Question

The values of $K_{sp}$ for $CuS,A{g}_{2}S$ and $HgS$ are ${10}^{-31},{10}^{-42}$ and ${10}^{-54}$ respectively. The correct order of their solubility in water is

• A. $A{g}_{2}S>HgS>CuS$
• B. $HgS>CuS>A{g}_{2}S$
• C. $HgS>A{g}_{2}S>CuS$
• D. $A{g}_{2}S>CuS>HgS$

Answer 1

The correct option is D $A{g}_{2}S>CuS>HgS$

It is given that the values of $K_{sp}$ for $CuS,A{g}_{2}S$ and $HgS$ are ${10}^{-31},{10}^{-42}$ and ${10}^{-54}$ respectively.

a. $CuS\leftrightharpoons C{u}^{2+}+S^{2-}$

$K_{sp}=s\times s$, if $s$ is the solubility of $CuS$.

Solubility of $CuS$ $=\sqrt{K_{sp}}=\sqrt{{10}^{-31}}=3.16\times {10}^{-16}M$

b. $A{g}_{2}S\leftrightharpoons 2A{g}^{+}+S^{2-}$

$K_{sp}=(2s{)}^2\times s$, if $s$ is the solubility of $A{g}_{2}S$.

Solubility of $A{g}_{2}S={\left(\frac{K_{sp}}{4}\right)}^{1/3}={\left(\frac{{10}^{-42}}{4}\right)}^{1/3}=6.3\times {10}^{-15}M$

c. $HgS\leftrightharpoons H{g}^{2+}+S^{2-}$

$K_{sp}=s\times s$, if $s$ is the solubility of $HgS$.

Solubility of $HgS$ $=\sqrt{K_{sp}}=\sqrt{{10}^{-54}}=1\times {10}^{-27}M$

Hence, the correct order of solubility in water is $A{g}_{2}S>CuS>HgS$.