Elementary Experimental Chemistry 9781487583026

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Elementary Experimental Chemistry

M. W. LISTER M.A., D.Phil.

W. A. E. McBRYDE M.A., Ph.D. FOURTH EDITION, REVISED

UNIVERSITY OF TORONTO PRESS

Copyright, Canada, 1959, by University of Toronto Press Printed in Canada Reprinted, 1960 Reprinted, 1964

Reprinted in 2018 ISBN 978-1-4875-8174-9 (paper)

CONTENTS General Introduction 1. Purification by Crystallization

1 4

2. Ionic Reactions

10

3. Displacement Reactions

18

4. Oxidation-Reduction Reactions

24

5. Double Salts

35

6. Complex Ions

39

7. Solubility

52

8. Solubility Product

58

9. Reversible Reactions and the Mass Law

62

10. Formula of a Salt Hydrate

73

11. Gas Law and Equivalent Weight of a Metal

75

12. Colloids

78

13. Acids and Bases

88

14. Calibration of Pipettes and Burettes

96

15. Acid-Base Titrations

98

16. Redox Titrations

103

17. Voltages of Electric Cells

110

18. Heat of Reaction

116

19. Rate of Reaction

121

20. Molecular Weights from the Depression of the Freezing Point

128

GENERAL INTRODUCTION SAFETY PRECAUTIONS

1. Use common sense and care at all times. Be sure to use the amounts and concentrations of reagents specified. Some experiments, which are safe in dilute solution, present certain hazards with more concentrated reagents. For instance, use of concentrated instead of dilute sulphuric acid can be dangerous. 2. Do not throw solids, broken glass, or other rubbish into the sinks; use the waste receptacles provided. In particular, broken glass must be put in a receptacle by itself, and, if dropped on the floor, it must be swept up. Pour waste liquids down the sink, and wash them down with water. 3. Certain organic liquids are inflammable, and these include alcohol, benzene, and ether. Never have the burner lit when these are being handled at your working place. A fire extinguisher should be used on any extensive fire. Find out where the nearest fire extinguisher is located, and be sure you know how to operate it. 4. Do not pipette certain liquids by direct suction by the mouth. These liquids include concentrated acid or alkali, or ammonia of concentration above about 0.5 molar. If a rubber tube is attached to the top of the pipette and suction is applied through this, the hazard in pipetting can be reduced. In pipetting up sodium hypochlorite, for instance, use this method. Be sure there is enough liquid present so that the tip of the pipette is never out of the liquid. 5. A first aid kit is available. If an accident occurs apply to the demonstrator immediately for treatment. Wash acid or alkali burns with water. Heat burns should be treated with ointment without washing. Cuts should be washed, disinfected, and bandaged. Remember that many laboratory chemicals are poisonous, and must be washed off if spilt on the skin. RULES ON THE USE OF THE BALANCE

1. You will be assigned to a particular balance. Use this one for all your weighings. The balance is a delicate instrument, and must be treated with care at all times. 2. Never transfer weights with your fingers: always use the forceps. 3. Never weigh any chemical directly on the pan. Always place it in a weighing bottle or on a watch glass. 4. Always turn the release knob slowly when releasing the balance beam, or when arresting it again. Never alter the weights or object on the pans, or move the rider, unless the beam is arrested. 5. If any chemical is spilt on the pans or floor of the balance, sweep it up immediately. 6. At the end of a weighing the balance must be left with (a) the beam arrested, (b) the weights in the box, or on the tray of fractional weights, (c) the rider on the hook, and (d) the door or window of the case closed.

2 RULES ON THE USE OF PIPETTES AND BURETTES

These rules outline the precautions that must be taken to obtain accurate results with pipettes and burettes. 1. Both pipettes and burettes must be clean before use. Clean them with soap solution, or with a "cleaning mixture" if this is provided. When they are clean, water drains down the walls in a continuous sheet without breaking into drops. 2. Rinse out pipettes or burettes with a little of the solution being measured. Do this two or three times, before finally filling them. This is, of course, only necessary if a new solution is being used. Pipettes

1. Suck up the liquid above the mark on the pipette. Put your first finger (not your thumb) on top of the pipette. Hold the pipette vertical, and let the liquid drain out until the bottom of the meniscus just rests on the mark. Remove any drop from the tip by touching it to the side of the flask. 2. Transfer the pipette to the flask or beaker being filled. Release the liquid. Keep the pipette vertical. Let the liquid drain out under gravity. Touch the tip of the pipette to the side of the flask, and hold it there for at least 10 seconds after all visible flow has finished. 3. Do not blow out the drop inside the tip. Remove the pipette with this drop still in it. 4. Never dip a pipette directly into a reagent bottle. Take some solution in a beaker or flask and pipette from that. Burettes

1. Be sure the stopcock is greased so that it does not leak. Use the special grease provided if it does leak; note that only a little grease is needed. 2. Never leave a funnel in the top of a burette during use. 3. After running out liquid, let the burette stand for about half a minute before reading it. This allows time for drainage. 4. It is easier to read a burette if a piece of white paper with a horizontal black rectangle on it is placed just behind the burette. Hold the paper so that the top of the rectangle is about 1 mm. below the meniscus. Read the bottom of the meniscus. 5. In all accurate work, read your burette to 0.02 ml. NOMENCLATURE

The development of inorganic chemistry has made it necessary from time to time to revise the system of naming compounds. Such revisions have mostly been concerned with complicated molecules, which will not be encountered in this book. However, even for simple compounds there is not complete uniformity. The following rules should suffice in naming any compounds in the following experiments.

3 1. In naming a salt the more positive constituent comes first: NaCl is sodium chloride, not chloride of sodium. 2. In a compound of two elements, the more negative element has its name changed to end in -ide: CaS is calcium sulphide. 3. Oxy-anions are distinguished by the endings -ate and -ite; the -ate ending always refers to the higher valency. Examples are: S04- - sulphate, SOa- sulphite; or Cl0 3- chlorate, Cl02- chlorite. Higher valencies than the -ate compound may have a prefix per- added; as in Cl04- perchlorate. Lower valencies than the -ite compound may have a prefix hypo- added, as in CIO- hypochlorite. 4. Complex anions are named so that the central element comes last. Thus ZnC1 4- - is the chlorozincate ion; SiF 5- - is the fluosilicate ion. In complex cations, at least in the metal ammonia complexes encountered in this book, two methods are used; thus Co(NH 3) 6+++ is either the cobalt-hexammine ion, or the hexammino-cobalt ion. The first alternative is used here. Other examples are: Ag(NHa)2+, silver diammine ion, or Cu(NH 3)4++, copper tetrammine ion. 5. The valencies of metallic elements, or at least of those metallic elements that have more than one important valency, are distinguished in two ways. Either the valency is given in Roman numerals after the element; or the valency is distinguished by -ic and -ous endings for higher and lower valencies respectively. According to the first system FeC12 is iron (II) chloride, and FeCla is iron (III) chloride; according to the second system FeCb is ferrous chloride and FeCla is ferric chloride. The first system is more flexible since more than two valencies may exist; for example for manganese there is Mn(II), Mn(III), and also Mn(IV) as in Mn02. However in some complicated compounds it is difficult to be sure what the valency is (unless other rules are agreed upon); and it has occasionally led to confusion in that the Roman numeral is taken to mean the number of other atoms attached; for example, manganese (II) oxide might mistakenly be interpreted as Mn02, while MnO is meant. In this book the -ic and -ous endings are used where these are very well established. The following list gives a number of equivalent names:

Cr+a Cr+ 2 Cu+ 2 Cu+ Fe+a Fe+ 2 Mn+a Mn+2 SnH Sn+ 2

= = = = =

chromic chromous cupric cuprous ferric ferrous manganic manganous stannic stannous

= = = = = = = =

chromium (III) chromium (II) copper (II) copper (I) iron (III) iron (II) manganese (III) manganese (II) tin (IV) tin (II)

In some cases, another system employing numerical prefixes has been established by long custom. Thus Mn0 2 has long been known as manganese dioxide, CrOa as chromium trioxide, and so on. For simple compounds, at least, these names are unambiguous, and they are occasionally used.

4

6. The number of atoms or radicles attached are generally distinguished by Greek prefixes; these are: mono = one; di = two; tri hepta = seven; octa = eight.

= three;

tetra

= four;

penta

= five;

hexa

= six;

Thus CuSO4.5H2O is copper (II) sulphate pentahydrate, or cupric sulphate pentahydrate. For well known compounds, these numbers are sometimes omitted, but if there is any uncertainty they should be included. The same applies to the valency numerals; thus "copper sulphate" would always be taken to mean the well known CuSO., and not the relatively rare Cu2SO4.

1.

PURIFICATION BY CRYSTALLIZATION INTRODUCTION

Many laboratory and industrial preparations of chemicals result in a product that contains appreciable quantities of impurities. There are various ways by which the pure substances desired may be isolated from these impure materials; one of the commonest methods is crystallization. When the solubility of a dissolved solid is decreased, for example, by cooling a warm concentrated solution, the solid is deposited as crystals. Usually small amounts of other dissolved substances remain in the solution when the principal substance crystallizes, and hence the latter is produced much purer. EXPERIMENTAL

1. Purification of Alum from a Mixture

The starting material for this experiment is a mixture containing Potassium alum (soluble in water) 85% Copper sulphate hydrate (soluble in water) 10% Charcoal (insoluble in water) 5% The alum is a colourless compound while copper sulphate is blue, and the same colours appear in the aqueous solutions of these substances. Charcoal is black. The object of the experiment is to recover pure alum from this mixture. The charcoal is separated from the other two substances because of its insolubility in water. Alum can be partially separated from copper sulphate by taking advantage of the large difference in its solubility in hot and cold water (see page 7). Some of the alum remains in solution with the copper sulphate, while the remainder crystallizes from solution reasonably pure. If you work carefully you should be able to recover as much as 80% of the alum by following the experimental directions given. (a) Weigh out 20 g. of mixture. Decide what weight of each mechanically mixed constituent is present.

4

6. The number of atoms or radicles attached are generally distinguished by Greek prefixes; these are: mono = one; di = two; tri hepta = seven; octa = eight.

= three;

tetra

= four;

penta

= five;

hexa

= six;

Thus CuSO4.5H2O is copper (II) sulphate pentahydrate, or cupric sulphate pentahydrate. For well known compounds, these numbers are sometimes omitted, but if there is any uncertainty they should be included. The same applies to the valency numerals; thus "copper sulphate" would always be taken to mean the well known CuSO., and not the relatively rare Cu2SO4.

1.

PURIFICATION BY CRYSTALLIZATION INTRODUCTION

Many laboratory and industrial preparations of chemicals result in a product that contains appreciable quantities of impurities. There are various ways by which the pure substances desired may be isolated from these impure materials; one of the commonest methods is crystallization. When the solubility of a dissolved solid is decreased, for example, by cooling a warm concentrated solution, the solid is deposited as crystals. Usually small amounts of other dissolved substances remain in the solution when the principal substance crystallizes, and hence the latter is produced much purer. EXPERIMENTAL

1. Purification of Alum from a Mixture

The starting material for this experiment is a mixture containing Potassium alum (soluble in water) 85% Copper sulphate hydrate (soluble in water) 10% Charcoal (insoluble in water) 5% The alum is a colourless compound while copper sulphate is blue, and the same colours appear in the aqueous solutions of these substances. Charcoal is black. The object of the experiment is to recover pure alum from this mixture. The charcoal is separated from the other two substances because of its insolubility in water. Alum can be partially separated from copper sulphate by taking advantage of the large difference in its solubility in hot and cold water (see page 7). Some of the alum remains in solution with the copper sulphate, while the remainder crystallizes from solution reasonably pure. If you work carefully you should be able to recover as much as 80% of the alum by following the experimental directions given. (a) Weigh out 20 g. of mixture. Decide what weight of each mechanically mixed constituent is present.

5

Have a supply (200 ml.) of boiling water in readiness. Add about 35 ml. of this to the mixture in a beaker. Stir for one minute then allow a few seconds for the suspended material to settle. In this interval pour about 50 ml. of boiling water through a fluted filter paper in a short-stemmed funnel. This heats the paper and funnel immediately prior to filtration. In your report explain why this is necessary. Discard this hot water. Decant the hot supernatant liquid from the beaker containing the mixture into the filter paper by pouring down a stirring rod. Remember that the tip of the funnel should touch the side of the receiving beaker. In this filtrate will be grown the first crop of alum crystals. Set this beaker aside, covered by a watch glass. (b) To the residue from the mixture in the beaker add 20 ml. of boiling water. Swirl for one minute, allow to settle, and filter as before through the same fluted filter paper into a fresh beaker. This is covered and set aside. (c) Repeat (b) with a second 20 ml. portion of boiling water, catching this filtrate in a third beaker. Cover this with a watch glass. These three filtrates are set away until the next class period, but before doing so they should be examined and compared for the intensity of their blue colour by placing the beakers on a sheet of white paper. Draw conclusions regarding the location and distribution of the copper sulphate and, by inference, the alum in these operations. (d) There may still remain some soluble substances on the fluted filter paper. These will represent a loss of alum in the over-all process. Add 2 ml. of water to the filter paper and catch 2 or 3 drops of the filtrate on a watch glass. Add 1 drop of potassium ferrocyanide, K4Fe(CN)&, solution; this provides a sensitive test for the copper sulphate. For comparison place 1 drop of 0.5% copper sulphate solution on another watch glass and add 1 drop of potassium ferrocyanide solution. The brown-coloured precipitate is copper ferrocyanide, Cu2Fe(CN) 8 • State whether all the copper sulphate was separated from the insoluble part of the mixture as a result of steps (a) to (c). (e) In the second laboratory period examine the three filtrates from the previous operations. In the first you should have a moderate crop of alum crystals. If you have not, or if too much liquid has evaporated so that the copper sulphate has crystallized out, consult the demonstrator for advice on how to proceed. Filter this first crop of alum crystals through a dry filter paper, folded as a cone and placed dry in the funnel. When all the mother liquor has drained, wash twice with 2 ml. portions of cold water. Now remove the filter paper from the funnel and open it out on a clean dry filter paper; this will absorb the greater part of the moisture. Scrape the damp alum crystals onto fresh sheets of dry filter paper and dry them as well as possible by pressing between dry sheets. When no further moisture can be removed this way dry the crystals in air for 15 minutes, transfer to a clean dry test-tube and stopper the tube. Explain in your report why a very small filter paper is desirable for filtering the crystals, and why so little wash water was used. (f) The mother liquor and washing from (e) and the filtrates from (b) and (c) are combined in one beaker. Add to this any crystals spilled or not transferred to the filter in (e). Evaporate by heating, but not boiling, until the first appearance of turbidity or until the volume is 10 ml. Remove from the heat, cover and let

6

cool. Crystals should appear on cooling; if they do not, consult the demonstrator. After the mixture has stood for twenty minutes, filter it and wash the crystals twice with 20 ml. portions of cold water. Open up the filter paper and dry the crystals as in (e). Record the volume of mother liquor and washing which is discarded. (g) Take one or two small crystals from each crop on separate watch glasses; add 2 drops of distilled water and 1 drop of potassium ferrocyanide solution. This is a test of the purity of the alum prepared (that is, how free it is from the soluble impurity copper sulphate). (h) Combine the two crops of crystals and weigh on a counterpoised filter paper. State your yield based upon the amount of alum originally in the mixture. Place your crystals in a clean test-tube, stopper, and label with your name, class, and the weight of contents. Hand in to a demonstrator. 2. Preparation of Potassium Nitrate from Sodium Nitrate

Weigh out 17 g. of sodium nitrate and 15 g. of potassium chloride. Dissolve the sodium nitrate in 35 ml. of hot water in a 100-ml. beaker, and add the potassium chloride. Allow the resulting solution to evaporate slowly just below its boiling temperature, with occasional stirring, until the volume has been reduced to 30 ml. Fit a filter paper into a short-stemmed funnel and pour through about 100 ml. of hot water to heat the paper and funnel. Place a clean beaker below the funnel, and quickly filter the hot solution of salts. It is essential that this filtration be carried out quickly with the solution still hot, or otherwise crystallization may begin in the stem of the funnel. Then allow the filtrate to cool, preferably by placing the beaker in a bath of cold water. This is most easily done by placing a stopper in the drain of a sink with an outlet at such a height as to maintain a depth of¾ to 1 inch, and then allowing cold tap water to flow through the sink by way of a length of rubber tubing. When the mixture is quite cool, filter through a dry filter paper, wash with two 5-ml. portions of cold distilled water, and remove the crystals and paper from the funnel. Dry the crystals partially by squeezing between paper towels or filter papers. Dissolve one crystal in 2 ml. of distilled water and add 1 ml. of silver nitrate. A white precipitate or cloudiness indicates the presence of some sodium chloride in the solid which is mainly potassium nitrate. Dissolve the remainder of the crystalline product in its own volume of boiling water and, if the solution is not clear, filter hot, catching the filtrate in a clean beaker. Allow the contents of the beaker to cool as before and, when cold, filter out, wash, and dry the potassium nitrate crystals. Test these in the same way for the presence of sodium chloride. 3. Preparation of Boric Acid Dissolve 15 g. of borax (about 0.05 gram formula weight of N a2B4O1. 5H2O) in 40 ml. of boiling water. Add 1 or 2 drops of methyl orange indicator to the solution, and then add concentrated (12 molar) hydrochloric acid to the boiling solution a few drops at a time, stirring thoroughly after each addition. When the

7

colour of the indicator changes to pink, indicating that sufficient acid has been added to convert the borax to boric acid, add 5 drops more hydrochloric acid, and then allow the resulting solution to cool. It is desirable to cool the resulting solution to 15° C or cooler, and this can best be achieved by sitting the beaker in a sink filled to a depth of¾ inch with cold running water. When the mixture is cold, filter; wash the crystals in the filter paper with two 5-ml. portions of cold distilled water. When the washings have been drained from the crystals, remove the filter paper and solid from the funnel and squeeze between paper towels or filter papers to remove as much as possible of the moisture. Transfer this damp first crop of crystals to a clean beaker and dissolve in about 50 ml. of boiling distilled water. If the solution is not perfectly clear filter it hot (use a short-stemmed funnel) into a clean beaker, and reheat the filtrate if necessary to dissolve any boric acid that has crystallized. Cover this solution and set away to cool until the next laboratory period. The following period filter this mixture, remove the crystals of boric acid from the filter paper, and spread out on a dry paper towel or filter paper to dry in the air. Dissolve a small amount of the product in distilled water and test for chloride by the addition of a few drops of silver nitrate solution. EXPLANATIONS

1. To understand the purificati9n of alum from copper sulphate by crystallization it is necessary to consider the solubilities of each salt in water at different temperatures. These are represented graphically in Figure 1. Observe that between goo (hot) and 20° (cold) the solubility of alum decreases about 30-fold, whereas that of copper sulphate decreases by less than a factor of 3. According to these figures 10 ml. of water at go 0 should be capable of dissolving 32 g. of

100.---.----.--,,--.....--,---,--..---,----,--,,--.....--,---,--..---.---r---,

80

COPPER SULPHATE POTASSIUM

~

l&I

a::

i a..

