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, K_{HCN} = 7E−10, and K_{HOH} = 1.2E−14, at 25°. Inserting these values into the equation [KCN] / ([HCN] × [KOH]) = K_{HCN} / K_{HOH} we have: (0.1 − ‹x›) / ‹x›^2 = 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, α_{1} and α_{2}, 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]) = K_{Acid} / K_{HOH}. It is clear, from the equation, that the weaker the acid of the salt (measured by the ionization constant K_{Acid}, 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).