Q. 624.0( 6 Votes )

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

Answer :

Given:

pH = 3.44

[C_{6}H_{5}NCl] = 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 [C_{5}H_{5}NCl]

C_{6}H_{5}NCl + aq ↔ C_{5}H_{5}NCl + H^{+}

Using the formula,

K_{h} =

⇒K_{h} =

As it is completely ionized,

∴ [C_{5}H_{5}NCl] = [H^{+}] = 3.63 × 10^{-4} M

As [C_{6}H_{5}NCl] = 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}

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