Calculate degree of dissociation of ${\mathrm{CH}}_3\mathrm{COOH}$ if $0.1\mathrm{M} {\mathrm{CH}}_3\mathrm{COOH}$ $({\mathrm{K}}_{\mathrm{a}}= {10}^{–5})$ has $\mathrm{pH} = 2$.

$$\begin{gathered} \underset{{\mathrm{C}}_0-\mathrm{x}}{{\mathrm{CH}}_3\mathrm{COOH}} \perp \underset{\mathrm{x}}{{\mathrm{CH}}_3{\mathrm{COO}}^{- }}+ \underset{{10}^{-2}}{{\mathrm{H}}^{+}} {\mathrm{K}}_{\mathrm{a}}={10}^{–5} \\ \mathrm{pH} = 2 \Rightarrow \left[{\mathrm{H}}^{+}\right] = {10}^{–2} \\ {\mathrm{K}}_{\mathrm{a}\left( {\mathrm{CH}}_3\mathrm{COOH}\right)} = \frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{CH}}_3{\mathrm{COO}}^{- }\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]} \\ {10}^{–5} = \frac{{10}^{-2}\times \mathrm{x}}{0.1-\mathrm{x}}\Rightarrow \mathrm{x} = {10}^{-4} \\ {\mathrm{\alpha }}_{{\mathrm{CH}}_3\mathrm{COOH}} = \frac{\mathrm{x}}{{\mathrm{C}}_0} = \frac{{10}^{-4}}{0.1} = {10}^{-3} \end{gathered}$$

If it is solution of only acetic acid then $\left[{\mathrm{H}}^{+}\right] = {\mathrm{C}}_0\mathrm{\alpha } = {10}^{–1} \times {10}^{–2} = {10}^{–3}$ but $\left[{\mathrm{H}}^{+}\right]$ is greater than this value i.e. some other source of ${\mathrm{H}}^{+}$ present.

$\mathrm{\alpha } = \sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{{\mathrm{C}}_0}} = \sqrt{\frac{{10}^{-5}}{{10}^{-1}}} = 0.01$