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

Answer: (D)

Solubility product $(K_{sp})$ of $CuS={10}^{-31}$ liability of $CuS=\sqrt{K_{sp}}=\sqrt{{10}^{-31}}$ $=3.16\times {10}^{-16}mol/lit$ Similarly, solubility of $A{g}_{2}S$ $=\sqrt[3]{\frac{K_{sp}}{4}}$ $(\because forA{g}_{2}S,4s^3=K_{sp})$ $=\sqrt{\frac{{10}^{-42}}{4}}$ $=6.3\times {10}^{-15}mol/litre$ Solubility of $HgS=\sqrt{K_{sp}}$ $=\sqrt{{10}^{-54}}$ $={10}^{-27}mol/lit$ Hence, the correct order of solubility is $A{g}_{2}S>CuS>HgS$