If the equilibrium constant of the reaction of weak acid HA with strong base is 10⁹, then pH of 0.1 M Na A is

$$\begin{gathered} \mathrm{H}A+{\mathrm{OH}}^{-}\rightleftharpoons {\mathrm{H}}_2\mathrm{O}+{\mathrm{A}}^{-} \\ \mathrm{K}=\frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]\left[{\mathrm{OH}}^{-}\right]} \\ \mathrm{HA}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{-} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ \therefore \frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{K}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={\mathrm{K}}_{\mathrm{w}} \\ {\mathrm{K}}_{\mathrm{a}}={\mathrm{K}}_{\mathrm{w}}\mathrm{K}={10}^{-14}\times {10}^9={10}^{-5} \\ \therefore {\mathrm{pK}}_{\mathrm{a}}=5 \\ {\mathrm{A}}^{-} \mathrm{solution} \mathrm{is} \mathrm{alkaline} \mathrm{due} \mathrm{to} \mathrm{hydrolysis}, \\ \therefore \mathrm{pH}=7+\frac{{\mathrm{pK}}_{\mathrm{a}}}{2}+\frac{\mathrm{logC}}{2} \\ =7+\frac{5}{2}+\frac{\log 0.1}{2}=9 \end{gathered}$$