JEE Advanced2013ChemistryIonic EquilibriumActual
The K sp of Ag 2 CrO 4 is 1 . 1 × 1 0 - 1 2 at 298 K . The solubility (in mol L - 1 ) of Ag 2 CrO ⁡ 4 in a 0 . 1 M AgNO 3 solution is
Options
- A1 . 1 × 1 0 - 1 1
- B1 . 1 × 1 0 - 1 0
- C1 . 1 × 1 0 - 1 2
- D1 . 1 × 1 0 - 9
Correct answer
B. 1 . 1 × 1 0 - 1 0
Step-by-step solution
Ag 2 CrO 4   s ⇌ 2 Ag +   aq + CrO 4 2 - 2 S   molar S   molar AgNO 3   aq ⟶ Ag +   aq + NO 3 -   aq 0 . 1   molar 0 . 1   molar Ag + = 2 S + 0 . 1   molar CrO 4 2 - = S  molar Here, S is the solubility of Ag 2 CrO 4 in aqueous AgNO 3 . ⇒ Ag + 2  CrO 4 2 - = 2 S + 0 . 1 2 S = 1 . 1 × 1 0 - 1 2 S ≪ 0 . 1 ⇒ 0 . 1 2 S = 1 . 1 × 10 - 12 = 10 - 2 × S S = 1 . 1 × 10 - 10   M