0.25M solution of pyridinium chloride was found to have a pH of 2.699. What is Kb for pyridine?

The Chemical formulae of pyridine and pyridinium ion are C₅H₅N and C₅H₅NH⁺ respectively.

C₅H₅NH⁺(aq) + H₂O(l) ⇌ C₅H₅N(aq) + H₃O⁺(aq) …. Ka = [C₅H₅N] [H₃O⁺] / [C₅H₅NH⁺]

At equilibrium :
pH = 2.699, hence [H₃O⁺] = 10⁻²·⁶⁹⁹ M = 0.002 M
Then, [C₅H₅N] = [H₃O⁺] = 0.002 M
and [C₅H₅NH⁺] = (0.25 - 0.002) M = 0.248 M

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]
Kb = [H₃O⁺] [OH⁻] [C₅H₅NH⁺] / [C₅H₅N] [H₃O⁺]
Kb = Kw × {[C₅H₅NH⁺] / [C₅H₅N] [H₃O⁺]}
Kb = (1 × 10⁻¹⁴) × (0.25 - 10⁻²·⁶⁹⁹) / (10⁻²·⁶⁹⁹)²
Kb = 6.2 × 10⁻¹⁰