Both the hydrogen and the hydroxide ions of water would disappear, and in approximately equal quantity, if the base and acid were approximately equally weak, and the ions would be regenerated from water ‹with no accumulation of either one to suppress the other›, as in the two previous cases considered. Under these circumstances, the decomposition by water ‹must proceed very much further than in the previous cases›. For instance, in the hydrolysis of potassium cyanide in 0.1 molar solution, at 25°, we find the concentration of the hydrogen-ion [H^{+}] reduced[373] from 1.1E−7, its [p185] value in pure water, to 1.1E−11, as a result of the accumulation of potassium hydroxide (the hydroxide-ion), and only ‹this small value› for [H^{+}] appears in the equation for the formation of the free acid, HCN (first equation, p. 182; ‹vide› the calculation, p. 183). But, in the present case, the factors [HO^{−}] and [H^{+}], in the equations on p. 184, maintain practically their original value, about the same as in pure water, and the formation of nonionized MeOH and HX must go correspondingly further to satisfy the constants K_{Base} and K_{Acid}. Just how far the action must proceed, can be formulated with the aid of the theory of ionization and the law of chemical equilibrium,[374] much in the same way as for the hydrolysis of potassium cyanide. The final equation, as developed by Arrhenius, reads: [Me^{+}] × [X^{−}] / ([HX] × [MeOH]) = α^2 [Salt]^2 / ([Acid] × [Base]) = (K_{Acid} × K_{Base}) / K_{HOH} = K, in which K_{Acid} and K_{Base} represent the ionization constants of the acid and the base, as given in the tables (pp. 104 and 106), and α is the degree of ionization of the salt. For the cyanide of a base, which is as weak a base as hydrocyanic acid is an acid, we find that the decomposition by water, at 25° in a 0.1 molar solution, must comprise 99.35%[375] of the salt, in order to establish equilibrium. In the case of potassium cyanide, in 0.1 molar solution, only 1.3% of the salt is decomposed (p. 182).