Question

The $K_{sp}$ value for some sulphides are $K_{sp}\left(B{i}_2{S}_3\right)=1.2\times {10}^{-98};$ $K_{sp}\left(A{g}_{2}S\right)=4\times {10}^{-48};$ $K_{sp}\left(NiS\right)=1.6\times {10}^{-25}$. An unknown cation $X^{+}$ was added to a solution of $N{a}_{2}S$. The precipitation of sulphide of the cation was found to occur before $NiS$ but after $A{g}_{2}S.$ The $K_{sp}$ value for the sulphide of cation $X^{+}$ is closest to:

• A. $1\times {10}^{-26}$
• B. $1\times {10}^{-44}$
• C. $1\times {10}^{-65}$
• D. $1\times {10}^{-95}$

Answer 1

Answer: (B)

$1\times {10}^{-44}$

For the sulphide of cation $X^{+}(i.e,X_{2}S)$ the solubility must be between those of $NiS$ and $A{g}_{2}S$.

Solubility of $NiS=(1.6\times {10}^{-25}{)}^{\frac{1}{2}}=4\times {10}^{-13}$.

Solubility of $A{g}_{2}S=(\frac{4\times {10}^{-48}}{4}{)}^{\frac{1}{3}}=1\times {10}^{-16}$.
Hence the solubility of the sulphide $X_{2}S$ must be between the two values, i.e., $1\times {10}^{-15}$.

$K_{sp}\left(X_{2}S\right)=(2\times {10}^{-15}{)}^2\times (1\times {10}^{-15})=4\times {10}^{-45}\approx 1\times {10}^{-45}$

The value is closest to $1\times {10}^{-44}$.

Option B is correct.