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Chemistry Question on Solutions

If the solubility product of CuS is 6×10–166 × 10^{–16}, calculate the maximum molarity of CuS in aqueous solution.

Answer

The correct answer is: 2.45×10−8molL−1.2.45 \times 10^{-8} mol L^{-1}.
Solubility product of CuS, Ksp=6×10−16K_{sp} = 6 \times 10^{-16}
Lets be the solubility of CuS in molL−1.mol L^{-1}.
CuS↔Cu2++S2−CuS ↔Cu^{2+} + S^{2-}
ss s s
Now, Ksp=[Cu2+][S2−]K_{sp}=[Cu^{2+} ][S^{2-}]
=s×s=s \times s
=s2= s^2
Then, we have, Ksp=s2=6×10−16K_{sp} = s^2=6×10^{-16}
⇒s=6×10−16⇒s=\sqrt{6\times10^{-16}}
=2.45×10−8molL−1= 2.45 \times 10^{-8}mol L^{-1}
Hence, the maximum molarity of CuS in an aqueous solution is 2.45×10−8molL−1.2.45 \times 10^{-8} mol L^{-1}.