Hydrolysis of Salt of Strong Base and Weak Acid (Anionic Hydrolysis)

Consider $NaOH(aq)+CH_3COOH(aq) \rightleftharpoons CH_3COONa(aq)+H_2O(l)$. $CH_3COONa$ dissociates completely, $CH_3COONa(aq) \rightarrow CH_3COO^-(aq)+Na^+(aq)$. $CH_3COO^-$ is the conjugate base of the weak acid $CH_3COOH$, so it does have a real tendency to pull a proton from water, $CH_3COO^-(aq)+H_2O(l) \rightleftharpoons CH_3COOH(aq)+OH^-(aq)$, while $Na^+$ has no such tendency toward $OH^-$; the result is $[OH^-]>[H^+]$, so the solution is basic (pH > 7) -- this is anionic hydrolysis.

Relating the hydrolysis constant Kh to Ka. For the hydrolysis equilibrium $CH_3COO^-+H_2O \rightleftharpoons CH_3COOH+OH^-$: $K_h=\dfrac{[CH_3COOH][OH^-]}{[CH_3COO^-][H_2O]}=\dfrac{[CH_3COOH][OH^-]}{[CH_3COO^-]}$ (water's concentration absorbed as before). Multiplying this by the acid's dissociation equilibrium $K_a=\dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}$ gives $K_h\cdot K_a=[H^+][OH^-]=K_w$, so $K_h=\dfrac{K_w}{K_a}$ -- a weaker parent acid (smaller $K_a$) gives a larger hydrolysis constant, i.e. more hydrolysis. As with Ostwald's law, $K_h$ can also be written in terms of the degree of hydrolysis $h$ and salt concentration $C$ as $K_h=h^2C$, giving $[OH^-]=\sqrt{K_h\cdot C}$. …