Given:

pH = 3.44

[C₆H₅NCl] = 0.02M

As we know that,

pH = -log[H⁺]

3.44= -log[H⁺]

By taking antilog of both the sides, we get

Antilog 3.44= [H⁺]

[H⁺] = 3.63 × 10^-4

Ionisation of pyridinium hydrochloride [C₅H₅NCl]

C₆H₅NCl + aq ↔ C₅H₅NCl + H⁺

Using the formula,

K_h =

⇒K_h =

As it is completely ionized,

∴ [C₅H₅NCl] = [H⁺] = 3.63 × 10^-4 M

As [C₆H₅NCl] = 0.02M (given)

∴ K_h =

⇒K_h =

⇒K_h = 6.6 × 10^-6

To calculate ionization constant, we apply the formula:

K_h = K_w / K_a

Or K_a = K_h / K_w

As the value of K_w is taken as 10^-14

K_h = 6.6 × 10^-6 (given)

∴ K_a =

⇒K_a = 1.51 × 10^-9

Thus, the ionization constant of pyridine is 1.51 × 10^-9

Note: K_h is the degree of hydrolysis of a salt which is defined as the fraction of a mole of the salt which is hydrolyzed. When the equilibrium is attained. Hence for a salt (acid and base), K_h = K_w / K_a