Hydrolysis of Salt of Weak Acid and Weak Base

Consider ammonium acetate, $CH_3COONH_4(aq) \rightarrow CH_3COO^-(aq)+NH_4^+(aq)$, where BOTH ions hydrolyse: $CH_3COO^-+H_2O \rightleftharpoons CH_3COOH+OH^-$ and $NH_4^++H_2O \rightleftharpoons NH_4OH+H^+$. Whether the resulting solution is acidic, basic or neutral now depends purely on which parent was the weaker: if $K_a>K_b$, the solution is acidic (pH < 7); if $K_a<K_b$, the solution is basic (pH > 7); if $K_a=K_b$, the solution is neutral. The relation between the two dissociation constants and the hydrolysis constant here is $K_a\cdot K_b\cdot K_h=K_w$, and the pH of the solution is given by $pH=7+\tfrac12(pK_a-pK_b)$ -- notably independent of the salt's concentration, unlike the single-hydrolysis cases above.

Example 8.8 – hydrolysis and pH of sodium acetate. Calculate (i) the degree of hydrolysis, (ii) the hydrolysis constant, and (iii) the pH of 0.1M $CH_3COONa$, given $pK_a=4.74$ for $CH_3COOH$. Since $CH_3COONa$ is the salt of a weak acid and strong base, this is anionic hydrolysis (Section 8.8.2 formulas apply). First $K_a=\text{antilog}(-4.74)=1.8\times10^{-5}$. (i) $h=\sqrt{K_w/(K_aC)}=\sqrt{10^{-14}/(1.8\times10^{-5}\times0.1)}=\sqrt{10^{-14}/1.8\times10^{-6}}\approx7.5\times10^{-5}$. (ii) $K_h=K_w/K_a=10^{-14}/1.8\times10^{-5}=5.56\times10^{-10}$. (iii) $pH=7+\tfrac12pK_a+\tfrac12\log_{10}C=7+\tfrac{4.74}{2}+\tfrac{\log_{10}0.1}{2}=7+2.37-0.5=8.87$. …