CHAPTER XI.

REACTIONS BETWEEN ACIDS AND SALTS.

Exp. 1. To the salt barium chloride add a little concentrated sulphuric acid. Arrange the experiment as shewn in fig. 20. Put about 3 or 4 grams of

the salt in the small flask, and pour the acid down the funnel tube. Allow the gas which comes off to pass into a little distilled water; the gas dissolves rapidly. When the reaction has moderated, heat the flask gently and continue to heat for 5 minutes or so.

Remove the exit tube from the water before you take away the lamp.

When the contents of the flask are cold add some water; shake up well; pour off the water; add more water, and shake up again; pour

Now prove

Fig. 20. off; continue to wash the white solid by decantation until the washings are no longer acid to litmus paper. that the white solid is barium sulphate (s. Chap. VII. Exp. 5). Prove that the aqueous solution of the gas produced in the reaction, (1) is acid to litmus, (2) gives the reaction with silver nitrate characteristic of a chloride, (3) neutralises caustic potash forming potassium chloride, (4) dissolves metals-say zinc and magnesium-with evolution of hydrogen. The liquid is an aqueous solution of hydrochloric acid.

The reaction which you have carried out is represented in an equation thus ;

By the interaction of a salt with an acid there have been produced, (1) a new salt, composed of the metal of the original salt and the radicle of the acid used; (2) a new acid, composed of the acid radicle of the original salt and the hydrogen of the acid used.

Exp. 2. In a flask, arranged with a cork and tubes as in Exp. 1, place 3-5 grams of the salt sodium bromide, and add a quantity of a very concentrated solution of phosphoric acid. Warm the flask, and lead the colourless acid-smelling gas which comes off into distilled water. Proceed as directed in Exp. 1. Prove that the water into which the gas was conducted contains an acid, by shewing that it turns blue litmus red, dissolves metals with evolution of hydrogen, and neutralises caustic potash; prove that the acid is hydrobromic acid, by adding silver nitrate and obtaining a very pale yellow pp. of silver bromide (AgBr), soluble with difficulty in a considerable quantity of ammonia solution. Prove that the solid obtained by evaporating the liquid in the flask to dryness and heating till acid fumes cease to be evolved is, (1) a compound of sodium, (2) a phosphate.

The reaction which occurred in the flask is represented in the following equation:

2NaBr + H2PO1 = Na2HPO + 2HBr

4

salt acid new salt new acid.

This reaction is analogous to that which occurred between sodium chloride and sulphuric acid.

Exp. 3. Conduct an experiment similar to 1 and 2, but put the salt potassium nitrate in the flask and add concentrated sulphuric acid.

Lead the gas evolved into water; prove that the solution contains nitric acid.

Evaporate the solution in the flask to dryness in the draught cupboard, heat the residue as long as acid fumes come off, and prove that the residual solid is potassium sulphate.

Nitric acid is colourless; you noticed a reddish-brown gas

in the flask while the salt and the acid were heated together. What is this gas?

Exp. 4. Heat in a test tube a little concentrated nitric acid and a little concentrated sulphuric acid; a reddish-brown gas is evolved. Put a cork with an exit tube into the test tube and lead a little of the gas into a solution of ferrous sulphate acidified with sulphuric acid; a brown-black liquid is formed (comp. test for nitric acid, Exp. 4, Chap. VII.).

Now lead some of the gas produced in Exp. 3 into a cold acidulated solution of ferrous sulphate.

From the result obtained you conclude that, probably, the gas obtained in Exp. 4 is the same as the reddish-brown gas produced along with nitric acid in Exp. 3. The gas in question is a mixture of nitric oxide (NO) and nitrogen dioxide (NO,).

The reaction which occurred in Exp. 3 may be thus represented in equations;—

(1)

4

3

2KNO2+ H2SO = K2SO, + 2HNO,
acid new salt new acid

salt

4

(2) 2HNO,+ H2SO1 = H2O.H2SO + 2NO + 30 (some of the NO and O combine to form NO2).

