A mixture contains \(1.0 \times 10^{-3} M \mathrm{Cu}^{2+}\) and $1.0 \times 10^{-3} M$ \(\mathrm{Mn}^{2+}\) and is saturated with 0.10\(M \mathrm{H}_{2} \mathrm{S} .\) Determine a pH where CuS precipitates but MnS does not precipitate. \(K_{\mathrm{sp}}\) for \(\mathrm{CuS}=8.5 \times 10^{-45}\) and $K_{\mathrm{sp}} \mathrm{for} \mathrm{MnS}=2.3 \times 10^{-13} .$

First, we'll write down the balanced chemical equations for both CuS and MnS precipitation: \[ Cu^{2+} + HS^- \rightleftharpoons CuS + H^+ \] \[ Mn^{2+} + HS^- \rightleftharpoons MnS + H^+ \] Since we're given the solubility product constants for both CuS and MnS, we can write the equilibrium expressions as follows: \[ K_{sp_{CuS}} = [Cu^{2+}][HS^-]/[H^+] \] \[ K_{sp_{MnS}} = [Mn^{2+}][HS^-]/[H^+] \] where \(K_{sp_{CuS}}\) and \(K_{sp_{MnS}}\) are the solubility products of CuS and MnS, respectively.

We are given that the concentration of H₂S is 0.10 M. Since the only source of HS⁻ is the dissociation of H₂S, we can write an equation for the concentration of HS⁻: \[ [HS^-] = K_a[H^+]/[H_2S] \] Because the main source of H⁺ would be neither CuS nor MnS, it would be from auto-ionization of water, so we could find relationship between HS⁻ and H⁺.

Since we want CuS to precipitate but MnS not to precipitate, we can set up the following inequality: \[ K_{sp_{CuS}} \geq [Cu^{2+}][HS^-]/[H^+] > K_{sp_{MnS}} \] Substituting the expression for the concentration of HS⁻ from step 2 into the inequality, we get: \[ K_{sp_{CuS}} \geq [Cu^{2+}]K_a[H^+]/[H_2S] > K_{sp_{MnS}} \]

Now, we can substitute the given values into the inequality: \[ 8.5 \times 10^{-45} \geq (1.0 \times 10^{-3})(K_a[H^+]/0.10) > 2.3 \times 10^{-13} \] We know the \(K_a\) for H₂S is 1.0 × 10⁻⁷ , so we can now separate the inequality and solve the equation for H⁺ concentration: \[ 8.5 \times 10^{-45} \geq (1.0 \times 10^{-3})(1.0 \times 10^{-7}[H^+]/0.10) \] \[ (1.0 \times 10^{-3})(1.0 \times 10^{-7}[H^+]/0.10) > 2.3 \times 10^{-13} \] Solving for [H⁺] in these inequalities, we get the range: \[ 8.5 \times 10^{-38} \leq [H^+] < 2.3 \times 10^{-6} \] Now we can calculate pH within this range. The pH where CuS precipitates, but MnS does not, would be the pH that corresponds to the higher [H⁺] value (to make sure MnS does not precipitate): \[ pH = -\log[H^+] = -\log(2.3 \times 10^{-6}) \approx 5.64 \] So, the pH where CuS precipitates but MnS does not precipitate is approximately 5.64.