Question

$$K_{sp}$$ of $CuS$ , $A{g}_{2}S$ and $HgS$ are ${10}^{-31}$ ,${10}^{-44}$ and ${10}^{-54}$ $mo{l}^2$ $litr{e}^{-2}$ respectively. Select the correct order for their solubility in water.
1. $A{g}_{2}S$>$HgS$>$CuS$
2. $HgS$>$CuS$>$A{g}_{2}S$
3. $HgS$>$A{g}_{2}S$>$CuS$
4. $A{g}_{2}S$>$CuS$>$HgS$

Answer 1

Hint: The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol $$K_{sp}$$. It is found out by using the solubility product of the components given.

Complete Step by step answer:
Equilibrium reaction for dissolution reaction of $CuS$ is:
$CuS\rightleftharpoons [C{u}^{+2}]+[S^{-2}]$
So, $$K_{sp}$$= $[C{u}^{+2}][S^{-2}]$ = ${10}^{-31}$
Hence, we get $S^2$ =${10}^{-31}$
Therefore S= 3.16 $\times {10}^{-16}$
Equilibrium reaction for dissolution reaction of $A{g}_{2}S$ is:
$A{g}_{2}S\rightleftharpoons 2A{g}^{+}+S^{-2}$
So, $$K_{sp}$$=$[A{g}^{+}{]}^2[S^{-2}]$ = ${10}^{-44}$
Hence, we get $(2S{)}^2(S)$ = ${10}^{-44}$
Therefore, 4$S^3$ = ${10}^{-44}$
So, S= 1.357$\times {10}^{-15}$
Equilibrium reaction for dissolution of $HgS$ is:
$HgS\rightleftharpoons H{g}^{+2}+S^{-2}$
So, $$K_{sp}$$=$[H{g}^{+2}][S^{-2}]$ =${10}^{-54}$
Hence, we get $S^2$= ${10}^{-54}$
Therefore, S= $1\times {10}^{-27}$

Therefore solubility of $CuS$ , $A{g}_{2}S$ and $HgS$ are 3.16$\times {10}^{-16}$ , 1.357$\times {10}^{-15}$and$1\times {10}^{-27}$. This implies that the highest solubility is that of $A{g}_{2}S$ and lower than this is of $CuS$. And lowest solubility is that of $HgS$ according to the values found. So the correct order according to the solubility in water is $A{g}_{2}S$>$CuS$>$HgS$.

So the correct option is 4th.

Note: The solubility product is a kind of equilibrium constant and its value depends on temperature. $$K_{sp}$$ Usually increases with an increase in temperature due to increased solubility. It should be kept in mind that solubility products represent the extent of dissolution of any compound.