Can \(\mathrm{Fe}^{2+}\) and \(\mathrm{Mn}^{2+}\) be separated by precipitating \(\mathrm{FeS}(\mathrm{s})\) and not \(\mathrm{MnS}(\mathrm{s}) ?\) Assume \(\left[\mathrm{Fe}^{2+}\right]=\left[\mathrm{Mn}^{2+}\right]=\) \(\left[\mathrm{H}_{2} \mathrm{S}\right]=0.10 \mathrm{M} .\) Choose a \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) that ensures maximum precipitation of \(\mathrm{FeS}(\mathrm{s})\) but not \(\mathrm{MnS}(\mathrm{s}) .\) Will the separation be complete? For \(\mathrm{FeS}, K_{\mathrm{spa}}=6 \times 10^{2} ;\) for \(\mathrm{MnS}, K_{\mathrm{spa}}=3 \times 10^{7}\).

The general form for a solubility product expression is \(Ksp = [M^z+][A^y-]\), where M is the cation, A is the anion, z and y are their charges. For FeS, \(Ksp = [Fe^{2+}][HS^-]\). Since H2S is a weak acid, the concentration of HS- can be approximated as concentration of H2S. Therefore, for FeS, \(Ksp = [Fe^{2+}][H2S]\). Similarly, for MnS, \(Ksp = [Mn^{2+}][H2S]\).

From the expressions above, it can be seen that the [H2S] needed to begin precipitation of each metal is just \(Ksp/M^{2+}\) . It is given in the problem statement that [Fe^{2+}] = [Mn^{2+}] = [H2S] = 0.1 M . We want to find a [H_{3}O^+] that ensures maximum precipitation of FeS and not MnS. Let's denote [H_{3}O^+] as h, and [HS^-] as hs. The equilibrium concentration of HS^- can be estimated using the equilibrium constant expression for the ionization of H2S in water, which is \(Kws = [HS^-][H_{3}O^+]/ [H2S]\). Substituting, we get hs = \(Kws \cdot [H2S]/ h\). As the [HS^-] need for precipitation is bigger for FeS, set hs to match the [HS^-] needed for precipitation for both ions, that is \(Ksp_{Fe}/[Fe^{2+}] = Ksp_{Mn}/[Mn^{2+}]\). Substitute into \(Kws \cdot [H2S]/ h = Ksp_{Fe}/[Fe^{2+}] \)Solving for h, we get, h = \(Kws \cdot [H2S] \cdot [Fe^{2+}]/Ksp_{Fe}\)

To evaluate if the separation is complete, compare hs needed for precipitation of FeS with that of MnS. If hs needed for MnS precipitation is larger than that of FeS, then separation is complete. On the other hand, if hs needed for MnS precipitation is smaller than that required for FeS, then separation will not be complete.