The percentage of aniline $\left({\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\right)$ which forms anilinium ion $\left({\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_3^{+}\right)$ in 0.25 M aqueous solution is $\left({\mathrm{K}}_{\mathrm{b}} \mathrm{of} {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2={10}^{-10}\right)$

$$\begin{gathered} \mathrm{Aniline} \mathrm{is} \mathrm{a} \mathrm{weak} \mathrm{base}. \\ \mathrm{\alpha }=\sqrt{\frac{{\mathrm{K}}_{\mathrm{b}}}{\mathrm{C}}} \\ \mathrm{\alpha }=\sqrt{\frac{{10}^{-10}}{0.25}} \\ \mathrm{\alpha }=2\times {10}^{-5} \\ \mathrm{Percentage} \mathrm{of} \mathrm{ionization} = 2\times {10}^{-5}\times 100 \\ \mathrm{Percentage} \mathrm{of} \mathrm{ionization} =2\times {10}^{-3} \end{gathered}$$