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Chemistry Question on Equilibrium

The Ksp of Ag2CrO4K_{sp} \, of \, Ag_2CrO_4 is 1.1×10−121.1 \times 10^{ - 12 } at 298 K. The solubility (in mol/L) of Ag2CrO4Ag_2 CrO_4 in a0.1M AgNO3a 0.1 M \, AgNO_3 solution is

A

1.1×10−111.1 \times 10^{ - 11}

B

1.1×10−101.1 \times 10^{ - 10}

C

1.1×10−121.1 \times 10^{ - 12}

D

1.1×10−91.1 \times 10^{ - 9}

Answer

1.1×10−101.1 \times 10^{ - 10}

Explanation

Solution

PLAN In presence of common ion (in this case Ag+Ag^+ ion) solubility of
sparingly soluble salt is decreased.
Let solubility of Ag2CrO4Ag_2 CrO_4 in presence of 0.1 M
AgNO3=xAgNO_3 = x
Ag2CrO4⇌2Ag++CrO42−Ag_2CrO_4 \rightleftharpoons 2 Ag^+ + CrO_4^{ 2 - }
AgNO3⇌Ag++NO3−AgNO_3 \rightleftharpoons Ag^+ + NO_3^-
Total [ Ag+ Ag^+ | = (2v + 0.1) M = 0.1 M
as x <<< 0.1 M
[CrO42]=x MCrO_4^2 ] = x \, M
Thus, [Ag−]2 [CrO42]=Ksp[ Ag^ - ]^2 \, [ CrO_4^2 ] = K_{ sp }
(0.1)2 (x)=1.1×10−12(0 . 1)^2 \, (x) = 1.1 \times 10^{ - 12 }
x = 1. 1 ×10−10\times 10^{ - 10} M