Question

Solubility products of $CuI$ and $ Ag_{2}CrO_{4} $have almost the same value $\left(\sim4\times10^{-12}\right).$ The ratio of solubilities of the two salts$\left(CuI:Ag_{_2CrO_4}\right)$ is closest to

• A. 0.01
• B. 0.02
• C. 0.03
• D. 0.10

Answer 1

Answer: (B)

$CuI(s) \rightleftharpoons Cu^+(aq) + I^- (aq)$
$K_{sp} = [ Cu^+][I^-]$
Since, $K_{sp} = 4 \times 10^{-12}, x = 2 \times 10^{-6}\,M$
$Ag_2CrO_4(s) \rightleftharpoons 2Ag^+ (aq) + CrO^{2-}(aq)$
$K_{sp} = [Ag^+]^2[CrO_4^{2-}]$
If its solubility $ = y $, at eqbm., $[Ag^+] = 2y$
and $[CrO_4^{2-}] = y$
$\therefore K_{sp} = (2y)^2y = 4y^3$
Since, $K_{sp} = 4\times 10^{-12}$
$4y^3 = 4\times 10^{-12}$,
$y = 10^{-4}\,M$
$\therefore x : y = 2\times 10^{-6}; 10^{-4}$
$ = 2\times 10^{-2} = 0 . 02$