In a qualitative analysis, Bi3+ is detected by appearance of precipitate of BiO(OH)(s). Calculate pH when the following equilibrium exists at 298 K : BiO(OH)(s) BiO+(aq) + OH-(aq), K = 4 × 10-10 (Given : log 2 = 0.3010)
- 4.699
- 5.286
- 8.714
- 9.301
Answer
(4) 9.301
Solution
BiO(OH)(s) BiO+(aq) + OH-(aq)
Let solubility = S mol L-1 ⇒ [BiO+] = [OH-] = S
K = S2 = 4 × 10-10 ⇒ S = 2 × 10-5 M = [OH-]
⇒ pOH = −log(2 × 10-5) = 5 − 0.3010 = 4.699
∴ pH = 14 − 4.699 = 9.301
Answer: (4)