Step-by-Step Solution

1. Understand the Condition for Precipitation

   The precipitation of a sparingly soluble salt like Zinc Sulfide ($ZnS$) occurs when its ionic product ($Q_{sp}$) exceeds its solubility product constant ($K_{sp}$). The equilibrium for the dissolution of $ZnS$ is: $ZnS(s) \rightleftharpoons Zn^{2+}(aq) + S^{2-}(aq)$ The solubility product expression is: $K_{sp} = [Zn^{2+}][S^{2-}]$ To prevent precipitation, the ionic product must be less than or equal to the solubility product: $Q_{sp} = [Zn^{2+}]_{initial}[S^{2-}]_{solution} \le K_{sp}$ The minimum concentration of $H^+$ required corresponds to the threshold condition where the solution is just saturated, i.e., $Q_{sp} = K_{sp}$.

2. Calculate the Maximum Allowable Sulfide Ion Concentration ($[S^{2-}]_{max}$)

   We are given the initial concentration of $Zn^{2+}$ ions and the $K_{sp}$ for $ZnS$.

   • $[Zn^{2+}] = 0.05$ M
   • $K_{sp}(ZnS) = 1.25 \times 10^{-22}$

   Using the threshold condition, we can calculate the maximum concentration of sulfide ions, $[S^{2-}]_{max}$, that can be present in the solution without causing precipitation. $[Zn^{2+}][S^{2-}]_{max} = K_{sp}(ZnS)$ $0.05 \times [S^{2-}]_{max} = 1.25 \times 10^{-22}$ $[S^{2-}]_{max} = \frac{1.25 \times 10^{-22}}{0.05} = \frac{1.25 \times 10^{-22}}{5 \times 10^{-2}}$ $[S^{2-}]_{max} = 0.25 \times 10^{-20} = 2.5 \times 10^{-21} \text{ M}$

3. Relate Sulfide Ion Concentration to Hydrogen Ion Concentration

   The sulfide ions are produced from the dissociation of hydrogen sulfide ($H_2S$). The overall dissociation equilibrium is: $H_2S(aq) \rightleftharpoons 2H^+(aq) + S^{2-}(aq)$ The overall dissociation constant ($K_{net}$) is given by: $K_{net} = \frac{[H^+]^2[S^{2-}]}{[H_2S]} = 1 \times 10^{-21}$ This equation shows that the concentration of sulfide ions, $[S^{2-}]$, is controlled by the concentration of hydrogen ions, $[H^+]$. An increase in $[H^+]$ will decrease $[S^{2-}]$ (Le Chatelier's principle).

4. Calculate the Minimum Required Hydrogen Ion Concentration ($[H^+]_{min}$)

   To prevent precipitation, we need to ensure that $[S^{2-}] \le [S^{2-}]_{max}$. The minimum $[H^+]$ required is the concentration that keeps the sulfide ion concentration exactly at its maximum allowable value, $[S^{2-}]_{max}$.

   Rearranging the $K_{net}$ expression to solve for $[H^+]^2$: $[H^+]^2 = \frac{K_{net} [H_2S]}{[S^{2-}]}$ For the minimum $[H^+]$, we use the maximum $[S^{2-}]$: $[H^+]_{min}^2 = \frac{K_{net} [H_2S]}{[S^{2-}]_{max}}$ We are given that the solution is saturated with $H_2S$, so $[H_2S] = 0.1$ M.

   Substituting the known values: $[H^+]_{min}^2 = \frac{(1 \times 10^{-21}) \times (0.1)}{2.5 \times 10^{-21}}$ $[H^+]_{min}^2 = \frac{0.1 \times 10^{-21}}{2.5 \times 10^{-21}} = \frac{0.1}{2.5}$ $[H^+]_{min}^2 = \frac{1}{25} = 0.04$

   Now, we take the square root to find the minimum molar concentration of $H^+$: $[H^+]_{min} = \sqrt{0.04} = 0.2 \text{ M}$

Conclusion

The minimum molar concentration of $H^+$ required to prevent the precipitation of $ZnS$ is 0.2 M.