A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine
Given, pH = 3.44
We know that
pH = -log [H+]
∴ [H+] = 3.63x10-4
Then Kh = ( 3.63x10-4 )2 / 0.02 (∵ Concentration is 0.02M)
⇒ Kh = 6.6 x 10-6
Now , Kh = Kw / Ka
⇒ Ka = Kw / Kh
= 10-14 / 6.6 x 10-6
= 1.51 x 10-9
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 62: A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of py....
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