What is the pH of a 1 M ${\mathrm{CH}}_3\mathrm{COOH}$ a solution $K_a$ of acetic acid = $1.8 \times {10}^{-5} ?$

$\mathrm{K} = {10}^{-14} {\mathrm{mol}}^2 {\mathrm{litre}}^{-2}$

${\mathrm{CH}}_3{\mathrm{COO}}^{-} + {\mathrm{H}}_2\mathrm{O} \rightleftharpoons {\mathrm{CH}}_3\mathrm{COOH} + {\mathrm{OH}}^{-}$

$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right] = \mathrm{c} \times \mathrm{h}; \mathrm{h} = \sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}} \times \mathrm{c}} = \sqrt{\frac{{10}^{-14}}{1.8 \times {10}^{-5} }\times 1} \\ =2.35 \times {10}^{-5} \\ \therefore \mathrm{pOH}=4.62; \mathrm{pH} = 9.38 \approx 9.4 \end{gathered}$$