H 2 SO4 appear. Treat by (a) or (b). (a) If there is no residue or precipitate evaporate carefully to dryness. If there is still no residue, or only an insignificant one, the material con- tained only silica or silicate and may be dis- carded. (b) If there is a residue or precipitate pour the con- tents of the crucible into 15 cc. of water, rinsing out the crucible with the resulting solution. Boil the mixture gently as long as any of the residue seems to dissolve. Filter and treat the filtrate by (10). Wash the residue with iN. H 2 SC>4, rejecting the washings, and treat by (7). (6) Treatment with Na 2 CO 3 Solution. Mix the resi- due obtained in (2, c) or (4, a) with 10 parts of solid Na 2 COs and 20 cc. of water and boil the mixture for about five minutes. Filter and reject the filtrate (see Discussion 8). Wash the residue and dissolve it by adding 6N. HC1 until the solution remains acid; then add just 5 cc. more of the acid and 10 cc. of water. Filter if necessary and treat the filtrate by (20). If there is a residue undissolved by this treatment treat it by (7). 26 CATIONS (METAL IONS) (7) Fusion with Na 2 CO 3 . Mix the residue from (2, c), (4, b) or (6) with ten times its weight of Na 2 CO 3 in a plati- num or nickel crucible (see Discussion 9) and heat over a very hot flame until complete fusion takes place. If necessary to secure a clear fusion, add 0.1-0.3 gram of NaNOs (see Discussion 10). Cool, place the crucible and contents in a dish and add carefully 6N. HC1 until the solution is acid. Evaporate to dryness and heat to 120- 130 to render the silica insoluble. Add just 5 cc. of 6N. HC1 (see Discussion 3) and 10 cc. of water and heat to boiling. Filter to remove silica and treat the filtrate by (20). DISCUSSION 1. Difficultly soluble solids are more easily brought into solution if they have previously been reduced to a fine powder. This is usually done by grinding the solid in a porcelain or agate mortar. In the case of very hard substances, such as certain minerals and rocks, a heavy porcelain or iron mortar should be used to reduce them to small particles. 2. The preliminary tests with water and dilute acids should be carried out, since they furnish important indi- cations as to the nature of the constituents present and often enable the analyst, especially the beginner, to obtain a solution more quickly and easily. 3. In order that the solution may have the proper acid concentration for the precipitation of Group II metals, just 5 cc. of 6N. acid should be present. If, therefore, the sample can be dissolved by the use of 5 cc. of 6N. acid, a considerable saving of time will be obtained. 4. A mixture of HC1 and HNO 3 (3:1), known as aqua regia, is a very powerful oxidizing agent and is often used very effectively as a solvent. It should be used, however, only in case dilute acids or cone. HC1 prove ineffective, since compounds comparatively soluble in these reagents may be rendered insoluble by its oxidizing DISCUSSION (i-io) 27 action; e.g., antimony and tin compounds may be oxidized to insoluble antimonic oxide (Sb 2 O5) and metastannic acid (H 2 SnO3)n respectively. The action of I2N. HC1 is noted, therefore, before the HNO 3 is added. Such substances as MnO 2 and PbO 2 are reduced and dissolved by I2N. HC1. Hot I2N. HC1 slowly dissolves such oxides as Sb 2 O5, SnO 2 , Fe 2 O 3 and A1 2 O 3 . Upon the addition of cone. HNO 3 , gold, platinum and HgS are dissolved. Silver compounds are changed by HC1 to AgCl, somewhat soluble in concentrated acid but left almost completely in the residue on the subsequent addition of dilute HC1. 5. A residue undissolved by aqua regia may consist wholly of silica or silicates. The H 2 F 2 treatment decom- poses most silicates with the formation of SiF 4 , a gas insoluble in cone. H 2 SO 4 , and hence volatilizes. The treatment is, therefore, very effective for decomposing rocks, ores or other substances which might contain silica or silicates. The residue undissolved by H 2 SO 4 may contain the sulphates of barium, lead, strontium, calcium and chro- mium. It may also contain bismuth as basic sulphate and antimony as Sb 2 Os along with undecomposed AgCl. 6. Most alloys are attacked by cone. HNO 3 , all of the elements present going into solution except antimony, tin, carbon and silicon. Antimony, tin and silicon are oxidized to Sb 2 Os, (H 2 SnO 3 ) ra and H 2 SiO 3 respectively, all of which form white amorphous precipitates. Certain alloys, especially those containing iron and alu- minium, are more readily attacked by HC1. Treatment of the residue with HC1 and aqua regia not only brings these alloys into solution but tends to dissolve the oxides of tin and antimony formed by HNO 3 . The HNO 3 treatment is made first in order to remove any silver or lead which would be precipitated as chlorides if aqua regia were used. 7. Certain kinds of organic matter, such as sugars, 28 CATIONS (METAL IONS) tartaric acid, etc., prevent the precipitation of aluminium and chromium hydroxides. Large quantities of organic matter of any kind interfere in precipitations, filtrations, etc. Therefore, if organic matter is present it should be removed before beginning the systematic analysis. Organic matter may be removed by the H 2 SO4 and HNOs treat- ment as outlined, or by ignition. The latter, however, is inadvisable in a systematic analysis since such substances as mercury and arsenic are volatilized thereby. 8. The sulphates of barium, lead, strontium, calcium and bismuth are converted into carbonates by boiling with Na2COs. A second treatment is sometimes necessary to convert all the barium into the carbonate. The car- bonates are readily dissolved in HC1. Anhydrous chromic sulphate is converted into the hydroxide by boiling with Na2COs. The hydroxide is soluble in HC1. AgCl is only slightly attacked by the Na2COs solution. 9. Most substances are decomposed and rendered soluble by fusion with Na2COs, the basic elements form- ing carbonates and the acidic elements forming sodium salts. In some cases, however, the carbonate is decom- posed with the formation of the oxide or even the metal. For this reason care must be exercised in the use of a plat- inum crucible, since these metals readily alloy with the platinum. Substances which might contain any of the metals in Groups I and II should not be fused with an alkali flux in platinum. Although a nickel crucible may be used, the latter is attacked to such an extent that a subsequent test for this element or for the alkali metals is rendered unreliable. 10. The addition of NaNO 3 to the Na 2 CO 3 fusion serves to oxidize certain substances not acted upon by the Na2COa alone. Sulphides are oxidized to sulphates, chromium compounds to chromates and manganese com- pounds to manganates. If the fusion is to be made in platinum the quantity of NaNOs added should be as small as possible. PRELIMINARY EXPERIMENTS 29 Group I Prelminary Experiments Bi+ ++, Sb+++, Pb++, Hg+, Ag+ In connection with the following experiments study Table I. Experiment i. Introduce into separate test-tubes 5 cc. portions of the test solutions containing the above ions. Test for acidity with litmus paper and if not already acid make distinctly acid with 6N. HNOs. In each case dilute with water to a volume of 10 cc. and add NEUCl as long as a precipitate continues to form. Decant, or filter if necessary, and treat the precipitates as directed in Exp. 2. Write equations. NOTES. Water, though very slightly ionized, reacts with salts of the less basic metals to form oxy-compounds or even acids. This reaction is called hydrolysis. The ionization of water is prevented in large meas- ure by the presence of strong acids or strong bases. Bi and Sb are both weakly basic in character and hence tend to react with the water. This tendency is aided either by a high concentration of their salts or by a low concentration of acid. The addition of NH 4 C1 as directed above, there- fore, may or may not give a precipitate, since the neutral chlorides are soluble. (See Introduction 14). If hydrolysis occurs, however, the change may be considered to proceed as follows: BiCl s -> Bi(OH) 2 Cl-> BiOCl Experiment 2. To each of the above precipitates (Exp. i) add 5-10 cc. of cold 2N. HC1 and mix thoroughly. Note the solution of BiOCl and SbOCl. (Difference, separation of Bi and Sb.) Filter off the solution from the PbCl 2 and saturate the nitrate with H 2 S. Saturate the solutions containing Bi + + + and Sb + + + with H 2 S. Write all equations. Experiment 3. Add to the precipitates of PbCl 2 , Hg 2 Cl 2 and AgCl (Exp. 2) 10 cc. of boiling water. Note the solution of PbCl 2 . (Difference, separation of Pb.) 30 CATIONS (METAL IONS) Experiment aa. Divide the PbCl 2 solution (Exp. 3) into two parts and add K 2 Cr 2 O? to the one and H2SO4 to the other. Note the color and nature of the precipi- tates formed. Write equations. Which of the above reagents should give the more satisfactory test for Pb + +. Experiment 4. Decant the liquid from the residues of Hg 2 Cl 2 and AgCl (Exp. 3) and add NH 4 OH as long as the reaction seems to continue. Note the solution of AgCl. (Difference, separation of Hg.) NOTES. In the presence of NH 4 OH, Hg 2 Cl 2 is changed by auto- oxidation to white HgNH 2 Cl and black, finely divided Hg. A part of the mercury in its reduction to the metallic state oxidizes the remainder to the bivalent condition. The compound HgNH 2 Cl may be considered to be formed from HgCl 2 by replacing a chlorine atom by the univalent radical (NH 2 ) . The reaction proceeds as follows : Hg 2 Cl 2 + 2 NH 4 OH-+ HgNH 2 Cl +Hg +NH 4 C1 + 2H 2 O. A solution of NH 4 OH contains a considerable proportion of NH 3 molecules which unite readily with Ag + to form the complex ion Ag(NH 3 ) 2 + . Therefore, when AgCl is treated with NH 4 OH the Ag+ is removed by the formation of the complex Ag(NH 2 ) 2 + which shifts the equilibrium toward the formation of more Ag + (Le Chatelier's Prin- ciple) with the result that the AgCl is dissolved. The reaction proceeds as follows: ^ Ag(NH 3 ) 2 Cl Experiment 5. Acidify the solution of Ag(NH 3 ) 2 Cl (Exp. 4) with HNOs- A white precipitate of AgCl is obtained, owing to the removal of NHs in the reversible action above. OUTLINE OF ANALYSIS (TABLE I) 31 TABLE I OUTLINE FOR THE SYSTEMATIC SEPARATION AND DETECTION OF GROUP I Ions present in acid solution Reagent Bi+++ Sb+++ Pb++ Hg+ Ag+ I NH 4 C1 BiOCl (?) SbOCl (?) PbCl 2 H g2 Cl 2 AgCl 2 HCHaN) BiCl 3 SbCl 3 PbCl 2 Hg 2 Cl 2 AgCl H 2 S B 2 iS 3 * Sb 2 S 3 * 1 1 ~T 3 H 2 (Hot) (a) K 2 Cr 2 O 7 ?l*tl< Qrt&ny^ PbCl 2 (a) PbCr0 4tf , Hg 2 Cl 2 AgCl 1 (&) H 2 S0 4 4 NH 4 OH ( fe ) PbS0 4 HgNH 2 Cl + Hg Ag(NH 3 ) 2 Cl 5 Hn0 3 B^ c /C. AgCl * May be added to Group II ppt. or tested according to method outlined in Group II. ANALYSIS Group I , Pb++, Hg+, Ag+ (10) Precipitation. To about 25 cc. of the solution of the substance acidified with HNOs (see Discussion n) add NIHUCl solution as long as a precipitate continues to form. Mix thoroughly and allow the mixture to stand for two or three minutes. Filter and treat the precipitate by (n). Reserve the filtrate for analysis of Group II. (20). (n) Separation of Bismuth and Antimony. Pour re- peatedly through the filter (10) a cold 10 cc. portion of 2N. HC1. Treat the residue by (12). Dilute the filtrate with an equal volume of water and saturate with H 2 S. A brown precipitate indicates the presence of bismuth. An orange-red precipitate indicates the presence of antimony and the absence of bismuth. Confirmatory tests may be made according to the methods outlined in Group II, or the precipitate may be added to that of the Group II x sulphide (20) and analyzed according to the general scheme. (12) Separation and Detection of Lead. Pour re- peatedly through the filter (n) a 10 cc. portion of boiling ;, 32 CATIONS (METAL IONS) water. Cool, acidify with HC 2 H3O 2 and add a slight excess of K 2 Cr 2 O7 solution. The formation of a yellow crystalline precipitate shows the presence of lead. If lead has been found, wash the residue left by the hot water, and treat it by (13). (See Discussion 14.) (13) Detection of Mercury. Pour repeatedly through the filter (12) a 5-10 cc. portion of NH 4 OH. A black residue left on the filter paper shows the presence of mercury. Treat the filtrate by (14). (14) Detection of Ag. Acidify the filtrate (13) with HNOs. The formation of a white curdy precipitate undis- solved by the HNOs shows the presence of Ag. (See Dis- cussion 15.) DISCUSSION 11. The chlorides of lead, silver and mercury (ous) are insoluble in acid of moderate concentration but tend to redissolve in strongly acid solutions, owing to the forma- tion of complex ions such as AgQ 2 ~. If the solution to be tested is strongly acid with an unknown acid it should be nearly neutralized with NH 4 OH before the addition of NFUCl. Complete neutralization causes the precipi- tation of metal hydroxides and oxy-compounds which may not readily dissolve on the subsequent addition of acid. It is preferable, therefore, not to completely neutralize at this time. 12. Owing to the rather high solubility product of PbCl 2 it is often incompletely precipitated in Group I. Its solubility, however, is considerably lessened by the addi- tion of a large excess of NHUCl becau.se of the common ion effect. The mass-law equation representing the solu- bility product of PbCl 2 is C Pb ++ C 2 c r =K. Since the prod- uct of the concentrations of its ions must under all con- ditions have a definite value it is readily seen that an increase of Cl~ must result in a decrease of Pb ++ . The Pb ++ , however, can decrease only by forming non-ionized PbCl 2 , and if the solution is saturated with respect to DISCUSSION (11-15) 33 PbCl 2 molecules, precipitation must take place. The reac- tion may be represented by the following equilibrium: PbCl 2 ^ PbCl 2 Pb++ +2C1- 13. As was pointed out in the preliminary experi- ments, the precipitation of bismuth and antimony in this group is due to hydrolysis. Application of the Law of Mass Action to this reaction shows that their precipitation may be practically or wholly prevented not only by an excess of acid but also by an excess of chloride ion. For the solubility product of BiOCl we have the expression C B i +++ -C Cci~=K. The excess of H + from the acid unites with the O to form the very slightly ionized H 2 O and so reduces the concentration of O . This in turn tends to reduce the ion product and hence causes the Bi to remain in solution. An excess of Cl~ would tend to shift the equilibrium toward the formation of non- ionized Bids and so reduce the concentration of Bi + ++ . The result again is to reduce the ion product and prevent precipitation. The same reasoning applies to the formation of SbOCl. (The student should make the application.) 14. The solvent action of hot water on PbCl 2 is some- what slow, hence the necessity of pouring the water repeatedly through the filter containing the mixed chlorides. Any undissolved PbCl 2 left on the filter may react with the NH 4 OH, subsequently added for the separation of mercury and silver, to form a basic chloride, PbOHCl. This causes a turbidity in the filtrate containing the dis- solved silver. The presence of this, however, does not interfere with the test for silver, since it is readily dis- solved by the HNO 3 . 15. The presence of a large quantity of mercury renders the test for silver less delicate, owing to the reduc- ing action of free mercury on the silver salt. Hg+2AgCl^HgCl 2 +2Ag The silver thus reduced is insoluble in NIHUOH and 34 CATIONS (METAL IONS) is, therefore, left in the residue mixed with the and free mercury. In case the quantity of silver is relatively small as compared to the mercury it might be completely reduced and hence remain in the residue. If a heavy black residue is obtained on the addition of NH 4 OH and silver is not found in the nitrate, the residue should be tested for silver as follows: Dissolve the residue in a small quantity of aqua regia, dilute with water and filter off the dissolved mercury. Leach the residue with NH 4 OH to dissolve any AgCl, and test the filtrate for silver as outlined in Analysis (14). GROUP II Preliminary Experiments Group II, Copper Division Hg ++ , Pb ++ , Bi +++ , Cu++, Cd+ + In connection with the following experiments study Tables II and III. Experiment 6. Dilute separate 5 cc. portions of each of the test solutions containing the above ions with an equal volume of water and saturate with H^S. Allow the precipitates to settle; then decant off the supernatant liquid (filter if necessary) and wash each of the precipitates by decantation with about 10 cc. of water in order to remove the remaining acid. Write equations. Experiment 7. To each of the above precipitates (Exp. 6) add 10 cc. of 2N. HNO 3 and heat to boiling. Note the insolubility of HgS. (Difference, separation of Hg.) Write equations. NOTE. If too strong HN0 3 is added or if the boiling is long continued the HgS is largely converted into a white difficultly soluble substance of the probable formula 2HgS-Hg(N0 3 )2 while some may be completely dissolved as Hg(N0 3 ) 2 . PRELIMINARY EXPERIMENTS 35 Experiment 7a. Pour off the liquid above the residue of HgS (Exp. 7) and add 5 cc. of aqua regia (3 parts of I2N-HC1 to i part of I6N-HNO3) and warm gently until solution is complete. Evaporate almost to dryness, add 5-10 cc. of water and then stannous chloride (SnCy drop by drop. Write equations. NOTES. The failure of the HgS to dissolve in 2N. HN0 3 is due not only to the small number of its ions in a saturated solution but also to the slow removal of the sulphide ion by oxidation with the dilute HNO 3 . Aqua regia is a much stronger oxidizing agent. It readily oxidizes the sulphide ion and so brings the mercury into solution. HgCl 2 is reduced to white Hg 2 Cl 2 by the action of SnCl 2 . This is then further reduced to free Hg by an excess of SnCl 2 , hence the change in color from white to gray. Experiment 8. To each of the solutions containing Pb++, Bi +++ , Cu ++ , and Cd++ (Exp. 7) add i cc. cone. H 2 SO4 and evaporate until the white fumes of SOs begin to appear. Cool and add 10 cc. of water. Note the fact that PbSCU remains insoluble. (Difference, separation of Pb.) Write equations. NOTES. PbS0 4 is somewhat soluble in HN0 3 . In order to effect a complete precipitation of lead it is necessary to remove the HNO 3 . This may be done by evaporation since the boiling-point of HN0 3 (120.5) is so much lower than that of the H 2 SO 4 (338). The appearance of the white fumes of SO 3 is evidence of the complete removal of HN0 3 . Bi 2 (SO 4 ) 3 ordinarily dissolves on the addition of water but if much bismuth is present a coarsely crystalline precipitate of (BiO) 2 S0 4 may separate out slowly when cold, and more quickly when heated. Experiment 8a. Decant or filter off the clear liquid from the insoluble PbSCX and dissolve it by shaking with 10 cc. of ammonium acetate (NH 4 C2H 3 O2). Acidify the solution with H^HsCb and add a few drops of K^Q^Oz. The precipitate is PbCrO4. Write equations. NOTE. PbS0 4 is soluble in NH 4 C 2 H 3 2 owing to the formation of the very slightly ionized Pb(C 2 H 3 O 2 ) 2 . It should be remembered that most salts are highly ionized. (See table, Introduction, 9.) 