When the cyanide is dissolved in water, we must obtain, for the same reasons as were developed in the discussion of the hydrolysis of sodium chloride, a little nonionized potassium hydroxide, from the union of potassium ions with hydroxide ions, formed by the water. Potassium hydroxide being a strong, easily ionizable base, there will be only a slight tendency towards this union. Hydrocyanic acid, on the other hand, is an exceedingly weak acid. The value of its ionization constant KHCN = [H+] × [CN−] / [HCN] is only 7E−10, as compared with a similar ratio approximating 1 for potassium hydroxide ([K+] × [HO−] / [KOH] = 1; see the tables, p. [104] and p. [106] and see pp. [106]–7). The hydrogen-ion, formed from the water, must therefore combine with cyanide-ion, to form nonionized hydrocyanic acid, much more completely than the hydroxide-ion combines with potassium-ion. With the disappearance of the ions of water, in this case notably of its hydrogen ions, more water must ionize to satisfy the ionization constant [p181] for water (p. [176]), and the formation of hydrocyanic acid will continue, towards the satisfying of its own constant. It is important to note that, for the reasons given, the hydrogen-ion of water is used up to a far greater extent than is the hydroxide-ion; the latter therefore accumulates, and this accumulation results in the formation of smaller and smaller concentrations of the hydrogen-ion, by the water. Since [H+] × [HO−] = 1.2E−14 (at 25°; p. [104]), as [HO−] grows larger, [H+] must grow proportionally smaller. The suppression of the hydrogen-ion by the accumulation of the hydroxide ion will, ultimately, make [H+] so small, that the equilibrium ratio [H+] × [CN−] / [HCN] will equal the equilibrium constant. Since the union of the hydrogen-ion with the cyanide-ion, to form little ionized hydrocyanic acid, is the main moving cause for the changes, the latter will then come to a standstill and equilibrium will be established. The net result of the action of water on potassium cyanide may be said to consist in the formation of practically nonionized hydrocyanic acid and the liberation of (chiefly) ionized potassium hydroxide, until all the equilibrium constants of the system are satisfied. We note that potassium cyanide solution must react strongly alkaline (exp.) and that a free acid (e.g. HCN) may well exist in the presence of a free base (e.g. KOH), provided the acid is present in a nonionized, and therefore chemically inactive, condition (inactive as an acid).
Ignoring the (practically) unimportant formation of small quantities of nonionized potassium hydroxide, we may summarize the action in a single equation, which shows the main action:
CN− + HOH ⇄ HCN + HO−.
Whereas water, as an acid and as a base, is so exceedingly weak, that it can form but traces of its own salts, sodium hydroxide and hydrochloric acid, when acting on sodium chloride and competing for the base with such a strong acid as hydrochloric acid and for the acid with such a strong base as sodium hydroxide (see p. [179]), the result, evidently, is quite different when water competes for a base with so weak an acid as hydrocyanic acid. In this case, we note that a considerable quantity of (ionized) potassium hydroxide, the salt of water in its rôle of an acid, is formed as a result of the action of water on potassium cyanide. [p182]
The theory of ionization, with the aid of the law of chemical equilibrium, gives us the means for accurately defining the relative concentrations of the products, in the final condition of equilibrium.[368] For the weak acid, hydrocyanic acid, we have the condition of equilibrium
[H+] × [CN−] / [HCN] = KHCN = 7E−10.
The symbols [H+], [CN−] and [HCN] denote the final concentrations for the condition of equilibrium, indicated in the equations on p. [180]; in such a mixture [H+] is not equal to [CN−], as it is in pure solutions of hydrocyanic acid in water. [CN−], representing the total concentration of the cyanide-ion, is very much larger than [H+], since the salt, potassium cyanide, produces the cyanide-ion in large concentrations.
For water, we have [H+] × [HO−] = KHOH = 1.2E−14, at 25°. Here, again, the symbols represent the final, total concentrations of the ions in the mixture and [HO−] is much larger than [H+], since hydroxide-ion is formed in large quantities, as described above.
Combining the two equations, we have:
[CN−] / ([HCN] × [HO−]) = KHCN / KHOH = KHydrolysis.