The pH of 0 . 1 M solution of acetic acid will be [degree of dissociation of acetic acid is 0 . 0132 ]

Reaction:-

$$\begin{gathered} {\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COO}}^{-} + {\mathrm{H}}^{+} \\ \mathrm{c}(0.1) 0 0 \\ \mathrm{c}-\mathrm{c\alpha } \mathrm{c\alpha } \mathrm{c\alpha } \end{gathered}$$

$\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]$

Degree of dissociation = $\mathrm{\alpha }=0.0132$

$\left[{\mathrm{H}}^{+}\right]=\mathrm{c\alpha }=0.1\times 0.0132=0.00132$

$\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=-\log 0.00132=2.88$

Hence, the $\mathrm{pH}$ of solution is $2.88$.