JEE Advanced | 2013Ionic EquilibriumHard
Question
The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in mol/L) of Ag2 CrO4 in a 0.1 M AgNO3 solution is
Options
A.1.1 × 10-11
B.1.1 × 10-10
C.1.1 × 10-12
D.1.1 × 10-9
Solution
Ag2CrO4 = 2Ag+(aq) + CrO42- (aq)
(2s + 0.1) s
Ksp = [Ag+]2 [CrO42-]
Ksp = (2s + 0.1)2 (s)
2s < < 0.1
⇒1.1 × 10-12 = (0.1)2 × s
s = 1.1 × 10-10
(2s + 0.1) s
Ksp = [Ag+]2 [CrO42-]
Ksp = (2s + 0.1)2 (s)
2s < < 0.1
⇒1.1 × 10-12 = (0.1)2 × s
s = 1.1 × 10-10
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