\(\mathrm{pH}=3.44\), i. e., \(\log \left[\mathrm{H}^{+}\right]=-3.44=-4.56\)
Taking Antilog
\(\begin{aligned}
&\therefore\left[\mathrm{H}^{+}\right]=3.63 \times 10^{-4} \\
&\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}^{+} \mathrm{HCl}^{-}+\mathrm{aq} \rightleftharpoons \mathrm{C}_5 \mathrm{H}_5 \mathrm{NCl}+\mathrm{H}^{+} \\
&\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{C}_5 \mathrm{~N}_5 \mathrm{~N}^{+} \mathrm{Cl}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}^{+} \mathrm{HCl}^{-}\right]} \\
&=\frac{\left(3.63 \times 10^{-4}\right)\left(3.63 \times 10^{-4}\right)}{2 \times 10^{-2}}=6.59 \times 10^{-6} \\
&\mathrm{pK}_{\mathrm{a}}=-\log \left(6.59 \times 10^{-6}\right)=6-0.8187=5.18 \\
&\mathrm{pK}_{\mathrm{b}}+\mathrm{pK}_{\mathrm{b}}=14 \quad \therefore \mathrm{pK}_{\mathrm{b}}=14-5.18=8.82 \\
&-\operatorname{log~\mathrm {K}_{\mathrm {b}}}=8.82 \text { or } \log \mathrm{K}_{\mathrm{b}}=-8.82 \therefore \mathrm{K}_{\mathrm{b}},=1.514 \times 10^{-9}
\end{aligned}\)