If $H_2S$ is passed in a solution of $0.1 \ M \ Zn^{2+}$ and $0.01 \ M \ Cu^{+}$ and the concentration of $S^{2-}$ is made $8.1 \times 10^{-31} \ M$. Which precipitation of $ZnS$ and $CuS$ will take place? Given: $K_{sp}(ZnS) = 3.0 \times 10^{-23}$ and $K_{sp}(CuS) = 8.0 \times 10^{-34}$.

(B) The ionic product $(Q_{sp})$ for $ZnS$ is calculated as: $Q_{sp}(ZnS) = [Zn^{2+}][S^{2-}] = (0.1)(8.1 \times 10^{-31}) = 8.1 \times 10^{-32}$.
Since $Q_{sp}(ZnS) < K_{sp}(ZnS)$ $(8.1 \times 10^{-32} < 3.0 \times 10^{-23})$,$ZnS$ will not precipitate.
The ionic product $(Q_{sp})$ for $CuS$ is calculated as: $Q_{sp}(CuS) = [Cu^{+}][S^{2-}] = (0.01)(8.1 \times 10^{-31}) = 8.1 \times 10^{-33}$.
Since $Q_{sp}(CuS) > K_{sp}(CuS)$ $(8.1 \times 10^{-33} > 8.0 \times 10^{-34})$,$CuS$ will precipitate.
Therefore,only $CuS$ will precipitate.