An acidified solution of 0.05 M $$Zn^{2+}$$ is saturated with 0.1 M $$H_2S$$. What is the minimum molar concentration (M) of $$H^+$$ required to prevent the precipitation of ZnS?
Use $$K_{sp}(ZnS) = 1.25 \times 10^{-22}$$ and overall dissociation constant of $$H_2S, K_{NET} = K_1K_2 = 1 \times 10^{-21}$$.

For ZnS to remain dissolved, the ionic product $$[Zn^{2+}][S^{2-}]$$ must not exceed its solubility product $$K_{sp}$$.

Given $$[Zn^{2+}] = 0.05\ \text{M}$$ and $$K_{sp}(ZnS)=1.25\times10^{-22}$$, the maximum permissible sulphide-ion concentration is

$$[S^{2-}]_{\text{max}} = \frac{K_{sp}}{[Zn^{2+}]} = \frac{1.25\times10^{-22}}{0.05} = 2.5\times10^{-21}\ \text{M}$$

In the acidified medium, the equilibria

$$H_2S \rightleftharpoons H^{+}+HS^{-}\qquad K_1$$
$$HS^{-} \rightleftharpoons H^{+}+S^{2-}\qquad K_2$$

combine to give the overall relation

$$K_{NET}=K_1K_2 =\frac{[H^{+}]^{2}[S^{2-}]}{[H_2S]}$$

Rearranging, the sulphide-ion concentration in the solution is

$$[S^{2-}] = \frac{K_{NET}\,[H_2S]}{[H^{+}]^{2}}$$

The solution is saturated with $$0.1\ \text{M}\ H_2S$$ and $$K_{NET}=1\times10^{-21}$$, so

$$[S^{2-}] = \frac{(1\times10^{-21})(0.1)}{[H^{+}]^{2}} = \frac{1\times10^{-22}}{[H^{+}]^{2}}$$

To avoid precipitation, this value must not exceed $$2.5\times10^{-21}\ \text{M}$$:

$$\frac{1\times10^{-22}}{[H^{+}]^{2}}\;\le\;2.5\times10^{-21}$$

$$[H^{+}]^{2}\;\ge\;\frac{1\times10^{-22}}{2.5\times10^{-21}} = 0.04$$

$$[H^{+}] \;\ge\; \sqrt{0.04} = 0.20\ \text{M}$$

Therefore, the minimum hydrogen-ion concentration required to prevent the precipitation of ZnS is

$$[H^{+}]_{\text{min}} = 0.2\ \text{M}$$