Question
On decreasing the pH from 7 to 2 , the solubility of a sparingly soluble salt ( MX ) of a weak acid ( HX ) increased from 10 - 4   mol   L - 1 to 10 - 3   mol   L - 1 . The pK a of HX is
JEE Advanced
Chemistry
Ionic Equilibrium
2023
JEE Advanced 2023 (Paper 1)
JEE Advanced Chemistry Question (2023) — Solution
Options
- A. 3
- B. 4
- C. 5
- D. 2
Answer
B. 4
Step-by-step solution
At pH = 7 , i.e., the solubility in water is S , The solubility product, K sp = S 2     - - - - - > 1 Assume the solubility of sparingly soluble salt is at   pH = 2 MX ⇌ M ⊕ + X ⊖                         S 1               S 1 - x The weak dissociation reaction is as follows, X ⊖ + H ⊕ ⇌ HX S 1 - x     10 - 2             x Assume the acid dissociation constant is K a . Now, 1 K a = HX H + X - K sp = S 1 S 1 - x 1 K a = S 1 H + S 1 - x assume S 1 ≃ x S 1 - x = K sp S 1 1 K a = S 1 2 H + K sp ⇒ S 1 2   = 10 - 2 K sp K a - - - - - - - > 2 Now from 1 and 2, S 1 2 S 2 = 10 - 2 K a ⇒ K a = 10 - 2 × 10 - 8 10 - 6 ⇒ K a = 10 - 4 ⇒ pK a = 4
Related: Chemistry — Ionic Equilibrium · All PYQ Banks