The conclusions may be summarized in the statement that the salts of strong bases with weak acids are more or less decomposed by water (hydrolyzed) and the resulting solutions must react alkaline. We find, as a matter of fact, that aqueous solutions of potassium cyanide, sodium carbonate, sodium sulphide, borax (see the table, p. [104]), all react strongly alkaline to litmus (exp.). Conversely, it may be said, that if the sodium or potassium salt of an acid dissolves in water with a decidedly alkaline reaction, it is the salt of a weak, poorly ionized acid.[370] [p184]
Action of Water on a Salt of a Strong Acid with a Weak Base.
For MeX + HOH ⇄ MeOH + HX, where MeOH is a weak base and HX a strong acid, we have as before:[371]
[Me+] / ([H+] × [MeOH]) =
[Salt] / ([Acid] × [Base]) = KBase / KHOH.
Action of Water on a Salt of a Base and an Acid, Both of which are Weak.
Like all salts, such a salt, say MeX, would ionize very readily, when dissolved in water (the few exceptions to readily ionizable salts are not under consideration), and, in this case, both the positive and the negative ions would have to combine respectively with the hydroxide and the hydrogen ions of water to form the nonionized weak base and the nonionized weak acid, and satisfy two very small constants, KBase and KAcid:
[Me+] × [HO−] / [MeOH] = KBase
and [H+] × [X−] / [HX] = KAcid.
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 KBase and KAcid. 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−] | = | α2 [Salt]2 | = | KAcid × KBase | = K, |
| [HX] × [MeOH] | [Acid] × [Base] | KHOH |