NCERT Class XI Chemistry Equilibrium Solutions

Question : 62

Total: 73

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

Solution:

Pyridinium hydrochloride is a salt of weak base and strong acid.
Therefore, pH = 7 −

(log X + pKb)
where Kb = dissociation constant of pyridine
3.44 = 7 -

(log 0.02 + pKb)
- 3.56 = −

(−1.70+pKb) ⇒ - 7.12 = 1.0 - pKb
pKb = 1.70 + 7.12 = 8.82,
Kb = antilog (– 8.82) = 1.513 × 10–9

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