How much water must be added to 300 mL of 0.2 M solution of ${\mathrm{CH}}_3\mathrm{COOH}\left({\mathrm{K}}_{\mathrm{a}}=1.8\times {10}^{-5}\right)$ for the degree of ionization  $\left(\mathrm{\alpha }\right)$ of the acid to double?

$$\begin{gathered} \because \mathrm{\alpha }\text{ is negligible w.r.t. }1 \\ {\mathrm{K}}_{\mathrm{a}}={\mathrm{C}}_1{\mathrm{\alpha }}_1^2={\mathrm{C}}_2{\mathrm{\alpha }}_2^2 \\ \therefore {\mathrm{C}}_2={\mathrm{C}}_1{\left(\frac{{\mathrm{\alpha }}_1}{{\mathrm{\alpha }}_2}\right)}^2 \\ =0.2\times \frac{1}{4}=0.05 \\ {\mathrm{C}}_1{\mathrm{V}}_1={\mathrm{C}}_2{\mathrm{V}}_2 \\ \Rightarrow 300\times 0.2=0.05\times {\mathrm{V}}_2 \\ \Rightarrow {\mathrm{V}}_2=1200 \mathrm{mL} \end{gathered}$$

Volume of H₂O added =1200 - 300

                = 900mL