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

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,

Kh =

Kh =

As it is completely ionized,

[C5H5NCl] = [H+] = 3.63 × 10-4 M

As [C6H5NCl] = 0.02M (given)

Kh =

Kh =

Kh = 6.6 × 10-6

To calculate ionization constant, we apply the formula:

Kh = Kw / Ka

Or Ka = Kh / Kw

As the value of Kw is taken as 10-14

Kh = 6.6 × 10-6 (given)

Ka =

Ka = 1.51 × 10-9

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

Note: Kh 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), Kh = Kw / Ka

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