36 CATIONS (METAL IONS) Experiment 9. To each of the solutions containing Bi+ ++, Cu+ + , and Cd + + (Exp. 8) slowly add NH 4 OH to distinctly alkaline reaction. Note the formation of a permanent white precipitate of BiOOH. The precipitates of Cu(OH) 2 and Cd(OH) 2 redissolve on the addition of an excess of the reagent. (Difference, separation of Bi.) Write equations. NOTE. According to the Law of Mass Action the limit of solubility of any acid, base or salt is reached when the product of the concentra- tion of its ions equals a certain maximum, called the solubility product. (Cone, of pos. ions X Cone, of neg. ions = a const., solubility product.) If NH 4 OH is added to a bismuth salt, precipitation of BiOOH begins when enough OH~ have been added to reach the solubility product for BiOOH. The greater the concentration of OH~, i.e., the greater the excess of NH 4 OH, the smaller must be the concentration of the Bi + + + and hence the more complete the precipitation. It will be recalled, however, that an addition of an excess of NH 4 OH dissolves both Cu(OH) 2 and Cd(OH) 2 . This is explained by the fact that the Cu ++ and Cd + + unite with HN 3 to form the complex ions Cu(NH 3 ) 4 ++ and Cd(NH 3 ) 4 + + (See Introduction 15.) This removes the Cu ++ and Cd ++ as such and prevents their precipitation. The Bi + ++ , incapable of forming a complex ion with NH 3 , must be removed by precipitation as indicated above. Experiment pa. Decant or filter off the clear liquid from the precipitate of BiOOH and dissolve it in a few cc. of 6N. HC1. Evaporate to about I cc. and pour into a large volume of warm water. Filter off the white milky precipitate of BiOCl and pour over the precipitate on the filter paper a freshly prepared solution of sodium stannite (Na2$nO2). (See Appendix i.) The black residue is finely divided bismuth. Write equations. NOTE. The solution of BiCl 3 is evaporated to remove the excess of HC1 in order to promote hydrolysis. (See Preliminary Exp. i, also Introduction, 14.) Experiment 10. Divide each of the solutions contain- ing copper and cadmium (Exp. 9) into two parts. Acidify PRELIMINARY EXPERIMENTS 37 the first portion of each with H^HsCb and add a few drops of K 4 Fe(CN) 6 . Note the color of the precipitates formed. (Difference, detection of Cu.) Write equations. To the second portion containing copper add KCN (Care, Poison) until the blue color just disappears; then add an equal volume to the second portion containing cadmium. Saturate each with H^S. Note the formation of insoluble CdS. (Difference, detection of Cd.) NOTE. Excess KCN reacts with salts of copper and cadmium to form the complex ions Cu(CN) 2 ++ and Cd(CN) 4 ++ . The Cu(CN) 4 + + first formed immediately decomposes into Cu(CN) 2 + and C 2 N 2 . The C 2 N 2 reacts with NH 4 OH to form NH 4 CNO and other more or less com- plex substances. The failure to obtain a precipitate with H 2 S is evidence that not enough Cu + or Cu ++ is present to reach the solubility product for Cu 2 S or CuS and shows that the complex ion Cu(CN) 2 + is only very slightly dissociated. On the other hand, the complex ion Cd(CN) 4 + + must be dissociated to a considerable extent since it produces sufficient Cd ++ to reach the solubility product for CdS. The reactions proceed as follows: 2 Cu(NH 3 ) 4 SO 4 + 4 KCN -* 2 KCu(CN) 2 +C 2 N 2 + 4 NH 3 +K 2 SO 4 Cd(NH 3 ) 4 S0 4 +4KCN^ K 2 Cd(CN) 4 + 4 NH 3 +K 2 SO 4 CATIONS (METAL IONS) TABLE II t OUTLINE FOR THE SYSTEMATIC SEPARATION OF GROUP II (SEPARATION INTO Cu AND SN DIVISIONS) Ions present in O.3N. HC1 solution As + Sn++ No. Reagent Hg++ Pb++ Bi + Cu++ Cd++ or + + Sb + or As + + Sn + + i H 2 S HgS PbS Bi 2 S 3 CuS CdS As 2 S 3 As 2 S5 Sb 2 S 3 SnS or SnS 2 2 (NH 4 ) 2 Sx HgS PbS Bi 2 S 3 CuS CdS (NH 4 ) 3 AsS 4 (NH 4 ) 3 SbS 4 (NH 4 ) 2 SnS 3 3 6N. HC1 1 1 I I 1 As 2 S 5 Sb 2 S 5 SnS 2 I I 1 ANALYSIS Group II Hg++, Pb++, Bi + , Cu++, Cd++, As + , As + + , Sb + , Sn++, Sn + - (20) Precipitation. Dilute the filtrate (10) which should contain just 5 cc. of 6N. acid (see Discussion 16) to 100 cc. and saturate with H 2 S. Filter, heat the filtrate to boiling and again saturate with I-bS. If no further precipitate forms reserve the solution for analysis of Group III (50) and treat the precipitate by (21). If a further precipitate forms (see Discussion 17) evaporate the mixture to dryness, moisten the residue with cone. HC1 and evaporate to dryness again to remove all HNO 3 . Add 10-15 cc. of 6N. HC1, heat to boiling and pass H 2 S into it for five to ten minutes. Filter, unite the precipitate to that above and treat by (30). Reserve the filtrate for analysis of Group III (50). (21) Separation of the Copper and Tin Divisions. Transfer the precipitate (20) to an evaporating dish, add 1015 cc. of (NHU^Sz reagent, cover the dish with a watch glass and warm the mixture for about five minutes. (Do not boil.) (See Discussion 18.) Dilute with an equal DISCUSSION (16-20) 39 volume of water and filter. A second treatment with (NH^Sz reagent should be made if the residue is large and much was extracted by the first treatment. Wash the residue with hot water (see Introduction 3, /) and treat it by (30). Treat the filtrates separately by (22). (22) Precipitation of the Tin Division. To the first filtrate obtained on treating the Group II sulphides with (NH 4 )2Sz reagent (21) add 6N. HC1 with frequent stirring until the solution remains milk-white from the separation of finely divided sulphur. Allow the mixture to stand for one or two minutes to coagulate the precipitate. The presence of the tin division is indicated by the presence of a flocculent yellow or orange precipitate. (See Discussion 19.) Treat the second filtrate obtained in (21) in the same way, and if a flocculent precipitate forms, unite it with the one obtained above. Filter and wash the precipitate, using suction to dry it as thoroughly as possible. (See Discussion 20.) Reject the filtrate and treat the precipitate by (40). DISCUSSION 1 6. The precipitation of the sulphides of the metals of both Groups II and III is determined very largely by the acid concentration of the solution. From the stand- point of the Law of Mass Action and the Ionic Theory two factors must be taken into consideration in the pre- cipitation of the metal sulphides, viz., the value of the solubility product and the concentration of the sulphide ion. The solubility product varies with the nature of the sulphide and with the temperature. The order of pre- cipitation with H 2 S from cold HC1 solution of decreasing acid concentration is about as follows: As ++ , As + , Hg++, Cu++, Sb +++ , Bi +++ and Sn++, Cd++, Pb + + and Sn++, Zn++, Fe++, Ni ++ , Co++, Mn++. According to Henry's Law of the Solubility of Gases, the solubility and therefore the concentration of H 2 S will have a constant value at a definite temperature and 40 CATIONS (METAL IONS) pressure. H^S ionizes to a slight extent into H + and HS~ and to a much less extent into 2H+ and S = . Since the S = is the active agent in the precipitation under dis- cussion only the latter ionization need be considered here. The expression of the Law of Mass Action for IHUS is C 2 H + -C S = - - =K In a saturated solution, however, C H2 s is a constant; hence simplifying the above equation gives C 2 H + -C s ==K.C H2 s=K From this it is readily seen that any increase in the con- centration of H + must result in a corresponding decrease in S = . But in order that a sulphide, e.g., CdS, may precipitate, its solubility product, C C d ++ -C s == =Kcds, must be reached. Therefore, if the acid concentration is large the concentration of S = may become so small that its product with C C d ++ will not equal the solubility product for CdS. On the other hand, if the acid concentration is small the concentration of S = may become so large that its product with C Zn ++ will equal the solubility product for ZnS, in which case ZnS would be precipitated in Group II. In order to make a complete separation between Groups II and III, the concentration of S = must be so regulated that there will be sufficient to reach the solubility product of the most soluble of the sulphides of Group II but not sufficient to reach the solubility product of the least soluble of Group III. (See table, Introduc- tion n). The necessity of following the directions very accurately is therefore evident. 17. When H 2 S is passed into a cold acid solution con- taining an arsenate, reaction takes place only very slowly between the two substances. Part of the arsenic is pre- cipitated as As2S 5 and a part is reduced to the trivalent state and precipitated as As2Ss. When the solution is heated the reduction takes place somewhat more rapidly. The most favorable conditions for the reduction and pre- DISCUSSION (16-20) 41 cipitation of arsenates with H 2 S is from a hot solution having a comparatively high concentration of acid. 1 8. In the separation of arsenic, antimony and tin from the copper division, advantage is taken of the acid character of the elements of the tin division and their ability to form sulpho-salts with (NH 4 ) 2 Sz. This reagent, however, has a slight solvent action on the sulphides of copper and mercury. If (NH 4 ) 2 Sz, to which about 5 per cent NaOH has been added, is used, the solvent action on the sulphides of copper and mercury is very much reduced and a more satisfactory separation is obtained. Excessive heat not only increases the solvent action on HgS and CuS but decomposes the reagent with precipitation of free sulphur. 19. If the separation of arsenic, antimony and tin from the copper division has been complete the presence of the tin division is shown by a flocculent yellow or orange precipitate on acidification of the (NH^S* solution with HC1. However, if the precipitate obtained is brown (indicating copper) dark gray (indicating mercury), or of unpronounced yellow or orange, so as to make doubtful the presence of the tin division, much time can often be saved by proceeding as directed below. Small amounts of copper which otherwise might be overlooked may also be detected. Heat the precipitate with 10-20 cc. of NH 4 OH almost to boiling for five minutes, and filter. This treatment dissolves all but the copper and free sulphur. The residue, therefore, should be tested for copper according to (31) (34) and (36) unless copper has already been found. Pass H 2 S into the filtrate for fifteen to twenty seconds to pre- cipitate any HgS and change the partially sulphurated acids (NH 4 )3AsO 3 S, (NH 4 )3AsO 2 S 2 , etc., into the fully sulphurated form (NH 4 )sAsS 4 , etc. Filter if necessary and acidify the filtrate with HC1. Filter off the precipi- tated sulphides, wash, dry by suction and treat by (40). 20. If a suction pump is not available the precipitate may be dried satisfactorily by pressing the filter containing it between several thicknesses of clean filter paper. 42 CATIONS (METAL IONS) I H Q Q fc 5 1 3 <% 5 u uu I S 3 SI <-s U CU PL, ! ii c| OUTLINE OF ANALYSIS (TABLE III) 4? ANALYSIS Group II, Cu Division (30) Separation of Mercury. Transfer the residue (21) to a porcelain dish and add 10-20 cc. of 2N. HNOs and heat to boiling. Boil gently for two to three minutes, not longer. (See note, Preliminary Exp. 7). Filter, wash the residue and treat it by (31). Treat the filtrate by (32). (31) Confirmatory Test for Mercury. Transfer the residue (30) undissolved by HNOs to a porcelain dish and add 5-10 cc. of aqua regia (see Preliminary Exp. 7a). Warm gently till solution is complete, then evaporate almost to dryness, dilute with 5-10 cc. of water, filter and add to the clear filtrate some SnCl 2 solution, at first 1-2 drops then 2-3 cc. (see Discussion 22). The formation of a white precipitate which turns gray on the addition of excess SnCb shows the presence of mercury. (See Discussion 23.) (32) Separation of Lead. To the filtrate obtained in (31) add 2-3 cc. of cone. H 2 SO4, transfer to a porcelain dish and evaporate until the dense white fumes of SOs appear. Cool, pour the mixture into a small beaker containing 1015 cc. of water and rinse out the vessel with a portion of the solution formed, in order to be sure that all of the solid is transferred. Allow the mixture to stand four to five minutes. The formation of a fine, white, crystalline precipitate indicates the presence of lead. The precipitate may be more readily distinguished if the liquid is given a slight whirling motion, so that the pre- cipitate will collect on the bottom of the beaker toward the center. Filter, treat the precipitate by (33) and the filtrate by (34). (33) Confirmatory Test for Lead. Dissolve the pre- cipitate of PbSC>4 (32) by pouring a 10-20 cc. portion of ammonium acetate (NH 4 C 2 H 3 O2) solution repeatedly through the filter. To the filtrate add a few drops of 44 CATIONS (METAL IONS) K 2 Cr 2 O 7 and 3-5 cc. of HC 2 H 3 O 2 . A yellow precipitate is PbCrO 4 (see Discussion 24). (34) Separation of Bismuth. To the filtrate obtained in (32) add NH 4 OH until, after shaking, a distinctly alka- line reaction is obtained. (Test with litmus.) Shake to coagulate the precipitate of BiOOH, and filter. Wash the precipitate and treat it by (35). Treat the filtrate by (36). (35) Confirmatory Test for Bismuth. Pour through the filter containing the precipitate of BiOOH (34) a cold freshly prepared solution of Na 2 SnO 2 (see Appendix I). The formation of a black residue shows the presence of bismuth. (See Discussion 25.) (36) Detection of Copper. If the filtrate from the BiOOH (34) is deep blue, copper is shown to be present. If, however, it is colorless or nearly so, about one-fourth of the solution should be acidified with HC 2 H 3 O 2 and a few drops of K 4 Fe(CN) 6 added. The formation of a red precipitate or coloration shows the presence of copper. Treat the remainder of the solution by (37) or (38.) (37) Detection of Cadmium. To the remainder of the NH 4 OH solution obtained in (36) add KCN solution (care, poison) until the blue color just disappears; if the solution is colorless add only a few drops. Pass H 2 S into the colorless solution for about half a minute. The formation of an immediate yello>w precipitate (see Dis- cussion 26) shows the presence of cadmium. (38) Optional Method for the Detection of Cadmium. Acidify the remainder of the NH 4 OH solution (36) with H 2 SO 4 , add a few iron nails or some iron filings and boil for a short time. Filter, and unless the solution is still acid make it just acid with 6N.H 2 SO 4 and pass H 2 S into it. The formation of a yellow precipitate shows the pres- ence of cadmium. Discussion 21. If the elements of the copper division are present in large quantity, small quantities of tin may remain DISCUSSION (21-27) 45 undissolved by the (NH 4 ) 2 Sz treatment. This may also occur when small quantities of cadmium and stannous tin are present together. Any tin sulphide, either SnS or SnS2, remaining in the copper division, will be con- verted by the HNOs (30) into insoluble metastannic acid, H 2 SnOs. This is practically unaffected by the aqua regia or bromine water used in (31). In order to recover any tin that may be left in the copper division the residue obtained by treatment with aqua regia (31) should be treated as follows: If it is dark- colored, showing incomplete removal of HgS, digest with bromine water in order completely to remove the HgS, filter and dissolve the residue in a small quantityof(NH 4 ) 2 Sa; reagent. Dilute, filter if necessary, and add the solu- tion to the main ammonium sulphide solution obtained in (21). 22. The addition of SnCl 2 to a solution of HgCl2 causes an immediate reduction to the white Hg 2 Cl2. Further addition of SnCl 2 carries the reduction to free mercury, which imparts a gray appearance to the pre- cipitate. The latter reduction is hindered very materially and may be almost prevented if excess aqua regia has not been completely removed before making the test. Bromine water may be used in place of the aqua regia. Excess bromine, however, must be removed before the addition of SnCk. 23. If the separation of mercury is incomplete and a large amount of CuS is left undissolved by the HNOs treatment (30), a white precipitate of CuCl will separate out on the addition of SnCl2. This, however, does not turn gray with excess SnCl 2 . 24. The confirmatory test for lead should always be made, since a precipitate with H 2 SO4 may consist of (BiO) 2 SO 4 or BaSO 4 . The (BiO) 2 SO 4 is coarsely crystal- line, dissolves in NH 4 C 2 H 3 O 2 and gives a yellow color with K 3 Cr 2 O7. The precipitate, however, differs from PbCrO* in that it dissolves readily in HC 2 H 3 O 2 . The 46 CATIONS (METAL IONS) BaSO 4 resembles PbSO 4 in appearance but is insoluble in NH 4 C 2 H 3 O2. 