:::E

ALUM

60

40

~

0 1---L..-..JL--...&.......&-...____.__..__.__.___.____,_ _._......._....___.__.._-' 100 200 0 300 GRAMS SOLUTE PER 100 GRAMS WATER FIGURE

1

8

alum and 5.5 g. of copper sulphate hydrate. The same volume of water at 20° can dissolve only 1.1 g. of alum, but 2.1 g. of copper sulphate. It is found that in cases such as this the solubility of one solute is not greatly altered by the presence of a second solute, and so we are treating these solubilities as independent of each other. Thus, if we dissolve the 17 g. of alum and 2 g. of copper sulphate contained in 20 g. of mixture in 10 ml. of water at 80°, and then allow the solution to cool to 20°, the cold water can still hold the copper sulphate in solution but cannot hold all the alum. Accordingly, alum is obtained free of copper sulphate on cooling. The test for copper sulphate (strictly speaking, for soluble cupric salts) is the formation of brown cupric ferrocyanide. This salt is very insoluble and so dark in colour that its formation reveals the presence of very minute amounts of copper salts. One drop of 0.5% copper sulphate solution contains about 6Xl0-6 g. of copper. 2CuSO4

+ K4Fe(CN)e -

2K2SO4

+ Cu2Fe(CN)e.

2. The conversion of sodium nitrate and potassium chloride to obtain potassium nitrate and sodium chloride is made possible by the individual solubilities of these salts and by the way in which the solubilities change with temperature. The process taking place can be represented as NaNOa

+ KCI -

KNOa

+ NaCl.

However, in view of the fact that these four salts are all completely ionized (see Experiment 2) their solution contains the ions potassium (K+), sodium (Na+), chloride (CI-), and nitrate (NOa-); and the only factor that would suggest that the reaction described by the above equation had occurred would be the appearance of crystals of either KNO 3 or NaCl. In dilute solutions all four ions can remain together in solution, and there is no reason to speak of a reaction occurring at all. As the solution is made more concentrated, however, a time is reached when saturation occurs, and following that the growth of crystals. The question of which salt will crystallize from solution is determined by the solubilities of the four individual salts whose ions are in solution. These solubilities are shown in Figure 2. Observe that at 100° NaCl is the least soluble, while at 0° KNO 3 has the smallest solubility. Moreover, the solubility of NaCl is relatively unaffected by changing the temperature from 100° to 0° C. When water is evaporated from a hot solution of equivalent amounts of these salts, the least soluble will be the first salt to crystallize and this will be NaCl. If the evaporation is carried far enough to cause a moderate amount of this salt to crystallize from solution, and then these crystals removed by filtration, the solution will retain more K+ ions than Na+, and more NOa- ions than CI-. Now when the solution is cooled it soon becomes saturated with and subsequently deposits crystals of KNOa. During the cooling a small amount of NaCl may also crystallize owing to the slight decrease in solubility of this salt in the temperature interval. Hence the first crop of KNO 3 crystals may not be pure, and should be purified by recrystallization. The principle involved in this purification is the same as that in the purification of alum discussed above.

9

100

I I I I I I

I

I

80

:u : ::.c I

I

u

• 60 Ill 0

I I

0

I

2

a: ~ a:

I

I

'

I

"' 40 a.

I

I

Ill

:::i:

I I

Ill

I-

I

I I

I

0

10

20

I

I

I

I

I

I

0")

/


l2

+ 2 e-

(to electrode).

There is a complication, however, because elementary iodine combines with an iodide ion to form a so-called complex ion:

h

+ 1-- 13-.

It is this ion, l3-, called the tri-iodide ion, that is responsible for the brown colour observed on the filter paper. The fact that this colour appears at the positive pole is evidence that the iodide ions carry a negative charge. At the negative electrode the probable reaction is:

2 e-

+ 2 H2O ---+ 2 OH- + H2.

It would be more logical to expect the potassium ion, K+, to be discharged at the negative electrode. However, in Experiment 3 we shall see that chemically active metals like potassium or sodium have a strong tendency to lose electrons, K---+K+

+e-

and conversely it is difficult to get the ions of these metals to accept electrons from an electrode. In aqueous solution a water molecule is a more ready acceptor of an electron than a potassium ion. 3. The results of the tests described in part 3 can be grouped into various types. (a) No precipitate is formed. This corresponds to the blanks in the table above. For instance if 0.1 M copper nitrate is mixed with sodium sulphate solution, no precipitate is formed. In this case you are adding a solution containing Cu++ and NO 3- ions to one containing Na+ and SO 4- ions. The resulting solution contains Cu++, Na+, NO 3- and SO4- ions; that is, a mixture of what was put in, and no reaction has occurred. It is misleading to write Cu(NO3)2

+ Na2SO4 µ

CuSO.

+ 2 NaNO 3

since this implies the reversible formation of copper sulphate and sodium nitrate molecules in solution. In certain instances such molecules are formed to a greater or lesser extent, but for the most part the resulting solution contains merely a mixture of the ions of the constituent solutions. (b) A precipitate is formed with the simplest formula: for instance, silver nitrate and sodium sulphide; the precipitate is Ag~. The formulae of these precipitates can all be deduced from the fact that the resulting molecule is electrically neutral, so that if it results from Ag+ ions and s- ions, 2 Ag+ are needed to neutralize one s--, and the formula is Ag2S. For this reaction the following equation is frequently written:

2 AgNO3

+ Na~---+ Ag~ + 2 NaNO3.

This is satisfactory provided it is clearly understood what the symbols mean. On the left-hand side we have solutions providing Ag+, Na+, NO 3- and S-; on the right-hand side we have a precipitate of Ag~, and Na+ and NO 3- ions left in solution. Therefore what really occurs is:

15 2 Ag+ + 2 NOa- + 2 Na+ + s - - Ag2S + 2 Na+ + 2 NO 3and cancelling the ions that occur on both sides: 2 Ag+.+ s - -

Ag2S.

This is the equation of the reaction that really occurs. So far as the Na+ and NO 3- are concerned nothing happens: they remain in solution though the other constituents of that solution are changed. The Na+ ions were initially in the presence of water and s - ions; finally they are in the presence of water and NOa- ions. It does not follow that the Ag+ ands- ions remain as ions in the precipitate. Of the various precipitates obtained in this experiment some probably still contain ions, some probably not. This can only be discovered by further examination of the precipitates by more advanced methods (for instance by X-ray diffraction) and will not be considered further immediately. However, in Experiment 6 on complex ions, we shall come on some further evidence on this point. It can be assumed that the precipitates all have the simplest formulae as described in this section except in the following particular cases. In certain cases other or subsequent reactions occur and these may be divided as follows. (c) Loss of water. Most hydroxides can lose water to give the corresponding oxides. If XOH is any hydroxide, the reaction would be 2 XOH µ X2O + H2O,

This equilibrium may be predominantly over to one side or the other depending on the particular hydroxide being considered, the temperature, and the amount (concentration) of water present. Even in the presence of much excess water some hydroxides lose water very largely to give precipitates of the oxides. The actual precipitates obtained are: Hydroxides Cu(OH)2, Zn(OH)2, Mg(OH)z, Pb(OH)2, Mn(OH)2*, Fe(OH)2*. heat Oxides

i

CuO, Ag2O, HgO.

It will be seen that oxides are most readily formed by the most noble, or least chemically active, metals. It must be admitted that the distinction is not quite so clear cut since partial loss of water may occur; similarly if silver oxide is shaken with water, the water becomes distinctly alkaline, showing the presence of at least traces of AgOH in solution. (d) Other interfering reactions. It frequently happens that when the sodium salt of a weak acid, such as carbonic acid, is added to a solution of a salt, the hydroxide is precipitated. For instance, aluminium nitrate and sodium carbonate give a precipitate of aluminium hydroxide, not aluminium carbonate. This sort of reaction is called hydrolysis, and is dealt with in Experiment 13. However, it should be noted that in one of the tests in this experiment hydrolysis results; this is the reaction between magnesium nitrate and sodium sulphide, when a precipitate of magnesium hydroxide is formed. The reaction is: Mg(NOs)z + 2 Na2S + 2 H2O -

Mg(OHh + 2 NaHS + 2 NaNOa

16 or in ionized form :

+

+

SH20 µ HSOH-. Mg++ 2 OH- - Mg(OH)2.

+

In one other test another type of reaction occurs because one ion is capable of oxidizing the other. This type of reaction is dealt with in Experiment 5, and is also fairly common. Copper (II) ions can oxidize iodide ions so that the reaction between copper nitrate and potassium iodide is 2 Cu(NOah

+ 5 KI -

2 Cul

+ Kia + 4 KNOa

or in ionized form: 2 Cu++

+ 5 1- -

2 Cul

+ Ia-.

The precipitate is copper (I) iodide (cuprous iodide), while the solution contains the brown tri-iodide ion Ia-. Oxidation also occurs when iron (II) and manganese (II) hydroxides are exposed to the oxygen of the air 4 Fe(OH)2 4 Mn(OHh

+ 02 -

4 FeO. OH

+ 2 H20.

+ 02 - 4 MnO.OH + 2 H20.

The products are hydrated ferric and manganic oxides. These last reactions go at a moderate rate, so that the colour changes gradually over a considerable time. Pure ferrous hydroxide is nearly white; ferric oxide (hydrated) is brown; there is an intermediate compound which is dark green and contains both iron (II) and iron (III). The tests performed in part 3 of this experiment will have provided some indications of which salts are soluble and which are not. The following rules (sometimes called solubility rules) are of use in remembering which salts are soluble. They must not be treated as rigid laws, admitting of no exception; they are more in the nature of guides to what is usually found. Also it must be realized that the distinction between soluble and insoluble is not sharply defined. Observed solubilities extend continuously from very soluble down to very insoluble, and there is no sharp break at any point. The following rules apply to water at room temperature as solvent. (i) Nitrates, perchlorates, and acetates are usually soluble. (ii) Most chlorides are soluble; exceptions are those of silver, lead, copper (I), and mercury (I). Bromides and iodides usually resemble the chlorides, but fluorides may be quite different. (iii) Most sulphates are soluble; exceptions are calcium, strontium, barium (Group IIA), and lead. (iv) Salts of sodium and potassium (Group IA generally) are soluble. So are ammonium salts. (v) Hydroxides are usually insoluble, except those of Group IA, and to a lesser extent those of Group IIA. (vi) Sulphides are insoluble, except those of Group IA. (vii) Carbonates are insoluble, except those of Group IA. 4. If barium sulphate is added to a cold sodium carbonate solution very little barium carbonate will be formed. Not only is barium sulphate more insoluble

17 in water than barium carbonate, but it is probable that the carbonate solution never reaches the inside of the barium sulphate grains. In molten sodium carbonate (the mixture with potassium carbonate is used to give a lower melting point), not only are the solubility relations changed, but the higher temperature leads to a more vigorous attack. Also the concentration of carbonate ion is much higher so that the equilibrium BaSO4

+ COa- µ

BaCOa

+ SO4-

is much further to the right. On washing out the melt with water, barium carbonate and sulphate are both left on the filter. Dilute hydrochloric acid dissolves barium carbonate (but not barium sulphate), giving a solution of barium chloride. Barium chloride is almost insoluble in concentrated hydrochloric acid; this is an example of an effect which is dealt with in Experiment 8. Consequently concentrated hydrochloric acid can be used to precipitate barium chloride. Note that the total effect of these reactions has been to prepare a soluble barium salt from an insoluble one, and is a reaction which could not have been carried out in aqueous solution. 5. The formulae of the compounds used in part 5 are: ethyl alcohol CH 3 • CH 2• OH; ethyl iodide CH 3 .CH 2 I; and diphenyl sulphide (C 6H 6)2S. It will have been found in part 3 that Hg++ ions precipitate with OH-, J-, and s- ions. Therefore no precipitate would indicate that these compounds in solution do not form these ions; that is, they are bound together by covalent links. Most compounds of carbon (organic compounds) are held together by covalent links. QUESTIONS

1. Suppose you had a solution of copper nitrate into which two electrodes connected to a source of direct current were dipping. To which pole would the ions travel when the current flowed through the solution? 2. State which ions will be in solution, and their approximate concentrations in moles per litre, when the following are mixed. Assume that all the compounds dissolved in water are completely dissociated into ions. (a) 0.1 mole cupric nitrate, 1 litre of water. (b) 0.1 mole cupric nitrate, 0.1 mole sodium hydroxide, 1 litre of water. (c) 0.1 mole cupric nitrate, 0.1 mole sodium sulphide, 1 litre of water. (d) 0.1 mole barium nitrate, 0.2 mole sodium sulphate, 1 litre of water. (e) 0.1 mole silver nitrate, 0.1 mole sodium sulphide, 1 litre of water. 3. Predict, with the aid of the common solubility rules, whether the following compounds are likely to be soluble or insoluble in water: (a) barium bromide, (b) ammonium fluoride, (c) cobalt sulphate, (d) nickel sulphide, (e) zinc hydroxide, (J) caesium carbonate, (g) strontium sulphate, (h) silver iodide, (i) copper acetate, (j) rubidium hydroxide. 4. Given solutions containing soluble salts of the three metals named in each case, suggest procedures for separating the metals by selective precipitation of suitable compounds: (a) silver, copper, and magnesium. (b) zinc, lead, and barium.

18 (c) mercury, barium, and manganese. 5. Would you expect to be able to obtain a quantitative yield of barium chloride from barium sulphate by the method of part 4? Give reasons. 6. Why can barium chloride not be obtained by adding barium sulphate to a solution of hydrochloric acid?

3.

DISPLACEMENT REACTIONS INTRODUCTION

When a piece of zinc is placed in a solution of hydrochloric acid, hydrogen gas is liberated, and the zinc is said to displace hydrogen from hydrogen chloride, forming zinc chloride. In the same way zinc can displace copper from copper chloride when placed in an aqueous solution of the latter. In this experiment examples of this sort of reaction will be studied, and evidence adduced that groups of elements may be arranged in series on the basis of the ability of each to displace others in the series. EXPERIMENTAL

1. Displacements Involving Acids or Water (a) Obtain a small piece of sodium metal. As it is stored sodium frequently becomes coated over by a layer of oxide; you may cut through a piece with a knife or spatula to expose a fresh area of the metallic surface. Drop the metal onto about 50 ml. of water in a 250-ml. beaker. DO NOT PUT YOUR FACE CLOSE TO THE BEAKER WHILE THE REACTION IS PROCEEDING. When the sodium has dissolved test the solution with red litmus paper. (b) Set up a series of test-tubes, one row to contain 10 ml. each of distilled water and the other 10 ml. each of 1.0 M sulphuric acid. Clean two I-inch pieces of magnesium ribbon with emery cloth and add one to each of water and sulphuric acid. Observe the rates of reaction. After the tube containing water has stood for some time, add a drop of bromthymol blue indicator (in its yellow form), boil the solution, and keep it hot for several minutes. (c) Prepare two pieces, about 1 inch square, of amalgamated aluminium foil by cleaning the surface lightly with emery paper, and then rubbing on this clean surface a few drops of mercuric chloride solution. Use a piece of filter paper or a small cloth for this. Rub the surface for about one minute, and then wash it thoroughly with distilled water. Immediately add the foil, one piece to distilled water and the other to the molar sulphuric acid solution. Add other pieces of untreated foil to different tubes containing water and acid. Observe and compare the rates of reaction in all cases. (d) Place 2 pieces of iron (for example, a I-inch nail) in water and sulphuric acid in the same way, and observe the extent and rate of reaction. (e) Repeat with two pieces of lead foil. Allow the lead to stand a week.

18 (c) mercury, barium, and manganese. 5. Would you expect to be able to obtain a quantitative yield of barium chloride from barium sulphate by the method of part 4? Give reasons. 6. Why can barium chloride not be obtained by adding barium sulphate to a solution of hydrochloric acid?

3.

DISPLACEMENT REACTIONS INTRODUCTION

When a piece of zinc is placed in a solution of hydrochloric acid, hydrogen gas is liberated, and the zinc is said to displace hydrogen from hydrogen chloride, forming zinc chloride. In the same way zinc can displace copper from copper chloride when placed in an aqueous solution of the latter. In this experiment examples of this sort of reaction will be studied, and evidence adduced that groups of elements may be arranged in series on the basis of the ability of each to displace others in the series. EXPERIMENTAL

1. Displacements Involving Acids or Water (a) Obtain a small piece of sodium metal. As it is stored sodium frequently becomes coated over by a layer of oxide; you may cut through a piece with a knife or spatula to expose a fresh area of the metallic surface. Drop the metal onto about 50 ml. of water in a 250-ml. beaker. DO NOT PUT YOUR FACE CLOSE TO THE BEAKER WHILE THE REACTION IS PROCEEDING. When the sodium has dissolved test the solution with red litmus paper. (b) Set up a series of test-tubes, one row to contain 10 ml. each of distilled water and the other 10 ml. each of 1.0 M sulphuric acid. Clean two I-inch pieces of magnesium ribbon with emery cloth and add one to each of water and sulphuric acid. Observe the rates of reaction. After the tube containing water has stood for some time, add a drop of bromthymol blue indicator (in its yellow form), boil the solution, and keep it hot for several minutes. (c) Prepare two pieces, about 1 inch square, of amalgamated aluminium foil by cleaning the surface lightly with emery paper, and then rubbing on this clean surface a few drops of mercuric chloride solution. Use a piece of filter paper or a small cloth for this. Rub the surface for about one minute, and then wash it thoroughly with distilled water. Immediately add the foil, one piece to distilled water and the other to the molar sulphuric acid solution. Add other pieces of untreated foil to different tubes containing water and acid. Observe and compare the rates of reaction in all cases. (d) Place 2 pieces of iron (for example, a I-inch nail) in water and sulphuric acid in the same way, and observe the extent and rate of reaction. (e) Repeat with two pieces of lead foil. Allow the lead to stand a week.

