Question

If the solubility product of $CuS$ is $6 \times 10^{-16}$, the maximum molarity of $CuS$ in aqueous solution will be

• A. $3.45 \times 10^{-5}\, mol^{-1}$
• B. $2.45 \times 10^{-8}\, mol^{-1}$
• C. $2.45 \times 10^{8}\, mol^{-1}$
• D. $3.45 \times 10^{5}\, mol^{-1}$

Answer 1

Answer: (B)

Maximum molarity of $CuS$ in aqueous solution = Solubility of $CuS$ in mol $L^{-1}$
If $S$ is the solubility of $CuS$ in mol $L^{-1}$, then
$CuS\rightleftharpoons Cu^{2+} + S^{2-}, K_{sp} = [Cu^{2+}][S^{2-} = S \times S = S^{2}$
$\therefore S^{2} = 6 \times 10^{-16}$
or $S = \sqrt{6 \times 10^{-16}} = 2.45 \times 10^{-8}\, mol \,L^{-1}$