25. If, owing to occlusion or incomplete washing of the H2S precipitate, an incomplete separation from Group III is obtained, the addition of NH 4 OH for the separation of bismuth will cause the precipitation of Fe(OH)3 or other hydroxides of Group III. These differ from BiOOH in that none of them are reduced by Na2SnC>2 by short contact in the cold. 26. An immediate yellow precipitate with H^S shows the presence of cadmium. When much copper is present and the solution is saturated with H^S a deep yellow color soon develops and an orange-red precipitate of (CSNH 2 )2 may separate out on standing, owing to a reaction between the H 2 S and C 2 N 2 set free by the reduction of Cu(CN) 2 . 27. If, owing to previous errors in analysis, a black precipitate (due to HgS, PbS, etc.) is obtained in the final test for cadmium with I-bS, it should be thoroughly washed and the cadmium dissolved out by boiling the precipitate with 15 cc. of 1.2 N. H2SO 4 and filtering. After it has been diluted with two to three times its volume of water the CdS if present may be precipitated by saturating with H 2 S. Preliminary Experiments Group II, Tin Division + + + + + ++ + + As + , As ++ , Sb + , Sn++, Sn + - In connection with the following experiments study Tables II and IV. Experiment n. Dilute separate 5-cc. portions of the test solutions of each of the above ions with an equal volume of water, and saturate cold with H 2 S. If a precipitate does not form heat to boiling and saturate again with PRELIMINARY EXPERIMENTS 47 H 2 S. Note the difference in action between As + and As ~ l ~ + . Write equations. Experiment 12. Decant or filter the liquid from the precipitates (Exp. n) and dissolve each, by warming if necessary, with about 5 cc. of (NH^S* reagent. Do NOT boil NOTES. The solubility of the sulphides of arsenic, antimony and tin is doubtless due to their acidic character and tendency to form sulpho- salts. This is in direct contrast to the metals of the copper division of Group II in that the latter are all basic in character and have very little tendency to form sulpho-salts. The use of (NH 4 ) 2 Sa; instead of (NH 2 ) 4 S is made necessary only in the cases of Sb 2 S 3 and SnS. Antimony and tin must be oxidized to the higher valence since the sulpho-salt of the lower valence is apparently incapable of existence. The following reactions for the solution of As 2 S 3 and As 2 Ss are as follows: S^ 2 (NH 4 ) 3 AsS 4 + (6;x; -s)S As 2 S 5 +6(NH 4 ) 2 Sx-> 2(NH 4 ) 3 AsS 4 + (6JC - 3 )S Write the corresponding equations for the solution of Sb 2 S 3 , SnS and SnS 2 noting the fact that in the sulpho-salt arsenic and antimony are quinquivalent while tin is quadrivalent. Experiment 13. Since the two solutions of arsenic are now identical, (NH^AsS^ as are also those of tin, (NH 4 ) 2 SnS 3 , one of each may be discarded. Dilute the remaining solutions of arsenic, antimony and tin (Exp. 12) with an equal volume of water and acidify with 6N. HC1. Note the precipitate formed in each case. Write equations. NOTE. When (NH 4 ) 2 S X is treated with an acid, considerable sulphur is apt to be set free. This gives a white, milky appearance to the mixture and may prevent the detection of small amounts of the sulphides. Dilu- tion with water obviates this difficulty to some extent. Experiment 14. To each of the precipitates of Sb 2 S 5 and SnS 2 (Exp. 13) dried by suction or by pressing 48 CATIONS (METAL IONS) between filter paper, add just 10 cc. of I2N. HC1. Place the test-tubes containing the 'mixtures in a beaker of water and heat to boiling. Boil gently for ten minutes. Note the solution of Sb2S 5 and SnS2. (Difference, sepa- ration of As.) Write equations. NOTES. If the boiling is carried out so that the bubbles rise but slowly from the solutions, practically no As 2 S 6 is dissolved. However, if the boiling is too vigorous some of the arsenic may go into solution. This is due to removal of H 2 S and consequent shifting of equilibrium. Sb 2 S 5 dissolves in i2N. HC1 somewhat slowly, especially when large amounts are present. It dissolves with the formation of SbCl 3 , the anti- mony being reduced by the H 2 S to the trivalent condition, Experiment i4a. Without filtering, add small particles of KClOs or NaClOa, a few at a time, to the As2S 5 and stir till solution is complete. Evaporate to 1-2 cc. Make distinctly alkaline with NH 4 OH and add 1-2 cc. of magnesia mixture (MgCl2-NH 4 Cl-NH 4 OH). If no pre- cipitate forms add about one-fourth its volume of NIrUOH and rub the inside of the tube with a glass rod to hasten precipitation. NOTE. The action of HC1 on KC10 3 or NaC10 3 is to liberate chlorine and C1O 2 , which remove the S =by oxidation, causing the arsenic to go into solution. It is dissolved as HsAs0 4 instead of AsCl 5 , very little of which remains as such even in very concentrated HC1. MgNHiAsCX is somewhat soluble even in cold water; therefore it is necessary to have a rather concentrated solution. It also has a tendency to hydrolize into NH 4 OH and MgHAs0 4 , a soluble salt. This tendency is counteracted by the addition of a large excess of NH 4 OH. It has an- other peculiarity, which is not uncommon, in that it is inclined to super- saturate. This may be overcome and precipitation effected, by vigorous shaking, or by producing a rough surface in contact with the liquid. Experiment 15. Dilute the solutions of antimony and tin (Exp. 14) to 50 cc. and transfer one- third of each to a third vessel. Heat the remaining portions to 90 and saturate with FbS. Note the precipitation of Sb2Ss. (Difference, separation of Sb.) Write equations. Experiment isa. Dilute the solution in the third PRELIMINARY EXPERIMENTS 49 vessel (Exp. 15) to 50 cc. and saturate with H 2 S. Note the color of the precipitate. Without filtering, boil to expel H 2 S. Add bromine water and heat till solution is complete, then add 5 gms. of solid oxalic acid (H 2 C 2 O4) and saturate with H 2 S. Note the precipitation of Sb 2 S 3 . (Difference, separation of Sb.) Write equations. NOTES. There is enough difference in the solubility products of Sb 2 S 3 and SnS 2 for the careful worker to make a complete separation of antimony and tin according to Exp. 15. However, if the directions for temperature and acid concentration are not carefully observed, a pre- cipitate containing both antimony and tin may be obtained (Exp. 150), in which case the separation can be made according to Exp. 1 50. The bromine water is used to oxidize the S = and so hasten the solu- tion. The same results may be obtained more slowly by evaporation to concentrate the acid. The failure of SnS 2 to precipitate in the presence of a large excess of H 2 C 2 O 4 is doubtless due to the formation of a complex which reduces the + + concentration of Sn + + . Experiment 16. Dilute the solution containing tin (Exp. 15) to 50 cc. and saturate cold with H 2 S. Note the precipitation of SnS 2 . Without filtering, add 1-2 gms. of granulated test lead and boil two to three minutes. Cool and filter, allowing the filtrate to run into a solution of HgCl 2 . The precipitate is Hg 2 Cl 2 +Hg. + + NOTE. The action of lead is to reduce the Sn ++ to Sn ++ , which in turn will reduce HgCl 2 to HgiCl 2 or even to free mercury. Stannic tin has no effect on HgCl 2 ; hence the necessity for reduction before the final test. 50 CATIONS (METAL IONS) TABLE iy OUTLINE FOR THE SYSTEMATIC SEPARATION AND DETECTION OF GROUP II (SN DIVISION) No. Reagent As 2 S 5 SbjS, SnS 2 i I2N. HC1 As 2 S 5 SbCl 3 SnCl 4 HC1 KC1O 3 H 3 AsO 4 Magnesia mixture MgNH 4 AsO 4 6N. HC1 H 3 AsO 4 H 2 S As2S 3 - 5 2 H 2 S 1 Sb 2 S 3 SnCl 4 3 HCl-Br 2 SbCl 3 SnCl 4 4 H 2 C 2 4 i H 2 S Sb 2 S 3 5 6 Pb HgCl 2 i- SnCl 2 SnCl 4 (Hg 2 Cl 2 ) ANALYSIS Group II, Tin Division (40) Separation of Arsenic. Transfer the precipitated sulphides (22), dried by suction or by pressing between filter paper, to a test-tube, and add just 10 cc. of I2N. HC1. (See Discussion 28.) Place the test-tube in a beaker of water and heat until the contents of the tube just begin to boil. Keep the water at this temperature for about ten minutes, occasionally stirring the contents of the tube with a glass rod. When the reaction is complete and no further solution takes place, dilute with 5 cc. of water and filter, allowing the filtrate to run into a 5o-cc. graduate. Wash the residue with 5-10 cc. of water, catching the wash water in the graduate with the filtrate. Test the residue for arsenic by (41) and treat the filtrate by (43). (41) Detection of Arsenic. Punch a hole through the filter and wash the residue of As 2 S 5 into a test-tube with 5-10 cc. of 6N. HC1. Warm the mixture gently, adding solid KC1O 3 or NaClO 3 , one crystal at a time, until the OUTLINE OF ANALYSIS (TABLE IV) 51 arsenic is dissolved. Make the solution just alkaline with NH 4 OH, then add about one- third its volume of I5N. NH 4 OH and about 0.5 cc. of magnesia mixture (MgCl 2 -NH 4 Cl-NH 4 OH). Shake the contents vigor- ously and if no precipitate forms rub the walls of the tube with a glass rod and allow it to stand for some time. (See note, Preliminary Exp. I4a). The formation of a white crystalline precipitate indicates the presence of arsenic. Filter and confirm the arsenic by (42). (42) Confirmatory Test for Arsenic. Dissolve the precipitate of MgNH 4 AsO 4 (41) by pouring 5-10 cc. of 6N. HC1 repeatedly through the filter. Heat the solu- tion to boiling and pass H 2 S into it for five to ten minutes. The formation of a white precipitate, changing to yellow, shows the presence of arsenic. (See Discussion 17.) (43) Detection of Antimony. To the filtrate (40) add water just sufficient to make a total volume of 50 cc. ; transfer to a small flask and heat to about 90. Pass H 2 S into the hot solution for about five minutes, keeping the temperature at about 90 by placing the flask in a beaker containing boiling water. The formation of an orange-red precipitate shows the presence of antimony. (See Discussion 29.) Filter while hot, add 5 cc. of water, heat to 90 and again saturate with H 2 S to remove all antimony. Filter if a precipitate forms, and treat the filtrate by (45). (44) Confirmatory Test for Antimony. If the precipitate (43) is not orange, and the presence of antimony is doubt- ful, boil the mixture without filtering until the precipitate has redissolved. Add 5-10 gms. of solid oxalic acid (H2C 2 O 4 ) and pass H 2 S into the hot solution. If antimony is present a bright red precipitate will be formed. Filter and test the filtrate for tin by (46). (45) Detection of Tin. Dilute the filtrate from (43) to 70 cc., cool and pass in H 2 S for ten minutes. The formation of a yellow precipitate indicates the presence of tin. 52 CATIONS (METAL IONS) (46) Confirmatory Test for Tin. Boil the mixture (45) or the filtrate from (44) with 1-2 gms. of granulated test lead for two to three minutes. Cool the mixture and filter into a solution of mercuric chloride (HgCl 2 ). The presence of tin is shown by the formation of a white precipitate, which may turn gray if much tin is present. (See Discussion 22.) DISCUSSION 28. The separation of arsenic, antimony and tin depends on the relative solubility of their sulphides in HC1. It is necessary, therefore, to follow the directions very carefully in order to secure a satisfactory separation. The solubility product of As 2 Ss is considerably smaller than that of Sb 2 S 5 or SnS2 ; hence it is stable in a much higher concentration of HC1. It is practically insoluble in hot I2N. HC1 unless the H 2 S formed in the equilibrium As 2 S 5 + ioHCl^ 2(AsCl 5 )+5H 2 S is expelled by too vigorous boiling. The hot acid, however, dissolves the tin very readily and the antimony more slowly. In the presence of much antimony some may be left undissolved even in the strong acid, but the amount extracted will never be so small as to escape detection. Also, if the acid becomes much diluted, considerable Sb 2 S5 will be left undissolved. The undissolved portion may be sufficient to give the residue an orange color, but it will not interfere with the test for arsenic. 29. When both antimony and tin are present and the acid concentration is not sufficiently high or the solution has been allowed to cool, some tin may be precipitated with the antimony, in which case the color of the H 2 S precipitate will usually be brown. 30. The magnesia mixture used in the detection of arsenic contains a large excess of NH4 + for the purpose of reducing the OH~ concentration. This prevents the PRELIMINARY EXPERIMENTS 53 precipitation of Mg(OH) 2 by the NH 4 OH. See Intro- duction 13, also Appendix I.) 31. Test lead is used for the reduction of tin, rather than zinc or iron, since lead does not reduce tin to the metallic state. If the stronger reducing agent is used the tin will be left in the residue and will have to be dis- solved in HC1 before being added to the HgCb solution. GROUP III Preliminary Experiments Group III, Aluminium Division + + + + Al + , Cr + , Zn++ In connection with the following experiments, study Tables V and VI. Experiment 17. Introduce into separate test-tubes 5-cc. portions of the test solutions containing the above ions. Make each distinctly alkaline with NH 4 OH, then saturate with H^S. Heat nearly to boiling to coagulate the precipitates. Write equations. NOTES. In the presence of water A1 2 S 3 and Cr 2 S 3 are completely hydrolized (see Introduction 14); hence the hydroxides formed on the addition of NH 4 OH remain unchanged when H 2 S is introduced. Zn ++ forms with NH 3 the complex ion Zn(NH 3 ) 4 ++ . (See Intro- duction 15.) This removes enough Zn ++ to prevent the precipitation of Zn(OH) 2 but not enough to prevent precipitation of ZnS with HgS. Experiment 18. Decant off the clear liquid from the precipitates of A1(OH) 3 , Cr(OH) 3 and ZnS (Exp. 17) and dissolve in a slight excess of HC1. Add NaOH slowly to alkaline reaction, and then about I gram of solid Na2C>2, a little at a time. (Caution: Do not carry the Na2O 2 on paper. Use a dry watch glass. Why?) Boil to expel excess Na2<32. Write equations. 54 CATIONS (METAL IONS) NOTE. It should be recalled (see note, Preliminary Exp. 9, also Introduction 15) that a more complete precipitation does not necessarily follow the addition of a reagent in excess of that necessary to reach the solubility product for the compound concerned. The solution of A1(OH) 3 , as well as that of chromium and zinc, in an excess of NaOH presents another example, but of quite a different type. All three belong to the class known as amphoteric compounds (see Introduction 16). In the presence of excess NaOH, therefore, they form the soluble sodium salts NaAlO 2 , NaCrO 2 and Na 2 Zn0 2 . Na 2 O 2 oxidizes the NaCrO 2 to the more readily soluble Na 2 CrO 4 , but is without effect on NaA10 2 or Na 2 ZnO 2 . Why? Experiment 19. Acidify the solutions of aluminium, chromium and zinc (Exp. 18) with HNOa and add NH 4 OH to alkaline reaction. Note the precipitation of Al(OH)s. (Difference, separation of Al.) Write equations. NOTE. The action of Na 2 O 2 in Exp. 18 oxidized the chromium to the more soluble chromate, a compound which does not possess ampho- teric properties; hence on the addition of HNO 3 it does not react as a base to form a chromium salt. On the other hand, the aluminium and + + zinc return to the positive radical as Al + and Zn ++ which react with the NH 4 OH subsequently added, as outlined in Exp. 17. Experiment ipa. Filter off the precipitate of A1(OH) 3 (Exp. 19) and wash with water. Dissolve by pouring 5 cc. of 6N. HNO 3 through the filter. Add 4-5 drops of N. 100 Co(NOs)2 and evaporate nearly to dryness. Soak up the liquid with a small piece of filter paper. Roll it up, wind a platinum wire around it in the form of a spiral and heat in a flame till all the carbon is burnt off. A blue residue is characteristic of aluminium. NOTE. The blue substance formed by the interaction of Co(N0 3 ) 2 and A1(NO 3 ) 3 is a compound of CoO and A1 2 O 3 , probably Co(AlO 2 ) 2 , though the exact composition has not been determined. In carrying out the reaction care must be exercised in the addition of Co(NO 3 ) 2 . The aluminium must be in excess, otherwise the blue color is hidden by the black cobalt oxide. Furthermore, the presence of sodium or potassium salts causes the mass to fuse and thus interferes with the test. They may be removed by washing the NH 4 OH precipitate with water. PRELIMINARY EXPERIMENTS 55 Experiment 20. Acidify the solution of chromium and zinc (Exp. 19) with HC2H3O2, and add BaCU solu- tion. Note the precipitation of BaCrOi. (Difference, separation of Cr.) Write equations. Experiment 2oa. Filter off the precipitate of BaCrCX (Exp. 20) and dissolve it in a very little 6N. HNOs. Dilute with 9-10 volumes of water and to a portion of it in a test-tube add about 2 cc. of ether and i cc. of H 2 O 2 , and shake. NOTE. The blue color in the ether layer is a perchromic acid, probably H 3 Cr0 7 . It is very unstable, decomposing into oxygen and a chromic salt. Excess H 2 2 or acid accelerates this decomposition. Experiment 21. Saturate the zinc solution (Exp. 20) with H2S. The white flocculent precipitate is ZnS. Filter off the precipitate and dissolve it by pouring 5 cc. of 6N. HNO 3 repeatedly through the filter. Add 4-5 drops of N. 100 Co(NOs)2 and evaporate in a porcelain dish almost to dryness. Neutralize with Na2COs solu- tion and add a slight excess. Evaporate to dryness and ignite gently till the purple color of the cobalt disappears. Allow the residue to cool. The green color is due to a compound of the oxides of cobalt and, zinc, probably CoZnO 2 . NOTE. ZnS is somewhat more flocculent when precipitated from a warm solution. A white, milky-looking precipitate does not show the presence