But more than one salt is sometimes obtained by the reaction between a specified acid and the same metallic oxide or hydroxide. Exp. 4. You are given two exactly equal quantities of a solution of oxalic acid in water, and a quantity of a solution of Fig. 17. potassium hydroxide*. To one of the quantities of oxalic acid add the potash solution very cautiously, from a graduated tube with a stopcock (a burette) (Fig. 17) until the liquid is exactly neutral (8. Exp. 2, Chap. VII.). Read off, on the burette, the number of cubic centimetres of potash solution added. To the other quantity of oxalic acid add exactly half as much potash as you have added to the first quantity of the acid. Evaporate both solutions over low flames until a drop taken out on a rod solidifies on cooling; allow to cool; collect, dry, and recrystallise the solids from as little water as possible; collect, dry, and recrystallise each again. Prove that each solid is (1) a compound of potassium, (2) an oxalate. The production of a white pp. (calcium oxalate) when calcium chloride is added to a solution, which pp. is insoluble in acetic acid, is a test for the of an oxalate in that solution. presence Dissolve the solids in distilled water, noting that one is much more soluble than the other, and test each solution with blue and red litmus; the oxalate of potassium obtained by exactly neutralising oxalic acid with potash is neutral to litmus; the other oxalate of potassium turns blue litmus red; this is often expressed by saying that an aqueous solution of the salt has an acid reaction. Quantitative analyses of these oxalates of potassium shew that the composition of the neutral salt is expressed by the formula K,C,O,, and the composition of the other salt is expressed by the formula KHCO A salt may then be formed from an acid by replacing a part, or the whole, of the hydrogen of the acid by a metal. * Note to Demonstrator. About 10 grams oxalic acid in 250 c.c. water; 100 c.c. is enough for each experiment: about 20 grams caustic potash in 250 c.c. water; 50 c.c. for each experiment. Exp. 5. You are given two exactly equal quantities of a solution of potash, in basins, and a quantity of a solution of sulphuric acid. You are also told the weight of each basin. Fill a burette with the acid; run the acid into one of the quantities of potash until the liquid is exactly neutral; to the other quantity of potash add exactly twice as much sulphuric acid as was required to neutralise the first (equal) quantity of potash. Evaporate each solution to dryness on a water-bath; then place each basin on a sand-tray, heap the sand round the basins, and heat them to about 200o for some time. Any free sulphuric acid present will be volatilised at this temperature. Now allow the basins to cool, wipe them carefully, and weigh them. Deduct the weight of each basin from the weight of basin + salt; the difference is the weight of salt obtained. In one case you obtain a considerably greater weight of salt than in the other. If it is now assumed-and this has been experimentally proved--that sulphuric acid is completely volatilised at a moderate red heat, it follows that you have obtained two distinct potassium sulphates. The reasoning is as follows: A certain mass, say x grams, of potash (in solution) was exactly neutralised by sulphuric acid; the solution was evaporated; the solid obtained was heated to moderate redness, and weighed; it weighed, say, x' grams. The same mass (x grams) of potash was neutralised by sulphuric acid, as much sulphuric acid as was used to neutralise the potash was added over and above that required for the neutralisation; the solution was evaporated; the solid obtained was heated to moderate redness, and weighed; it weighed considerably more than x' grams (say " grams). (The ratio of x' to x" should be nearly 1 to 1.28.) But, from the conditions, neither solid obtained could be a mixture of sulphuric acid with potash, or a mixture of potassium sulphate with sulphuric acid; therefore each solid was a sulphate of potassium; therefore there are two potassium sulphates. Confirm this result by dissolving each salt in water and adding a little blue litmus; in one case the litmus is turned red, in the other it remains unchanged. *Note to Demonstrator. About 20 c.c. of a normal solution of potash, as free as possible from carbonate; and about 100 c.c. of a normal solution of sulphuric acid. Quantitative analysis shews that the compositions of the two salts are expressed by the formulæ K,SÓ, and KHSO, respectively. You again conclude that a salt may be formed from an acid by replacing a part, or the whole, of the hydrogen of the acid by a metal (comp. Exp. 4). Exp. 6. You are given