99 Percentile Qs Bank for JEE MainChemistryIonic Equilibrium
Determine the solubility of silver chromate at 298 K given its K sp value is 1.1 × 10 -12 .
Options
- A6.5 × 10 -5
- B2.4 × 10 -2
- C3.6 × 10 -3
- D8.9 × 10 -4
Correct answer
A. 6.5 × 10 -5
Step-by-step solution
Ag 2 CrO 4 ⇌ 2Ag + + CrO 4 2 - ; K sp = 1.1 × 10 - 12 S 2S S K sp = Ag + 2 · CrO 4 2 - K sp = [2S] 2 · [S] = 4S 3 S 3 = K sp 4 = 1.1 × 10 - 12 4 ⟹ S = 6.53 × 10 - 5