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


[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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