of zinc. Sulphur often separates out as a white, m'lky precipitate, especially on standing, or if the current of H 2 S is long continued. It is often necessary, therefore, to make the confirmatory test. The addition of too much Co(NO 3 ) 2 must be avoided. (See note, Exp. 56 CATIONS (METAL IONS) TABLE V OUTLINE FOR THE SYSTEMATIC ANALYSIS OF GROUP III (SEPARATION INTO AL AND FE DIVISIONS) Ions present in acid solution No. Reagent A1+ + + Cr+ + + Zn+ + i NH 4 OH (A1OH), Cr(OH) 3 Zn(NH) 4 + + 2 H 2 S (A1OH), Cr(OH) 3 ZnS 3 HC1 A1C1 3 CrCl 3 ZnCl 2 4 HNO 3 Aldi CrCl 3 ZnCl 2 5 NaOH NaAlO 2 NaCrO 2 Na 2 ZnO 2 6 Na 2 O, NaAlO 2 Na 2 CrO 4 Na 2 ZnO 2 No. Mn + + Fe+ + Co+ + Ni+ + i Mn(OH) 2 Fe(OH) 2 Co(NH 3 ) 4 + + Ni(NH 3 ) 4 + + 2 MnS FeS CoS NiS 3 MnCl 2 FeCl 2 CoS* NiS * 4 MnCl 2 FeCl 3 CoCl 2 NiCl 2 5 Mn(OH) 2 Fe(OH) 3 Co(OH) 2 Ni(OH) 2 6 MnO(OH) 2 Fe(OH) 3 Co(OH) 3 Ni(OH) 2 _ 3 1 1 1 ! * See Note, Exp. 23. ANALYSIS Group III Al + , Cr + , Zn++, Mn++, Fe++, Co++, Ni++ Precipitation and separation of the aluminium and iron divisions. (50) Precipitation. Boil the filtrate from Group II (20) till the H 2 S is expelled. Test the vapor with filter paper wet with lead acetate [Pb(C2HsO2)2] solution. Add NH 4 OH in slight excess, and after shaking note whether a precipitate is formed. (See discussion 32.) DISCUSSION (32-37) 57 Without filtering add 2-3 cc. more NH^OH and pass in H 2 S until, after shaking, the vapors blacken a piece of filter paper moistened with Pb(C2HsO2)2 solution. Heat the mixture nearly to boiling to coagulate the pre- cipitate. Filter and wash the precipitate with water containing about i per cent of (NH 4 ) 2 S. If filtration is slow the funnel should be kept covered with a watch glass to prevent oxidation to soluble sulphates. The filtrate should be colorless. (See Discussion 33.) Treat the pre- cipitate by (51) and reserve the filtrate for analysis of Group IV (80). (51) Separation of Aluminium and Iron Divisions. Transfer the Group III precipitate (50) with the filter if necessary, to a porcelain dish, add 5-20 cc. of 6N. HC1 and stir for one to two minutes in the cold. Heat the mixture to boiling and if a black residue still remains add a few drops of HNOs and boil again. Dilute with 5-10 cc. of water and filter off the residue of sulphur. Evaporate nearly to dryness to remove excess acid, dilute to 10-20 cc., and add NaOH to alkaline reaction, avoiding a large excess. If a very large precipitate forms add 1020 cc. more water. Cool the mixture and add 1-3 grams of solid Na 2 O 2 , a little at a time and with constant stirring. (See Discussion 35.) Add about 5 cc. of 3N. Na2CO 3 solution unless phosphate or the alkaline earth metals are known to be absent. (See Discussion 37.) Boil to decompose excess Na 2 O2; cool, dilute with an equal volume of water and filter with the aid of suction if possible. (See Discussion 36.) Treat the filtrate by (60) and the residue by (70). DISCUSSION 32. The H 2 S is removed and the effect produced by NH 4 OH alone is noted, in order to obtain information regarding the presence of aluminium or other insoluble hydroxides. NaOH and Na2O 2 , used later in the analysis, often contain small quantities of silica and aluminium 58 CATIONS (METAL IONS) which tend to make the detection of aluminium more dif- ficult. Any information gained 'at this point, therefore, may be of considerable importance. 33. Ammonium monosulphide ( (NH^S) is some- times used instead of H 2 S in the precipitation of Group III. When this is done some NiS may be dissolved and pass into the filtrate, giving it a brown or nearly black color. The use of H 2 S as directed in (50) almost completely pre- vents this. If, however, (NH 4 ) 2 S is used and the filtrate is brown or nearly black, indicating that some NiS has been dissolved, it may be precipitated by boiling the solu- tion for a few minutes. It should be filtered off and added to the main precipitate. 34. The presence of a considerable amount of ammo- nium salts not only lessens the solubility of A1(OH)3, but prevents the precipitation of Mg(OH) 2 in this group. A sufficient quantity of ammonium salts is usually formed by the neutralization of the acid already in the solution. The presence of ammonium salts decreases the ionization of NH 4 OH, owing to the common ion effect. (See Intro- duction 12.) The solubility product for the hydroxides of Group III and also for Mg(OH)2 is so small that even a slight excess of NH 4 OH causes precipitation. In the presence of much ammonium salts, however, the concentra- tion of OH~ is so reduced that its product with that of certain of the metal ions, e.g., Mn ++ or Mg ++ , does not reach the solubility product for those compounds. In the cases of ferric, aluminium and chromium ions the solubility product (C M + + + *C 3 oH~) is so small that the presence of large amounts of ammonium salts does not appreciably affect their solubility. On the other hand, zinc, nickel and cobalt ions unite with NHs to form complex ions similar to those of silver and copper. (See note, Exp. 9 also Introduction 15.) 35. Na 2 O 2 is a very unstable substance and decomposes slowly, even in a cold solution, oxygen being given off. In a hot solution the decomposition may become explosive. DISCUSSION (32-37) 59 The peroxide should therefore be added in very small quantities to a cold solution. This may be easily done by transferring a little of it to a dry watch glass and sprinkling it into the solution, stirring constantly. The reaction is known to be complete when a steady stream of gas is evolved after the mixture has been thoroughly stirred. If much chromium is present the Na2O 2 should be added until the green chromic salt has been entirely changed to the yellow chromate. The solution should be diluted before filtering in order to prevent disinte- gration of the filter paper by the strong alkali. 36. The separation of the aluminium and iron divisions by means of NaOH, Na2O 2 and Na2CO 3 is very satisfactory except in the case of zinc. When much of the iron division is present 1520 mgs. of zinc may be carried down in the precipitate, so that provision must be made in the iron division for its detection. 37. It should be remembered that the phosphates of barium, strontium, calcium and magnesium are insoluble in alkaline solution, and hence may have precipitated along with Group III, in case phosphate was present in the original material. Na2CO,3 is used as a reagent in the separation of the aluminium and iron divisions to insure complete precipitation of these elements since their hydrox- ides are somewhat soluble even in strong NaOH solution. ZnCOs, though insoluble in dilute solutions of Na2COs alone, is soluble in the presence of NaOH, owing to the formation of the zincate ion (ZnO 2 ~). Na2CO 3 also serves to decompose the chromates of the alkaline earths and so prevent the precipitation of chromium. 60 CATIONS (METAL IONS) TABLE VI ( OUTLINE OF THE SYSTEMATIC SEPARATION AND DETECTION OF GROUP III (Al DIVISION) No. Reagent NaAlO, Na 2 CrO 4 Na ZnO 2 i 2 3 4 5 HNOi NH 4 OH HNO 3 Co(N0 3 ) 2 HC 2 H 3 O 2 BaCl 2 HNO 3 Ether H,0 2 H 2 S HNO 3 Co(N0 3 ) 2 Na 2 C0 3 A1(N0 3 ) 3 M(OH), Co(A10 2 ) 2 (?) Na 2 Cr 2 O 7 Na 2 CrO 4 Na 2 Cr 2 O 7 BaCrO 4 Zn(N0 3 ) 2 Zn(NH 3 ) 4 (N0 3 ) 2 1 Zn(N0 3 ) 2 ZnCl 2 I ZnS Zn(NO,), CoZnO 2 (?) 1 H 2 Cr0 4 H 3 Cr0 7 (?) 1 ANALYSIS Group III, Aluminium Division (60) Separation of Aluminium. Acidify the nitrate from (51) with i6N. HNO 3 avoiding a large excess; add NH-iOH just to alkaline reaction and then 2-3 cc. more. (See Discussion 38.) The formation of a white flocculent precipitate shows the presence of aluminium. (See Dis- cussion 39.) Heat the mixture almost to boiling to coagu- late the precipitate, filter and wash thoroughly with hot water. Unless the precipitate is white and flocculent the confirmatory^fcifct should be made. Treat the filtrate by (62). (61) Confirmatory Test for Aluminium. Dissolve the precipitate (60) in 5 cc. of 6N. HNO 3 ; add 4-5 drops of N. 100 Co(NO3)2 or less if the precipitate was small, evaporate almost to dryness and add 1-2 cc. of water. Absorb the solution in a small piece of filter paper, wind OUTLINE OF ANALYSIS (TABLE VI) 61 a platinum wire around it in the form of a spiral and ignite the paper in a flame until the carbon is completely burned. A blue residue shows the presence of aluminium. (See note, Exp. iga.) (62) Detection of Chromium. To the filtrate form (66) add HC2H 3 O2 slowly until the solution is just acid. (Test with litmus.) If the solution is colorless, chromium is absent and the solution should be tested for zinc accord- ing to (64). If the solution is at all yellow, add about IO cc. of BaCl 2 solution and heat the mixture to boiling. If a yellow precipitate forms, chromium is present. Unless the precipitate is distinctly yellow the confirmatory test should be made. (See Discussion 40.) Treat the filtrate by (64). (63) Confirmatory Test for Chromium. Dissolve the precipitate (62) by pouring repeatedly through the filter a cold lo-cc. portion of o.6N. HNOs. To the solution contained in a test-tube add 1-2 cc. of ether and i cc. of 3 per cent H 2 O 2 solution. Shake the mixture. A blue color appearing in the ether layer shows the presence of chromium. (See note, Exp. 2Oa.) (64) Detection of Zinc. Warm the HC2H3O 2 solution (62) or the filtrate from (62) to about 50 and saturate in a small flask with H 2 S. If a flocculent precipitate does not form at once, cork the flask and allow it to stand five to ten minutes. The formation of a white flocculent precipitate shows the presence of zinc. Unless the pre- cipitate is white and flocculent the confirmatory test should be made. (65) Confirmatory Test for Zinc. Filter off the pre- cipitate obtained in (64) and dissolve it by pouring re- peatedly through the filter a 5-cc. portion of 6N. HNO 3 . To the resulting solution add 4-5 drops of N. 100 Co(NO3) 2 , or less if the precipitate was small. Evaporate the mix- ture almost to dryness in a porcelain dish, neutralize with Na 2 COa solution and add 0.5 cc. in excess. -Evapo- rate to dryness and ignite gently until the purple color 62 CATIONS (METAL IONS) due to the Co(NO 3 )2 disappears. (See Discussion 41.) The appearance of a green color shows the presence of zinc. DISCUSSION 38. In the separation of aluminium a large excess of NH 4 OH must be avoided, since it tends to redissolve the A1(OH) 3 by forming NH 4 AlOo. A moderate excess, how- ever, must be present in order to keep the zinc in solution. 39. Since aluminium and silica are often present in the Na2O2 and NaOH used as reagents in the separation of the aluminium and iron divisions, a blank test should be made for these impurities by treating 10-15 cc - of water, to which has been added about the same quantities of these two reagents as was used in the regular analysis, by (60) and comparing the precipitate formed with that obtained in the regular analysis. The confirmatory test should always be made in case the NH 4 OH precipitate is small, in order to avoid mistaking silicic acid (H^SiOs) for A1(OH) 3 . 40. If, owing to careless manipulation in filtering and washing the second or third group precipitates, and sul- phides have been oxidized to sulphates, the addition of BaCb for the separation of chromium may give a white or pale yellow precipitate composed largely of BaSO 4 . This not only makes it necessary to confirm the presence of chromium but renders the confirmatory test less delicate owing to the difficulty of extracting the barium chromate (BaCrO 4 ) from the BaSO 4 with acid. The color of the solution after treatment with NaOH and Na2C>2 for the separation of aluminium and iron divisions should be noted. A yellow color, changing to orange on the addition of acid, indicates chromium. 41. The confirmatory test for zinc is of value only when the IrbS precipitate is small and finely divided or when the presence of foreign materials causes it to be PRELIMINARY EXPERIMENTS 63 dark-colored. In the hands of the careful worker the test is quite satisfactory and will detect 0.5 mg. of zinc. If, owing to too much heat, the residue becomes black, it should be dissolved in a few drops of HNOa and evaporated almost to dryness, and the test should be repeated with the exception that no more Co(NO3)2 should be added. Prelminary Experiments Group III, Iron Division Mn++, Fe++, Co++ In connection with the following experiments study Tables V and VII. Experiment 22. To separate 5-cc. portions of the test solutions containing the above ions, add NH 4 OH to alka- line reaction and saturate with FhS. Note the color of the precipitates formed. Write equations. Experiment 23. Decant or filter off the liquid from the precipitates (Exp. 22) and dissolve by the addition of HC1. Add a few drops of HNO 3 if HC1 fails to effect a solution. Write equations. NOTE. CoS and NiS are dissolved very slowly by HCl but much more rapidly in aqua regia. The oxidizing action of aqua regia removes the S = , thus shifting the equilibrium and allowing the cobalt and nickel to pass into solution. Experiment 24. To the clear solutions (Exp. 23) add NaOH to alkaline reaction and then about I gram of solid Na2C>2, a little at a time. Note any changes in the precipitates caused by the Na2C>2. Boil to expel excess Na2C>2. Write equations. NOTE. Na 2 2 oxidizes Mn(OH) 2 to the less soluble MnO(OH) 2 , Fe(OH) 2 to Fe(OH) 3 and Co(OH) 2 to Co(OH) 3 . Ni(OH) 2 is only parti- ally oxidized to Ni(OH) 3 . All these compounds are insoluble in an excess of NaOH, differing in this respect from the hydroxides of the aluminium division. 64 CATIONS (METAL IONS) Experiment 25. Decant or filter off the liquid from the precipitates (Exp. 24) and add 5-10 cc. of HNO 3 . Note the fact that iron, cobalt and nickel hydroxides readily dissolve. Heat the mixture containing the manganese nearly to boiling and add H^Oo, a few drops at a time, until solution is complete. Write equations. NOTE. MnO(OH) 2 is not affected by HN0 3 alone, but if reduced to a lower oxide it is dissolved as Mn(N0 3 ) 2 . The reduction is readily brought about with H 2 2 , oxygen being evolved. Experiment 26. Heat the solutions (Exp. 25) to boil- ing and add gradually 1-2 grams of solid KC1O 3 or NaClOa. Boil for a minute or two. Note the precipitation of MnO2. (Difference, separation of Mn.) Write equations. Experment 26a. Filter off the liquid from the MnC^ (Exp. 26), add 5-10 cc. of i6N. HNOs and 1-2 grams of Boil for a minute or two and set aside till the settles. The purple color of the solution is due to HMnO4. Write equations. Experiment 27. To the solutions of iron cobalt and nickel (Exp. 26) slowly add NH 4 OH to distinctly alkaline reaction. Note the precipitation of Fe(OH)s. (Differ- ence, separation of Fe.) NOTE. The Co(OH) 2 and Ni(OH) 2 at first formed dissolve in an excess of NH 4 OH with the formation of the complex ions Co(NH 3 ) 4 + + and Ni(NH 3 ) 4 ++ . (See Introduction 15.) Experiment 2ya. Decant or filter the liquid from the Fe(OH)a precipitate (Exp. 27) and dissolve it in a small quantity of 6N. HC1. Divide the solution into two parts and add K 4 Fe(CN) 6 to the one and KCNS to the other. Write equations. NOTE. It is important to use HC1 as the solvent for Fe(OH) 3 - HN0 3 must not be used since the subsequent reaction with KCNS is rendered much less delicate in its presence, N0 2 , a common impurity in HNO 3 , also giving a red color with KCNS. The test is extremely delicate, and PRELIMINARY EXPERIMENTS 65 when only a faint red color is produced the acids used should be tested for iron. The red color is due to the formation of non-ionized ferric thiocyanate [Fe(CNS) 3 ] and the test is therefore rendered more delicate + + when an excess of the reagent is used. K 4 Fe(CN) 6 gives with Fe + a deep-blue precipitate of Fe 4 (Fe(CN) 6 ) 3 (Prussian blue). Experiment 28. Saturate the solutions of cobalt and nickel (Exp. 27) with H 2 S. Decant or filter the liquid from the precipitates of CoS and NiS and dissolve by the addition of HC1 and a few drops of HNOa. Evaporate to 1-2 cc. and just neutralize with NaOH. Add HC 2 H 3 O2 to very faintly acid reaction, and a considerable excess of potassium nitrite (KNp2). Note the precipitation of potassium cobaltinitrite (K 3 Co(NO2)6). (Difference, detec- tion of Co.) NOTE. In very faintly acid reaction, cobalt produces with KN0 2 a yellow crystalline precipitate of K 3 Co(NO 2 ) 6 ; hence the necessity of neutralizing the strong acid with NaOH and subsequently rendering the solution acid with the weak HC 2 H 3 O 2 . The precipitate may form rather slowly. The reaction preceeds as follows: Co(NO 3 ) 2 +7KNO 2 +2HC 2 H 3 O 2 ->K 3 Co(NO 2 ) 6 -i-NO+2KC 2 H30 2 +2KN03 A part of the KN0 2 is used in oxidizing the cobalt to the trivalent con- dition. This then unites with the excess KN0 2 to form the complex salt K 3 Co(N0 2 ) 6 . The K 3 Co(N0 2 ) 6 is somewhat soluble in water but much less so in KN0 2 solution, owing to the common ion effect of K + . Experiment 29. To the solution containing nickel (Exp. 28) add NH 4 OH to faintly alkaline reaction, heat to boiling to drive off excess NHs and then add a few drops of dimethyl-glyoxime. Note the precipitation of nickel dimethyl-glyoxime [NiKCHs^CiN^Hh]. (Difference, detection of Ni.) 66 CATIONS (METAL IONS) TABLE VII OUTLINE FOR THE SYSTEMATIC SEPARATION AND DETECTION OF GROUP III (Fe DIVISION) No. Reagent MnO(OH) 2 Fe(OH) 3 Co(OH) 3 Ni(OH) 2 2 3 4 5 6 7 8 HNO 3 H 2 2 KC10 3 HN0 3 Pb0 2 *NH 4 OH HC1 (o) K 4 Fe(CN) 6 (b) KCNS H 2 S HC1 KC1O 3 NaOH HC2H 3 2 KNO 2 NH 4 OH (CH 3 ) 2 C 2 (NOH) 2 MnO(OH) 2 Mn(NO 3 ) 2 MnO-2 Fe(N0 3 ) 3 Fe(N0 3 ) 3 Fe(N0 3 ) 3 Fe(OH) 3 FeCl 3 Fe 4 (Fe(CN) 6 ) 3 Co(N0 3 ) 2 Co(N0 3 ) 2 Co(N0 3 ) 2 Co(NH 3 ) 4 (NO 3 ) 2 J CoCl 2 KsCo(N0 2 ) 6 I Ni(N0 3 ) 2 Ni(N0 3 ) 2 Ni(N0 3 ) 2 Ni(NH 3 ) 4 (N0 3 ) 2 NiS NiCl 2 K 4 Ni(N0 2 ) 6 Ni(CH3) 2 C 2 N 2 O 2 H) 2 HMnO 4 Fe(CNS) 3 * If phosphates may be present, the procedure should follow Table VIII. TABLE VIII OUTLINE FOR THE SYSTEMATIC SEPARATION AND DETECTION OF Fe, Co AND Ni WHEN PHOSPHATES MAY BE PRESENT Ions present in HNO 3 solution Reagent po 4 Fe+++ Co++ NI++ i (a) (NH 4 ) 2 MoO 4 (b) HC1 + K4Fe(CN) 6 (c) (NH 4 OH + HC2H 3 2 (NH 4 C 2 H 3 2 + FeCl 3 2 NH4C2H 3 O 2 3 H 2 S (NH4)3P04(Mo0 3 ),2 Fe+++ Fe 4 (Fe(CN) 6 )3 Fe(C2H 3 2 ) 3 Fe(OH) 2 C 2 H 3 2 Co++ Co++ Co++ Co(C2H 3 02) 2 (?) CoS Ni++ Ni++ Ni++ Ni(C2H 3 02) 2 (?) NiS FeP0 4 i i OUTLINES OF ANALYSIS (TABLES VII, VIII) 67 ANALYSIS Group III, Iron Division (70) Precipitation of Manganese. Transfer the pre- cipitate obtained in (51) to a porcelain dish, together with the filter if necessary; add 5-20 cc. of 6N. HNOs, heat nearly to boiling and, if any of the precipitate remains undissolved, add slowly and with constant stirring 3 per cent H 2 O2 solution until the precipitate has completely dissolved. Filter to remove the paper and evaporate almost to dryness. Add 10-15 cc. of i6N. HNO 3 , heat to boiling, add 0.5-1 gram of powdered KC1O 3 and boil gently. If a large precipitate forms add 1-2 grams more KClOs, a small portion at a time. A dark brown or black precipitate shows the presence of manganese. Filter through an asbestos filter (see Discussion 42). Heat the filtrate to boiling and add 0.5 gram more of powdered KC1O 3 . Boil gently. If a precipitate forms add 1-2 grams more KClOs, heat to boiling and filter through the same filter. Wash the precipitate with a little i6N. HNOa which has been freed from the oxides of nitrogen by warming with a little KClOs. Treat the precipitate by (71) and the filtrate by (72). (71) Confirmatory Test for Manganese. Transfer the precipitate (70) to an evaporating dish, add 1-2 grams of lead dioxide (PbO 2 ) and 10-15 cc. of 6N. HNO 3 . Heat the mixture to boiling and boil for one to two minutes. Then pour into a test-tube and allow the PbO 2 to settle. A purple solution shows manganese to be present (72) Detection of Phosphate. Heat about one- tenth of the filtrate from (70) to boiling and pour it into about three times its volume of ammonium molybdate [(NH 4 )2MoO 4 ] reagent. (See Discussion 43.) The formation of a fine yellow precipitate shows the presence of phosphate. If phosphate is shown to be present treat a second one- tenth portion of the filtrate by (75). 