A portion of the new acid produced in the primary change reacts with some of the acid used to decompose the salt, and is thereby itself decomposed.

Exp. 5. Allow the salt sodium carbonate to react with nitric acid; prove that carbon dioxide is evolved and that sodium nitrate remains in the flask (evaporate the contents of the flask to dryness on a water-bath and heat at 100o until acid fumes cease to come off). The reaction is thus represented: Na,CO2+2HNO, = 2NaNO,

salt

3

acid new salt

+ CO. acidic oxide

+ H2O. water.

A new acid is not formed, but an acidic oxide and water are produced. Reasoning from the results of Exp. 1 and 2 we should have expected carbonic acid (H,CO,) to be produced in the present exp.; this acid, if it exists, is extremely unstable and is very easily decomposed into an oxide and water.

In Exp. 3 an acid was produced, part of which was decomposed under the experimental conditions; in the present Exp. an acid has probably been produced, but at the moment of its production it has been decomposed into an oxide and water.

Exp. 6. Allow moderately concentrated sulphuric acid to react with a little potassium iodide; lead the gas produced into water; prove (1) that the solution contains an acid; (2) that this acid is hydriodic, by adding silver nitrate and obtaining a pale yellow pp. of silver iodide (AgI) insoluble in ammonia solution. Prove that the residue in the flask, after removing the excess of sulphuric acid by evaporation and heating, is potassium sulphate.

Notice that a greyish violet solid is formed in the neck of the flask. Remove a little of this solid to a test tube: prove that it is slightly soluble in water; and very soluble in alcohol, giving a brown liquid, a drop or two of which added to a very little starch, which has been boiled with a good deal of water and allowed to cool, produces a deep blue colour (so-called iodide of starch). These reactions prove that the solid is iodine.

In this reaction a new acid (hydriodic) has been produced; but part of this acid interacts with the excess of sulphuric acid present to produce iodine &c. Thus

In this chapter you have learned that the normal products of the reaction between an acid and a salt are, (1) a new salt, composed of the metal of the original salt and the radicle of the acid used; and (2) a new acid, composed of the acid radicle of the original salt and the hydrogen of the acid used. You have learned that secondary changes frequently occur, generally resulting in a change of part, or even the whole, of the new acid produced into less complex compounds or into its elements.

Incidentally, you have learned tests for an iodide and a

bromide.

Exp. 7. You are given the salt ethylic acetate (CH.CH2O); determine whether the reaction which occurs when this salt is heated with hydrobromic acid follows the course of the normal reaction between an acid and a salt. You are told that ethylic bromide is a liquid about 1 times heavier than, and insoluble in, water; you are also told that the test for acetic acid consists in exactly neutralising with ammonia and adding ferric chloride solution, when a red colour (due to formation of ferric acetate) is produced.

XI.

Reference to "ELEMENTARY CHEMISTRY"; Chaps. IX. and

CHAPTER XII.

CLASSIFICATION OF ELEMENTS.

The elements chlorine, bromine, and iodine are placed in the same class.

Exp. 1. Mix about 2 grams of potassium chloride with about 3 grams of manganese dioxide; place the mixture in a flask fitted with a cork, exit tube and funnel tube, as shewn in fig. 20, (p. 53); let the exit tube pass into a dry jar; set the apparatus in the draught place; pour a little concentrated sulphuric acid diluted with its own weight of water down the funnel tube into the flask, and warm gently.

A greenish-yellow gas collects in the jar. This gas is chlorine.

Exp. 2. Conduct an experiment similar to Exp. 1, but use potassium bromide in place of potassium chloride.

A reddish-brown gas, condensing to dark-red drops, collects in the jar. This liquid is bromine.

Exp. 3. Place a mixture of about 2 grams potassium iodide and 3 grams manganese dioxide in a small retort

Fig. 21.

arranged so that the beak passes into a small dry flask (Fig. 21). Add some sulphuric acid and warm.