$A$ $0.02 \ M$ solution of pyridinium hydrochloride has $pH = 3.44$. Calculate the ionization constant of pyridine.

Given: Concentration of salt $(C)$ = $0.02 \ M$,$pH = 3.44$.
Step $1$: Calculate $[H^{+}]$.
$[H^{+}] = 10^{-pH} = 10^{-3.44} = 3.63 \times 10^{-4} \ M$.
Step $2$: Calculate the hydrolysis constant $(K_h)$.
For a salt of a weak base and strong acid,$[H^{+}] = \sqrt{K_h \times C}$.
$K_h = \frac{[H^{+}]^2}{C} = \frac{(3.63 \times 10^{-4})^2}{0.02} = \frac{1.3177 \times 10^{-7}}{0.02} = 6.59 \times 10^{-6} \approx 6.6 \times 10^{-6}$.
Step $3$: Calculate the ionization constant of pyridine $(K_b)$.
$K_h = \frac{K_w}{K_b} \Rightarrow K_b = \frac{K_w}{K_h}$.
$K_b = \frac{1.0 \times 10^{-14}}{6.6 \times 10^{-6}} = 1.515 \times 10^{-9}$.