Derive the equation for the ionization constant $K_a$ of a weak acid $HX$.

(N/A) weak acid $HX$ is partially ionized in an aqueous solution. The equilibrium is expressed as:
$HX_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^{+} + X_{(aq)}^{-}$
Initial Concentration $(M)$: $C, 0, 0$
Change in concentration: $-C\alpha, +C\alpha, +C\alpha$
Concentration at equilibrium: $C(1-\alpha), C\alpha, C\alpha$
where $\alpha$ is the degree of ionization.
The equilibrium constant expression is:
$K = \frac{[H_3O^{+}][X^{-}]}{[HX][H_2O]}$
Since $[H_2O]$ is constant in dilute solutions,we define the acid dissociation constant $K_a$ as:
$K_a = K[H_2O] = \frac{[H_3O^{+}][X^{-}]}{[HX]}$
Substituting the equilibrium concentrations:
$K_a = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)}$
$K_a = \frac{C^2\alpha^2}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$
This is the expression for the ionization constant of a weak acid.