Question

Derive an expression for the hydrolysis constant of sodium acetate.

Answer 1

$$Hydrolysis\ of\ Sodium\ acetate\ -\\CH_3COO^-+ H_2O \iff CH_3COOH+OH^-$$

equilibrium constant equation -

$$K_e=\frac{[CH_3COOH][OH^-]}{[CH_3COO^-][H_2O]}$$

Concentration of water is very large and is regarded as practically constant so it will assumes the

form -

$$K_h=\frac{[CH_3COOH][OH^-]}{[CH_3COO^-]}\ \ \ \ .......(1)$$

but,

$$K_w=[H^+][OH^-]\ \ \ \ ........(2)$$

For the dissociation of weak acid acetic acid,

$$CH_3COOH\iff H^++CH_3COO^-$$

acid dissociation constant, $K_a$ is expressed as

$$K_a=\frac{[H^+][CH_3COO^-]}{[CH_3COOH]}\ \ \ \ \ .......(3)$$

Dividing (2) by (3)

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

or,

$$K_h=\frac{K_w}{K_a}$$