19

(f) Repeat with two pieces of copper. (g) Repeat with two small pieces of calcium. (h) Add a piece of silver (for instance, a dime) to molar sulphuric acid in a test tube or beaker. In all the above cases, if the metal did not completely dissolve, close the tube loosely with a cork or plug of cotton wool, and allow to stand until the next laboratory period. Compare the rates of reaction in all cases, noting especially if the reaction of metal is faster with sulphuric acid or with water. Compare also the rates with which the various metals react with water or with sulphuric acid. Give equations for all cases where a reaction definitely occured. 2. Displacements Between Different Metals The relative activity of two metals can be found (in certain cases) from the fact that, if a more active metal is added to a solution of a salt of a less active one, the more active metal will go into solution and deposit the less active one. The reaction will be of the following type:

X

+

YNOa -

Y

+

XNOa,

where X is the more active, and Y the less active, metal. The limitation on this method is that, if both metals are much more active than hydrogen, then hydrogen will be evolved from the water; and little or no free metal will be deposited from solution. The net reaction would then be:

For this experiment you are provided with powdered metallic zinc, lead, and copper; and with solutions of zinc nitrate, lead nitrate, copper nitrate, and silver nitrate. If you need silver metal, use a dime (to be provided by the student). Determine the order of activity of the four metals, zinc, lead, copper, and silver, as far as you can, by performing experiments of the following sort. In a test-tube place a few grains of one metal powder and 5 ml. of a solution of the nitrate of another metal. Observe whether any new metal is deposited from the solution. The reactions in many cases are slow; therefore, place the tubes containing the reaction mixtures in a beaker of boiling water, and leave them there not less than 15 minutes. Remove the tubes and allow their contents to cool. Test 2 ml. portions of each solution by adding them to 2 or 3 ml. of (i) aqueous ammonia, and (ii) potassium chromate solution (slightly acidified). The purpose of these tests is to ascertain which metallic ions are in solution after sufficient time has elapsed for any reaction to take place. For the reaction X

+ YNOa -

Y

+ XNOa

proof of the presence of X ions in solution is proof of the reaction having taken place as written; the presence of Y ions in solution shows that no reaction has occurred. Sometimes you may see evidence for the presence of both ions in solution; this presumably indicates that a reaction has taken place but incompletely.

20 The observations characteristic of each ion are tabulated in Table I. TABLE I Ion-Test With Copper (Cu++)

Ammonia (NH8)

Pot. chromate (K2Cr0,)

Dark blue solution Cu(NH,),++ White precipitate Pb(OH)2 Colourless solution Ag(NHa)2+ Colourless solution Zn(NH3),++

Lead (Pb++) Silver (Ag+) Zinc (Zn++)

No precipitate Yellow precipitate Red precipitate No precipitate

By a suitable selection of experiments of this sort, place these elements in order of their decreasing chemical activity. 3. Displacements Between Halogens and Their Ions (a) Add 5 ml. of aqueous chlorine solution to 5 ml. of 0.1 M sodium bromide solution in a test-tube. Then add about 1 ml. of carbon tetrachloride, and shake briefly. (b) Add 5 ml. of aqueous bromine solution to 5 ml. of 0.1 M potassium iodide solution. Then add 1 ml. of carbon tetrachloride, and shake briefly. Note and account for the colour changes, giving equations for all reactions observed. What is the role of the carbon tetrachloride? EXPLANATIONS

The term displacement applied to these reactions emphasizes the molecular nature of reactants and products; thus for Zn

+ 2 HCl --+ H2 + ZnCl2

the zinc takes the place of hydrogen in its association with chloride. Viewed as a reaction among fully ionized substances, however, the reaction should be written Zn

+ 2 HaO+ --+ H2 + Zn++ + 2 H20

with the chloride fully ionized before and after reaction. In aqueous solutions the hydrogen ion combines with a molecule of water to give the ion H.H20+ = HaO+. From this point of view, the reaction amounts to a transfer of positive charge from hydrogen ions to zinc ions. According to our conception of atomic structure, however, the transfer of charge takes place by the exchange of electrons (negatively charged); thus we have

2 e-

Zn --+ Zn++ + 2 e+ 2 HaO+ --+ H2 + 2 H20.

The exchange of electrons in this case takes place at the boundary between the solid and the solution; no electrons are set free. Thus for a zinc atom to part with 2 electrons and form a zinc ion there must be two hydrogen ions in solution to receive these electrons. Processes such as the above in which electrons are produced or consumed are often called half-reactions, and if these are added together (as written above) their sum (in which the electrons cancel out) gives the displacement reaction. We shall see in Experiment 4 that oxidation and reduction

21 processes can be regarded in the same way as producers and consumers of electrons, and described by half-reactions; thus displacements come to be regarded as special cases of oxidation-reduction reactions. No half-reaction can in practice be isolated in a displacement reaction, because free electrons cannot be produced there must always be a pair of complementary half-reactions. However if we set up the apparatus shown in Figure 3,

Cu

POROUS PLUG

GALVANOMETER FIGURE

3

we have a chemical cell, a source of electrical energy. The galvanometer indicates a current (flow of electrons) from one pole of the cell to the other, and a suitable instrument can measure the voltage (electromotive force) developed between the poles. The source of the electron flow is considered to be half-reactions at the electrodes. We could have the following half-reactions in one direction or the other: Zn µ Zn++ (in solution) + 2 eCu µ Cu++ (in solution) + 2 e-. The electrons will flow along the wire from one metallic electrode to the other depending on which of these half-reactions can go forward at the expense of the other. Experimentally it is found that the electron flow is from zinc to copper; thus the galvanometer shows the zinc terminal to be negative and the copper positive. Thus what happens when current flows is the following: Zn - Zn++ + 2 e2 e- + Cu++ - Cu Total reaction: Zn+ Cu++ -

Zn+++ Cu

22

The relative ease with which this occurs can, in fact, be measured by the voltage or electromotive force between the electrodes; this aspect of half-reactions is gone into more fully in Experiment 17. In general, for any cell of this sort the electrons will move along the wire to the less active metal, which therefore becomes the positive pole. When, as in part 2, one metal is immersed in a solution of the ions of another, the electron transfer can take place directly at the metal surface. If X and Y are the two metals, the reaction will be X

+ Y+-x+ + Y.

This reaction will occur if Xis more active than Y. By experiments of this sort it is possible to arrange metals in a series starting with the most active one and continuing to the least active one. Any metal in this series, which is sometimes called the electrochemical series, is less active than the metals which precede it, and more active than those which come after it. In part 2 a few experiments of this sort should enable you to place copper, lead, silver, and zinc in order of activity. There is one difficulty in applying this method to all metals, which is that the solution contains a small concentration of hydrogen ions from ionization of the water. If the metal being used is much more active than hydrogen, hydrogen may be evolved, instead of the other metal being precipitated out of solution. This sort of reaction is illustrated in part 1, where it will be found that a metal such as sodium can produce hydrogen from water: 2 Na+ 2 H20 -

2 NaOH + H2.

Sodium can react directly with water. Other less active metals only react at all fast with solutions of acids, that is, they react with hydrogen ions; for example, Fe + 2 H 30+ -

Fe++ + 2 H20 + H2.

It is probably incorrect to say that a metal reacts only with hydrogen ions and not at all with water; however, the reaction with hydrogen ions is very much faster. Reactions of this sort can also be formally divided up into two halfreactions: Fe - Fe++ + 2 e2 HaO+ + 2 e- - 2 H20 + H2 Fe + 2 HaO+ -

Fe++ + 2 H20 + H2

In aqueous solutions, hydrogen ions are combined with water, and have the formula HaO+. Thus it is possible to place hydrogen in the electrochemical series; metals above hydrogen will dissolve readily in dilute acids, while those below it will not. The actual speed of solution of a metal in water or acid may be much influenced by another phenomenon, which is independent of its activity. A layer of insoluble oxide or hydroxide may form round the metals and prevent actual contact between the liquid and the metal. Aluminium is the best example of this. It is reactive enough to react with water, but it forms a thin coherent coating of its insoluble oxide, so that no further attack occurs; hence its possible use in kitchen-ware. If a coherent coating is prevented by amalgamation, the metal is attacked. The process of amalgamation consists in the formation of a layer of aluminium dis-

23

solved in mercury spread over the surface of the aluminium. The mercury itself is formed by one of these displacement reactions, 3 HgCl2 + 2 Al -

3 Hg+ 2 AIC1 3•

The aluminium oxide (or hydroxide) cannot form a continuous layer over the liquid mercury film, as the film can move and break the layer. Iron in pure water, out of the air, forms only a thin coating of Fe(OHh; air, however, oxidizes this and gives the familiar rusting of iron. Magnesium also gives a fairly insoluble hydroxide, but this does not form such an impermeable coating as does aluminium oxide, so that though it may slow down the reaction it does not stop it. Although the reaction of magnesium with sulphuric acid is far quicker than with water, owing to the much higher concentration of hydrogen ions in the former, you will probably find the reverse to be true for calcium. This can be explained by the formation around the calcium of a deposit of sparingly soluble calcium sulphate, and this interferes with the access of hydrogen ions to the surface of the metal. In the same way lead in sulphuric acid is somewhat protected by the nearly insoluble lead sulphate; lead will displace hydrogen much more easily from perchloric acid (HCl04) since lead perchlorate is soluble. In part 3 are two experiments which show that the halogens can be placed in order of activity, as can the metals. Since chlorine is more active than bromine, we get the reaction: which can also be considered as a transfer of electrons: Cl2 +

2 Br 2 e- -

Cl2 + 2 Br -

Br2 + 2 e2 CI2 CI- + Br2

In this case the most active halogen is the one that combines with electrons most readily; with metals, the most active metal loses electrons most readily. In part 3 the carbon tetrachloride is merely added to show which free halogen is present. It does not react, but dissolves the halogens to give solutions of the following colours: chlorine, pale yellow; bromine, orange brown; iodine, purple. From these experiments it is possible to place chlorine, bromine and iodine in order of activity; it would be found that fluorine is the most active of all the halogens. QUESTIONS

1. From the observations in these experiments (a) place the following metals in order of decreasing chemical activity: Cu,

Fe, Ag, Na, Al, Mg, Ca, Pb. (b) arrange the halogens in order of decreasing activity. 2. Metallic sodium, exposed to air, quickly becomes coated over with a layer of oxide. Why does this not protect the metal from attack by water? 3. Predict what would happen if you dropped a piece of sodium into a solution of magnesium chloride. 4. Explain clearly why magnesium dissolves more rapidly in molar sulphuric acid than in water.

24 5. Suggest reasonable explanations, in terms of the principles illustrated in this experiment, for the following practical applications of chemistry: (a) Hydrochloric acid is used by a tinsmith to clean oxide film from copper without dissolving the metal. (b) The tarnish on silver (silver sulphide) may be removed by holding the tarnished article in contact with metallic aluminium under a solution of household soda.

4.

OXIDATION-REDUCTION REACTIONS INTRODUCTION

A large number of reactions are grouped together in the category of oxidationreduction (redox) reactions. The two processes of oxidation and reduction occur simultaneously, and these may be regarded as involving either the transfer of oxygen atoms or of electrons from one element to the other. The element being oxidized may combine with oxygen or with a greater proportion of oxygen; this can be associated also with the loss of electrons by that element. EXPERIMENTAL

1. Reduction of a Metallic Oxide by Natural Gas

Set up the apparatus as shown in Figure 4.

8 FIGURE 4

24 5. Suggest reasonable explanations, in terms of the principles illustrated in this experiment, for the following practical applications of chemistry: (a) Hydrochloric acid is used by a tinsmith to clean oxide film from copper without dissolving the metal. (b) The tarnish on silver (silver sulphide) may be removed by holding the tarnished article in contact with metallic aluminium under a solution of household soda.

4.

OXIDATION-REDUCTION REACTIONS INTRODUCTION

A large number of reactions are grouped together in the category of oxidationreduction (redox) reactions. The two processes of oxidation and reduction occur simultaneously, and these may be regarded as involving either the transfer of oxygen atoms or of electrons from one element to the other. The element being oxidized may combine with oxygen or with a greater proportion of oxygen; this can be associated also with the loss of electrons by that element. EXPERIMENTAL

1. Reduction of a Metallic Oxide by Natural Gas

Set up the apparatus as shown in Figure 4.

8 FIGURE 4

25 Place about 1 g. of the metallic oxide at the bottom of a l"X6" Pyrex testtube, and attach the rubber stopper and tubes as shown. Ask a demonstrator to check your assembly to be sure it is gas tight. Then turn on the gas, and after about 10 seconds light the jet A. It is important to sweep the tubes out with gas before lighting the jet in order to remove any air; otherwise a backfire would result. Adjust the gas flow through the tube to give a flame at A about ½inch long. Then heat the metallic oxide at B, gently at first, and then as strongly as possible in a Bunsen flame for a few minutes. Note any change in the appearance of the oxide, and whether drops of water condense in the cooler part of the test-tube. Allow the tube to cool before turning off the jet at A. Examine the residue in the tube for evidence of any metal formed. Use successively (a) cupric oxide, (b) magnesium oxide. Record all observations, and give an equation for any reaction proposed to account for what you saw.

2. Reduction of a Metallic Oxide by Charcoal (a) Grind together in a mortar about 1 g. of lead monoxide and 1 g. of charcoal. Place the mixture in a crucible and cover with its lid. Heat the mixture with a moderate Bunsen flame for two or three minutes, then allow to cool. When cool, remove the lid from the crucible, and examine the contents for evidence of reduction to the metal. (b) Repeat using aluminium oxide in place of lead oxide.

3. Reduction of a Metallic Oxide by Aluminium This experiment will be demonstrated. It consists of the reduction of chromic oxide or ferric oxide by powdered aluminium. The reaction is described in textbooks as the "thermite" or "Goldschmidt" process for the reduction of certain oxides. 4. The Oxidation States of Iron (a) Put about 2 ml. of 0.1 M ferrous sulphate solution in a test-tube, and add 5 ml. of the sodium hypochlorite solution provided. This solution contains sodium hypochlorite and sodium hydroxide. Note and account for any reaction. (b) Put 20 ml. of ferric chloride solution (0.1 M) and 5 ml. of cone. HCl into a small conical flask. Add several pieces of zinc, and add another piece from time to time so that there is always undissolved zinc in the flask. At the start, and at intervals of 5-10 minutes, take a drop of the solution in the flask, and mix it with 5 ml. of 0.1 M potassium thiocyanate solution in a test tube. Note and explain the appearance of the solution in the flask, and the relative intensity of the colours in the test tubes of thiocyanate. (c) In a test-tube place 1-2 ml. of 0.1 M ferric chloride solution and 10 ml. of the sodium hypochlorite sodium hydroxide solution. Heat cautiously to nearly boiling and keep hot for 1-2 minutes. Cool under the tap, and filter through glass wool. Use only enough glass wool to make a small plug at the bottom of your funnel. Add a few ml. of barium chloride solution to the filtrate. Account for all your observations.

+

26 5. The Oxidation States of Chromium (a) Place a mixture of 0.5 g. of chromic oxide, Cr2O3, and 6-8 pellets of sodium hydroxide (about 1 g.) in a crucible, and heat gradually until the sodium hydroxide melts. Sprinkle in gradually about 1 g. of sodium peroxide until the melt no longer has a green tinge; the colour should be yellowish brown. SAFETY NOTE: SODIUM PEROXIDE MUST NEVER BE PUT

ON ANY

INFLAM-

MABLE MATERIAL SUCH AS PAPER; IT MUST NEVER BE STORED IN YOUR LOCKER; IT MUST BE KEPT DRY, AND MUST NOT BE STORED IN THE AIR FOR MORE THAN A SHORT TIME BEFORE USE.

You

MUST TAKE CARE TO PROTECT YOUR FACE AND

ESPECIALLY YOUR EYES FROM SPLASHES OF THE MOLTEN MATERIALS HANDLED IN THIS SECTION.

Allow the crucible to cool, and place it in a beaker containing enough water to cover the crucible. Warm the water in the beaker, and stir to dissolve the contents of the crucible. Filter the resulting solution, and record its colour. To a 5 ml. portion of this solution add dilute nitric acid in small portions until the resulting solution is just acid to litmus. Describe and account for any change in colour. Then add a few ml. of lead nitrate solution. Describe and account for what is observed. (b) Place 5 ml. of 0.1 M sodium or potassium chromate solution in each of two test-tubes, adding to each 1 or 2 ml. of dilute sulphuric acid. To one of these solutions add, with gentle shaking to keep the contents of the tube mixed, 0.2 M ferrous sulphate solution until no further change in colour is visible. Keep this tube for part (d). Prepare a solution of sodium sulphite by placing about ¾inch depth of the solid in a test-tube and dissolving in sufficient water to fill the tube ¾full. Add this solution in small portions to the second chromate solution until no further change is seen. Keep this tube for part (c). Account for your observations. (c) Boil this second solution from part (b) to expel SO2. Add solid sodium carbonate in small portions until all of the acid is neutralized, as shown by the appearance of a precipitate of chromic hydroxide. Then add sodium hypochlorite solution, mixing and heating the contents of the tube until all of the precipitate has dissolved. Note and account for the colour of the solution. (d) In a 1" X6" test-tube dissolve 0.5 g. of potassium dichromate in 10 ml. of water (or use 10 ml. of a 5% solution) and add 5 ml. of concentrated hydrochloric acid. Drop in one or two pieces (2 or 3 grams) of zinc. Close the tube with cotton wool or a loosely fitting cork. Leave for at least fifteen minutes, adding more zinc, if necessary, to maintain a continual evolution of hydrogen. Meanwhile, dissolve 4.5 g. of sodium acetate trihydrate in 4 ml. of water in a small flask or another 1" X6" test-tube. Compare the colour of the chromium solution with that retained from part (b). When the solution is blue, filter it quickly through a small plug of glass wool, wedged in the apex of the cone of a funnel, into the prepared solution of sodium acetate. It is preferable to arrange for the tip of the funnel to dip below the surface of the sodium acetate solution.

