The expression of degree of hydrolysis of strong conjugate base such as acetate ion is given by the expression

where K_a is the ionization constant of conjugate acid of the base.

$$\begin{gathered} {\mathrm{Ac}}^{-} +{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{HAc}+{\mathrm{OH}}^{-} \\ \mathrm{c}(1-\mathrm{\alpha } ) \mathrm{c\alpha } \mathrm{c\alpha } \end{gathered}$$

$K_{\mathrm{b}}=\frac{\left[\mathrm{HAc}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{Ac}}^{-}\right]}=\frac{\left(c\alpha \right)\left(c\alpha \right)}{c(1-\alpha )}=\frac{c{\alpha }^2}{1-\alpha }$

Assuming$\alpha \ll 1,K_{\mathrm{b}}=c{\alpha }^2\text{ or }\alpha =\sqrt{K_{\mathrm{b}}/c}\text{. }$But $K_{\mathrm{b}}=K_w/K_{\mathrm{a}}$ where ${\mathrm{K}}_{\mathrm{a}}$ is the ionization constant of acetic acid .Hence $\alpha =\sqrt{K_{\mathrm{w}}/K_{\mathrm{a}}c}$.