Question
In a qualitative analysis Bi ^ 3+ 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 : 2 = 0.3010)
NEET
Chemistry
Ionic Equilibrium
2026
NEET 2026 (Cancelled)
NEET Chemistry Question (2026) — Solution
Options
- A. 8.714
- B. 4.699
- C. 5.286
- D. 9.301
Answer
D. 9.301
Step-by-step solution
The equilibrium reaction is given by: BiO(OH)(s) BiO ^ + (aq) + OH ^ - (aq) The equilibrium constant expression is: K = [ BiO ^ + ][ OH ^ - ] Let the solubility of BiO(OH) be s. Then [ BiO ^ + ] = s and [ OH ^ - ] = s. Substituting the values into the equilibrium expression: s^2 = 4 10^ -10 s = 2 10^ -5 M Therefore, the concentration of hydroxide ions is: [ OH ^ - ] = 2 10^ -5 M Calculating the pOH: pOH = - [ OH ^ - ] = - (2 10^ -5 ) = 5 - 2 Given 2 = 0.3010: pOH = 5 - 0.3010 = 4.699 The pH of the solution at 298 K is: pH = 14 - pOH = 14 - 4.699 = 9.301 Answer: 9.301
Related: Chemistry — Ionic Equilibrium · All PYQ Banks