A 0.02 M solution of pyridinium hydrochloride has pH = 3.44.

$\mathrm{pH}=3.44$

We know that,

$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

$\therefore\left[\mathrm{H}^{+}\right]=3.63 \times 10^{-4}$

Then, $K_{h}=\frac{\left(3.63 \times 10^{-4}\right)^{2}}{0.02}$

$(\because$ concentration $=0.02 \mathrm{M})$

$\Rightarrow K_{h}=6.6 \times 10^{-6}$

Now, $K_{b}=\frac{K_{w}}{K_{a}}$

$\Rightarrow K_{a}=\frac{K_{w}}{K_{h}}=\frac{10^{-14}}{6.6 \times 10^{-6}}$

$=1.51 \times 10^{-9}$