Q.Derive an expression for Ostwald's dilution law.

A weak acid $HA$ is only partially dissociated: $HA+H_2O \rightleftharpoons H_3O^++A^-$, with dissociation constant $K_a=\dfrac{[H^+][A^-]}{[HA]}$ (writing $H_3O^+$ as $H^+$ for convenience); the analogous weak-base constant is $K_b=\dfrac{[B^+][OH^-]}{[BOH]}$.

Ostwald's dilution law relates $K_a$ to the degree of dissociation $\alpha$ (the fraction of moles that dissociate at equilibrium) and the concentration $C$. Working through acetic acid, $CH_3COOH \rightleftharpoons H^++CH_3COO^-$, with equilibrium concentrations $(1-\alpha)C$, $\alpha C$, $\alpha C$: $K_a=\dfrac{(\alpha C)^2}{(1-\alpha)C}=\dfrac{\alpha^2 C}{1-\alpha}$. Since a weak acid dissociates only slightly, $\alpha \ll 1$ so $(1-\alpha)\approx1$, giving the simplified law $K_a \approx \alpha^2 C$, i.e. $\alpha=\sqrt{K_a/C}$ -- as dilution increases (C falls), $\alpha$ rises, which is Ostwald's dilution law in words: dilution increases the degree of dissociation of a …