The percentage of pyridine $\left({\mathrm{C}}_5{\mathrm{H}}_5 \mathrm{N}\right)$ that forms pyridinium ion$\left({\mathrm{C}}_5{\mathrm{H}}_5{ \mathrm{N}}^{+}\mathrm{H}\right)$ in a $0.10\mathrm{M}$ aqueous pyridine solution $\left({\mathrm{K}}_{\mathrm{b}}\right.$ for $\left.{\mathrm{C}}_5{\mathrm{H}}_5 \mathrm{N}=1.7\times {10}^{-9}\right)$  is

$\begin{array}{l}{\mathrm{C}}_5{\mathrm{H}}_5 \mathrm{N}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{C}}_5{\mathrm{H}}_5\mathrm{NH}+{\mathrm{OH}}^{-}\\ 0.10\mathrm{M}\end{array}$

$\mathrm{\alpha }=\sqrt{\frac{{\mathrm{K}}_{\mathrm{b}}}{\mathrm{C}}}=\sqrt{\frac{1.7\times {10}^{-9}}{0.10}}=1.30\times {10}^{-4}$

$\therefore \mathrm{Percentage} \mathrm{of} \mathrm{pyridine} \mathrm{that} \mathrm{forms} \mathrm{pyridinium} \mathrm{ion}$

$=1.30\times {10}^{-4}\times 100=0.013\%$