Equilibrium Ques 99
99. The solubility products of $CuS, Ag_2 S$ and $HgS$ are $10^{-31}, 10^{-44}, 10^{-54}$ respectively. The solubilities of these sulphides are in the order
[1997]
(a) $Ag_2 S>HgS>CuS$
(b) $Ag_2 S>CuS>HgS$
(c) $HgS>Ag_2 S>CuS$
(d) $CuS>Ag_2 S>HgS$
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Answer:
Correct Answer: 99.(b)
Solution:
- (b) For Binary salts like $CuS$ & $HgS$, solubilty,
$s=\sqrt{K _{s p}}$
$\Rightarrow s _{CuS}=\sqrt{10^{-31}}=3.16 \times 10^{-16}$,
$s _{HgS}=\sqrt{10^{-54}}=10^{-27}$
For $Ag_2 S \to \underset{2 s}{2 Ag^{+}}+\underset{s}{S^{2}}$
$K _{s p}=4 s^{3}$ or $s _{Ag_2 S}=\sqrt[3]{\frac{K _{s p}}{4}}=\sqrt[3]{\frac{10^{-44}}{4}}$
$ =\sqrt[3]{0.25 \times 10 \times 10^{-45}}=13.54 \times 10^{-16} $
$\therefore \quad$ The order is $Ag_2 S>CuS>HgS$
BETA
