Question

Solubility products of $CuI$ and $A{g}_{2}Cr{O}_4$have almost the same value $\left(\sim 4\times 10^{-12}\right).$ The ratio of solubilities of the two salts$\left(CuI:A{g}_{{}_{2}Cr{O}_4}\right)$ is closest to

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

Answer 1

Answer: (B)

$CuI(s)\rightleftharpoons C{u}^{+}(aq)+I^{-}(aq)$
$K_{sp}=[C{u}^{+}][I^{-}]$
Since, $K_{sp}=4\times 10^{-12},x=2\times 10^{-6}M$
$A{g}_{2}Cr{O}_4(s)\rightleftharpoons 2A{g}^{+}(aq)+Cr{O}^{2-}(aq)$
$K_{sp}=[A{g}^{+}{]}^2[Cr{O}_4^{2-}]$
If its solubility $=y$, at eqbm., $[A{g}^{+}]=2y$
and $[Cr{O}_4^{2-}]=y$
$\therefore K_{sp}=(2y{)}^{2}y=4y^3$
Since, $K_{sp}=4\times 10^{-12}$
$4y^3=4\times 10^{-12}$,
$y=10^{-4}M$
$\therefore x:y=2\times 10^{-6};10^{-4}$
$=2\times 10^{-2}=0.02$