68 CATIONS (METAL IONS) If phosphate is absent treat the remaining nine- tenths by (73). (73) Separation of Iron in the Absence of Phosphate. If phosphate was shown to be absent add NH.tOH to the remaining nine-tenths of filtrate (70) until it is distinctly alkaline and then 4-5 cc. in excess. The formation of a reddish-brown precipitate shows the presence of iron. Filter and wash the precipitate. Treat the filtrate by (77) and the precipitate by (74). (74) Confirmatory Test for Iron. Dissolve the pre- cipitate (73) by pouring a 5-10 cc. portion of 6N. HC1 repeatedly through the filter. To the resulting solution add a few drops of potassium ferrocyanide [K 4 Fe(CN) 6 ] solution. A deep-blue precipitate (Prussian blue, Fe 4 (Fe(CN) 6 )3) shows the presence of iron. (75) Detection of Iron in the Presence of Phosphate. If phosphate was shown to be present evaporate about one-tenth of the filtrate (70) to dryness, add 1-2 cc. of I2N. HC1 and evaporate to dryness again. Dissolve the residue in 5-6 cc. of 6N. HC1 and add 5 cc. of potassium thiocyanate (KCNS) solution. A red coloration shows the presence of iron. Treat the remainder of filtrate (70) by (76). (76) Removal of Phosphate. To the remainder of the filtrate from (70) add NH 4 OH slowly and with fre- quent stirring, until the precipitate formed just fails to redissolve. The solution should not be alkaline. If, owing to the addition of too much NH 4 OH, it has become alkaline or a large precipitate separates, H^HaCb should be added to distinct acid reaction. Add 15 cc. of 3N. NH 4 C2H3O2 solution and unless the mixture has a brownish-red color add ferric chloride (FeCls) solution drop by drop, until a brownish-red color is produced. (See Discussion 44.) Transfer the mixture to a 250 cc. flask, add water to make a total volume of about 100 cc. and boil for five minutes, adding more water if a large precipitate separates. Allow the mixture to stand one OUTLINES OF ANALYSIS (TABLES VII, VIII) 69 to two minutes, filter while still hot and wash with hot water. To the filtrate add 10 cc. of 3N. NH^H.sCb solution and boil. If a precipitate separates filter it off through a separate filter and reject. Make the filtrate alkaline with NH 4 OH and treat it by (77). (77) Separation of Zinc, Cobalt and Nickel. Pass H 2 S into the filtrate (73) or (76) until, after shaking, the vapors blacken filter paper moistened with lead acetate [Pb(C2H 3 O2)2] solution. The formation of a black pre- cipitate indicates the presence of nickel or cobalt or both; a white flocculent precipitate indicates zinc. Filter, wash the precipitate with water containing a few drops of (NH 4 )2S and treat it by (77a) unless zinc has already been found; otherwise treat it by (78). Treat the filtrate by (80) if phosphate or much chromium has been found. Otherwise reject it. (See Discussion 45.) (77a) Separation of Zinc from Cobalt and Nickel. If zinc has already been found this procedure may be omitted and the precipitate from (77) treated directly by (78). If zinc has not already been found, transfer the precipitate (77), with the filter if necessary, to an evaporating dish, add 10-30 cc. of iN. HC1 and stir the cold mixture for about five minutes. Filter, wash the residue and treat it by (78). Boil the filtrate until the H 2 S is completely expelled. Make the solution alkaline with NaOH. To the cold solution add, with constant stirring, about I cc. of powdered Na2O2, a little at a time. Boil to decompose excess Na2C>2, cool and filter. Unite the residue with that undissolved by the iN. HC1 above and treat by (78). Acidify the filtrate with HC 2 H 3 O2 and test it for zinc according to (64). (See Discussion 46.) (78) Detection of Cobalt. Transfer the combined resi- dues from (77a) to a porcelain dish and add 5-15 cc. of 6N. HC1. Heat the mixture nearly to boiling, and while hot sprinkle into it a little powdered KC1O 3 . When the reaction is complete and only a residue of sulphur remains, filter and evaporate the filtrate almost to dryness. Take 70 CATIONS (METAL IONS) up the residue with 1-2 cc. of water and make the solution just neutral with NaOH. (Test with litmus.) Add 2-3 cc. of 6N. HC2H 3 O2 and then 3-5 cc. of potassium nitrite (KNO2) solution and let the mixture stand, with occasional shaking, for fifteen to twenty minutes. (See Discussion 47.) The formation of a yellow crystalline precipitate shows the presence of cobalt. Filter and treat the nitrate by (79). (79) Detection of Nickel. Dilute the filtrate from (78) to about 25 cc., add NH 4 OH until the solution is just alkaline. Heat to boiling to expel excess NH 3 and then add a few drops of dimethyl-glyoxime [(CH 3 CNOH) 2 ] solution, and if a red precipitate or coloration does not form at once allow the mixture to stand five to ten minutes. A red precipitate shows the presence of nickel. (See Discussion 48.) DISCUSSION 42. Owing to the oxidizing action of strong HNO 3 , the precipitate of MnO2 cannot be filtered through an ordinary filter paper. An asbestos filter may be prepared by placing a small pinch of glass wool in the neck of a funnel, tamping it gently with the finger and pouring over it a suspension of fine asbestos fiber in water, enough to make a layer 2-3 mm. thick. 43. If the solution containing the ammonium molyb- date is heated much above 60 C., white insoluble molybdic acid (MoO 3 ) may separate out and render the test for phosphate much less delicate. The yellow precipitate is (NH 4 )3PO4-i2MoO 3 and is most readily formed at about 60 in the presence of a large excess of (NH 4 ) 2 MoO4 and HNO 3 . 44. The separation of ferric iron and phosphate from the bivalent elements is based on the following facts: The solubility product for FePO 4 is much smaller than that DISCUSSION (42-48) 71 for the phosphates of the bivalent elements. In a boiling acetic acid solution containing a large excess of the acetate ion, insoluble basic ferric acetate [Fe(OH) 2 C2H 3 O2] is formed. In order to issure complete removal of the phos- phate and prevent its interaction with the bivalent elements -f + excess Fe + is necessary. This is shown by the brownish- red color of Fe(C2HsO2)3. The separation is satisfactory and complete if accurately carried out. If the solution is allowed to become alkaline much of the iron will pre- cipitate ^s Fe(OH)3 while the bivalent elements may form insoluble phosphates or hydroxides. On the other hand, if the solution is too acid the FePO4 will not be precipi- tated owing to the formation of HPO4 = and the consequent + + reduction of the ion product C Fe + -C P0 ^. If there is a tendency toward the formation of colloidal FePCU its coagulation may be greatly promoted by boiling. 45. The solution obtained after the removal of zinc, cobalt and nickel may contain some or even all of the alkaline earth metals. In the presence of phosphate, barium, strontium, calcium and magnesium may be pre- cipitated as phosphates along with the metals of Group III. In the presence of much chromium, magnesium may be precipitated completely as Mg(CrO2)2. 46. Although cobalt and nickel sulphides are soluble in HC1 the reaction is comparatively slow even in moder- ately concentrated solutions. Therefore, when a mixture of the sulphides of zinc, cobalt and nickel is treated with a cold iN. solution of HC1 the greater portion (80-90 per cent) of the cobalt and nickel remains undissolved, while all of the zinc passes into solution. The subsequent treatment of the solution with NaOH and Na2O 2 gives a very satisfactory separation since the cobalt and nickel are present in such small quantities that only an insig- nificant amount of zinc is carried down with them. 47. It has been mentioned (see note, Exp. 28) that 72 CATIONS (METAL IONS) when KNO2 is added to an acid solution containing cobalt and nickel a part of the HNC>2 formed oxidizes the cobalt + + from the cobaltous (Co 4 " 4 ") to the cobaltic (Co + ) state. The cobaltic ion in turn unites with the nitrite ion to form + + the complex ion Co(NO2)e + In the presence of HC 2 H 3 O2 the solubility product for K 3 Co(NO 2 )6 is soon reached, resulting in the precipitation of potassium cobalt- initrite [K 3 Co(NO2)6]. Nickel is not readily oxidized by + + HNO2 to the nickelic (Ni + ) state although it does form the complex ion Ni(NO2)e ++ . The potassium salt K4Ni(NO2)e is fairly soluble, however, so that little or no nickel is precipitated with the cobalt. Since K 3 Co(NO 2 )6 is readily soluble in strong acids the necessity of neutral- izing the HC1 and subsequently acidifying with the weaker HC2H 3 O 2 is apparent and since both cobalt and nickel form complex ions with NC>2~ a large excess of KNO2 must be added. 48. Dimethyl-glyoxime is a weak monobasic organic acid having the formula (CH 3 ) 2 C2(NOH) 2 . It reacts with the nickel ion (Ni + + ) according to the following equation : C H 3 C NOH CH 3 C NOH HON C CH 3 Ni+++2 : : : + 2 H+ CH 3 C NOH CH 3 C NO Ni ON C CH 3 Nickel dimethyl-glyoxime is least soluble in a neutral or weakly acid solution, is very voluminous and has an intense red color. It is capable, therefore, of detecting very small quantities (o.i mg.) of nickel. In the presence of cobalt, dimethyl-glyoxime produces a brown coloration which deepens as the amount of cobalt is increased. It is advisable, therefore, to remove the cobalt before making the test for nickel. PRELIMINARY EXPERIMENTS 73 GROUP IV Preliminary Experiments Ba ++ , Sr++, Ca+ + In connection with the following experiments study Table IX. Experiment 30. To separate 5-cc. portions of the test solutions containing the above ions add NH 4 OH, drop by drop, till its odor persists after shaking. Heat to boiling and then add ammonium carbonate [(NH 4 )2CO 3 ] in slight excess. Allow the precipitates to settle, decant off the liquid and dissolve the precipitates in HC2H 3 O 2 . Write equations. Experiment 31. Divide the solutions (Exp. 30) into two portions each. To one add an equal volume of satu- rated calcium sulphate (CaSO4) solution. If a precipitate does not form immediately heat to boiling. Note the character of the precipitates and the relative time for precipitation. Write equations. Experiment 32. Determine the color of the flame produced by barium, strontium and calcium by introducing a little of each (second portion, Exp. 31) on a clean platinum wire into the colorless flame of the Bunsen burner. (Cau- tion: the platinum wire should be held only in the outer cone of the Bunsen flame.) Experiment 33. To the acetic acid solutions of the above metals (second portion, Exp. 31) diluted to about 50 cc. add potassium dichromate (K^C^O?) solution. Note the precipitation of BaCrO4. (Difference, separation of Ba.) Write equations. NOTE. The solubility of the chromates of barium, strontium and calcium in 100 cc. of water at 10 is 0.38, 120 and 400 mgs. respectively From this it is evident that in comparatively dilute solutions practically no strontium or calcium will be precipitated. From more concentrated solutions, however, some strontium may separate out as SrCr0 4 . 74 CATIONS (METAL IONS) Experiment 34. To each of the solutions containing strontium and calcium (Exp. 33) add NH 4 OH to alkaline reaction, heat to boiling and add (NH^oCOs in excess. Note the precipitation of SrCOs and CaCOs. Filter, wash the precipitates and dissolve them by pouring a small quantity of hot HC2H 3 O 2 repeatedly through the filter. Write equations. Experiment 35. Dilute the solutions (Exp. 34) to 2025 cc., heat to boiling and add ammonium sulphate [(NH 4 ) 2 SO 4 ] solution. Boil for two to three minutes and filter. To the filtrates add NH 4 OH to alkaline reaction and then ammonium oxalate [(NH 4 )2C2O 4 ] solution. Note the precipitation of CaC 2 O 4 . (Difference, detection of Ca.) Write equations. NOTES. The solubility of the sulphates of strontium and calcium in 100 cc. of water at 18 is n and 200 mgs. respectively. The correspond- ing solubility of their oxalates is 4.6 and 0.56 mgs. From this it will be noted that a complete separation of strontium and calcium cannot be obtained by means of their sulphates, but it is also evident that there will remain in the nitrate enough CaSO 4 to be easily detected with (NH 4 ) 2 C 2 4 . The solubility of the salts of barium, strontium and calcium in mgs. per 100 cc. of water at 18 is shown in the following table: C0 3 Cr0 4 S0 4 C 2 4 Ba 2-3 0.38 0.23 8.6 Sr i . i 120. II . 4.6 Ca i TI 4OO 2OO o 56 OUTLINES OF ANALYSIS (TABLE IX) TABLE IX OUTLINE FOR THE SYSTEMATIC SEPARATION AND DETECTION OF GROUP IV Ions present in acid solution ; ?J Reagent Ba+ + Sr+ + Ca+ + I NH 4 OH + (NH 4 )2C03 BaCO 3 SrC0 3 CaCO 3 2 ^ HC 2 H 3 O 2 (a) CaSO 4 (sat. soln.) Ba(C 2 H 3 O 2 ) 2 BaSO 4 Sr(C 2 H 3 2 ) 2 SrSO 4 Ca(C 2 H 2 O 2 ) 2 Ca(C 2 H 3 O 2 ) 2 (&) K 2 Cr 2 7 BaCrO 4 SrCrO 4 CaCrO 4 4 NH 4 OH + (NH 4 ) 2 C0 3 1 SrCO 3 CaCO 3 5 6 HC 2 H 3 O 2 (a) CaSO 4 (sat. soln.) Sr(C 2 H 3 2 ) 2 SrS0 4 Ca(C 2 H 3 O 2 ) 2 Ca(C 2 H 3 2 ) 2 (&) (NH 4 ) 2 S0 4 SrSO 4 CaSO 4 7 (NH 4 ) 2 C 2 4 1 CaC 2 O 4 1 ANALYSIS Group IV Ba++, Sr++, Ca+ + (80) Precipitation. Evaporate the filtrate from Group III (50) to 15-20 cc. Filter off any free sulphur. (See Discussion 49.) Heat to boiling (see Discussion 50), add (NH^COs as long as a precipitate continues to form, and allow the mixture to stand about ten minutes. A white crystalline precipitate indicates the presence of Group IV. Filter, treat the precipitate by (81) and the filtrate by (82). (See Discussion 51.) (81) Detection of Barium and Strontium. Dissolve the precipitate (80) by pouring a hot lo-cc. portion of 6N. HC2HsO2 repeatedly through the filter. To a small portion of the resulting solution add an equal volume of saturated CaSCU solution. The immediate formation of a white precipitate shows the presence of barium. A white precipitate which forms only slowly or on warming shows the absence of barium and the presence of strontium. 