Highly selective Backlog Qs for JEE MainChemistryIonic Equilibrium
A 0 . 02   M solution of pyridinium hydrochloride has pH = 3 . 44 . Calculate the ionization constant of pyridine .
Options
- A0 . 8 × 10 - 11
- B1 . 5 × 10 - 9
- C2 . 6 × 10 - 7
- D3 . 4 × 10 - 8
Correct answer
B. 1 . 5 × 10 - 9
Step-by-step solution
C 5 H 5 N · HCl + H 2 O ⇌ C 5 H 5 N + Cl - + H 3 O + K a = C 5 H 5 N + Cl - H 3 O + C 5 H 5 N · HCl pH = 3 . 44 = - log H 3 O + log H 3 O + = - 3.44 = 4 ¯ .56 H 3 O + = antilog 4 ¯ .56 = 3.63 × 10 - 4 M H 3 O + = C 5 H 5 NCl - = 3 . 63 × 10 - 4 Hence ,   K a = 3 .63 × 10 - 4 × 3 .63 × 10 - 4 0 .02 = 6 .588 × 10 - 6 pK a = - log   K a = - log   6 . 588 × 10 - 6 pK a = 6 + ( - 0 . 8187 ) = 5 . 18 pK a + pK b = 14 pK b = 14 - 5 . 18 = 8 . 82 -