Question

The $ K_{sp} $ of $ Ag_2CrO_4 $ is $ 1.1 \times 10^{ - 12 } $ at $298\, K$. The solubility (in mol/L) of $Ag_2 CrO_4 $ in $ a 0.1 \,M \, AgNO_3 $ solution is

• A. $ 1.1 \times 10^{ - 11} $
• B. $ 1.1 \times 10^{ - 10} $
• C. $ 1.1 \times 10^{ - 12} $
• D. $ 1.1 \times 10^{ - 9} $

Answer 1

Answer: (B)

Let solubility of $Ag_2 CrO_4$ in presence of $0.1 \,M$
$ AgNO_3 = x $
$ Ag_2CrO_4 \rightleftharpoons \underset{2x}{2 Ag^+} + \underset{x}{CrO_4^{ 2 - }} $
$ AgNO_3 \rightleftharpoons \underset{0.1}{Ag^+ }+ \underset{0.1}{ NO_3^-} $
Total [ $ Ag^+ ] = (2x + 0.1) M = 0.1\, M$
as $x < < < 0.1\, M$
[$CrO_4^2 ] = x \, M $
Thus, $ [ Ag^ +]^2 \, [ CrO_4^{2-} ] = K_{ sp } $
$ (0 . 1)^2 \, (x) = 1.1 \times 10^{ - 12 } $
$\because x = 1. 1 \times 10^{ - 10} \, M$