27

Note the appearance of the precipitate formed. Prepare a fresh plug of glass wool in a clean funnel, and quickly filter the mixture, discarding the filtrate. Observe the colour of the insoluble residue. 6. The Oxidation States of Manganese (a) Mix equal weights (about 1½ g.) of manganese dioxide and sodium nitrate. Add the mixture to 15-20 pellets of sodium hydroxide in a crucible and heat cautiously till molten. Keep the mixture just molten for 2-3 minutes, and then allow to cool. Put about 100 ml. of water in a 250 ml. beaker, and dissolve about 10 pellets of sodium hydroxide in it. Then place the cool crucible in the water, and dissolve out most of the solidified mass in the crucible. As soon as possible filter this solution, and treat the filtrate in the following ways. In a small beaker mix about 10 ml. each of the filtrate and sodium hypochlorite solution. Warm to nearly boiling, and note any reaction. To the rest of the filtrate, add about 1 g. of barium chloride dissolved in a few ml. of water. Note any precipitate. Filter, wash, and dry. (b) In each of three test-tubes place 10 ml. of 0.02 M potassium permanganate solution. To the first add about 5 ml. of dilute sulphuric acid; to the second an equal volume of water; and to the third an equal volume of 1.0 M sodium hydroxide. To each add sodium sulphite solution (see 5 (b)); add about 5 ml. to the first two tubes (acid and neutral) until no further colour change takes place; to the alkaline tube add the solution a few drops at a time. Note any colour changes.

7.

The Oxidation States of Vanadium

A solution of 0.05 M pervanadyl chloride, VO2Cl, containing about 1 M HCl, is provided. (a) To 1-2 ml. of the pervanadyl chloride solution in a test-tube, add a few pellets of sodium hydroxide until no further colour change occurs. Allow each pellet to dissolve before adding the next. Now add 1-2 ml. of barium chloride solution. Note the colour of the precipitate. (b) To 5 ml. of the pervanadyl chloride solution in a test-tube, add about as much anhydrous sodium sulphite as could be picked up on a dime. Boil the mixture gently for 10-15 seconds and note its colour. Cool the solution, and add sodium hydroxide pellets until no further colour change occurs. Divide the solution equally between two test-tubes. To the first add 1-2 ml. of barium chloride solution. To the second add 2-3 ml. of sodium hypochlorite solution, and allow to stand for a few minutes; then add 1-2 ml. of barium chloride solution. (c) In a I-inch test-tube put 15 ml. of the pervanadyl chloride solution, add about 2 ml. concentrated HCl, then several pieces of zinc. Close the tube with a glass wool plug. Allow to stand until the colour is pale purple; if necessary add more HCl or zinc so that gas is continuously evolved. Pour off 2-3 ml. of the solution so formed, and add to it a few drops of copper chloride solution (about 0.1 M). Note what occurs. Pour off another 2-3 ml. of the solution, and add 1-2 ml. of the stannic chloride solution (in dilute HCl), warm to nearly boiling, and note the colour change. Now add a few drops of copper chloride solution.

28 Pour off 3 ml. of the pale purple solution prepared above, and mix it with 1 ml. of the original pervanadyl chloride solution. Note the colour. Pour off another 3 ml. of the pale purple solution, and add an equal volume of chlorine water. After a few minutes warm the solution gently, and note the colour. In every case note any colour changes or precipitates, and write a balanced chemical equation for all the reactions. 8.

Strength of Oxidizing or Reducing Agents

It is not found that any oxidizing agent will oxidize any reducing agent. Some substances are powerful oxidizing agents which will effect many oxidations, while others will only oxidize relatively few compounds. The same applies to reducing agents. The following experiments illustrate qualitatively the relative strengths of the hydrogen halides (HCl, HBr, and HI) as reducing agents. (a) Place 1 spatula point of sodium chloride in a dry test-tube and add about ½ ml. of concentrated sulphuric acid. Note whether there is any sign of free chlorine as judged by the yellowish green colour of chlorine. Repeat with sodium bromide, and then with potassium iodide. Note whether free bromine (a deep orange red), or free iodine (almost black with a purple vapour) is produced in appreciable amounts. (b) In each of 3 test-tubes place 5 ml. of a 0.1 M solution of sodium nitrite, and 5 ml. of 0.1 M solution of either sodium chloride, sodium bromide, or potassium iodide. Add to each 2 ml. of carbon tetrachloride. This will dissolve any free halogen that is formed. Now add 2 ml. of 30% acetic acid to each. Shake and observe any change. If no change is noted, add about 5 ml. of dilute hydrochloric acid. Shake and observe any change. EXPLANATIONS

A large number of reactions, of which the above are examples, are classed as oxidation-reduction (redox) reactions. In any such reactions an oxidizing substance (which is then reduced) and a reducing substance (which is then oxidized) must be present. In deciding which is the oxidizing and which the reducing agent, and in balancing the chemical equations we can use a number of different, though essentially equivalent, methods. Oxidizing Agents Provide oxygen. 2. Transfer of electrons. Combine with electrons. 3. Change of oxidation Oxidation state goes from higher to lower value. state. Method

1. Transfer of oxygen.

Reducing Agents Combine with oxygen. Provide electrons. Oxidation state goes from lower to higher value.

To balance equations by method 1, the reducing agent molecules are made to gain as many atoms of oxygen as are lost by the oxidizing agent molecules. Not all oxidation-reduction reactions can be dealt with by method 1, as no oxygen may be involved anywhere in the reactions. It was, however, in reactions involving oxygen that this type of reaction was first distinguished, which accounts for the name. An example is

29 + CuO ---+ Cu + H2O, H2 oxidizing reducing agent agent In method 2, the transfer of electrons may be real enough or it may be only formal. For instance in the preceding example it is only in a formal sense that the hydrogen provides electrons to the copper. However if we take another example from part 4 (b), the reaction is 2 Fe+++ + Zn ---+ 2 Fe++ + Zn++.

It is evident that the zinc loses and the ferric ions gain electrons. The reaction can be broken up formally into two parts: Zn ---+ Zn++ + 2 e2 Fe+++ + 2 e- ---+ 2 Fe++ Total: 2 Fe++++ Zn---+ 2 Fe+++ Zn++ To balance the chemical equations, two half-reactions of this sort must be combined so that the electrons cancel. It was shown that the displacement reactions of Experiment 3 involve electron transfers, and so are special examples of redox reactions. Any reaction used in an electric cell (for example, a dry cell or storage battery) transfers electrons around the external circuit. For instance in a storage battery Pb + SO4-----+ PbSO4 + 2 e- ( -

J

pole)

PbSO4 + 2 H2O + SO4- +- PbO2 + 2 H2SO4 + 2 e- ( + pole)

electrons round circuit

The current is carried through the solution by H 3O+ and S0 4- moving from one pole to the other. Closely related to this last method is method 3, which is in some ways the most general. The oxidation state of an atom in a molecule or ion is assigned a number, an integer, which may be positive, negative, or zero. They are assigned in accordance with the following rules. 1. The oxidation state of an uncombined element is zero. 2. The oxidation state of a monatomic ion equals its charge; for example Fea+ : oxidation state + 3. Cl- : oxidation state -1. 3. In a molecule oxygen is assigned the oxidation state -2 (corresponding formally to the oxide ion o-), and hydrogen + 1 (corresponding formally to the hydrogen ion H+). The sum of the oxidation states of all atoms in a molecule is zero; for example,

H2

2(

HNOa:

+ 1)

H (+ 1)

+ +

s (+ 6) N (+ 5)

+ +

04 4( - 2) Oa 3( - 2)

= 0. 0.

30 The oxidation state of sulphur in sulphuric acid (and all sulphates) is therefore + 6; of nitrogen in nitric acid (and all nitrates) is + 5. K

Mn

04

K2

Cr

(+ 1) + (+ 7) + 4( - 2) = 0 2( + 1) + (+ 6) + 4( The value + 1 is assigned to potassium as it is present in K+ ions.

04 2)

= 0.

There are some compounds which are not covered by these simple rules and which need others, but with the exception of sodium peroxide we have not met them yet, so they will not be considered here. The oxidation of sodium peroxide can be dealt with by method 1. Having assigned the oxidation states, and noted (as above) that oxidizing agents go from higher to lower oxidation states, and reducing agents from lower to higher, we can find the equation for any oxidation-reduction reaction by the further rule that: The total change in oxidation state of the oxidized atoms is equal and of opposite sign to the total change in oxidation state of the reduced atoms. It is understood that a change of, for example, 2 in 3 atoms counts as a total change of 6. To take the simple example used above:

Oxidation states

2 Fe+++ + Zn ---+ 2 Fe++ + Zn++ 2( + 3)------2( + 2) 0------ + 2

Total change - 2 Total change + 2

It is evident that a change of oxidation state of 1 is equivalent to the transfer of one electron. It is for this reason that the total changes in oxidation states must balance, since we must not be left with any electrons unaccounted for. We can thus attack the hydrogen copper oxide reaction by the three methods. 1.

H2 + gains 1 oxygen

CuO ---+ loses 1 oxygen

H20

+

2.

H2 ---+ 2 H+ + 2 eCuO + 2e- ---+O- + Cu Then 2 H+ + o- ---+ H20

3.

H2 CuO ---+ H20 Cu 0-------2( 1) +2------0

+

+

+

Cu

Oxidation Reduction

change + 2 }Oxidation change - 2 states

The following are some more complicated examples: (a) Potassium permanganate and sodium sulphite in acid solution 2 KMnO, + 5 Na2SOa + 3 H2SO,---+ 2 MnSO, + K2SO, + 5 Na2SO, + 3 H20 2( + 7) - - - - - - - - - - - - - - - - - 2( + 2) Change - 10 5( + 4) - - - - - - - - - - - - - - - - 5( + 6) Change + 10 The manganese and sulphur atoms change their oxidation states. This equation was written in terms of the molecules involved; it could better be written as a reaction of ions:

31 Oxidation: SOa- + 3 H2O -so,-+ 2 HaO+ + 2 eReduction: MnOr + 8 HaO+ + 5 e- - Mn++ + 12 H~O Then 2 MnOr + 5 SOa- +6 HaO+ - 2 Mn++ + 5 SO,- + 9 H2O These two equations are essentially equivalent. (b) Ferric chloride and 2 FeCla + 3 NaOCl + 2( + 3) - - - - - 3( + 1) -

sodium hypochlorite- part 4 (c). 10 NaOH - 2 Na2FeO, + 9 NaCl + 5 H~O - - - - - - - 2( + 6) Change + 6 - - - - - - - - - - - - - 3( - 1) Change - 6

The iron and chlorine change their oxidation states. Note that of the nine NaCl formed only three come from hypochlorite. In general the steps to follow in balancing equations of this sort are, first to decide which atoms change their oxidation states and by how much, then by balancing these changes to get the correct ratio of oxidizing agent to reducing agent, and lastly to fill in the subsidiary molecules such as sodium hydroxide or water. This same reaction may be represented by ionic half-equations Fe+++ + 8 OH- - FeO,- + 4 H2O + 3eOxidation: Reduction: OCI- + H2O + 2 e- CI+ 2 OHThen 2Fe+++ + 3 OCI-+ 10 OH--2 FeO,-+ 3CI-+5 H2O The following notes apply to the actual reactions carried out in this experiment. 1. The illuminating gas in Toronto is natural gas of which the principal constituent is methane. It probably reduces copper oxide by a reaction such as: It is unlikely that any reduction of magnesium oxide will occur at all. Manufactured illuminating gas contains a high proportion of hydrogen, which will more easily reduce numerous metallic oxides at elevated temperatures. 2. Lead oxide can be reduced by carbon to give lead and carbon monoxide at about 550°. The crucible in which the reduction is carried out must be kept covered during the heating, otherwise the charcoal just burns in the air. Aluminium oxide is much more difficult to reduce. The ease of reduction of oxides can be related to the activity of the metal found in the last experiment; active metals are not readily obtained by reduction of their compounds. 3. The reaction between aluminium and chromium or iron oxide is another instance of a displacement reaction, comparable with those studied in Experiment 3. Aluminium is more active than chromium, thereby displacing the latter from its oxide:

4-7. The reactions in these parts illustrate the various oxidation states of the metals iron, chromium, manganese, and vanadium. You should make a point of locating these four elements in the Periodic Table, and of finding out from textbooks something of the origin of the variability in oxidation state exhibited by these and related elements.

32 Table I will assist you m identifying the products formed m the various experiments. TABLE I Element Iron

Chromium

Manganese

Vanadium

Oxidation State

Name

Typical Formulae

Colour purple, red usually yellow-brown solutions, but the ion Fe+++ is pale violet pale green solutions

+6 +3

ferrates ferric or iron (Ill)

Na2Fe0,, BaFeO, (insol.) FeCla, FeO.OH (insol.)

+2 +6 +3 +2 +7 +6 +4 +3 +2 +s

ferrous or iron (II) dichromates chromates chromic or chromium (III) chromous or chromium (II) permanganates manganates managanese dioxide manganic or manganese (III) manganous or manganese (II) pervanadyl vanadates

FeSO,, Fe(OH)2 (insol.)

+4 +3 +2

vanadyl hypovanadates vanadic vanadous

CrCl2, Cr( C2 H 30 2)2 (insol.)

orange in solution yellow purple or dark green solutions blue solutions

KMnO, K2MnO,, BaMnO,(insol.) Mn02 (insol.)

purple solutions green solutions brown-black

MnO.OH (insol.)

brown

MnSO,

pale pink in solution

V02CI N a3VO, and some other types VOCl2 Na2V,O, VCla VCl2

yellow white or pale yellow

K2Cr201 K2CrO,, BaCrO, (insol.) CrCl.a, Cr(OH)a (insol.)

deep blue brown blue-green lavender

4. The oxidations of hypochlorite encountered in this and subsequent sections involve the ion CIO- or OCI-, in which chlorine is in the oxidation state of + 1; it is reduced to chloride Cl- where the oxidation state is - 1.

2 e- + OCI- + H2O - Cl-+ 2 OH- (in alkali). In 4 (b) the ferric chloride is slowly reduced; the amount remaining can be determined by the red colour it gives with potassium thiocyanate, KCNS. The equation for the oxidation to ferrate in 4 (c) has already been given. 5. The oxidation of chromic oxide by sodium peroxide involves an equation that is best balanced by considering the oxygen atoms transferred: Cr2Oa + 3

Na2O2-► 2

Na2CrO4 + Na2O.

The dichromate ion is formed when solutions of chromates are acidified: 2 CrO4- + 2 HaO+ -

Cr2O1- + 3 H2O.

The reduction of this ion, as performed in 5 (b), requires an acidic medium: Cr2O1- + 14 HaO+ + 6 e- - 2 Cr+++ + 21 H2O. The reducing agents used in this part were ferrous or sulphite ions Fe++ - Fe+++ + eSOa- + 3 H2O - SO4- + 2 HaO+ + 2 e-.

33

In part 5 (d), zinc will eventually reduce dichromate to chromous ions (Cr++) which are a bright blue colour. The reaction is fairly slow. The formation of chromous ions can be seen by the change in colour of the solution, and made further manifest by their reaction with sodium acetate; in this reaction a red precipitate or colour of chromous acetate is seen. Any chromic ions still in this solution will form a lilac-coloured precipitate of chromic hydroxide with the sodium acetate solution. The chromous acetate precipitate is crystalline and will be retained on a coarse filter such as glass wool, while the chromic hydroxide will pass through as a non-crystalline gelatinous precipitate. The chromous compounds are very easily oxidized even by air. 6. The oxidation of manganese dioxide in 6 (a) involves the reduction of sodium nitrate to sodium nitrite. Mn02

+

NaNOa

+4

L

+

2NaOH

+5 oxidation:

-1- increase of 2

-

Na2Mn04

+

NaN02

+6

+

H20

+3

I

- - Reduction: decrease of 2

The product sodium manganate can be recognized by its intense green colour in solution. It is, however, not stable in solution unless a moderate concentration of sodium hydroxide is also present. In neutral or acidic solution it decomposes thus: 3Na2MnO, + 2H20 -2NaMn04 + Mn02 + 4NaOH. Such a reaction, in which one element undergoes simultaneous oxidation and reduction, is called a disproportionation. Barium manganate is blue-grey in colour and insoluble. In 6(b) the course of the reduction of permanganate is seen to be influenced by the acidity of the medium. The relevant half-reactions are 5 e- + MnO,- + 8 HaO+ 3 e- + Mn04- + 4 HaO+ e- + MnO,- -

Mn++ + 12 H20 Mn02 + 6 H20 MnO,-

(acidic) (neutral) (alkaline)

The sulphite in all cases is oxidized to sulphate. 7. The chemistry of vanadium is comparatively complicated. In addition to forming compounds in four different oxidation states, it forms both cationic and anionic species in the higher two oxidation states. Finally, vanadates (VOi-) which are colourless in solution, may upon mild acidification form orange-coloured divanadates (H 3V201-), by a reaction comparable to the formation of dichromates from chromates (see above) 2 VOt- + 5 HaO+ -

HaV201- + 6 H20.