76 CATIONS (METAL IONS) If no precipitate forms, both barium and strontium are absent. (See Discussion 51 and 52.) If barium is shown to be present, treat the remainder of the solution by (83) ; if barium is absent and strontium is shown to be present, treat by (85) ; if both barium and strontium are absent treat by (86). (82) Detection of Traces of Barium and Calcium. To a portion of the nitrate (80) add 1-2 cc. of (NH 4 ) 2 SO 4 and warm. A white precipitate (turbidity) shows the presence of barium. To the remainder of the nitrate (80) add 1-2 cc. of (NH 4 ) 2 C 2 O 4 . A white precipitate (tur- bidity) shows the presence of calcium. Reunite the two portions, filter off any precipitate and treat the filtrate for Group V by (90). (See Discussion 51.) (83) Separation of Barium. Dilute the remainder of the solution (81) to about 75 cc. and add 5 cc. of NH 4 C 2 H3O 2 . Heat to boiling, and while hot add K 2 Cr 2 O 7 in slight excess. A large excess should be avoided. (See note, Exp. 33.) Filter while hot and treat the filtrate by (84). A confirmatory test for barium may be made as follows: Dissolve the precipitate of BaCrO 4 in dilute HC1; add about 0.5 cc. of alcohol (C 2 H 5 OH) and boil until the yellow color entirely disappears and the solution becomes green. Neutralize with NH 4 OH and filter off the Cr(OH) 3 . Acidify the filtrate with HC1 and evaporate nearly to dry- ness. Introduce a little of this solution on a clean platinum wire into the colorless flame of the Bunsen burner. A green color shows the presence of barium. (84) Precipitation of Strontium and Calcium. To the filtrate (83) add NH 4 OH to alkaline reaction; heat to boiling and add (NH 4 ) 2 COa solution in excess. Allow the mixture to stand five to ten minutes and filter. Reject the filtrate. Dissolve the precipitate in a few cc. of 6N. HC 2 HsO 2 and test for strontium as outlined in (81). (85) Separation of Strontium. Dilute the solution (84) to about 20 cc.; heat to boiling and add (NH 4 ) 2 SO 4 DISCUSSION (49-53) W solution, as long as a precipitate continues to form. Filter and treat the nitrate by (86). A confirmatory test for strontium may be made as follows: Add to the precipitate of SrSO 4 about ten times its volume of solid Na2COs and 10 cc. of water, and boil three to five minutes. Filter, dissolve the residue in 2-3 cc. of HC1 and introduce a small portion of the solution on a clean platinum wire into the colorless flame of the Bunsen burner. A deep red color shows the presence of strontium. (Caution: Do not confuse with calcium, yellowish-red color.) (86) Detection of Calcium. To the filtrate (85) add NH 4 OH to alkaline reaction, and then (NH 4 ) zC&i. A white crystalline precipitate shows the presence of calcium. A confirmatory test for calcium may be made by moistening the CaC2O4 precipitate with a little cone. HC1 and introducing a small portion of it on a clean platinum wire into the colorless flame of the Bunsen burner. A yellowish-red color shows the presence of calcium. DISCUSSION 49. The solution to be tested for Group IV should be clear and colorless and should contain only a moderate excess of NH 4 C1. Any yellow or brown color due to undecomposed (NH 4 ) 2 S or small quantities of nickel or chromium, not precipitated in Group III, will usually be removed by the evaporation as directed in (80). NiS is slightly soluble in excess (NH 4 ) 2 S, forming a brown solution. A large excess of NH 4 C1 will hold chromium in solution as CrCls^NHs, a red solution. The filtrate from Group III may also contain small quantities of aluminium, owing to its amphoteric (see Introduction 16) nature. Evaporation decomposes (NH 4 ) 2 S and volatilizes excess NH 3 which causes the precipitation 78 CATIONS (METAL IONS) of any NiS and A1(OH) 3 that may be present. If the solution is still colored from chromium it should be evapo- rated to dryness, taken up in 10 cc. of water and filtered to remove excess ammonium salts and Cr(OH) 3 . 50. Group IV should be precipitated from a hot solu- tion in order to insure complete separation from magnes- ium, which forms a double salt, MgCO 3 - (NH 4 ) 2 CO-4H 2 O, with (NH 4 ) 2 CO3 only moderately soluble in cold water. It is much more soluble in hot water. The solution should not be boiled after the addition of the precipitating agent (NH 4 ) 2 CO3, owing to the fact that it decomposes quite readily when heated. From a cold solution the carbon- ates are precipitated in a fine state of division. Heat favors the formation of larger particles. 51. Owing to the appreciable solubility of the alkali earth carbonates (see table following note, Exp. 35), traces of these metals may remain in the filtrate after precipitation with (NH 4 ) 2 CO3. The insolubility of their phosphates makes it necessary to remove them before testing for magnesium in Group V. 52. The method of analysis as outlined does not permit a complete separation of barium, strontium and calcium; yet it enables the analyst to determine with a rather high degree of accuracy their presence or absence. The principle involved in the detection of barium and strontium is based on the relative solubility of their sulphates. (See table following notes, Exp. 35.) A saturated solution of CaSO4 produces enough SO 4 = to cause an immediate precipitation of BaSO 4 , but owing to the larger solubility product of SrSO 4 the solution is saturated with it only slowly or after heating. The difference in time required for the precipitation of BaSO 4 and SrSO 4 under the con- ditions stated is quite marked, so that it is comparatively easy to distinguish between barium and strontium. 53. The test for calcium assumes the absence of strontium and barium and the test for strontium assumes the absence of barium; hence, if barium is shown to be PRELIMINARY EXPERIMENTS 79 present it must be removed before making the tests for strontium or calcium. Precipitation as BaCrO 4 effect- ively removes the barium, while little or no SrCrO 4 or CaCrO 4 are precipitated. (See table following notes, Exp. 35-) The separation of strontium and calcium with (NH 4 )2SO 4 is not complete. Some of the calcium may be precipitated with the strontium and some of the stron- tium will remain in the solution, though not enough to interfere with the test. Enough calcium will always remain in the nitrate, however, to give a precipitate with (NH 4 ) 2 C2O 4 , owing to the very slight solubility of CaC2O 4 . V Preliminary Experiments Mg++, NH 4 + , K+, Na+ In connection with the following experiments study Table X. Experiment 36. To 5 cc. of the test solution contain- ing Mg ++ add an equal volume of NH 4 C1. Make dis- tinctly alkaline with NH 4 OH and add sodium phosphate (Na 2 HPO 4 ) solution. The precipitate is MgNH 4 PO 4 . To a second 5-cc. portion of the test solution add NH 4 OH to alkaline reaction. Note the precipitation of Mg(OH)2. Now add, without filtering an excess of NH 4 C1. Why does not Mg(OH)2 precipitate in Group III? Explain. Write equations. Experiment 37. Evaporate a small quantity of NH 4 C1 to dryness in a porcelain dish and heat the residue. To a second portion in a test-tube add NaOH. Note the odor of the gas evolved and its effect upon a piece of moist red litmus paper. What is the gas evolved? Why can this reaction be used as a test for NH 4 + ? 80 CATIONS (METAL IONS) Experiment 38. To separate i-cc. portions of the test solutions of Mg ++ , K+ and Na+ add 10-15 drops of perchloric acid (HC1O4). Allow the mixtures to stand for a short time. Note the precipitation of KCIO-*. (Dif- ference, detection of K.) Experiment 39. To separate i-cc. portions of the test solutions of K + and Na + add an equal volume of alcohol and then about i cc. of fluoboric acid (HBF 4 ). Note the precipitation of KBF4. (Difference, separation of K.) Write equations. Experiment 40. Repeat Exp. 39, substituting fluo- silicic acid (H 2 SiF 6 ) for the HBF 4 . Note the character of the precipitates formed. Why must potassium be removed before testing for sodium? Write equations. Experiment 41. Determine the color of the flame produced by potassium and sodium by introducing a salt of each on a clean platinum wire, into the colorless flame of the Bunsen burner. OUTLINES OF ANALYSIS (TABLE X) 81 TABLE X OUTLINE FOR THE SYSTEMATIC SEPARATION AND DETECTION OF GROUP V No. Reagent Mg+ + NH 4 + K+ Na+ i 2 3 4 Na 2 HPO 4 NH 4 OH NaOH (a) HC1O 4 (b) HBF 4 H 2 SiF 6 MgNH 4 PO 4 NH 4 + NH,*t K+ K+ KC1O 4 Na+ Na+ NaClO 4 NaBF 4 Na 2 SiF 1 KBF 4 i ANALYSIS Group V Mg++, NH 4 +, K+, Na+ (90) Detection of Magnesium. Evaporate the filtrate from (82) in a porcelain dish until the ammonium salts begin to crystallize out. Filter, and to one-third of the filtrate add 1-2 cc. of Na 2 HPO 4 and enough i$N. NH 4 OH to make one-third the total volume. Shake the mixture vigorously for two to three minutes, and if a precipitate does not form, rub the walls of the vessel with a glass rod (see note, Exp. I4a) and allow it to stand for some time. A white crystalline precipitate is MgNH 4 PO 4 . (See Dis- cussion 54.) Acidify the remaining two-thirds of the filtrate above with HC1 and treat by (91). If the precipitate of MgNH 4 PO 4 is of doubtful form it should be filtered, dissolved off the filter with a small quantity of HC2HsO2 and reprecipitated by the addition of NH 4 OH and a small quantity of NaHPO 4 . (91) Detection of Potassium. Transfer the solution (90) to a porcelain dish and evaporate to dryness. Place the dish on a clay triangle and ignite till the ammonium salts are completely expelled. (See Discussion 55.) Dis- 82 CATIONS (METAL IONS) solve the residue in 5-6 cc. of water, transfer about I cc. of the resulting solution to a wa,tch glass and add a few drops of perchloric acid (HC1O 4 ). The formation of a white crystalline precipitate shows the presence of potas- sium. Confirm by means of the flame test. (See Pre- liminary Exp. 41.) Treat the remainder of the solution by (92) if potassium was found, otherwise by (93). (92) Removal of Potassium. To the remainder of the solution (91) add an equal volume of alcohol and an excess of fluoboric acid (HBF 4 ). Filter off the precipitate of KBF 4 and treat the filtrate by (93). (93) Detection of Sodium. To the nitrate (92) add 1-2 cc. of fluosilicic acid (H 2 SiF 6 ) and allow the mixture to stand a few minutes. (See Discussion 58.) A white gelatinous precipitate shows the presence of sodium. Confirm by means of the flame test. (See Preliminary Exp. 41.) (94) Detection of Ammonium. The test for NH 4 + must be made on a small portion of the original substance. Introduce 3-4 cc. of the original substance, if it is a liquid, or about o.i gm. if it is a solid, into a small beaker, and add NaOH till the mixture is distinctly alkaline. Test the vapors with a piece of red litmus paper placed on the under side of a watch glass covering the beaker. If no change is observed in the litmus, warm the mixture gently, but do not boil. If the red litmus turns blue ammonium salts are present. DISCUSSION 54. Since there are no characteristic color reactions that can be used for the detection of magnesium it becomes necessary to use considerable care in manipulation in order to secure a precipitate that can be recognized and depended upon. MgNH 4 PO 4 is a white crystalline sub- stance soluble in HC 2 H 3 O 2 , but insoluble in NH 4 OH. Practically all the metals, except alkali metals, form. DISCUSSION (54-58) 83 phosphates insoluble in NH^OH, hence the necessity of having a clear solution entirely free from metals of the previous groups. The presence of NH^Cl is necessary to prevent the precipitation of Mg(OH) 2 , a white flocculent precipitate. (See Discussion 34.) 55. Since the presence of ammonium salts interferes with the subsequent tests for potassium and sodium, it is necessary that their removal be complete. Ammonium compounds react with both HC1O4 and H^SiFe to form the corresponding ammonium salts, which are only slightly soluble under the conditions of the experiments. In order to insure the complete volatilization of the ammonium salts all parts of the dish should be well heated, though it must not be heated nearly to redness since both KC1 and NaCl are somewhat volatile at that temperature. 56. In testing for potassium and sodium by means of the flame, a faint yellow coloration will almost invari- ably be obtained, owing to the fact that traces of sodium are nearly always present in the reagents previously used. A fleeting yellow tinge should not be taken as evidence of the presence of sodium, but the yellow color should be distinct and persistent. 57. The color of the flame may be used to detect potassium in the presence of considerable amounts of sodium, if a blue glass is used to cut out the yellow sodium rays. With small amounts of potassium, however, this test is not always satisfactory. 58. Fluosilicic acid (H 2 SiF 6 ) reacts slowly with glass, so that on long standing in a glass vessel it will usually produce a precipitate even in the absence of sodium. Since the Na2SiF 6 precipitate is semitransparent, the reaction should be carried out in a clear test-tube or on a watch glass. 84 CATIONS (METAL IONS) QUESTIONS FOR REVIEW 1. Name the group reagents and the compounds pre- cipitated by each. 2. Why is a precipitate sometimes obtained on the addition of water before adding the group reagent? Give examples. 3. Why do bismuth and antimony sometimes pre- cipitate in Group I ? 4. What is the effect of adding NH 4 OH to AgCl, Hg 2 Cl2? Write equations. 5. Can HC1 be substituted for NH 4 C1 HNO 3 in the precipitation of Group I? Explain. Could NaCl be used? 6. Explain what is meant by the following terms: hydrolysis, reagent, precipitate residue, filtrate, colloidal solution. 7. Define ion, acid, base, salt, oxidation, reduction. 8. State the law of mass action. 9. What is meant by common ion, solubility product, ionization constant, complex ion? 10. How may 2N. HC1 be made from 6N. HC1? 11. Explain by means of the solubility-product principle why PbCb is less soluble in a solution containing NH 4 C1 than in pure water. 12. Explain by means of the law of mass action why AgCl is soluble in NH 4 OH and is reprecipitated on the addition of an acid. 13. Why is it necessary to have a definite concentration of acid for the precipitation of Group II? What is the effect if the acid concentration is too great, or if it is too small? Explain by means of the solubility-product prin- ciple. 14. What precautions must be taken in the treatment of the Group II precipitate with (NH 4 ) 2 S 2 reagent! Why? 15. Why is it necessary to test for lead and mercury in both Groups I and II? 1 6. What are the colors of the following compounds, QUESTIONS FOR REVIEW 85 precipitated in Group II: HgS, Pbs, Bi 2 S 3 , CuS, CdS, As 2 S3, Sb 2 S 3 , SnS, SnS 2 ? 17. Why is it necessary to evaporate till white fumes appear in the separation of lead in Group II? 1 8. What is the effect of adding NH 4 OH to a solution of CuSO4 and Bi 2 (SO4)2? Write equations. Explain. 19. What is the purpose of KCN in the detection of cadmium? Write equations involved. 20. Explain by means of the solubility-product prin- ciple why CuS, which is only slightly soluble in hot 2N. HC1, dissolves readily in hot 2N. HNO 3 . 21. What effect does temperature have on the complete- ness with which Group II sulphides may be precipitated? Explain. 22. What precautions are necessary to insure complete precipitation of arsenic in Group II? Explain. 23. Given a solution which is known to contain no other metals than those given below, outline a method of analysis that will necessitate no unnecessary steps: (a) Lead, mercury, cadmium. (b) Copper, arsenic, cadmium. (c) Bismuth, cadmium, antimony, tin. 24. What is the purpose of H 2 C 2 O4 in the detection of antimony? 25. Show by a series of equations the changes through which arsenic goes when H 2 S is passed through a dilute HC1 solution of H 3 AsO 4 . 26. Explain by means of the solubility-product prin- ciple why PbSC>4 may be dissolved in NH4C 2 H 3 O 2 solution. 27. Should PbCrO4 dissolve in NH4C 2 H 3 O 2 solution? Explain. Why does PbCrCU precipitate from the same solution that dissolves PbSC>4? 28. How is Na 2 SnO 2 prepared? Write equations showing all the changes that take place. 29. Explain by means of the law of mass action why the addition of HC1 to a solution of (Na4) 2 SnS 3 pre- cipitates SnS 2 . 86 CATIONS (METAL IONS 30. In the separation of As2S5 from 80285 and SnS2 with I2N. HC1 why does more As2S 5 dissolve if the solu- tion is allowed to boil? 31. Why use just 10 cc. of I2N. HC1 in the treatment of the sulphides of the tin division? 32. What is the principle on which the separation of antimony and tin, by precipitation with H 2 S, is based? Explain by means of the solubility-product principle. 33. What is the confirmatory test for tin? What is the precipitate formed? Write all equations involved. 34. If phosphate is present, what elements may be precipitated on the addition of the third group reagent? 35. Why is it possible to separate the Al division from the Fe division with NaOH and Na2O 2 ? 36. What elements are affected by the Na2C>2? Write equations. 37. Why is Na2COs added in the separation of the Al and Fe divisions? 38. Why is HNO 3 added before NH 4 OH in the sepa- ration of aluminium? 39. What is the purpose of H2O2 in dissolving the Fe division ? 40. If it is necessary to use HNOs to dissolve Group III what conclusion can be drawn? 41. Since nickel and cobalt sulphides are not precipi- tated in acid solution why is it necessary to use aqua regia to dissolve them? 42. Given a solution which is known to contain no other metals than those given below, outline a method of anlysis that will necessitate no unnecessary steps; (a) Aluminium, chromium, iron. (b) Chromium, manganese, cobalt, nickel. (c) Zinc, manganese, iron, nickel. 43. Why is it necessary to test for zinc in the Fe division? 44. If phosphate is present in an unknown, in what QUESTIONS FOR REVIEW 87 solutions is it necessary to test for the metals of Group IV? Why? 45. What is the purpose of the following reagents in the analysis of Group IV; CaSO 4 , K 2 Cr 2 O 7 , (NH 4 ) 2 SO 4 , (NH 4 ) 2 C 2 4 +? 46. Could K 2 CrO 4 be used instead of K 2 Cr 2 O? in the precipitation of barium and lead? What is the relation between K2CrO 4 and K 2 Cr 2 O?? 47. Why is it necessary to add (NH 4 ) 2 SO 4 and (NH 4 ) 2 C 2 O 4 to the filtrate from Group IV before testing for Group V? 48. Why does Mg(OH) 2 not precipitate in Group III or Group IV? Explain by means of the solubility-product principle. 49. Why not evaporate to dry ness and drive off all ammonium salts before testing for magnesium? 50. Why is it necessary to test the original material for NH 4 + ? 51. Why remove ammonium salts before testing for potassium and sodium? 