Exp. 6. Place about 50 c.c. of dilute sulphuric acid in a beaker, add some hydrochloric acid*, heat to boiling, and then slowly run in a solution of barium chloride from a burette so long as, but no longer than, a pp. of barium sulphate is produced. In order to determine when the reaction is completed, shake the contents of the beaker vigorously after the addition of each few c.c. of the BaCl,Aq, allow the pp. to settle, and then add 1 or 2 drops of the BaCl,Aq; when only a very slight pp. is produced, add 1 or 2 drops more BaCl,Aq, shake well, allow to settle, pour a very little of the liquid through a filter (made of good paper), and add one drop of the BaCl,Aq to the clear filtrate; should the filtrate remain quite clear the reaction is completed, should a cloudiness be produced, run a drop or two of the BaCl, Aq into the beaker, and repeat the filtration &c. of a very little of the liquid; continue thus until the reaction is as nearly as possible completed, but do not add any excess of BaCl,Aq.

When the reaction is complete, prove that the filtrate from the pp. of barium sulphate does not give the reactions for a compound of barium, or for sulphuric acid; prove also that the pp. is a barium compound and that it is a sulphate. The reaction which has occurred is represented in an equation thus ;

BaCl,Aq+ H2SO ̧Aq = BaSO ̧ + 2HClAq. The foregoing Exp. proves that when a given quantity of barium chloride in solution has been wholly decomposed by dilute sulphuric acid the only compound of barium produced is the sulphate which is precipitated, and that the sulphuric acid employed has been itself wholly decomposed. By making the Exp. more accurate it is possible to prove the correctness of the equation given above.

Exp. 7. You are given aqueous solutions of barium chloride and sulphuric acid of stated strengths, i.e. the mass of each compound in a specified volume of the solution is stated. Measure off accurately a certain volume of the barium chloride solution, say 50 c.c.; calculate exactly what volume of the acid solution contains that mass of sulphuric acid which is shewn by the equation in Exp. 6 to be necessary for the completion of the reaction; add exactly this volume of the sulphuric acid solution; allow the pp. to settle completely; pour

* The pp. of BaSO4 settles better in the presence of HClAq.

the clear liquid through a filter, and prove the absence of a barium compound and of sulphuric acid, and the presence of hydrochloric acid, in the filtrate.

In the following Exps. solid compounds are produced, but they interact with the other products of the change to produce the original substances; hence it is necessary to add an excess of one of the interacting compounds in order to completely decompose the whole of the other compound in the original solution. Sometimes (as in Exp. 9) a very large excess of one of the interacting compounds is required to complete the change, even when that change results in the formation of a solid body.

Exp. 8. To a solution of calcium chloride add a solution of ammonium carbonate until the reaction is completed, i. e. until addition of a drop of ammonium carbonate solution fails to produce any pp. of calcium carbonate. Now filter, and prove that the filtrate contains a considerable quantity of ammonium carbonate. Hence some of the ammonium carbonate required to complete the chemical change has itself remained unchanged. The reaction may be represented by such an equation as the following:

CaCl,Aq + (NH ̧),CO ̧Aq + ∞ (NH ̧) ̧CO2 = CaCO ̧ + 2NH,ClAq +(NH,),CO Aq:

x must be made fairly large to complete the reaction.

Exp. 9. To six equal quantities of a solution of bismuth chloride in dilute hydrochloric acid add quantities of water, in the ratio 1:2:46:8: 10*. In each case bismuth oxychloride is precipitated. Filter off each pp. and prove (by adding H,SAq) that each filtrate contains bismuth chloride in solution. The reaction may be expressed thus:

2

BiCl + BiCl + H2O + Aq = BiOC1 + 2HClAq + ≈ BiCl ̧Aq ; the smaller the value of x relatively to that of Aq the greater is the quantity of BiCl, decomposed; but traces of BiCl, remain unchanged even when x is several hundred times less than Aq.

Exp. 10. You are given solutions of ferric chloride and potassium sulphocyanide. To a few drops of one add a little

3

* About 10 c.c. of a solution of 10 grams BiCl, in 500 c.c. HCIAq may be used: the smallest quantity of water may be about 10 c.c.

of the other; a deep red colour is produced, due to formation of ferric sulphocyanide. The reaction may be thus expressed: Fe ̧Cl. Aq + 6KCNSAq + æKCNSAq = Fe ̧ (CNS) ̧Aq + 6KClAq + KCNSAq. If x is made very large the reaction approaches completion.