The differences in colour among the various compounds in the different oxidation states make it possible to follow the course of the reactions examined in this part. In 7 (b) the reduction by sulphite forms a vanadyl salt, which is converted by alkali to a hypovanadate. e- + V02+ + 2 HaO+ - VO+++ 3 H20. 4 VO++ + 10 OH- - V,Ou- + 5 H20.

34 A possible half-reaction for the re-oxidation of a hypovanadate solution might be: V,O 9-

+ 14 OH- -

4 VO,-.

+ 7 H2O + 4e-.

In 7 (c) reduction by zinc should yield a vanadous salt (cf. 5 (d)). Bivalent vanadium is a powerful reducing agent, capable of reducing copper salts to the metal, and susceptible of rather easy oxidation. Different products of oxidation of bivalent vanadium are obtained depending on the oxidizing agent used. Thus stannic chloride, being a mild oxidant, converts V++ to v+++; while the original pervanadyl chloride converts it to VO++; and chlorine should regenerate the vanadium to the +5 oxidation state, although the oxidation tends to be slow. Half-reactions for conversions involving the vanadyl or pervanadyl ion require hydrogen ion to be included in order to provide or consume oxygen atoms. Thus, for example: eVO2+ 2 HaO+ -vo++ 3H2O. eVO++ 2 HaO+ - V+++ 3 H2O.

+ +

+ +

+ +

8. The displacement reactions of Experiment 3, part 3, showed that the halogens could be arranged in order of decreasing activity as chlorine, bromine, iodine. This amounts to the fact that the process 2 e-

+ X2 - 2 X-

proceeds with greater ease as we pass from iodine to chlorine (as X); or conversely that iodide is a better reducing agent than chloride, chlorine a better oxidizing agent than iodine. This order is shown again in the experiments of this part. Chloride ion does not reduce sulphuric acid, but bromide and iodide do. Moreover iodide appears under comparable conditions to bring about a more drastic reduction than bromide. The reactions involved are

+ + +

+ + +

+ + +

(i) 2 HX H2SO, - X2 2 H2O SO2. (ii) 8 HX H2SO, - 4 X2 4 H2O H2S. (iii) 6 HX H2SO, - 3 X2 4 H2O S. Bromide generally achieves reaction (i) while iodide often reduces by (ii) or (iii). The oxidation of the halide ion by nitrite or nitrous acid involves the halfreaction: eNO22 HaO+ - NO 3 H2O. The strength of nitrite as an oxidant can be regulated by the hydrogen ion concentration of the solution; the more acidic the solution the more readily the halfreaction above will take place, and hence the more powerfully the solution acts as an oxidant. The acetic acid used in the experiment provides low concentration of H 3O+, while hydrochloric acid provides a much higher concentration. A weaker reducing agent will therefore be able to reduce nitrite if the concentration of hydrogen ions is large (that is, in the presence of hydrochloric acid). The carbon tetrachloride is added simply to show whether free halogen is formed.

+

+

+

QUESTIONS

1. Name the oxidizing agent, and give the chemical equation, for the following reactions:

35

(i) Calcium + aluminium fluoride ---+ calcium fluoride + aluminium. (ii) Silver nitrate + lead ---+ lead nitrate + silver. (iii) Magnesium + hydrochloric acid ---+ magnesium chloride + hydrogen. (iv) Aluminium + nickel oxide---+ aluminium oxide + nickel. (v) Sodium ferrate + sodium sulphite+ sulphuric acid---+ sodium and ferric sulphates + water. (vi) Hydrogen peroxide + ferrous sulphate + sulphuric acid ---+ ferric sulphate + water. (vii) Hydrogen peroxide+ sodium hypochlorite---+ sodium chloride+ oxygen + water. (viii) Di-sodium phosphite + sodium dichromate + sulphuric acid---+ sodium and chromic sulphates + phosphoric acid + water. (ix) Sodium permanganate + sodium oxalate + sulphuric acid ---+ sodium and manganous sulphates + carbon dioxide + water. (x) Sodium hypovanadate + sodium hypochlorite + sodium hydroxide---+ sodium vanadate + sodium chloride + water. 2. Which of the following are redox reactions? Indicate which elements change their oxidation states: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x)

K2S04 + Al2 (S04)a + 24 H20---+ 2 KAI (S04)2.12 H20, Na2B401 + 2 HCI + 5 H20---+ 4 HaBOa + 2 NaCl. Cb + 2 NaCI02---+ 2 NaCl + 2 CI02. Na2Fe04 + 3 FeS04 + 4 H2S04 ---+ 2 Fe2 (S04)a + Na2S04 + 4 H20, 2 N02 + 2 NaOH---+ NaN02 + NaN0 3 + H20. 4 AgNOa + K2Cr201 + H20---+ 2 Ag2CrO, + 2 KNOa + 2 HNOa, NHa + HCI ---+ NH4Cl. 4 Zn + 10 HNOa ---+ NH4NOa + 4 Zn (NOa)2 + 3 H20. FeS04 + 6 KCN ---+ K4Fe (CN)e + K2SO,. 3 Na2S20a + 2 NaN02 + 6 HCI---+ Na2SeOe + 2 N02 + 6NaCl + 3 H20.

3. Discuss the correlation between position of an element in the activity series (as determined in Experiment 3) and ease of reduction of its oxide by carbon or illuminating gas. 4. Select from the following unbalanced ionic equations the half-reactions corresponding to oxidation and reduction, and show how these may be combined to give balanced equations: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x)

Cr04-- + CI- + HaO+ ---+ Cb + Cr+++ + H20. l2 + SOa- + H20 ---+ 1- + S04- + HaO+. Cr201-- + S02 + HaO+ ---+ Cr+++ + S04- + H20. Cb + h + H20 ---+ CI- + IOa- + HaO+. SOa-- + S- + HaO+---+ S + H20. Fe+++ + Ti+++ + H20 ---+ Fe++ + TiO++ + HaO+. VO+++ Sn+++ HaO+---+ v+++ +Sn+++++ H20. Cu + NOa- + HaO+ ---+ Cu++ + N02 + H20, Mn04- + H202 + HaO+ ---+ Mn02 + 02 + H20. HC02- + PtCie- + OH- ---+ COa- + Pt + CI- + H20.

36

5.

DOUBLE SALTS INTRODUCTION

This experiment consists of the preparation of several compounds belonging to a category known as double salts. These compounds are formed when two simple salts crystallize together in definite, simple molecular proportions. The compounds are only known in the solid state, for in solution they generate the separate ions of the component simple salts. The crystal form of a double salt need not be the same as that of either simple salt from which it is composed. Double salts must be distinguished from complex salts, which are illustrated in Experiment 6. EXPERIMENTAL

1. Common Alum, KAl(SO4)2.12 H~O Aluminium foil is provided for this preparation in sheets 6 inches X 6 inches (average weight 1.55 g.). Crumple one sheet and place in a flask, add 40 ml. of 2 M potassium hydroxide solution. Describe and explain what happens. After activity has subsided there will remain in the flask an insoluble residue due to other elements in the commercial aluminium alloy; filter this out and catch the filtrate in a 250 ml. beaker. Calculate the volume of dilute (3M) sulphuric acid that will contain twice the number of moles of KOH added, and add this quantity of the acid to the beaker while slowly stirring its contents. A considerable amount of a white precipitate (what is it?) will appear at first, but as the acid is all added this begins to dissolve. (Why?) Heat the solution nearly to boiling or until the precipitate has all or nearly all dissolved. A small residue may still remain undissolved after this heating in which case filter the mixture and catch the filtrate in a second clean beaker. (What salts do you think are now present in this solution?) Heat this solution gently to evaporate water from it, but do not let it boil, until the volume has been reduced to about 25 ml. Cover the beaker with a watch glass and set away to cool, preferably until the next laboratory period. Identify the crystals that form. Filter these out through a dry paper and wash with two 5 ml. portions of cold water. Remove the crystals and squeeze them dry between layers of filter paper. Leave the crystals to dry in the air for half an hour, and then weigh them. Place the specimen of crystals in a test-tube, label, and hand in. In your report show the necessary equations for this preparation of potassium aluminium sulphate. Calculate the weight of alum that could in theory be produced from 1.55 g. of aluminium, and suggest reasons why you got less than this amount. 2. Chrome Alum, KCr(SO4)2.12 H2O Weigh out 6 g. of potassium dichromate, grind fine in a mortar, and place in a 100-ml. beaker. Add 35 ml. of 2.5 M sulphuric acid; warm to dissolve. Do not allow the beaker to become hotter than can be held in the hand. Add, with stirring and in several small portions, 5 ml. of ethyl alcohol. Place the beaker in a dish or sink of cold water to keep its contents from becoming too

37 hot (not over 40°). Note the odour of acetaldehyde resulting from the oxidation of the alcohol, and the change in colour of the solution. When the reaction has ended there should be a dark solution in the beaker. If any white crystalline solid separates, set the beaker to heat over a gentle flame and swirl until the solid has dissolved. Set the beaker away, preferably for a week, to permit crystals of chrome alum, which are purple in colour, to form. Filter these, and dry them on a few sheets of filter paper. 3. Copper Ammonium Sulphate, Cu(NH4h(S04)2. 6 H20 Weigh out 25 g. of copper sulphate pentahydrate, 13½ g. of ammonium sulphate, and place them in a 250 ml. beaker. Add 100 ml. of water and about ½ml. (10 drops) of reagent dilute sulphuric acid. Warm the mixture, with stirring, until the crystals are dissolved. Filter hot into a second 250 ml. beaker, cover with a watch glass, and allow to stand for a week. Pour the liquid off the crystals and dry them between filter papers. Transfer to a test-tube, stopper, label, and hand in. Compare their colour with that of copper sulphate pentahydrate.

4. Potassium Cupric Chloride, K2CuCI,. 2 H20 Of the preparations in this experiment the following is the most difficult, but is included in order to illustrate the conversion of a double salt to a complex salt, and the interrelationship of the two. Weigh out 10 g. of cupric oxide and add to 50 ml. of 6 M hydrochloric acid (1 :1 dilution of the concentrated acid). Heat to dissolve. In a second beaker add 15 g. of potassium chloride to 30 ml. of water, and heat to dissolve. Mix the two solutions and set the resulting solution away to evaporate slowly at room temperature. It may be necessary to push the crust of salt crystals that forms around the edge of the liquid into the solution from time to time. When about 10 ml. of liquid remains, filter the mixture, and immediately dry the crystals by pressing between sheets of dry filter paper. Place the dried crystals in a test-tube, rejecting any that are brown, and stopper to keep these from effiorescing. The compound sought in this preparation is a clear turquoise blue in colour. It is not deliquescent, unlike cupric chloride, CuC'2. 2 H20. However, when heated, or exposed to dry air it will lose water and turn brown. The brown material is a mixture of potassium chloride (white) and potassium cuprichloride, KCuCh, (reddish-brown). The latter compound is representative of complex salts, and examples of these are prepared and studied in Experiment 6. Select one or two blue crystals, and heat them, in a Pyrex test-tube, to the point of decomposition to this brown mixture. 5. Magnesium Ammonium Phosphate, MgNH4PQ4.6 H20 To 25 ml. of 0.4 M magnesium chloride solution in a 100 ml. beaker add 1 ml. of dilute hydrochloric acid and 20 ml. of a 0.2 M solution of disodium hydrogen phosphate, Na2HP0 4. Now add slowly, with stirring, sufficient ammonia solution to yield a final solution which tests alkaline to litmus, and then 5 ml. more. Cool the beaker containing this mixture in a dish of cold water or, better, ice water for about an hour, filter, and transfer the solid to the filter paper. Wash with 10 ml. of a solution prepared by tenfold dilution of the reagent

38

ammonia solution, then twice with 10 ml. of 95% ethyl alcohol, and once with 10 ml. of ether. CAUTION: NO FLAMES NEAR. Describe the product formed. To show that the crystals contain the ammonium ion, transfer a little to a Pyrex test-tube, add two spatula points of solid calcium hydroxide, and warm. Demonstrate the production of ammonia gas by (i) moistened red litmus paper, (ii) a glass rod which has been dipped in cone. hydrochloric acid. EXPLANATIONS

The name alum is commonly applied to the salts prepared in parts 1 and 2. It is used in a generic sense for double salts of the type XY(SO 4)2.12 H 2O where Xis a univalent ion like K+, Na+, NH 4+; and Y is a tervalent ion like Al+++, Fe+++, Cr+++. Common alum is usually the potassium aluminium salt; when aluminium is replaced by iron we have ferric alum, and when replaced by chromium we have chrome alum. These compounds are all isomorphous; that is, their crystals are almost identical, and a supersaturated solution of one alum can be seeded successfully by a crystal of another. In solution the crystal forms the ions x+, y+++, and sO4-. The dissolving of aluminium in potassium hydroxide solution is explained on the basis of aluminium hydroxide being amphoteric. Thus aluminium hydroxide (also zinc hydroxide) will dissolve in solutions containing a high concentration of hydroxide ions. The phenomenon is discussed in Experiment 6. Upon withdrawal of hydroxide ions from such a solution aluminium hydroxide is precipitated. Al(OH).HaO+ -Al(OH)a 2 H2O.

+

+

In part 2, dichromate is reduced by alcohol to yield chromic ions:

Cr201-

+ 3 C2H5OH + 8 HaO+ -

3 CHaCHO

+ 2 Cr+++ + 15 H2O.

You will be quickly aware of the pronounced odour of acetaldehyde which is volatile above 22° C. Heat is evolved as the above reaction takes place, and should be dissipated to avoid the temperature of the solution rising above 40°. The reason for preventing this rise in temperature concerns a complication that may befall the chromic ion. This will be better understood after you have studied Experiment 6; in essence the purple-coloured chromic ion may in warm sulphuric acid solutions be converted to a dark green complex ion containing chromium and sulphate radicals. Cr(H2O)s+++ purple

+ SO4- -

Cr(H2O)GSO,+ green

+ H2O.

From the green solution chrome alum is not readily crystallized; from the purple colour of the crystals it is inferred that the ion Cr(H2O)G+++ is present in these. The compound prepared in part 3 is one of a class of similar double salts known as "picromerites." Another common compound of this group is Mohr's salt, ferrous ammonium sulphate, Fe (NH,)2(SO4)2.6 H2O. The formation of crystals of magnesium ammonium phosphate in part 5

39 required that the solution contain the three ions Mg++, NH 4 + and PO?-. The Mg++ ions are supplied by the magnesium chloride. The NH 4+ ions are produced by the neutralization of hydrochloric acid by ammonia NHa

+ HaO+ --+ NH,+ + H2O.

Because phosphoric acid is a weak acid and dissociates according to the following equations, but only slightly, H2O H2O H2O

+ +

+

HaPO, +=± HaO+ H2PO,- +=± HaO+ HPO,- +=± HaO+

+ +

+

H2PO,HPO,POt-

acidic solutions of phosphates contain scarcely any free phosphate ions, Poi-. However, as phosphate solutions are made alkaline the hydroxide ion causes a shift in the above equilibria in accordance with Le Chatelier's principle, and as a result the solution becomes populated by moderate concentrations of free phosphate ion. Hence, the addition of ammonia to the solution in this preparation serves to make available two of the three ions necessary for the formation of the product. QUESTIONS

1. Outline a scheme for the preparation of ferrous ammonium sulphate crystals, Fe(NH,h(SO4)2,6 H2O, starting with pure iron wire. 2. Outline a scheme for the preparation of ferric alum crystals, KFe(SO4)2,12 H2O, starting with pure iron wire. 3. Outline a scheme for the preparation of aluminium sulphate crystals, Al2(SO,)a.l8 H2O, from common alum. 4. If you had no copper oxide, but plenty of pure metallic copper, how could you modify the procedure of part 4 in order to prepare K 2CuCl 4 .2 H2O? 5. Name the ions present, and state their approximate molar concentrations, in solutions prepared by dissolving 0.1 gram-formula weight of each of the following compounds in water sufficient to make 1 litre of solution: (a) common alum, (b) copper ammonium sulphate, (c) potassium cupric chloride, (d) magnesium bismuth nitrate, 3 Mg(NOah,2 Bi(NOa)a.24 H2O,

6.

COMPLEX IONS INTRODUCTION

In this experiment you will encounter numerous reactions which require for their explanation the formation of what are called complex ions; there are also given preparations of several crystalline compounds in which these complex ions are believed to be present. It is difficult to lay down a hard and fast rule to distinguish what are usually called complex ions from simple ions, and indeed not all authors are agreed upon what constitutes a complex ion. Perhaps the best distinction is to say that simple ions comprise

39 required that the solution contain the three ions Mg++, NH 4 + and PO?-. The Mg++ ions are supplied by the magnesium chloride. The NH 4+ ions are produced by the neutralization of hydrochloric acid by ammonia NHa

+ HaO+ --+ NH,+ + H2O.

Because phosphoric acid is a weak acid and dissociates according to the following equations, but only slightly, H2O H2O H2O

+ +

+

HaPO, +=± HaO+ H2PO,- +=± HaO+ HPO,- +=± HaO+

+ +

+

H2PO,HPO,POt-

acidic solutions of phosphates contain scarcely any free phosphate ions, Poi-. However, as phosphate solutions are made alkaline the hydroxide ion causes a shift in the above equilibria in accordance with Le Chatelier's principle, and as a result the solution becomes populated by moderate concentrations of free phosphate ion. Hence, the addition of ammonia to the solution in this preparation serves to make available two of the three ions necessary for the formation of the product. QUESTIONS

1. Outline a scheme for the preparation of ferrous ammonium sulphate crystals, Fe(NH,h(SO4)2,6 H2O, starting with pure iron wire. 2. Outline a scheme for the preparation of ferric alum crystals, KFe(SO4)2,12 H2O, starting with pure iron wire. 3. Outline a scheme for the preparation of aluminium sulphate crystals, Al2(SO,)a.l8 H2O, from common alum. 4. If you had no copper oxide, but plenty of pure metallic copper, how could you modify the procedure of part 4 in order to prepare K 2CuCl 4 .2 H2O? 5. Name the ions present, and state their approximate molar concentrations, in solutions prepared by dissolving 0.1 gram-formula weight of each of the following compounds in water sufficient to make 1 litre of solution: (a) common alum, (b) copper ammonium sulphate, (c) potassium cupric chloride, (d) magnesium bismuth nitrate, 3 Mg(NOah,2 Bi(NOa)a.24 H2O,

6.