52. What is the purpose of HBF 4 in the analysis of Group V? 53. What salts are most suitable for use in flame tests? Why? 54. If an unknown is soluble in water, and phosphate is found, what metals will it be unnecessary to test for? PART III ACIDS In most cases the acid ions (anions) exist in solution as compound radicals composed of two or more elements held together and acting as a single substance. In this respect they differ from the metal ions (cations) which usually exist in solution as simple radicals. The most important anions consisting of a single element are the halides (Cl~, Br~, I~, F~) and sulphide (S = ). Because of this difference, more care must be exercised not only in the preparation of the solution for the detection of acids but also in the treatment during analysis in order to pre- vent decomposition, oxidation or reduction. The anions for whose separation and detection pro- vision is made in this scheme of analysis are : Group I. Anions whose silver salts are insoluble in cold dilute HNO 3 . Ferrocyanide, Fe(CN)e~ ~~ Sulphide, S~ Ferricyanide, Fe(CN) 6 ~ Iodide, I~ Thiocyanate, CNS~ Bromide, Br~ Cyanide, CN~ Chloride, Cl~ Group II. Anions whose salts decompose on boiling in acid solution and give characteristic volatile oxides. Carbonate, CO 3 ~ Thiosulphate, S 2 O 3 ~ Sulphite, SO 3 ~ Nitrite, NO 2 ~ Group HI. Anions whose silver salts are soluble in acid but insoluble in warm neutral solution. Arsenite, AsO 3 Arsenate, AsO 4 Oxalate, C 2 O 4 ~ Phosphate, PO 4 - Chromate, CrO 4 ~ Tartrate, C 4 H 4 Oe 88 THE SYSTEMATIC ANALYSIS 89 Group IV. Anions whose silver salts are soluble. Sulphate, SO 4 ~ Acetate, C 2 H 3 O 2 - Borate, BO 3 - Nitrate, NO 3 - Fluoride, F~ Provision was made for the detection of silicate (SiO 3 ~~) in the course of the analysis for metal ions. (See (5) ). THE SYSTEMATIC ANALYSIS ANIONS Since any interfering cations will be removed during the course of the analysis, the solution as prepared for the metal analysis may be used for the analysis of anions if the substance has been dissolved in water or cold dilute HNO 3 . If the substance is a liquid or solution treat directly by (no); if a solid treat by (100). PREPARATION OF SOLUTION (100) Treatment of a Solid. To about 0.5 gram of the finely powdered substance add 10 cc. of cold water and mix thoroughly. Filter and wash the residue with 5 cc. of cold water, catching the wash water in the vessel containing the nitrate. Acidify the filtrate with HNO 3 , noting if a gas is evolved (see Discussion 60), and treat by (no). Treat the residue by (101). (101) Treatment with Dilute HNO 3 . Pour repeatedly through the filter containing the residue undissolved by cold water, a 5-cc. portion of cold 2N. HNO 3 , noting if a gas is evolved. (See Discussion 59.) If a residue remains, punch a hole through the filter and wash it into a test-tube with 5 cc. of 6N. HNO 3 . Warm the mixture as long as any of the solid seems to. dissolve. Dilute with 5 cc. of water, cool and filter. Treat the combined filtrates by (no). Treat the residue by (102). 90 ACIDS (102) Treatment with Na 2 CO.3. If there is a residue undissolved by dilute HNO 3 , the anion constituents may be dissolved by treatment with Na 2 CO 3 solution, as directed in (6) or by fusion with Na 2 CO 3 as directed in (7). In either case, however, 6N. HNO 3 should be sub- stituted for HC1, and the solution boiled to drive off the CO2 before treatment with AgNO 3 by (no). DISCUSSION 59. A careful study of the solubility (see table, Appen- dix, III) and acidity of the material to be analyzed, in connection with the metal ions found, will often permit the omission of certain steps in the system of analysis for anions. Thus, if the substance contains silver and is soluble in water or dilute acid, it is unnecessary to test for Group I. Barium or lead and sulphate cannot exist together in a neutral water-soluble material. Group II anions will not be present in a strongly acid solution. A study of the nature and color of the group precipi- tates may also permit the omission of certain steps; e.g., Ag 2 S is black; Ag 3 Fe(CN) 6 , Ag 2 CrO 4 and Ag 3 AsO 4 are dark red; Agl, Ag 3 PO 4 and Ag 3 AsO 3 are yellow; AgBr is yellowish-white, while the remaining silver salts of Groups I and III are white. 60. When a strong mineral acid is added to a carbonate, the carbonic acid formed (H 2 CO 3 ) immediately decom- poses and, owing to its slight solubility, CO 2 , is evolved. Nitrites, sulphites and thiosulphates also tend to decom- pose in acid solution with the formation of N 2 O 3 , SO 2 , and SO 2 +S respectively. The solubility of N 2 O 3 and SO 2 is so great, however, that enough will always remain in solution for their detection according to the methods outlined in Group II. If the evolution of a gas is noted during the treatments as directed in (100) or (101), and carbonate is not found in the solution as prepared, a small portion of the orginal unknown should be tested for carbonate as directed in (130). OUTLINES OF ANALYSIS (TABLE XI) 91 TABLE XI OUTLINE FOR THE SYSTEMATIC SEPARATION AND DETECTION OF GROUP I (DIVISION A) Ions present in dil. HNO 3 solution No. Reagent Fe(CN) 6 = Fe(CN) 6 = CNS- CN- i i 2 3 4 5 6 7 AgN0 3 NaCl HC1 (NH 4 ) 2 MoO 4 Zn(N0 3 ) 2 HC1 FeSO 4 FeCl 3 NaOH FeS0 4 HC1 Ag 4 Fe(CN) 6 Ag 3 Fe(CN) 6 Ag( Na( Na( Zn( J Fe( , :NS :NS :NS CNS) 2 , CNS) 3 AgCN NaCN NaCN Zn(CN) 2 Fe(CN) 2 Na 4 Fe(CN) 6 Fe 4 (Fe(CN) 6 ) 3 Na 4 F Na(l (Fe(( 2Mo e(CN) 4 tfoO 2 ) 3 - :N),),. 2.3 Na 3 Fe(CN) 6 Na 3 Fe(CN) 6 Zn 3 (Fe(CN) 6 ) 2 H 3 Fe(CN) 6 Fe,(Fe(CN) 6 ) 2 ! ANALYSIS Group I Fe(CN) 6 = Fe(CN) 6 =, CNS-, CN-, S= I~, Br~, Cl- (no) Precipitation. To about 25 cc. of the cold solu- tion acidified with HNOs (see Discussion 60) add, slowly and with constant stirring, AgNO 3 solution in excess. Filter, divide the precipitate into two portions, treat one by (in) and the other by (120). Reserve the filtrate for analysis of Group II (130). Division A (in) Solution of the Cyanogen Compounds. Suspend one portion of the precipitate (no) in 5 cc. of water, add 5 cc. of the NaCl reagent and mix thoroughly. Filter, 92 ACIDS reject the residue of silver halides (see Discussion 61) and treat the nitrate by (112). '(See Discussion 59.) (112) Detection of Ferrocyanide. To the nitrate (in) add ammonium molybdate ( (NH 4 )2MoO 4 ) reagent in excess. A red flocculent precipitate shows the presence of ferrocyanide. (See Discussion 62.) Filter and treat the nitrate by (113). (113) Precipitation of Ferricyanide. To the nitrate (112) add a slight excess of Zn(NO)a solution. Allow the mixture to stand a few minutes. A white precipitate indicates ferricyanide. If the precipitate is colloidal or finely divided shake the mixture with a pinch of asbestos fiber to coagulate the precipitate and filter. Treat the precipitate by (114) and the filtrate by (115). (114) Confirmatory Test for Ferricyanide. Dissolve the precipitate (113) by pouring a 3-5 cc. portion of 6N. HC1 repeatedly through the filter. Dilute the solu- tion with an equal volume of water and add about i cc. of FeSCU solution. A deep blue precipitate shows the presence of ferricyanide. (115) Detection of Thiocyanate and Cyanide. To the filtrate (113) add about i cc. of FeCl 3 solution. A red coloration shows the presence of thiocyanate. Make the solution just alkaline with NaOH and add a few drops of FeSCU solution. Heat the mixture to boil- ing and boil gently for one to two minutes. Acidify with HC1 to dissolve the hydroxides of iron. A blue residue insoluble in HC1 shows the presence of cyanide. (See Discussion 63.) DISCUSSION 61. The solution of the cyanogen compounds by means of NaCl is based on the relative solubility of their silver salts. The lower solubility product of AgCl causes a displacement of the equilibrium toward the formation of more AgCl and a consequent decrease of Ag+, with the DISCUSSION (61-63) 93 result that the ion product of the cyanogen compounds of silver is reduced below their solubility-product value and they pass into solution as sodium salts. In the pres- ence of a small amount of HC1 the reaction is rapid and complete. The sulphide and other halides are unaffected, owing to the fact that their solubilities are less than that of AgCl. (See Introduction 12.) 62. The precipitation of ferrocyanide with (NHU^MoC^ reagent is complete only when a considerable excess of (NH 4 )2MoO4 is present. Unless the ferrocyanide is com- pletely removed it will be precipitated with ferricyanide and interfere with the test for that acid. The heavy red precipitate of ferrocyanide is a complex salt of molybdenum with K 4 Fe(CN) G , of the probable formula K 2 (MoO 2 Fe(CN) 6 ) 3 2MoO 3 2oH 2 O, and is somewhat soluble in excess of K4Fe(CN)e. (Note. The student will learn to distinguish between ferrocyanide and ferricyanide by noting that a ferrocyanide contains a complex ion consisting of CN~ and ferrous iron while a ferricyanide contains a complex ion consisting of CN~ and ferric iron.) 63. When a simple cyanide is boiled with NaOH and FeSCU, Na4Fe(CN)e is formed. This reacts in acid solu- tion with FeCls to form Fe4(Fe(CN) 6 )3 (Prussian blue). Very small quantities of this are not readily detected in the presence of much thiocyanate, but if the solution is filtered a blue precipitate may easily be detected on the filter paper. 94 ACIDS TABLE XII OUTLINE FOR THE SYSTEMATIC SEPARATION AND DETECTION OF GROUP I (DIVISION B) Ions present in dil. HNOs solution No. Reagent s- I- Br~ Cl - l i AgN0 3 Ag 2 S Agl AgBr AgCl 2 HNO 3 Ag 2 SO 4 Agl AgBr AgCl HC1 H 2 S0 4 1 i 1 BaCl 2 BaSO 4 3 Zn H L SO 4 1 HI HBr HC1 4 K*Cr,O 7 (cold) I 2 HBr HC1 5 K 2 Cr 2 O 7 (hot) T Br 2 HC1 KI KBrOO 1 6 KMn0 4 I ' Cl, KI KC1 ANALYSIS Division B (120) Detection of Sulphide. To the second portion (no) add 5 cc. of 6N. HNO 3 and heat to boiling to oxidize the sulphide to sulphate. Cool and filter. Treat the residue by (121). Evaporate the filtrate nearly to dry- ness, add 1-2 cc. of 6N. HC1, filter and add BaCl 2 to the clear filtrate. A white precipitate of BaSO 4 shows the presence of sulphide. (See Discussion 64.) (121) Solution of the Halides. If any of the cyanogen compounds have been found they should be destroyed (see Discussion 65) by placing the residue (120) in a por- celain dish and igniting to dull redness. After the dish and contents have cooled, add a small piece of granulated zinc and 5 cc. of 6N. H2SO4. By means of a glass rod loosen any particles adhering to the dish in order to give them free access to the reducing agent. When the reaction is complete and only a black spongy residue OUTLINES OF ANALYSIS (TABLE XII) 95 remains, filter, reject the residue of metallic silver and treat the filtrate by (122). (122) Detection of Iodide. To the cold filtrate (121) add 5 cc. of 6N. H2SO4 and 10 cc. of water. Transfer the solution to a test-tube and add about i cc. of carbon tetrachloride (CCU) and a few drops of K^C^O? solution. Shake the mixture thoroughly. The presence of iodide is shown by a violet color in the CCU layer. If iodide is found filter and repeat the treatment with CCU and K^C^OT until the CCU layer shows no further coloration. Treat the filtrate by (123). (123) Detection of Bromide. Transfer the filtrate (122) to a small flask fitted with a delivery tube as shown in Fig. 4. Place I cc. of KI solution and about 10 cc. of water in the receiver and connect to the delivery tube as shown. Heat the contents of the reaction flask to boiling and boil for about one minute. Remove the receiver, add i cc. of CCU and shake the contents vigorously. A pur- ple coloration in the CCU layer shows the presence of bromide. If bromide is found the boiling should be continued until no further test is obtained with KI solution and CCU. Care should be taken to keep the concentra- tion of the acid in the reaction flask near its original value by FIG. 4. replacing with water the loss due to evaporation. (See Discussion 66.) If the yellow color should disappear from the solution in the reaction flask a few more drops of K 2 Cr 2 O7 solution should be added. (124) Detection of Chloride. To the contents of the reaction flask from which all bromide has been removed 96 ACIDS (123), add a few drops of KMnCX solution and repeat the boiling, at the same time catching the evolved gas in KI solution as outlined in (123) and testing it with CCU as before. A purple color in the CCU layer shows the pres- ence of chloride. DISCUSSION 64. By the action of boiling HNOs, according to the equation 3H 2 S +8HN0 3 -+ 3H 2 S0 4 +8NO +4H 2 O, sulphides are oxidized to sulphates, which may be detected by the white precipitate of BaSCU produced by BaCl 2 . Excess HNOs should be removed from the solution, how- ever, in order to prevent the possible precipitation of Ba(NOs) 2 , a white crystalline precipitate, on the sub- sequent addition of BaCl 2 . HC1 is added to remove any silver ion that may be in the solution. The presence of sulphide may be due not only to its presence in the original material but also to the decompo- sition of thiosulphate. When AgNOa is added to an acid solution containing a thiosulphate, decomposition of the thiosulphate takes place with the formation of insoluble Ag 2 S. When much thiosulphate is present the decomposition is readily detected by more or less rapid change of color in the precipitate from white through varying shades of gray to black. 65. The reduction of the silver halides to metallic silver and the halogen acid by means of zinc and H 2 SO4 is complete only in the absence of the cyanogen compounds. If any of these compounds are found, therefore, they must be removed before the treatment with zinc and H 2 SO 4 . They are readily destroyed by a dull red heat. A higher temperature might volatilize some silver halide. 66. The separation and detection of the halides is based on their relative resistance to oxidation by K 2 Cr 2 O7 DISCUSSION (64-66) 97 in acid solution. In a cold I-5N. H 2 SO4 solution K 2 Cr 2 O 7 readily oxidizes HI. HBr is unattacked by the dichromate in acid concentrations as high as 3N. if the solution is kept cold, but is oxidized when the solution is boiled. Under these conditions HC1 remains unchanged but is readily oxidized by the more powerful oxidizing action of KMnO 4 . It is also slowly oxidized by R^C^Oy if the acid con- centration is allowed to become much greater than 3 N. Hence the necessity of adding water, to keep the solution near its original volume when testing for bromide, is evident. 98 ACIDS TABLE XIII OUTLINE FOR THE SYSTEMATIC SEPARATION AND DETECTION OF GROUP II Ions present in dil. HNO 3 solution No. Reagent C0 3 = S0 3 = S 2 3 = NO 2 - i 2 ^ Ca(OH) 2 HNO 3 HgNO 3 CaCO 4 | CaSO 3 H 2 SO 3 H,SO 4 (Hg) CaSO 3 H 2 S0 3 H 2 SO 4 (Hg) Ca(N0 2 ) 2 HNO 2 HN0 2 4 FeSO 4 i I I FeSO 4 -NO C0 3 =, ANALYSIS Group II SO 3 =, S 2 O 3 =, NO 2 - (130) Separation of Group and Detection of Carbon- ate. Transfer the nitrate (no) to a small flask connected with a delivery tube and receiver as shown in Fig. 4. About 10 cc. of lime-water (Ca(OH) 2 ) should be placed in the receiver so that the delivery tube reaches below the surface of the liquid. Heat the solution in the flask to boiling and boil for about one minute. If a milky white precipitate forms in the receiver (see Discussion 67) car- bonate is shown to be present. Filter and treat the fil- trate by (131). Reserve the solution in the flask for analysis of Group III (140). (131) Detection of Sulphite. Acidify the filtrate (130) with HNO 3 and add a few drops of mercurous nitrate (HgNO 3 ). A gray precipitate shows the presence of sul- phite. (See Discussion 68.) Filter and treat the filtrate by (132). (132) Detection of Nitrite To the filtrate (131) add about i cc. of FeSO 4 solution. The appearance of a brown coloration shows the presence of nitrite. (See Discus- sion 69.) DISCUSSION (69-69) 99 DISCUSSION 67. When CO 2 is passed through lime-water, insoluble CaCOs is first formed. Excess of CO 2 , however, reacts with the normal carbonate to form soluble Ca(HCO3). In the detection of carbonate the formation of a white precipitate of CaCOs, which may or may not redissolve, is evidence of the presence of carbonate. 68. Mercurous nitrate (HgNOs) is very readily reduced to Hg by the action of free H 2 SO3, but is unaffected by HNO 2 . The reaction proceeds as follows: 2HgNO 3 +H 2 SO3+H 2 O^2Hg+H 2 SO4+2HNO 3 It has been mentioned (see Discussion 64) that Ag 2 S 2 Os decomposes even in acid solution with the formation of Ag 2 S. The reaction is as follows: Ag 2 S 2 O 3 +H 2 O-+ Ag 2 S +H 2 SO 4 The decomposition, however, is comparatively slow, and some undecomposed H 2 S 2 C>3 may be left in the filtrate from Group I. Boiling the solution to expel Group II decomposes this with the formation of SO 2 and free sul- phur. The detection of both sulphide and sulphite, there- fore, may have been due to the presence of thiosulphate in the original material. In this case a portion of the origi- nal solution should be tested for sulphide, thiosulphate and sulphite as follows: To the solution made slightly acid with HC 2 H 3 O 2 add ZnSCU solution. This will precipitate sulphide as white flocculent ZnS. Filter and treat the filtrate with Sr(NOa)a solution and allow the mixture to stand for several hours. The formation of a white precipitate (SrSOs) indicates the presence of sulphite. This may be confirmed by the property of its HC1 solution to decolorize iodine solution. The thiosulphate if present, remains in the filtrate from the SrSOs. It can be detected by acidifying with HC1 and warming, when sulphur will be deposited. 100 ACIDS 69. The test for nitrite depends on the formation of brown FeSO 4 -NO, according to the following equations: 3HN0 3 +3FeS0 4 +3HNO 2 - Fe 2 (SO 4 ) 3 +Fe(NO 3 ) 3 + 3 NO+ 3 H 2 FeS0 4 +NO->FeS0 4 -NO If much HgNOs was added in testing for sulphite, a white precipitate of Hg 2 SO 4 will be formed on the addition of FeSO 4 . 