Question

Using the following equilibria
CH₃COO⁻ + H₂O ⇔ CH₃COOH + OH⁻
and CH₃COOH ⇔ H⁺ + CH₃COO⁻
show that K_h = (K_w/K_a)

Answer 1

$K_h=\frac{[CH_3COOH]*[OH^-]}{[CH_3COO^-]}$

$K_a=\frac{[H^+]*[CH_3COO^-]}{[CH_3COOH]}$

$K_w/K_a=[H^+]*[OH^-]/\frac{[H^+]*[CH_3COO^-]}{[CH_3COOH]}=\frac{[H^+]*[OH^-]*[CH_3COOH]}{[H^+]*[CH_3COO^-]}=\frac{[CH_3COOH]*[OH^-]}{[CH_3COO^-]}=K_h$