COMPLEX IONS INTRODUCTION

In this experiment you will encounter numerous reactions which require for their explanation the formation of what are called complex ions; there are also given preparations of several crystalline compounds in which these complex ions are believed to be present. It is difficult to lay down a hard and fast rule to distinguish what are usually called complex ions from simple ions, and indeed not all authors are agreed upon what constitutes a complex ion. Perhaps the best distinction is to say that simple ions comprise

40 1. Ions containing only one atom; for example, Na+, Cl-, S-, etc. 2. Ions containing several atoms held together sufficiently strongly that these have virtually no tendency to break up into simpler ions; for example, SO,-, NOa-, CO3- , OH-, CN-, NH,+, etc. These ions are sometimes called compound ions to distinguish them from simpler monatomic ions. Compound ions can be broken down chemically by oxidation or reduction, or in some cases by hydrolysis, but they never undergo decomposition into simpler ions. For instance, the following reactions do not occur to a detectable extent:

SO,- - Ss+ + 4OOH- -o- + H+ CO3- - CH+ 3 O-

etc.

By contrast with this behaviour, other ions are known which can be formed from or break up into simpler constituents quite readily. These are known as complex ions, and examples of these include the following: FeCl,µFe++++ 4 Cl-. FeCNS++ µ Fe+++ + CNS-. HgI,µ Hg++ + 4 I-. The simpler constituents need not both be ions; one can be a neutral molecule. For example: The distinction between a compound ion and a complex ion is sometimes a little vague. It may become clearer to you, from a practical standpoint, after you have encountered a variety of complex ions in the experimental work that follows. EXPERIMENTAL

1.

Hydrates

(a) Place 2 or 3 moderate sized crystals of copper sulphate pentahydrate in a Pyrex test-tube and heat in a Bunsen flame until all the water has been removed. Note any change of appearance. Allow the tube to cool. Then add 2 or 3 drops of water to the residue in the test-tube. Hold the bottom of the tube to feel any temperature change. (b) Repeat with cobalt (11) chloride hexahydrate.

2. Ammines (a) Take 5 ml. of 0.1 M silver nitrate in a test-tube and add 1 drop of aqueous ammonia. Mix the contents of the tube and observe any precipitate. Now add ammonia drop by drop until the solution is clear, and then add about 2-3 ml. more ammonia. Divide the solution into two parts and to the first add 1 or 2 ml. of 0.1 M sodium chloride (or any other dilute chloride) solution. To the second add 1 or 2 ml. of 0.1 M potassium iodide solution. Note and account for any precipitate formed, and for the failure of any expected precipitate to appear. (b) Take 5 ml. of 0.1 M copper sulphate solution, and add aqueous ammonia gradually until the solution is clear. Now add 1-2 ml. of 0.1 M sodium sulphide

41

solution. Note and account for any precipitates formed or any changes in colour of the solution. (c) Repeat (b), using 5 ml. of 0.1 M zinc nitrate solution. (d) Take 5 ml. of 0.1 M cobalt (II) chloride solution. Add about 1 g. of solid ammonium chloride, and then aqueous ammonia until the solution is clear when mixed by shaking. Allow to stand in air for some time. Then add 1 or 2 ml. of sodium carbonate solution. Compare the result to that observed when sodium carbonate is added to the original cobalt chloride solution. Note any colour changes or precipitates. (e) Take 5 ml. of 0.1 M magnesium nitrate solution. Add 10 to 15 ml. of aqueous ammonia. Shake and note the appearance of the mixture. (f) Preparation of nickel hexammine perchlorate. The object of this and the next two parts is to show that these ammines are perfectly definite compounds, which may be isolated as crystalline solids in the same way as any other salt. Dissolve 2 g. of nickel chloride hexahydrate in 10 ml. water in a 100 ml. beaker. Add 30 ml. of aqueous ammonia. If the solution is not clear filter it. Now add 30 ml. of a 10% solution of ammonium perchlorate. A blue precipitate of nickel hexammine perchlorate, Ni(NH 3) 8++(Cl04-)2, will form. This can be crystallized from hot water as follows. Heat the mixture fairly slowly until it is just beginning to boil. Stir while heating to prevent bumping. When it is boiling very gently the nickel hexammine perchlorate should have dissolved. Remove the beaker to the bench, remove the stirrer, cover with a watch glass, and allow to cool. When quite cold, filter, and wash the crystals with two successive portions of 10 ml. of cold water containing 1 or 2 ml. of reagent ammonia solution. Then wash with 10 ml. of alcohol. Dry the product by squeezing between dry filter papers, and finally let it stand in the open to dry completely. When dry, transfer the product to a test-tube provided with a cork, label, and hand in to the demonstrator. (g) Preparation of copper tetrammine sulphate. Weigh out 4 g. of copper sulphate pentahydrate and powder it in a mortar. Add 15 ml. of aqueous ammonia to the powdered salt in a small beaker, and stir well. Notice whether the solution changes temperature. The copper sulphate should dissolve completely, or very nearly. Filter, and catch the filtrate in a 100 ml. beaker. Pour in very slowly 20 to 25 ml. of alcohol. Allow to stand for a short time; stir occasionally. Filter the mixture, and wash the crystals twice with 10 ml. of alcohol containing 1 ml. of reagent ammonia solution. Finally wash with 5 ml. of ether (No FLAMES NEAR), and immediately press between dry filter papers to dry. When nearly dried the cake should be broken up on a watch glass and allowed to dry further in air for a few minutes. Transfer the product to a test-tube provided with a cork, label, and hand in to the demonstrator. It is Cu(NH3)4S04. H20. (h) Preparation of cobaltic hexammine chloride. Place together in a small flask 6 g. of cobalt chloride hexahydrate, 4 g. of ammonium chloride, and 8 ml. of distilled water and warm to dissolve the salts. Add through a paper cone (to avoid contact with the neck of the flask) 0.5 g. of activated charcoal powder. Cool the solution in the flask in running tap water and add 12 ml. of concen-

42 trated aqueous ammonia. Being sure that the solution in the flask is still cold, add in small portions 12 ml. of 6% hydrogen peroxide solution, shaking the flask to agitate its contents between each addition. When all the peroxide has been added, warm the mixture to about 60° (as hot as you can bear with the hand) and maintain it at this temperature until the solution in the flask no longer appears pink. Remove the flask from the heat, cool under the tap and then in an ice-bath; during this cooling crystals are deposited. Filter the mixture when it is apparent that no more solid is separating, allowing the mother liquor to drain from the solid. Then transfer the damp solid to a 100 ml. beaker in which you have heated 50 ml. of water containing 2 ml. of concentrated hydrochloric acid. The cobalt salt should all dissolve, leaving the charcoal undissolved. Filter the hot solution, and collect the filtrate in a clean small beaker or flask. Add to the filtrate 25 ml. of concentrated hydrochloric acid, and then set to cool in an icebath. The product separates as golden-yellow crystals. Filter and wash with a little alcohol. Set the crystals to dry in air on a watch glass. Then weigh your product, and calculate the yield based on the cobalt chloride taken. Label and hand in. (i) Determination of composition of a complex ion. The following procedure enables the number of molecules of ammonia attached to the silver or copper ion to be measured. (i) Pipette into a small beaker 10 ml. of 0.1 M silver nitrate. Fill your burette with 0.2 M ammonia solution, and note the volume in your burette. Run in ammonia to the beaker with stirring until the precipitate, which first forms, just redissolves. Note the volume of ammonia used, and hence calculate the formula of the silver ammonia complex. (ii) Repeat using 0.1 M copper sulphate. Somewhat better results are obtained if a small amount of sodium carbonate is added before beginning the titration; do not add more than what can be carried on the tip of a wire spatula. The beaker should be observed in a good light, preferably against a dark background.

3. Halogen Complexes (a) Dissolve about 2 g. of copper (II) chloride dihydrate crystals in 2 ml. of water in each of two test-tubes. Dilute one solution with 20 ml. of water, and compare the colours of the solutions in the two tubes. Look through the solution lengthwise of the tube with a piece of white paper as background. To the undiluted solution add 4 ml. of concentrated hydrochloric acid. Note the colour. (b) Dissolve 2 g. of copper sulphate (or copper nitrate) in 6 ml. of water in a test-tube. Pour half of this solution into another test-tube, and into one of the tubes place solid sodium bromide up to a total of 2 g. Do not add the sodium bromide all at once, and dissolve the added salt as you go along. Compare the colours of the two solutions. Warm the tube if necessary in order to dissolve the last crystals of sodium bromide. Dilute the solution containing sodium bromide with 20 ml. of water and note the colour change. (c) Dissolve 1 g. of cobalt (II) chloride hexahydrate in 10 ml. of water, and pour half of the solution into a second test-tube. Heat one tube nearly to boiling and compare the colours of the hot and cold solutions. Cool the hot test-tube somewhat and add gradually 5 ml. of concentrated hydrochloric acid. Mix and

43

note the colour. Add 5 ml. of water to this solution, mix, and divide equally between two test-tubes. The colour should be pink with a purplish tinge. Heat one of these tubes nearly to boiling and compare the colour of the solution in it with that in the other tube. (d) To 5 ml. of 0.1 M ferric chloride solution in a test-tube add 0.2 M potassium fluoride solution until no further colour change is visible. Add a few drops of potassium thiocyanate solution, and note the resulting colour. Compare with the results of part 5 (a). Continue to add potassium fluoride solution up to a total of 10 ml. (e) Preparation of cuprous iodomercurate (Cu2HgJ4). Weigh out 0.5 g. of mercuric chloride, and dissolve in 20 ml. of water in a 50-ml. beaker. Add very gradually a solution made by dissolving 1½ g. potassium iodide in a little water. Note that the precipitate that first forms (cf. Experiment 2, part 3) redissolves in excess iodide solution. The solution should be clear; if any appreciable amount of mercuric iodide fails to redissolve upon stirring, add an extra crystal or two of potassium iodide. Do not, however, delay in doing this. The clear solution contains the iodomercurate ion, HgI 4- . In another 50 ml. beaker prepare a solution of 1 g. of copper nitrate trihydrate and 1 g. of sodium sulphite in 10 ml. of water. Add to this concentrated hydrochloric acid drop by drop until a clear solution results. This solution contains cuprous copper in the form of the chlorocuprite ion, CuCl2- (compare with part 3 (a)). Pour the second solution into the first, and allow the mixture to stand for a few minutes. Filter off the precipitate; wash it three times with 10-ml. portions of water, and finally with 10-15 ml. of alcohol. Dry the precipitate by opening out the filter paper on a watch glass, and placing this on a beaker of boiling water (boil about 50 ml. of water in a 250 ml. beaker). When the product is dry, break up any lumps, transfer it to a small test-tube provided with a cork and label, and hand it in. Place the loosely corked tube containing the copper (I) iodomercurate (II) in a 50 ml. beaker of water. The water surrounds the tube, but none should enter it. Heat the water in the beaker to boiling; notice any colour change. Allow to cool and note the colour. Place a small amount, about 0.1 g., of the copper (I) iodomercurate at the bottom of a 6-inch Pyrex test-tube closed with a plug of cotton wool. Hold the tube in a clamp in a nearly horizontal position, and heat the bottom in a Bunsen flame until yellow or red crystals appear in the cool part of the tube. Heat about¼ minute longer, then allow to cool. Where yellow crystals have formed on the side of the tube, press or scratch them with a glass rod. Note any change of colour. (f) As an alternative to the above preparation, the silver salt may be prepared. Add 4-5 ml. of 1.0 M silver nitrate to the solution prepared as in paragraph 1. Filter and wash the yellow precipitate in the same manner for the cuprous salt. This compound changes sharply on heating to an orange-red form.

4. Hydroxy Complexes The amphoterism of certain metallic hydroxides can be explained by the formation of complex ions of the type M (OH)z:-.

44 Set up four test-tubes; and add to the first 5 ml. of 0.1 M zinc nitrate, to the second 5 ml. of 0.1 M lead nitrate, to the third 5 ml. of 0.1 M aluminium nitrate, and to the fourth 5 ml. of 0.1 M magnesium nitrate. Add reagent sodium hydroxide gradually to each until no further change occurs. The removal of the simple ions to form a complex ion can be demonstrated in the case of lead by the following experiment. In a test-tube mix equal amounts (about 2 ml. each) of 0.1 M lead nitrate and 0.1 M sodium chromate. Run in reagent sodium hydroxide (1.7 M) with shaking and observe how much lead chromate remains undissolved. 5. Thiocyanate Complexes (a) To 5 ml. of 0.1 M ferric chloride solution add 1-2 ml. of potassium thiocyanate solution. Account for the colour produced. This reaction is studied in detail in Experiment 9. (b) To 2-3 ml. of 0.1 M cobalt nitrate solution add 5 ml. of potassium thiocyanate solution. Compare the colour of the solution with that of the original cobalt nitrate solution. Now add an equal volume of acetone to the solution containing cobalt and thiocyanate. Observe the colour in solution. Explain your observations. (c) Preparation of cobalt mercurithiocyanate (CoHg(CNS) 4). In some of its reactions the thiocyanate ion resembles the iodide ion. This preparation closely resembles that of cuprous iodomercurate (part 3 (e)). Weigh out 1 g. of mercuric nitrate and dissolve in 20 ml. of distilled water to which 2 drops of dilute nitric acid have been added. To this solution add 4 ml. of 2 M potassium thiocyanate solution. Identify the precipitate formed. Continue to add potassium thiocyanate solution until this precipitate just dissolves; the resulting solution contains the mercurithiocyanate (or thiocyanatomercurate) ion. Heat this solution to boiling and add to it 10 ml. of 0.25 M cobalt nitrate solution. Cool to room temperature. Filter, and wash the precipitate with three 10-ml. portions of distilled water followed by one 10-ml. portion of ethyl alcohol. Spread the crystals to dry on a large watch glass. Label and hand in your preparation as directed.

6. Cyano Complexes The preparation of cyanide complexes requires the use of potassium cyanide solutions with their attendant toxic hazard. Because of this hazard, certain preparations will be demonstrated only. (a) The formation of cyanonickelate ion, Ni(CN) 4- - , and cyanocuprite, Cu(CN)!-, will be demonstrated by the instructor. (b) To show the stability of the complex ferricyanide ion, Fe(CN)t, test for the presence of free ferric ion in a solution of potassium ferricyanide. Add 1-2 ml. of potassium thiocyanate solution to 5 ml. of 0.1 M potassium ferricyanide solution. Compare the colour of the solution with that observed for ferric thiocyanate (part 5 (a)).

45 7. Nitrite Complex Preparation of sodium cobaltinitrite. Weigh out 2 g. of cobalt chloride hexahydrate, 6 g. of sodium nitrite, and add them to 7 ml. of water in a small beaker. Warm gently until everything is dissolved. Allow to cool somewhat and add 5 ml. of 30% acetic acid. Allow to stand with occasional stirring until no more gas is evolved; avoid breathing this gas (nitrogen dioxide, NO 2) as it is poisonous. Add 15-20 ml. of 95% alcohol, and allow to stand for½ hour (if possible). Filter the mixture (do not throw away the filtrate), and wash the precipitate with a further 20 ml. of alcohol. Dry the precipitate by warming on a watch glass as in part 3 (e). When it is dry, break up any lumps, transfer the product to a small test-tube provided with a cork and label, and hand it in. The formula of the product is NaaCo(NO2)&. Place 5 ml. of the filtrate obtained above, which contains some sodium cobaltinitrite, in each of two test-tubes, and add 1-2 ml. of a solution of potassium chloride to one tube, and ammonium chloride to the other.

EXPLANATIONS

To the remarks made in the introduction concerning complex and compound ions, and the distinction between them, we may add the following. The stability of complex ions towards decomposition into constituent parts may be quite high, as with ferricyanide, for instance, or quite low, as with cobaltothiocyanate. We are scarcely justified in using stability alone as a criterion for complex ions. Another feature which may be helpful in defining a complex ion is that they can usually be prepared from or dissociate into ions or molecules which themselves are stable in aqueous solution. Thus an ion like iodomercurate is derived from mercuric and iodide ions; these are common stable ions. In spite of its great stability the ferricyanide ion can be related through its method of preparation into the simple common ions ferric and cyanide. On the other hand, decomposition of compound ions does not occur, but even if it did the fragments would not be ordinarily encountered ions stable in aqueous chemistry. No very simple rules can be laid down to decide just which complex ions will be formed, or how stable they will be. However, as the bonds in a complex ion are (usually) essentially covalent, it follows that, on the whole, they are formed most strongly by ions which most easily give covalent bonds, that is, ions from weak acids or weak bases. Thus the iron-cyanide complex (from the weak acid HCN) is stronger than the iron chloride complex (from the strong acid HCl); similarly the Na+ ion (from the strong base NaOH) gives very few complex ions. However, at best this is only a rough guide, and other factors can be important. As regards the formulae of the complex ions, the central atom in the complex is usually a metal atom (+charge) attached to several negative atoms or neutral groups or both. The number of groups so attached does not remain constant. The following notes refer to the specific experiments of corresponding numbering. 1. Many salts crystallize with water bound in the crystal. The mode of binding may vary, but in some cases we can distinguish definite units consisting of an

46 ion (usually positively charged) attached to several water molecules, and there is reason to suppose that these complex ions persist in solution. A typical example is the tetrahydrated copper (II) ion

The symbol ---t indicates that both binding electrons come from the same atom. This ion is present in copper sulphate pentahydrate; the water molecules lie at the corners of a square around the copper atom; the fifth molecule of water in CuSO 4 • 5 H 2O is bound differently. Similarly in cobalt chloride hexahydrate we have the ion ++

Nickel gives a similar ion Ni(H2O) 8++. It will be found that these ions are stable enough to evolve considerable heat when they are formed from the anhydrous salt and water. Other elements are known which give fairly definite hydrated ions, most frequently with 4 or 6 molecules of water attached to the central atom. It should be emphasized that what is usually called a copper ion Cu++ is really this complex ion Cu(H 2O) 4++; similarly, the cobalt ion in solution is really the Co(H2O)s++ ion. In fact it may generally be considered that any positive ions in solution are attached more or less loosely to some water molecules. These hydrated ions are considered complex rather than simple ions because, generally speaking, the water is sufficiently loosely held that it can be removed from the crystal on moderate heating (though this may lead to complete decomposition). Also, anhydrous salts can frequently be precipitated from solution. For instance, sodium sulphide precipitates CuS from copper solutions; in this case the water is separated from the copper atom and its bonds replaced by bonds to sulphur atoms. On the whole hydrates are usually rather unstable complexes. It should be added that, though these hydrates have been written with definite covalent bonds, some at least are held together by forces which, properly speaking, are not chemical bonds at all. 2. The ammonia molecule also is capable of attaching itself to a metal atom. The resulting complex, which contains the group M +-- NH 3 (M is a metal atom) is called an ammine. Thus, on adding ammonia to copper sulphate solution, the reaction occurs

47 HaN HaN -

! Cu i

++

NH 3

+ 4 H20,

NHa

pale blue

deep blue copper tetrammine ion

The copper tetrammine ion is the more stable so that the equilibrium is over to the right in spite of the larger number (concentration) of water molecules than of ammonia molecules present. It is usually, though not always, true that the ammine complex ion is more stable than the water complex ion. Ammines with different numbers of ammonia molecules are known. Their formulae can be found in various ways of which one of the simplest is illustrated in part 2(i). Small amounts of ammonia when added to silver nitrate solution, precipitate silver oxide 2 AgNOa + 2 NHa + H20 -

Ag20 + 2 NH4NOa.