Measure out 6 equal quantities—say 10 c.c. each—of the ferric chloride solution; add to the first a measured quantity xc.c. (say 10 c.c.) of the sulphocyanide solution, to the second add 2x c.c., to the third 3x c.c., to the fourth 6x c.c., to the fifth 8x c.c., and to the sixth add 10x c.c., of the sulphocyanide solution. The depth of colour increases from the first to the sixth; this indicates that the quantity of ferric sulphocyanide produced has increased as the relative mass of potassium sulphocyanide was increased.

In the following Exp. a chemical change is more or less completely reversed by altering the relative masses of the interacting bodies.

Exp. 11. To some solid natural antimony sulphide (Stibnite) add a fair quantity of concentrated hydrochloric acid; antimony chloride is formed and passes into solution, while sulphuretted hydrogen gas is evolved. Prove the latter fact by holding a piece of paper moistened with a solution of a salt of lead in the escaping gas; brown lead sulphide (PbS) is formed on the paper. Now pour a considerable quantity of

water into the vessel in which the reaction has occurred; orangered antimony sulphide is precipitated. The two changes may be thus represented :

concentrated acid

Sb2S2+6HClAq

3

2SbCl2+ 3H ̧S.

dilute acid

The

indicates the direction in which the change proceeds, according as much or little water is present.

In the following Exp. a chemical change is reversed by altering the medium in which the reaction occurs.

Exp. 12. Pass carbon dioxide into a saturated alcoholic solution of dry potassium acetate; potassium carbonate is formed and precipitated as a white solid, and acetic acid

remains in solution.

To an aqueous solution of potassium carbonate add acetic acid; potassium acetate is formed and passes into solution, and carbon dioxide is evolved as a gas.

Exp. 13. Recall Exp. 13 Chap. XIV. wherein nitric acid was prepared by the interaction of sulphuric acid with potassium nitrate. Place a small crystal of potassium nitrate in a porcelain basin, and a small crystal of potassium sulphate in another similar basin; to the first basin add some concentrated sulphuric acid, and to the second some concentrated nitric acid. Evaporate each to dryness, allow to cool, add more acid to each and again evaporate to dryness; repeat this treatment; and then prove that the salt in the first basin is potassium sulphate free from nitrate, and the salt in the second basin is potassium nitrate free from sulphate. The acid added in large excess has completely replaced the other acid from combination with the potash.

Reference to "ELEMENTARY CHEMISTRY." Chap. XII.

7

M. P. C.

CHAPTER XVII.

OXIDATIONS AND REDUCTIONS.

ANY chemical change which results in an increase in the ratio of the negative to the positive constituents of a compound is called an oxidation. Thus the change of FeO to Fe̱O̟, or of FeSO, to Fe,(SO), or of HgI to HgI,, is a process of oxidation. Any change which results in an increase of the ratio of the positive to the negative constituents of a compound is called a reduction. Processes of oxidation and reduction usually occur together; one part of the complete change is called an oxidation, and the other part is called a reduction.

Oxidation is sometimes effected by the direct addition of

oxygen.

Exp. 1. Cut a piece of sodium from the inside of a stick of this metal, and notice that the bright surface is at once covered with a nonon-lustrous film; this film is sodium oxide (NaO).

Exp. 2. Place a little very finely divided iron in a crucible, counterpoise the whole, then allow it to stand in the air for about an hour, and counterpoise again; there is no change of mass, nor is there any change in the appearance of the iron. Now heat the crucible over a Bunsen-lamp; the iron glows; allow to cool, and counterpoise again. There has been an increase in weight; the iron has been oxidised (to Fe,O) by the oxygen of the air.

Exp. 3. Place a very small piece of lead, and a very small piece of tin, on charcoal, and direct the outer flame, i.e. the oxidising flame, of a blowpipe on to each in succession; the lead is slowly oxidised to a yellowish film of lead oxide, and the tin to a nearly white film of tin oxide.