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] = K_{HCN} = 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^{−}] = K_{HOH} = 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^{−}]) = K_{HCN} / K_{HOH} = K_{Hydrolysis}. 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 α_{1}, and if we represent, similarly, the total concentration of potassium hydroxide by [KOH] and its degree of ionization by α_{2}, the equilibrium equation may be written: α_{1}[KCN] / ([HCN] × α_{2}[KOH]) = K_{HCN} / K_{HOH} = K_{Hydrolysis}. Since the degrees of ionization of the two strong electrolytes are practically the same, we have further simply [KCN] / ([HCN] × [KOH]) = K_{HCN} / K_{HOH} = K_{Hydrolysis}.