102 ACIDS < o 2 K q E u qq do q 4^ 5 9 } ^ J ^ ^ II 1 1 18 k u of o d {So cd c3 cj <2 III 2 o|_o'_|oo ||_ "i d d d < d ^ < ^; Jz; P g ^ ^ _, d 6 _^ ^^ ~ ^ 4^ ! o|o ^ _^ b/ Sibi c/3 cB* d 4 solution, may be taken as conclusive evidence of the absence of tartrate. If other organic matter may be present and the uranous salt is obtained in (149), a portion of the original material may be tested for tartrate as follows: Make the some- what concentrated solution slightly alkaline with NH 4 OH, add CaCb solution in excess, and allow the mixture to stand a short time. If a precipitate forms filter and digest the precipitate with cold NaOH solution. Dilute slightly, filter, and boil the filtrate. If a precipitate forms, filter while hot, wash the precipitate and transfer it to a test- tube. Add a drop of NH 4 OH and a little AgNO 3 solution, and warm. The presence of tartrate is shown by the formation of a black precipitate or silver mirror on the walls of the tube. OUTLINES OF ANALYSIS (TABLE XV) 107 TABLE XV OUTLINE FOR THE SYSTEMATIC SEPARATION AND DETECTION OF GROUP IV Ions present in neutral solution No. Reagent SO 4 = BO 3 = C 2 H 3 O 2 - NO 3 ~ i 2 3 4 5 HC1 BaCl 2 HC1 turmeric H 2 SO 4 CH 5 OH FeS0 4 H 2 S0 4 H 2 SO 4 BaSO 4 H 3 BO 3 Ba 3 (BO 3 )2 (?) Brownish- red color 1 HC 2 H 3 O 2 Ba(C 2 H 3 O 2 ) 2 Ba(C 2 H 3 2 ) 2 C- 2 ri5(_x 2 ri 3 O 2 I HNO 3 < FeSO 4 -NO ANALYSIS Group IV = C 2 H 3 2 -, NO 3 -, F- (150) Detection of Sulphate. Evaporate the filtrate (140) to 15-20 cc. Add slowly, and with frequent stirring just enough 6N. HC1 to precipitate all the silver; filter and reject the precipitate. To the clear filtrate add BaC^ solution in excess. The formation of a white crystalline precipitate shows the presence of sulphate. (See Dis- cussion 76.) Treat the filtrate by (151). (151) Detection of Borate. To just 5 cc. of the filtrate (150) add 5 cc. of I2N. HC1 and 2 drops of turmeric solu- tion. Allow the mixture to stand ten minutes. If borate is present the solution will assume a brownish-red color. (See Discussion 77.) (152) Detection of Acetate. Test a portion of the original substance for acetate as follows: If the substance is a solid add to a small quantity on a watch glass a few drops of cone. H2SO4. Mix with a glass rod and note 108 ACIDS the odor. If acetate is present in considerable quantity the odor of acetic acid will be apparent. If the substance is a liquid a portion of the slightly alkaline solution should be evaporated to dryness and tested with cone. H2SO4 as directed above. (See Discussion 78.) If snail quantities are suspected the following test should be made: To a small portion of the solid, obtained by evaporation if necessary, add about 0.5 cc. of alcohol and i cc of cone. H2SO4. Heat gently and note the odor. The ethylscetate (C2H 5 C2H 3 O2) formed gives a character- istic pleasant odor. (153) Detection of Nitrate. Test a portion of the original substance for nitrate as follows: Acidify the solution contained in a test-tube with 6N. H 2 SO4, and add an equa' volume of FeSC>4 solution. Pour about 2 cc. of cone. H 2 SO4 slowly down the sides of the tube so that it forms a layer at the bottom of the solution. If nitrate is present a brown ring will form in a short time at the juncture oi the two liquids. (See Discussion 79.) (154) Detection of Fluoride. In the absence of silica or silicates the etching test may be made as follows: Mix, with the aid of a piece of wood, about I gram of the powdered material in a lead dish or platinum crucible with enough cone. H 2 SO4 to form a thick paste. Cover the dish or crucible with a watch glass that has been coated on the convex side with a thin layer of paraffin through which characters have been scratched. Put a little water in the watch glass to prevent melting the paraffin, and warm th( dish or crucible gently, preferably over a water bath, for some time. Remove the watch glass, melt off the pararfin and note whether the parts exposed to the action oi the fumes have been attacked. If fluoride is present the glass will be dissolved off or etched in the places that were exposed to the fumes. (See Discus- sion 80.) In the presence of silica or silicates it is evident that the etching test cannot be used. If the material can be decom- DISCUSSION (76-80) 109 posed by cone. H 2 SO4 a portion of the powdered material may be mixed with cone. HoSCU in a test-tube and warmed gently while a drop of water is held on the end of a glass rod in the vapors. If fluoride is present the drop of water will become turbid. (See Discussion 80.) If the material cannot be decomposed with cone. H2SO4 it may be fused with 7-8 times its weight of a riixture of equal parts of sodium and potassium carbonate, the melt extracted with water, filtered, the filtrate acid fied with HC 2 H 3 O2 and the fluoride precipitated by the addition of CaCl 2 solution. The above test for fluoride may then be made on the dried precipitate. DISCUSSION 76. It will be recalled (see Discussion 68) th it one of the products obtained by the decomposition of Ag 2 S 2 O3 is H 2 SO4; hence if thiosulphate has been found, a portion of the original solution made acid with HC1 should be used for the sulphate test, instead of solution (150). 77. In order to make sure of the presence of borate when the color is slight, a blank test should be made for comparison. In order to be able to estimate th^ relative amount present the student should make comparison tests with known amounts of borate. In either case ae should take care to make all tests alike, using the same quantities of HC1, alcohol and turmeric, since the shade of color depends to a considerable extent upon the concentrations of these substances. 78. Although acetate, if present in the original solu- tion, would appear in the filtrate from the BaSO 4 (150) its detection in that solution would necessitate considerable evaporation which will ordinarily be avoided if the original substance is taken. 79. The test for nitrate is very delicate and accurate except in the presence of certain substances which cover up the ring or form a somewhat similar one. Chromates 110 ACIDS are reduced by FeSO 4 in acid solution giving a green color while ferro- and ferricyanides give a blue precipitate which makes the detection of the brown ring more difficult. Iodides and bromides are oxidized by the cone. H 2 SO 4 and produce a ring somewhat similar to the brown of the nitrate. Chromates may be removed by reduction to chromic salts with SO 2 and precipitation with NH 4 OH; the others may be removed by precipitation from slightly acid solution with Ag 2 SO4 solution. 80. Most fluorides are decomposed by cone. H 2 SO4 according to the following equation: CaF 2 +H 2 S0 4 -> CaS0 4 +H 2 F 2 The reaction proceeds more rapidly if the mixture is heated, owing to the more rapid removal of the volatile H 2 F 2 . The H 2 F 2 may be recognized by its ability to etch glass or dissolve silica or silicates. In either case volatile silicon tetrafluoride (SiF 4 ) is formed. When this is brought into contact with water the following reaction takes place: SiF 4 +4H 2 O-> 2H 2 SiF 6 +H 4 SiO 4 When silicates undecomposed by cone. H 2 SO 4 are fused with a mixture of sodium and potassium carbonates, the silicon reacts to form soluble Na 2 SiOs and any fluoride present forms soluble Na 2 F 2 . QUESTIONS FOR REVIEW 111 QUESTIONS FOR REVIEW 1. What ions are precipitated by AgNOs in acid solution? 2. Why should the solution be kept cold in precipitating Group I? 3. If a ferricyanide is present what other acids of Group I are not likely to be present? Why? 4. Although it is possible to have all metals in the same solution, this is not possible in the case of acids. Why? 5. Why will NaCl solution dissolve all the cyanogen compounds of silver but have no effect of the correspond- ing halides? 6. What is the effect of boiling Ag 2 S with HNO 3 ? Write equation. 7. What is the principle on which the detection of chloride, bromide and iodide is based? 8. Why is it necessary to remove all iodide before testing for bromide? Why remove all bromide before testing for chloride? 9. Given a solution which is known to contain no other anions than those given below, outline a method of analysis that will necessitate no unnecessary steps; (a) Ferrocyanide, thiocyanate, cyanide. (b) Sulphide, iodide, chloride. (c) Iodide, bromide, chloride. 10. In driving off the last traces of bromide before testing for chloride, why must the water be replaced as it evaporates? 11. Why is it unlikely that Group II acids will be found in a strongly acid solution? 12. What is the precipitate formed in the test for sul- phite? Write equation. 13. Write equations showing the action of HNO2 on FeSO 4 . 14. If, while the Group II anions are being distilled off a white precipitate should form in the receiver but should redissolve before the distillation is completed what conclu- 112 ACIDS sions couid be drawn ? Write equations showing the changes. 15. Why should Group III acids be precipitated from a warm solution? 1 6. What precautions are necessary in the precipitation of Group 1)1? 17. What is the effect of adding an excess of NaOH to the silver salts of Group III? 1 8. Why should a large excess of NaOH be avoided in the separation of arsenite from the remaining Group III acids? 19. If arsenate is found why must it be completely removed before testing for phosphate? 20. What is the confirmatory test for oxalate? Write equations showing all reactions involved. 2 1 . Whkt is the black precipitate formed on the addition of NaOH daring the precipitation of Group III? 22. Why must an excess of AgNOs be present before precipitating Group III? 23. Wh?t method may be used for collecting and filter- ing a colbidal or finely divided precipitate? 24. G : vcn a solution which is known to contain no other anions thar. those given below outline a method of analysis that will necessitate no unnecessary steps: (a] Nitrite, arsenite, arsenate. (b] Oxalate, chromate, tartrate. (c] Sulphite, oxalate, chromate. 25. What effect would the presence of shreds of filter paper in the H 2 SO4 solution have on the test for tartrate? 26. Why must all oxalate be removed before testing for tartrate? 27. Could Ba(NOs)2 be used in the place of HC1 and BaCU in the test for sulphate? 28. What acids interfere with the test for nitrate? How ma> they be removed? 29. Describe the test for fluoride. 30. If z : nc and barium were found in an unknown soluble in dilute HNOs, what acids would it be unneces- sary to test for? APPENDIX I. Preparation of Reagents Acids : Acetic, 6N. ; Mix 350 cc. of glacial acid with 650 cc. of water. Fluoboric; dissolve HsBOs in 48 per cent H2F2, using a lead or platinum dish, until a test portion will not give a precipitate with Pb(NO3)2 solution. After cooling add an equal volume of alcohol and just enough fluosilicic acid to precipitate any sodium which may be present as an impurity. Fluosilicic; pour 48 per cent H 2 F 2 upon sand in a lead dish or wax bottle until the sand is barely covered. Allow the mixture to stand for a few hours or until the free H 2 F 2 has entirely disappeared. Pour off the liquid and add an equal volume of alcohol. Hydrochloric, I2N.; use the C.P. acid of commerce (Sp.gr. 1.19), Hydrochloric, 6N.; mix equal volumes of I2N. HC1 and water. Hydrofluoric, 48 per cent; use the C.P. acid of com- merce. Nitric, i6N.; use the C.P. acid of commerce (Sp.gr. 1.42). Nitric, 6N.; mix 380 cc. of i6N. HNOs with 620 cc. of water. Perchloric, 6N.; mix 650 cc. of the 60 per cent C.P. acid with 350 cc. of water. Sulphuric, 96 per cent; use the C.P. acid of commerce (Sp.gr. 1.84). Sulphuric, 6N. ; mix i volume of the 96 per cent H 2 SC>4 with 5 volumes of water. 113 114 APPENDIX Bases: Ammonium hydroxide, I5N.;*. use the C.P. solution of commerce (Sp.gr. 0.90). Ammonium hydroxide, 6N.; mix 400 cc. of the I5N. NH 4 OH with 600 cc. of water. Calcium hydroxide, saturated solution; shake 5-10 grams of C.P. CaO with 1000 cc. of water until saturated, and filter. Sodium hydroxide, 6N. ; dissolve 250 grams of pure NaOH in enough water to make a total volume of 1000 cc. Salts: Ammonium acetate, 3N.; mix equal volumes of 6N. HC 2 H 3 O 2 and 6N.NH 4 OH, or dissolve 250 grams of C.P. NH4C2H3O2 in enough water to make a total volume of 1000 cc. Ammonium carbonate, 9N.; dissolve 250 grams of freshly powdered (NHU^COa in enough cold 6N. NH 4 OH to make 1000 cc. Ammonium chloride, iN.; dissolve 54 grams of NH 4 C1 in enough water to make a total volume of 1000 cc. Ammonium molybdate; dissolve 75 grams of C.P. (NH 4 )2MoO4 in 500 cc. of water and pour the solu- tion slowly into 500 cc. of 6N. HNOs. If a pre- cipitate should form, shake the mixture occasionally until solution is complete. Ammonium oxalate, o.5N. ; dissolve 35 grams of (NH4)2C 2 O4-H 2 O in enough water to make a total volume of 1000 cc. Ammonium polysulphide ; saturate 500 cc. of 6N. NH 4 OH with H 2 S gas, and add to this solution 500 cc. more 6N. NH 4 OH, 50 cc. of 6N. NaOH and 25 grams of flowers of sulphur. Digest for some hours and filter. Ammonium sulphate, iN.; dissolve 66 grams of (NH 4 )2SO 4 in 1000 cc. of water. PREPARATION FOR REAGENTS 115 Ammonium sulphide; saturate 500 cc. of 6N. NH 4 OH with H 2 S gas and then add 500 cc. more 6N. NH 4 OH. Barium chloride, iN. ; dissolve 120 grams of BaCl2-2H2O in 1000 cc. of water. Bromine water; use saturated solution. Calcium chloride, iN.; dissolve 150 grams of CaCl 2 - 6H2O in 1000 cc. of water. Calcium sulphate; use saturated solution. Cobalt nitrate, o.oiN. ; dissolve 1.5 grams of Co(NO3)2- 6H 2 O in 1000 cc. of water. Dimethylglyoxime, o.iN.; dissolve 12 grams of the solid in 1000 cc. of 95 per cent alcohol. Ferric chloride, iN.; dissolve 90 grams of FeCls- 6H2O in 1000 cc. of water. Ferrous sulphate, iN.; dissolve 140 grams of FeSO 4 - yH2O in 1000 cc. of o.6N. IHbSOi and keep in con- tact with iron nails. Hydrogen peroxide; 3 per cent solution. Lead acetate, o.2N. ; dissolve 38 grams of Pb( 3H 2 O in 1000 cc. of water. Mercuric chloride, o.iN.; dissolve 27 grams of in 1000 cc. of water. Mercurous nitrate, o.iN. ; dissolve 29 grams of HgNOs 2H 2 O in 1000 cc. of water. Magnesia mixture, iN. ; dissolve 100 grams of MgCb- 6H 2 O and 100 grams of NH 4 C1 in water, add 50 cc. of I5N. NH 4 OH and dilute to 1000 cc. Potassium cyanide, iN.; dissolve 65 grams of KCN in 1000 cc. of water. Potassium dichromate, iN.; dissolve 147 grams of K 2 Cr2O? in 1000 cc. of water. Potassium ferrocyanide, iN.; dissolve 105 grams of K 4 Fe(CN) 6 -3H 2 O in 1000 cc. of water. Potassium iodide, o.iN.; dissolve 17 grams of KI in 1000 cc. of water. 116 APPENDIX Potassium nitrite, 3N.; dissolve 250 grams of KNO 2 in 1000 cc. of water. Potassium permanganate, o.iN.; dissolve 16 grams of KMnC>4 in 1000 cc. of water. Potassium thiocyanate, iN.; dissolve 97 grams of KCNS in 1000 cc. of water. Silver nitrate, O.5N.; dissolve 85 grams of AgNOa in 1000 cc. of water. Sodium carbonate, 3N.; dissolve 159 grams of Na 2 COa in 1000 cc. of water. Sodium chloride, iN.; dissolve 58 grams of NaCl in 1000 cc. of water and add 5 cc. of I2N. HC1. Sodium phosphate, iN.; dissolve 120 grams of Na 2 HPO 4 -i2H 2 O in 1000 cc. of water. Sodium stannite; add NaOH, drop by drop, to a solu- tion of SnCl 2 until the precipitate of Sn(OH) 2 is just dissolved. The solution must be kept cold to prevent decomposition and oxidation to Na 2 SnC>3. Sodium stannite is unstable and must be prepared as needed. Stannous chloride, iN.; dissolve 113 grams of SnCl 2 - 2H 2 O in 170 cc. of I2N. HC1 and dilute to 1000 cc. Keep in bottles containing granulated tin. Turmeric; shake an excess of turmeric powder with 95 per cent alcohol and filter. Uranium acetate, o.iN.; dissolve 20 grams of UO 2 (C 2 H 3 2 ) 2 -2H 2 in 1000 cc. of water. Zinc nitrate. iN.; dissolve 148 grams of Zn(NOs) 2 - 6H 2 O in 1000 cc. of water. II. Test Solutions Solutions used in the preliminary experiments should contain equal amounts of the given ions per cc., in order that the student may learn to estimate more accurately TEST SOLUTIONS 117 the relative amounts of the constituents in an unknown. The solutions given below when dissolved in the pro- portions stated will contain 10 mgs, of the ion per cc. of solution. Ion. Salt ! Grams per liter Ion. Salt Grams per liter Ag AgNO 3 15-7 Cr CrCl 3 30 Hg HgNO 3 -2H 2 O 15- (*) Zn ZnCl 2 21 Pb Pb(N0 3 ) 2 16. (t) Mn Mn(NO 3 ) 2 -6H 2 O 52 Hg HgCl 2 13-5 Fe FeSO 4 -7H 2 O 49 (II) Bi Bi(NO 3 ) 3 -5H 2 23- U) Co CoCl 2 -6H 2 O 40 Cu CuCl 2 -2H 2 O 21 Ni NiCl 2 -6H 2 O 4i Cd CdCl 2 -2H 2 O 19-5 Ba BaCl 2 -2H 2 O 18 As NaAsO 2 17-3 Sr SrCl 2 -6H 2 O 30 As Na 2 HAsO 4 -7H 2 O 41 Ca CaCl 2 -6H 2 O 55 Sb SbCl 3 19- () Mg MgCl 2 -6H 2 O 84 Sn SnCl 2 -2H 2 O 16. () K KC1 19 Sn SnCl 4 -3H 2 O 22. () Na NaCl 25 Al A1C1 3 49- * Dissolve in 0.6 N. HNO33- f Double the amount for Group I experiments. J Dissolve in 3 N.HNO 3 . Dissolve in 350 cc. of 6N. HCl and dilute to i liter. f| Dissolve in 0.6 N. H 2 SO 4 and keep in contact with iron nails. 118 APPENDIX apuonjj ?'c3 1 rtrtaJcj oj rtrtrtrtaj apiuiojg apipioj ajo3a3a3o3o3o3 gpiqdinc * -- 1- ^tttttt r.T. L O .^M.-^.^H Cy Cu Cu C3 03.-, o3 o3 o3 o3 03 03 ^.-< water only. ; <* * s S-*?! s w 8 .2 S .s .S 8 ^ . 1111 i? ;? ?y a = soluble in acids. i = insoluble in water and ydrolizes completely in presence of w lightly soluble in water, soluble in aci soluble in water, soluble in acids. wa A LIST OF THE MORE COMMON ELEMENTS 119 IV. A LIST OF THE MORE COMMON ELEMENTS WITH THEIR ATOMIC WEIGHTS Name Symbol At. Wt. Name Symbol At. Wt. Aluminium Al 27 I Lithium . Li 6 Q Antimony Arsenic .... Sb As 120.2 7? Magnesium . . . Manganese Mg Mn 24-3 C4. Q Barium Ba 208 Mercury Hg 2OO.6 Boron Bromine Cadmium B Br Cd II 79-9 112.4 Molybdenum Nickel Nitrogen .... Mo Ni N 9 6 58.7 Calcium Chlorine .... Ca Cl 40 -it. tr Oxygen Phosphorus . P 16 Chromium Cobalt Cr Co 52 58 9 Potassium. . . . Silicon .... K Si 39-1 28 3 Coooer . . Cu 61 5 Silver Ag IO7 8 Fluorine Gold Hydrogen F Au H 19 197.2 i 008 Sodium Strontium. . . . Sulphur Na Sr s 23 87.6 -J2 Iodine Iron .... I Fe 126.9 55 8 Tin Uranium Sn . u II8.7 2^8 2 Lead Pb 207.2 Zinc Zn 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. RENEWALS ONLY TEL. NO. 642-3405 This book is due on the last dace stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. maTWaa fef AIM 1 K. *. J. KK'DLD MAR 1 71-9A*H1^ inw 9 CHA \ OUii Dpp'n ,-, (p "-V. U MAV 24747$, ADD rz nCX> Bfeiuillnj xf^ W> 1 3 1986 : RECEIVED JUL Z 1986 CIRCULATION DEPT. LD2 1 A-60m.3/70 IlniSSS^rfnSlStiiia (N5382slO)476-A-32 Berkeley GENERAL LIBRARY -U.C. BERKELEY BQOOB7bfl32