Assuming that the degree of hydrolysis is small, the pH of 0.1 M solution of sodium acetate $\left(K_{\mathrm{a}}^{\circ }=1.0\times {10}^{-5}\right)$ will be

The acetate ion undergoes hydrolysis as ${\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{OH}}^{-}$

$K_{\mathrm{h}}=\frac{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]}\equiv \frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]\left[{\mathrm{H}}^{+}\right]/\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}=\frac{K_{\mathrm{w}}}{K_{\mathrm{a}}}=\frac{1.0\times {10}^{-14}{\mathrm{M}}^2}{1.0\times {10}^{-5}\mathrm{M}}=1.0\times {10}^{-9}\mathrm{M}$also,$\left[{\mathrm{CH}}_3\mathrm{COOH}\right]=\left[{\mathrm{OH}}^{-}\right]$  Hence, $K_{\mathrm{h}}=\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}_0}\text{ or }\left[{\mathrm{OH}}^{-}\right]={\left(K_{\mathrm{h}}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}_0\right)}^{1/2}$  O$\begin{array}{r}\left[{\mathrm{OH}}^{-}\right]={\left[\left(1.0\times {10}^{-9}\mathrm{M}\right)\left(0.1\mathrm{M}\right)\right]}^{1/2}=1.0\times {10}^{-5}\mathrm{M}\\ \left[{\mathrm{H}}^{+}\right]=K_{\mathrm{w}}/\left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-14}{\mathrm{M}}^2/\left(1.0\times {10}^{-5}\mathrm{M}\right)=1.0\times {10}^{-9}\mathrm{M}\\ \mathrm{pH}=-\log \left(\left[{\mathrm{H}}^{+}\right]/\mathrm{M}\right)=9\end{array}$