two small basins and a quantity of a solution of hydrochloric acid. Counterpoise the basins one against the other. It is advisable to label the basins, that in the right hand pan R, that in the left hand pan L. To each basin add an equal quantity, say 20 c.c., of an aqueous solution of potash. Run the solution of hydrochloric acid given you, from a burette, into one of the basins until the liquid is neutral; to the other basin add twice as much of the same hydrochloric acid solution; evaporate the contents of both basins to dryness on a water-bath; continue to heat the solids obtained at 100° so long as any trace of acidsmelling fumes is given off; then allow the basins to cool; dry each carefully; place them on the pans of the balance, basin R on the right-hand pan, and basin L on the left-hand pan, and put the counterpoise used at the beginning of the experiment on the same pan as it was then placed on. The basins and their contents equilibrate each other. Hence you conclude that the two equal masses of potash have reacted each with the same mass of hydrochloric acid, and that the excess of hydrochloric acid added in one casei.e. the hydrochloric acid which did not interact with the potash has been removed by evaporation. Hence, probably, the same salt has been formed in each case. Dissolve the two solids in the basins in distilled water, and prove that both solutions are neutral to litmus. Prove also that both solutions contain (1) a compound of potassium, (2) a chloride (s. Exp. 4, Chap. VII.; and Exp. 8, Chap. VI.). The proof that both solids are the same salt could be rendered complete only by careful quantitative analyses. Such analyses have been made, and have proved that only one salt is obtained by the interaction of an aqueous solution of hydrochloric acid with potash, and that this salt has the composition KCl. The relation between the composition of an acid and that of a salt is, that the salt is a compound of a metal with the elements of the acid except the whole, or a part, of the hydrogen of the acid. The term acid radicle is generally used to denote the elements of an acid minus the replaceable hydrogen of the acid. You have now gained a fairly clear notion of the essential property of an acid, viz. that it is a compound of hydrogen which reacts with metals, metallic oxides, hydroxides, and carbonates, to form salts. You also know the relation between the composition of a salt and that of the acid from which the salt is obtained. You have learned that some acids react with potash to form more than one salt. You have experimentally examined the meaning of the phrase, replacing the whole or a part of the hydrogen of an acid by a metal. Incidentally you have become acquainted with tests by which the presence in solutions of (1) a compound of magnesium, (2) an oxalate, (3) a ferric compound, may be proved. You have also learned how to use a burette. Acids which react with potash or soda to form only one stable salt are called monobasic acids. Exp. 7. You are given carbolic acid; examine the reactions which occur between this compound and (1) sodium, (2) sodium oxide or hydroxide, (3) sodium carbonate; and determine from the results of your experiments whether the compound is or is not properly called an acid. Reference to "ELEMENTARY CHEMISTRY." Chap. IX. CHAPTER IX. CLASSIFICATION OF SALTS. EXPERIMENTS Conducted in Chap. VIII. shewed that hydrochloric acid reacts with potash to produce one salt, and that sulphuric acid and oxalic acid each reacts with potash to produce two salts. Aqueous solutions of one of the potassium sulphates, and one of the potassium oxalates, you prepared, shewed an acid reaction towards litmus, i. e. the solutions turned blue litmus red. The compositions of these salts of potassium, and their actions on litmus, are presented in the following table : Potassium sulphates. Potassium chloride. KCl. Neutral to litmus. K,SO. Neutral to litmus. KHSO, Acid to litmus. Potassium oxalates. K.C.O. Neutral to litmus. There is an evident connexion between the compositions of these salts and their neutrality or non-neutrality towards litmus. But all salts composed of a metal, an acid radicle, and hydrogen, do not shew an acid reaction towards litmus. Exp. 1. Add potassium carbonate to water until no more of the salt dissolves; now pass carbon dioxide into the liquid for some time. Crystals separate from the solution: collect these on a filter; wash them by pouring two or three successive very small quantities of cold water over them on the filter. Then dissolve the crystals in water and examine the action of the solution on litmus; it is slightly alkaline, i.e. it turns red litmus bluish. The compositions of the two |