The cyanide-ion, whose concentration is expressed by [CN−], is formed practically altogether by the ionization of potassium cyanide, which is an easily ionizable and almost entirely ionized salt; the hydroxide-ion, whose concentration is expressed by [HO−], is formed by the ionization of potassium hydroxide, which is an easily ionizable base, ionized to practically the same degree as is the potassium cyanide in the solution. If we represent the total concentration of the potassium cyanide, ionized and nonionized, at the point of equilibrium, by [KCN] and its degree of ionization by α1, and if we represent, similarly, the total concentration of potassium hydroxide by [KOH] and its degree of ionization by α2, the equilibrium equation may be written:
α1[KCN] / ([HCN] × α2[KOH]) = KHCN / KHOH = KHydrolysis.
Since the degrees of ionization of the two strong electrolytes are practically the same, we have further simply
[KCN] / ([HCN] × [KOH]) = KHCN / KHOH = KHydrolysis.
The mathematical equations give us a measure of the extent to which water must decompose or hydrolyze the salt in question, as expressed in the chemical equations (p. [180]). The extent of the hydrolysis, clearly, depends on the relative ionization constants of hydrocyanic acid and water, the two acids competing for the base.
From the known values of the constants, one may calculate that, at 25°, in a solution of 6.5 grams potassium cyanide in a liter (0.1 molar), almost 1.3% of the cyanide is decomposed into potassium hydroxide and hydrocyanic acid. Since every molecule of hydrolyzed salt forms one molecule of [p183] the hydroxide and one molecule of the acid, we may put [KOH] = [HCN] = x and [KCN] = 0.1 − x. The ionization constant, KHCN = 7E−10, and KHOH = 1.2E−14, at 25°. Inserting these values into the equation [KCN] / ([HCN] × [KOH]) = KHCN / KHOH we have: (0.1 − x) / x2 = 7E−10 / 1.2E−14. Here x = 0.0013. This is 1.3% of the 0.1 mole of cyanide used.
One may convince himself, as follows, that the constants are satisfied when the decomposition of the cyanide has proceeded to this point: the degrees of ionization of the potassium cyanide and potassium hydroxide, α1 and α2, may be taken as 85% (the same as the degree of ionization of the similar electrolyte KCl in 0.1 molar solution). Then [HO−] = 0.85 × 0.0013 = 0.0011; [CN−] = 0.85 × (0.1 − 0.0013) = 0.083; [H+] = 1.2E−14 / [HO−] = 1.1E−11. For [H+] × [CN−] / [HCN] we have then: 1.1E−11 × 0.083 / (0.0013) or 7E−10, the value for the ionization constant of hydrocyanic acid. It should be noted that, whereas in pure water at 25° [H+] = [HO−] = √(1.2E−14) = 1.1E−7, in the solution under consideration [HO−] has increased to the value 0.0011 and [H+] is only 1.1E−11.
The relation developed for the hydrolysis of potassium cyanide is a general one, holding for the hydrolysis of salts, of the type MeX, of a weak acid with a strong base. It may be expressed in general as follows: for the hydrolysis of a salt according to MeX + HOH ⇄ MeOH + HX, where HX is a weak acid and MEOH a strong base, we have:[369]
[Salt] / ([Acid] × [Base]) = KAcid / KHOH.
It is clear, from the equation, that the weaker the acid of the salt (measured by the ionization constant KAcid, the numerator on the right), the more will water, ceteris paribus, be able to drive it out of its salt and form its own salt, the base (the smaller the numerator on the right, the larger must be the denominator on the left).