As more ammonia is added this redissolves as the complex ammine, so that the total reaction is AgNOa + xNHa - Ag (NHa)xNOa. The object of the titration in this part is to find the value of x for silver nitrate, or copper sulphate. The end-point in the titration with copper sulphate is harder to fix, but corresponds reasonably closely with the composition of the complex shown above. The addition of sodium carbonate suggested in the experimental directions maintains the solution slightly alkaline, and ensures that no significant part of the added ammonia is converted to ammonium ions. Other ammine ions formed in part 2 are: Zn (NHa),++ Co (NHa)e+++ Ni (NHa)e++

zinc tetrammine ion cobalt (III) hexammine ion nickel hexammine ion

Note that with cobalt, the cobalt (II) hexammine ion is first formed, but this is oxidized by air or hydrogen peroxide to the ion derived from tervalent cobalt. 4 Co (NHa)e++ + 4 NH,++ 2 H20 + 02 - 4 Co (NHa)s+++ + 4 NH,OH. There is always some degree of reversibility in the reactions which give these ammines. Thus with copper there will be a small concentration of uncomplexed copper ions (that is, the simple hydrated ion) in equilibrium with the copper tetrammine ion; and these ions are enough to give a precipitate of the very insoluble copper sulphide (CuS) with sodium sulphide. Similarly with silver it will be found that at the concentrations used in this experiment, there are not enough simple silver ions left to precipitate silver chloride, whose saturated solution in water is 0.00001 M; but there are enough to precipitate silver iodide, whose saturated solution is 0.00000001 M.

48 It is often found, as might be expected, that the number of ammonia molecules that can be attached to a metal is the same as the number of water molecules that can be attached. This is not always so. 3. Just as NH 3 molecules can replace water attached to copper, so can chloride or bromide ions (parts (a) and (b)). The equilibria involved are HaN

! HaN-tCu+-NHa i

++

NHa

Cl

I I

Cl-Cu -Cl Cl yellowgreen

green

Copper chloride crystals are the non-ionic (uncharged) substance CuCl2.2H 20 in the middle of the above equilibria. The different complex substances above can be detected by their different colours in solution. The copper bromide complexes are similar, though differently coloured. Cobalt gives a somewhat different equilibrium. It is found with relatively few exceptions that when a cobalt (II) atom is surrounded by 6 other atoms it is pink, and when surrounded by 4 it is blue. The blue is a much more intense colour than the pink. In these experiments we have HCl Co(H20)e++ (Cl-)2 ~ (Ha0+)2 C0Cl,H20 pink blue Other examples are: Co(NHa)&++ pink; C0Br4 - blue. The pale blue CoCb obtained by heating CoCb.6 H20 is an exception as the Co and Cl atoms are arranged in the crystal so that 6 Cl atoms surround every Co atom. In (d) we have a complex between ferric ions and fluoride ions, whose formula in dilute solution is FeF++ Fe+++

+ F- µ

FeF++.

Presumably water molecules are also attached. In more concentrated solutions complexes with more fluoride ions are obtained, and eventually a solid such as N a 3FeF 6 can be isolated. Potassium thiocyanate is used in this part to test for free ferric ions; if they are present a red colour will result. When potassium iodide was added to a mercury (II) salt in Experiment 2, a red precipitate was formed. This is Hgl2, and it is made up of immense sheets of atoms arranged (not quite flat) thus

49

I

I

I

I

I I I I Hg-I-Hg-I-H g-I-Hg-I I I I I I I I I I I I I Hg-I-Hg-I-H g-I-Hg-I I I I I I I I I I I I I Hg-I-Hg-I-H g-I-Hg-I I I I I I I I I I I I I

Hg-I-Hg-I-H g-I-Hg-I

Half of the bonds that hold this structure together are supposed to be formed by sharing a pair of electrons, one of which is from the mercury and one from the iodine. In the other half of the bonds, both electrons come from the iodine. It can be seen that HgI4 groups occur as a basis of the structure, and they are held together because the iodine atoms have to be shared. When more iodide ions are added, we get separate HgI4-- groups Hgh

+ 2 KI -

(K+)2 HgI4--.

This is potassium iodomercurate, which is soluble, but the cuprous or silver salts are not. When copper (II) nitrate and sodium sulphite are mixed together with a little hydrochloric acid, a precipitate of cuprous chloride is formed:

2 Cu(NOah

+ Na2SOa + 2 HCl + H2O -

2 CuCl

+ Na2SO4 + 4 HNOa.

The cuprous chloride dissolves in more HCl as the complex ion, CuCb-. On mixing cuprous chloride and potassium iodomercurate a precipitate of cuprous iodomercurate is formed. Generally speaking, cuprous salts are unstable unless the copper is combined in some, probably unionized, compound such as Cu2HgI4 because in solution they largely decompose thus: 2 Cu+ -

Cu++

+ Cu (metal).

On warming, the Cu2HgI4 changes colour. This is because the atoms rearrange themselves in a different pattern in the crystal. Many other substances give more than one sort of crystal for the same reason. The change from one form to the other occurs at a definite temperature (71 ° C in this case), just as the change to a liquid, that is, melting, occurs at a definite temperature. On stronger heating Cu2HgI4 decomposes Cu2HgI4 -

2 Cul

+ Hgh.

The mercuric iodide distils up the tube (it boils at 354°C) and forms red or yellow crystals on the cold walls of the tube. The red and yellow forms are also due to

50 different atomic patterns in the crystal. The yellow is stable above 127° C, the red below; but the yellow form can be obtained at room temperature though it is unstable. If it is scratched or pressed it turns into the red form. 4. A number of hydroxides which are precipitated by sodium hydroxide dissolve in excess of this reagent, owing to the formation of complex hydroxy anions. For instance zinc hydroxide dissolves to give sodium zincate. Zn(OH)2 + 2 NaOH -

(Na+)2 Zn(OH),-.

Lead nitrate similarly gives first lead hydroxide, then sodium plumbite Na2Pb(OH),. With aluminium nitrate the product is so extremely soluble that it is not easy to separate sodium aluminate from solution for analysis. In solution there is probably an equilibrium involving the ions Al(OH) 4- and AI(OH) 6- 3 : AI(OH)a + OH- µ Al(OH) 4. AI(OH) 4 + 2OH- µ Al(OH) 6- : - The sodium salt of the latter ion has been prepared; it is OHoH -a (Na+)a

I/

HO-AI-OH H0/6H

Compounds of this sort are essentially similar in type to the halogen complexes of part 3; with some elements compounds of the same sort of formula are formed either with a halogen ion or a hydroxide ion: for example, NaaAIFe NaaAI(OH) 6 Na2SnCie Na2Sn(OH) 6

sodium sodium sodium sodium

fluoaluminate aluminate chlorostannate stannate

The formation of compounds like sodium aluminate is sometimes called amphoteric behaviour, implying that the element can act either as a base (to give salts like aluminium chloride or sulphate) or as an acid (to give sodium aluminate). However it is only in a rather artificial sense that aluminium hydroxide acts as an acid, with sodium hydroxide; what is formed is really a complex ion. These compounds differ from well defined oxy-ions such as chromates or permanganates, firstly in their tendency to dissociate into simple ions, and secondly in that they have OH groups and not oxygen atoms round the central metal atom. 5. The thiocyanate ion, CNS-, will combine with a number of metallic ions to give complex ions. These are not particularly stable in some cases, but because of their strong colours they have often been used for the identification of the simple metallic ion from which they are formed. The reaction with ferric ion leads chiefly to a cationic complex: Fe++++ CNS-~ FeCNS++.

In Experiment 9 this reaction is studied more fully. With cobalt ion a series

51 of complexes containing up to 4 thiocyanate ions per atom of cobalt have been found. The formation of the complex ion with 4 thiocyanate ions can be represented thus: Co+++ 4 CNS-;:::= Co(CNS),--. It is believed that this ion is a deep blue colour. The equilibrium in aqueous solution lies rather to the left; that is, only a small fraction of the cobalt forms the blue complex. However, if acetone or some similar substance is added to reduce the ionizing properties of water, the above equilibrium is shifted much more to the right, and the blue colour of the complex is rendered visible. The preparation of cobalt mercurithiocyanate involves the sequence of reactions: Hg++ + 2 CNS---+ Hg(CNS)2 Hg(CNSh + 2 CNS---+ Hg(CNS),Co++ + Hg(CNS),----+ CoHg(CNS) 4 6. Cyano complexes are derived from the cyanide ion, CN-. In spite of the extreme toxicity of the cyanide ion itself, some of these cyano complexes are so stable as to be almost harmless, showing that virtually no free CN- ions are present. However, acids will gradually liberate HCN gas from any of them. If cyanide ion is added to a nickel salt an apple-green precipitate of nickel cyanide is first formed, but this redissolves in excess cyanide to give an ambercoloured solution containing the cyanonickelate ion, Ni(CN).-. The corresponding reaction with copper is more complicated. A white precipitate is formed when cyanide is added to a cupric salt, but this is actually cuprous cyanide which is formed with the evolution of cyanogen gas (compare Cu++ + J-, Experiment 2, part 3) 2 Cu+++ 4 CN----+ 2 CuCN + C2N2.

This precipitate redissolves in excess cyanide to form a colourless solution containing the cyanocuprite ion, Cu(CN)!-. 7. Sodium cobaltinitrite is a good example of the sort of complex salts formed by a weak acid, in this case nitrous acid, HN02. Note that in this reaction cobalt (II) is oxidized to cobalt (III) by nitrous acid as well as forming a complex with it. The acetic acid is used to liberate nitrous acid from sodium nitrite. The over-all reaction is: C0Cl2 + 7 NaN0 2 + 2 CHaCOOH ---+ NaaCo(N02)a + NO + 2 NaCl + 2 CHaCOONa + H20. The nitrite ion, N0 2-, is (0 = N-O)-, and it could be attached to cobalt either through the nitrogen atom thus:

52 or through the oxygen atoms thus: Co-0-N = 0. It is probable that the first formula is correct. The cobaltinitrite ion is remarkable in giving slightly soluble potassium and ammonium salts. QUESTIONS

1. Write detailed structural formulae for the following compounds:

(a) potassium cobaltinitrite; (b) potassium manganicyanide (cyanomanganate (III)); (c) barium zincate; (d) cobalt (III) hexammine nitrate; (e) potassium thiocyanatomercurate. 2. What are the oxidation states of the elements in the following compounds? (a) KaCo(CN)e; (b) K2PtCle; (c) Ni(NHa)e(Cl04)2; (d) (NH4)aCo(N02)e; (e) K~n(OH)e. 3. Explain the difference between copper ammonium sulphate (prepared in Experiment 5) and copper tetrammine sulphate. 4. From observations in this experiment, deduce whether the following reactions evolve or absorb heat: (a) CuS04 (solid) 5 H20 - CuS04. 5 H20 (solid); (b) Co(H20)e++ 4 CI- - CoCI46 H20 (in solution); (c) Cu(H20)4++ 4 NH a - Cu(NHa)4++ 4 H20 (in solution); (d) Hgh (red) - Hgl2 (yellow). 5. Name a metallic element whose hydroxide dissolves in (a) excess ammonia and in excess sodium hydroxide; (b) in excess ammonia but not in excess sodium hydroxide; (c) in sodium hydroxide but not in ammonia; (d) in neither. Using this sort of behaviour how would you separate iron, nickel, and aluminium, starting with a solution of their ions?

+ + +

+

7.

+

SOLUBILITY

INTRODUCTION

The various factors which affect the rate and extent of dissolving of one substance (the solute) in another (the solvent) are examined in this experiment. Some factors may alter the rate of dissolving, but not the final concentration when no more will dissolve; a solution in which no more solute will dissolve is said to be "saturated," and its concentration is called the "solubility" of the solute. For instance, stirring increases the rate of dissolving but not the solubility. In various parts of this experiment the influence of stirring, size of the solute crystals, temperature, and the chemical nature of solute and solvent are examined.

52 or through the oxygen atoms thus: Co-0-N = 0. It is probable that the first formula is correct. The cobaltinitrite ion is remarkable in giving slightly soluble potassium and ammonium salts. QUESTIONS

1. Write detailed structural formulae for the following compounds:

(a) potassium cobaltinitrite; (b) potassium manganicyanide (cyanomanganate (III)); (c) barium zincate; (d) cobalt (III) hexammine nitrate; (e) potassium thiocyanatomercurate. 2. What are the oxidation states of the elements in the following compounds? (a) KaCo(CN)e; (b) K2PtCle; (c) Ni(NHa)e(Cl04)2; (d) (NH4)aCo(N02)e; (e) K~n(OH)e. 3. Explain the difference between copper ammonium sulphate (prepared in Experiment 5) and copper tetrammine sulphate. 4. From observations in this experiment, deduce whether the following reactions evolve or absorb heat: (a) CuS04 (solid) 5 H20 - CuS04. 5 H20 (solid); (b) Co(H20)e++ 4 CI- - CoCI46 H20 (in solution); (c) Cu(H20)4++ 4 NH a - Cu(NHa)4++ 4 H20 (in solution); (d) Hgh (red) - Hgl2 (yellow). 5. Name a metallic element whose hydroxide dissolves in (a) excess ammonia and in excess sodium hydroxide; (b) in excess ammonia but not in excess sodium hydroxide; (c) in sodium hydroxide but not in ammonia; (d) in neither. Using this sort of behaviour how would you separate iron, nickel, and aluminium, starting with a solution of their ions?

+ + +

+

7.

+

SOLUBILITY

INTRODUCTION

The various factors which affect the rate and extent of dissolving of one substance (the solute) in another (the solvent) are examined in this experiment. Some factors may alter the rate of dissolving, but not the final concentration when no more will dissolve; a solution in which no more solute will dissolve is said to be "saturated," and its concentration is called the "solubility" of the solute. For instance, stirring increases the rate of dissolving but not the solubility. In various parts of this experiment the influence of stirring, size of the solute crystals, temperature, and the chemical nature of solute and solvent are examined.

53 It is possible to prepare a solution that is more concentrated than a saturated one; such a solution is said to be supersaturated. A supersaturated solution will usually deposit crystals of solute easily, and so reduce its concentration to that of a saturated solution. Some of the properties of supersaturated solutions are examined in this experiment. A saturated solution is in equilibrium with the solute; hence the solute will dissolve in an unsaturated solution until it becomes saturated, or crystals will form in a supersaturated solution until it also becomes saturated. The same equilibrium concentration is reached from either side. EXPERIMENTAL

1. Rate of Dissolving

Select four crystals of disodium phosphate, Na2HP0 4 .12 H 20. These should be of about the same size, and not too small. Crush two of the crystals separately in a mortar. Put 10 ml. of water in each of four test-tubes, and add the crystals so as to obtain the following combinations: (a) large crystal, cold water; (b) powdered crystal, cold water; (c) powdered crystal, cold water, the tube being well shaken; (d) large crystal, hot water. Compare the times required for the crystal to dissolve, and hence deduce what conditions increase the rate of dissolving.

2. Effect of Temperature on Solubility (a) Place 10 ml. of water in each of two large (6" X 1") Pyrex test-tubes. Add sodium chloride to one tube, and sodium nitrate to the other, and place them in a large (400 or 600 ml.) beaker, half filled with water. Bring the water in the beaker to boiling, and stir the contents of the tubes with stirring rods. Keep the water just boiling, and be sure that its level does not fall below that of the liquid in the test-tubes. Continue to add solid to the two tubes, until some remains undissolved even after 2 to 3 minutes stirring. Grasp the hot tubes by a clamp or piece of paper, and pour off 5 ml. of the clear liquid above the crystals into separate dry test-tubes. The ordinary testtube holds about 20 ml., so the tubes will be one quarter full. Allow the tubes to cool, and note the quantity of crystals that form in each. (b) Put about 5 ml. of sodium chloride in a test-tube, and about 5 ml. of sodium nitrate in another. To each add an equal volume of water at room temperature, and shake well. Note whether there is any marked temperature change as the solids dissolve. Do this either with a thermometer, or by holding the tubes in the hand. Was heat evolved or absorbed in the dissolving process? Relate the results of (a) and (b) above by means of Le Chatelier's principle. Consult textbooks for the "Solubility Curves" of these two salts in water.

3.

