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
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
The product of reaction of I- with MnO4- in alkaline medium :-...Arsenic (III) sulphide form a sol with negative charge. Which of the following ionic substances will have lowest floceul...The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.2 M NaOH would be : (pKa for CH3COOH = ...Given Ag(NH3)2+$\rightleftharpoons$ Ag+ + 2NH3, Kc = 7.2 × 10–8 and Ksp of AgCl = 1.8 × 10–10 at 298 K. If ammonia is a...Pure ammonia is placed in a vessel at a temperature where its dissociation constant (α) is appreciable. At equilibr...