Supersaturation

(a) Place about 5 ml. of sodium thiosulphate pentahydrate crystals, Na2S20 3.5 H20, in each of two test-tubes, and stopper the tubes loosely with cotton wool. Heat the tubes cautiously until all the crystals melt, and then set

54 the tubes aside until they are completely cool. To one tube add a very small crystal of sodium thiosulphate; note any temperature change by holding the tube in the hand. To the other tube add a small (not more than 1 mm. long) crystal of copper sulphate; as before note any temperature change. Note whether crystals form in the tubes. Is the crystallization of sodium thiosulphate endothermic or exothermic? (b) Place 10 ml. of 0.2 M calcium chloride in each of two test-tubes, and add to each 10 drops of dilute sulphuric acid. Shake one tube very gently to mix the reagents, and then let it stand undisturbed. Shake the other tube vigorously. Note whether a precipitate forms, and compare the rates of formation of the precipitate in the two tubes. What is the precipitate? What is the slow process in the formation of this precipitate?

4.

Saturation as an Equilibrium

In each of the following three parts a cold saturated solution is prepared, (i) from a cold unsaturated solution, and (ii) from a hot saturated solution. (a) Put 50 ml. of water in a 100 ml. beaker, and warm the water until it is hot, but not uncomfortably hot, to the hand (45-50° C). Remove the Bunsen flame, and after about one minute, add 10-15 g. of sodium bicarbonate. Stir well for several minutes, and then let the solid settle to the bottom. Pipette out 5 ml. of the clear liquid above the solid, and run it into a 125 ml. conical flask. Add a few drops of "screened" methyl orange indicator, and titrate the solution with 0.2 M hydrochloric acid. Calculate the concentration of the sodium bicarbonate in g./litre. Cover the beaker with a watch glass, and leave till the next laboratory period. Put 50 ml. of cold water in another 100 ml. beaker, and add 10-15 g. of sodium bicarbonate. Note the time of the addition. Stir the mixture for about one minute, and then let the solid settle to the bottom. Pipette out a 5 ml. sample, and analyse it by titration as before. Continue to stir the solution, and take samples at about twenty-minute intervals, until three or four have been taken. Titrate each sample, and note the time when it was taken. Cover the beaker with a watch glass, and leave till the next laboratory period. At the next period, stir both solutions, allow to settle, take a 5 ml. sample from each, and analyse it by titration as before. Note the temperature of the two solutions. What conclusion can you draw from the final concentrations of the two solutions? Save one of the solutions for use in part 5. (b) Follow exactly the experimental directions in part 4(a), but add 15-20 g. of potassium bicarbonate, instead of sodium bicarbonate, both to the hot and to the cold water. In titrating the solution, use 0.3 M hydrochloric acid, instead of 0.2 M. All other details are the same. (c) Put 50 ml. of water in a 100 ml. beaker and warm the water till it is moderately hot (50-60° C). Add 10-15 g. of oxalic acid, and stir well for several minutes. Let the solid settle to the bottom.

55 Pipette out 5 ml. of the clear liquid above the solid, and run it into a 125 ml. conical flask. Add a few drops of phenolphthalein indicator and titrate with 0.3 M sodium hydroxide. Calculate the concentration of oxalic acid in g./litre. Cover the beaker with a watch glass and leave till the next laboratory period. Put 50 ml. of cold water in another beaker, and add 10-15g. of oxalic acid. Note the time of the addition. Stir for about one minute, and let the solid settle. Pipette out a 5 ml. sample, and titrate it with 0.3 M sodium hydroxide as before. Continue to stir the solution, and take samples at about twenty-minute intervals, until three or four have been taken. Titrate each sample, and note the time when it was taken. Cover the beaker with a watch glass, and leave till the next laboratory period. At the next period, stir both solutions, allow to settle, take a 5 ml. sample from each, and analyse it by titration as before. Note the temperature of the two solutions. What conclusion can you draw from the final concentrations of the two solutions? CAUTION: OXALIC ACID IS POISONOUS

5.

Effect of Solvent on Solubility

(a) If part 4 (a) or 4 (b) was done, take the beaker of saturated sodium or potassium bicarbonate solution, and pour some of the clear liquid into a test-tube until it is half full. Add an equal volume of 95% ethyl alcohol, and shake to mix. Let the precipitate settle, pipette out a 5 ml. sample of the clear liquid, and titrate it as in part 4. Calculate the concentration in g./litre. Both these salts are very nearly insoluble in alcohol. Is the solubility in approximately 50% alcohol greater or less than half the solubility in water? (b) Shake 5 ml. of water in a test-tube with sodium chloride until no more dissolves. Pour off some of the solution into another test-tube, and add an equal volume of 95% ethyl alcohol. (c) Dissolve as much benzoic acid, C 8H 6COOH, as could be picked up on a nickel in about 1 ml. of 95% ethyl alcohol. Add 10 ml. of water, and shake. (d) Put 3 ml. each of benzene (CAUTION: INFLAMMABLE) and water in a testtube, and shake well. Note whether two layers form. Now add 3 ml. of ethyl alcohol, shake well, and note the relative volumes of the two layers. Add another 3 ml. of alcohol, shake, and observe the volumes of the layers as before. Continue until only one layer results. Draw conclusions about the mutual solubilities of benzene, alcohol, and water.

6.

Partition Between Two Solvents

(a) Put in a test-tube 5 ml. each of bromine solution and carbon tetrachloride. In a second tube put about 1 ml. of bromine solution, 4 ml. of water, and 5 ml. of carbon tetrachloride. Shake both tubes. Note the distribution of the bromine in the various phases, as judged by its colour. (b) Put 10 ml. of 0.2 M potassium iodide in each of three test-tubes, and add 5 ml. of carbon tetrachloride to each. Add one small (about 1 mm. long) iodine crystal to the first tube; add three such crystals to the second tube, and six to

56 the third. Shake the tubes well. Note the distribution of the iodine in the various phases, as judged by the colour. EXPLANATIONS

The process of dissolving a solid in a liquid involves the separation of molecules or ions from the solid to go into the liquid. In crystallization the molecules of solute separate from the liquid, or solution, to form the solid. These two processes oppose each other; and depending on which is faster, solid will either dissolve or the crystals of solid will grow. When these processes occur at the same rate, a crystal of solute in contact with the solution will stay the same size, and the concentration of the solution will not change. Because of this lack of change, the solution and solute are in equilibrium, and the solution is said to be saturated. The concentration of a given saturated solution is called the "solubility" of that particular solute in that particular solvent. Various conditions will affect either the rate of dissolving, or the solubility, or both. For instance, if the solid is finely ground so that it has a large surface area, this will increase the rate of transfer to the liquid, that is, the rate of dissolving. This will also increase the rate of transfer in the opposite direction, so that the solubility is unchanged. Similarly, stirring increases the rate of dissolving by bringing less concentrated solution in contact with the solid; but when all the solution is saturated, this has no effect, and the solubility is not altered. Higher temperatures, by increasing the thermal motion of the molecules also speed up dissolving, and in general alter the solubility. These effects are demonstrated in the first part of this experiment. The effect of temperature on solubility may be predicted by Le Chatelier's theorem. For an equilibrium: Undissolved solute

~

dissolved solute - heat,

in which heat is absorbed as the solute dissolves, higher temperatures will favour the dissolving process, and the solubility will increase with increasing temperature. In part 2, the heat change when a salt dissolves and the effect of temperature on solubility are observed. Thus these can be related by Le Chatelier's theorem. Strictly speaking, the heat quantity that governs the effect of temperature on solubility is the heat absorbed or evolved when solute dissolves in a nearly saturated solution. This is not necessarily the same in magnitude, or even in sign, as when solute dissolves in pure solvent. However, with sodium nitrate and chloride the effect is that to be expected from the total heat change. Part 4 deals with saturation as an equilibrium. One characteristic of equilibrium is that the same equilibrium situation is reached from either side; in this case, the same equilibrium concentration is reached from more concentrated or less concentrated solutions. A solution that is more concentrated than a saturated one is called supersaturated; it is characterized by the fact that a crystal of solute will grow when placed in it. In the absence of such a nucleus for crystal growth, the supersaturated solution may persist unchanged for some time. This is because a very

57

small crystal, containing a few molecules, is less firmly held together than a large one; hence molecules do not so readily start the formation of a crystal, and it is easier for them to attach themselves to an already sizable crystal. The crystal nucleus, or "seed" crystal should be a crystal of the solute; other substances will usually fail to start crystal growth. There is a rough rule governing the mutual solubility of substances, namely that "like dissolves like." Thus many organic compounds are soluble in organic solvents of somewhat similar formula. There are exceptions to this rule: common salt is soluble in water in spite of a lack of similarity in formula and properties; however here some special reasons apply. Part 5 gives a few examples illustrating ease of dissolving, or the reverse. One additional effect is also encountered here and that is the effect when a mixed solvent is used, for instance sodium bicarbonate dissolving in a mixture of alcohol and water. The solubility in a mixture of 50% water and 50% alcohol is in general between the solubilities in pure water and pure alcohol, but it is not necessarily half-way between. Usually it is less than the average of the solubilities in the pure solvents. If two immiscible solvents are used (for example, water and carbon tetrachloride), and a solute that is soluble in both, a new situation arises. Two liquid layers are formed, and solute is dissolved in both. It is found that, at equilibrium, the ratio of the concentrations in the two layers is constant (or approximately so) for all concentrations; this ratio, of course, varies for different solvents or solutes. The ratio is the same as the ratio of solubilities in the two solvents. In the experiment with iodine (part 6 (b)) a further complication arises, because iodine dissolves in aqueous potassium iodide to form potassium tri-iodide

KI+ h = Kia. Iodine is very slightly soluble in pure water. In carbon tetrachloride it dissolves as iodine molecules, h. However, the ratio of concentrations in the two layers still remains approximately constant, since the total iodine dissolved in the aqueous potassium iodide is proportional to the part present as I2 molecules, and this in turn is in a constant ratio to the concentration of iodine in carbon tetrachloride. Hence there is a definite ratio between the total iodine concentrations in the two layers. QUESTIONS

1. Give definitions of the terms: solvent, phase, solubility. 2. Is it more common for the solubilities of solids in liquids to be increased or decreased when the temperature is raised? Compare the solubilities of gases in liquids. 3. Consult textbooks to discover two solids whose solubility in water decreases with rising temperature. 4. Define a supersaturated solution. 5. Give two general methods of preparing supersaturated solutions of solids in liquids; of gases in liquids. 6. What guess would you make about the solubility (i) of sodium chloride in benzene, (ii) of benzoic acid in benzene. Explain your reasoning.

58 7. What is meant by saying that a saturated solution is in equilibrium with the solute? What experimental evidence supports the view that an equilibrium does exist?

8.

SOLUBILITY PRODUCT INTRODUCTION

When a salt dissolves in water, the solution contains two (or sometimes more) different sorts of ions. If another salt that provides one of the same ions as the original salt is also dissolved in water, it will in general reduce the solubility of the original salt. This phenomenon is sometimes called the "common ion effect." Conversely if some compound is added which combines with, and thereby removes, one of the ions from the original salt, its solubility will be increased. These effects are illustrated qualitatively in the first part of this experiment. The extent to which the solubility changes can be measured, and is found to conform, at least approximately, to a simple rule: for a salt AX, the concentration of A+ multiplied by the concentration of x- is a constant, at any one temperature. The measurements made in the following experiment illustrate this. EXPERIMENTAL

1. Lead Chloride

Put into a test-tube about as much lead chloride as can be picked up on a dime. Add 10 ml. of water and shake well for several minutes. Filter the solution into a dry test-tube, and add an equal volume of dilute hydrochloric acid (about 3 M) to the filtrate. Note any precipitate. 2. Lead Oxalate and Sulphate Put 5 ml. of 0.1 M lead nitrate solution in each of two test-tubes. To the first add 5 ml. of 0.1 M sodium oxalate; to the second add 2-3 ml. of dilute (about 3 M) sulphuric acid. Shake to mix. Let the precipitates settle, and pour off most of the liquid above them. Now add 5 ml. of dilute (about 3M) nitric acid to each tube, and stir with a stirring rod. Does any appreciable amount of either precipitate dissolve? Account for any differences of behaviour of lead oxalate and lead sulphate.

3.

Precipitation of Hydroxides

Dilute 5 ml. of the ammonia solution (about 5 M) to 25 ml. in a graduated cylinder. Put 5 ml. of 0.1 M magnesium chloride into a test-tube, and 5 ml. of 0.1 M aluminium nitrate into another. Add 5 ml. of the diluted ammonia solution to each. Shake to mix; note any precipitates. To each tube add 1 g. of ammonium chloride, and shake well. Does any appreciable amount of either precipitate dissolve?

58 7. What is meant by saying that a saturated solution is in equilibrium with the solute? What experimental evidence supports the view that an equilibrium does exist?

8.

SOLUBILITY PRODUCT INTRODUCTION

When a salt dissolves in water, the solution contains two (or sometimes more) different sorts of ions. If another salt that provides one of the same ions as the original salt is also dissolved in water, it will in general reduce the solubility of the original salt. This phenomenon is sometimes called the "common ion effect." Conversely if some compound is added which combines with, and thereby removes, one of the ions from the original salt, its solubility will be increased. These effects are illustrated qualitatively in the first part of this experiment. The extent to which the solubility changes can be measured, and is found to conform, at least approximately, to a simple rule: for a salt AX, the concentration of A+ multiplied by the concentration of x- is a constant, at any one temperature. The measurements made in the following experiment illustrate this. EXPERIMENTAL

1. Lead Chloride

Put into a test-tube about as much lead chloride as can be picked up on a dime. Add 10 ml. of water and shake well for several minutes. Filter the solution into a dry test-tube, and add an equal volume of dilute hydrochloric acid (about 3 M) to the filtrate. Note any precipitate. 2. Lead Oxalate and Sulphate Put 5 ml. of 0.1 M lead nitrate solution in each of two test-tubes. To the first add 5 ml. of 0.1 M sodium oxalate; to the second add 2-3 ml. of dilute (about 3 M) sulphuric acid. Shake to mix. Let the precipitates settle, and pour off most of the liquid above them. Now add 5 ml. of dilute (about 3M) nitric acid to each tube, and stir with a stirring rod. Does any appreciable amount of either precipitate dissolve? Account for any differences of behaviour of lead oxalate and lead sulphate.

3.

Precipitation of Hydroxides

Dilute 5 ml. of the ammonia solution (about 5 M) to 25 ml. in a graduated cylinder. Put 5 ml. of 0.1 M magnesium chloride into a test-tube, and 5 ml. of 0.1 M aluminium nitrate into another. Add 5 ml. of the diluted ammonia solution to each. Shake to mix; note any precipitates. To each tube add 1 g. of ammonium chloride, and shake well. Does any appreciable amount of either precipitate dissolve?

59 In two separate test-tubes, place 5 ml. of 0.1 M magnesium chloride, and 5 ml. of 0.1 M aluminium nitrate. Add 1 g. of ammonium chloride to each, and shake till dissolved. Now add 5 ml. of diluted ammonia solution to each tube, and shake to mix. Note any precipitates. 4.

Measurement of Solubility Products

(a) Silver acetate. Fill a burette with 0.25 M silver nitrate solution, and another with 0.25 M sodium acetate solution. Set up seven dry test-tubes, and run in the following amounts of the solutions from the burettes. TABLE I Tube a b C

d e f

g

Silver Nitrate 15 ml. 13 11 9 7

5

3

Sodium Acetate 3 ml.

5

7 9 11 13 15

Shake till well mixed, put a cork in each tube and leave till the next laboratory period. At the next laboratory period analyse the clear solution in each tube for its silver content as follows. Pipette 5 ml. of the solution from tube (a) into a 125 ml. conical flask. Be careful not to suck up any silver acetate crystals into the pipette. Add to the flask about 1 ml. of ferric alum solution, and a few drops of dilute (3 M) nitric acid. Fill a burette with 0.02 M potassium thiocyanate, and use it to titrate the solution in the flask. As you add the potassium thiocyanate, swirl the solution around in the flask frequently to ensure thorough mixing. A white precipitate will form. The end-point is reached when the liquid in the flask is just coloured a faint reddish-brown throughout. Note the volume of potassium thiocyanate solution needed. Repeat this analysis on the other six tubes. Note the laboratory temperature. For each tube calculate the silver ion concentration, and, from the amounts put in, calculate the acetate ion concentration; hence calculate Kap• (b) Silver nitrite. Use exactly the same method as in part (a) for silver acetate, but use 0.2 M silver nitrate and 0.2 M sodium nitrite as the two solutions. (c) Silver sulphate. Use exactly the same method as in part (a) for silver acetate, but use 0.3 M silver nitrate and 0.3 M sodium sulphate as the two solutions. EXPLANATIONS

Some general aspects of solubility were dealt with in Experiment 7. In this experiment we are dealing with the special case of a compound which, on dissolving, gives two (or more) different sorts of ions in solution. For instance, silver chloride in solution is present as silver ions and chloride ions. It is found that, for such cases, another salt giving one of the same ions (for example, silver nitrate which also

60 gives silver ions for the example above) reduces the solubility of the original salt. The extent of this change can be measured, and it is found for a salt, AX, that in saturated solutions at any given temperature: [A+] . [X-J = constant. The constant, designated by K ep, is called the solubility product constant. For instance in saturated silver chloride: [Ag+] . [CI-] =

Kap•

Each compound has its characteristic Kap, which may vary with temperature. It should be added that in general this equation is approximately, but not precisely, obeyed. If several ions are produced, as in silver sulphate, Ag2S04, K •P is given by The square on (Ag+] arises because the solid contains two silver ions to each sulphate ion. Similarly for magnesium hydroxide, Mg(OHh: [Mg++] . [OH-]2 =

Kap•

The solubility product equation is a special case of the Mass Law (see Experiment 9), for in the equilibrium: AX (solid) ~A+

+ x- (both in solution)

the Mass Law treats the concentration of solids as constant, so [A+] . [X-J = constant.

K •P is a special sort of equilibrium constant. If A+ and x- ions are mixed in solution so that [A+]. [X-J

>

Kap

the solution will be supersaturated, and a precipitate will probably form. Precipitation will continue until [A+] . [X-J =

Ksp•