Given that $K_{\mathrm{a}1}=1.1 \times 10^{-7}, K_{\mathrm{a}2}=1.3 \times 10^{-13}$ for $\mathrm{H}_{2} \mathrm{S}$; $K_{\mathrm{sp}}(\mathrm{ZnS})=2.5 \times 10^{-22}$. Dissolve $0.01 \mathrm{mol}\ \mathrm{ZnS}$ in $1.0\ \mathrm{L}$ hydrochloric acid. The closest value for the minimum concentration of hydrochloric acid required is
已知 $\mathrm{H}_{2} \mathrm{S}$ 的 $K_{\mathrm{a}1}=1.1 \times 10^{-7}, K_{\mathrm{a}2}=1.3 \times 10^{-13} ; K_{\mathrm{sp}}(\mathrm{ZnS})=2.5 \times 10^{-22} $。将$0.01 \mathrm{mol}\ \mathrm{ZnS}$溶于$1.0\ \mathrm{L}$盐酸。所需盐酸的最低浓度最接近的值为
$0.06 \ \mathrm{mol/L}$
$0.07 \ \mathrm{mol/L}$
$0.08 \ \mathrm{mol/L}$
$0.09 \ \mathrm{mol/L}$
$0.10 \ \mathrm{mol/L}$
$0.11 \ \mathrm{mol/L}$
$0.06 \ \mathrm{mol/L}$
$0.07 \ \mathrm{mol/L}$
$0.08 \ \mathrm{mol/L}$
$0.09 \ \mathrm{mol/L}$
$0.10 \ \mathrm{mol/L}$
$0.11 \ \mathrm{mol/L}$
This is a very classic physical chemistry problem. The equation for the reaction is $\mathrm{ZnS+2H^+=Zn^{2+}+H_2S}$. It can be calculated that the equilibrium constant for this reaction is:

$K=\frac{[Zn^{2+}] [H_2S]}{[H^+] ^2}=\frac{[Zn^{2+}] [H_2S] [S^{2-}]}{[H^+] ^2[S^{2-}]}=K_{sp}(ZnS)\frac{[H_2S]}{[H^+] ^2[S^{2-}]}=\frac{K_{sp}(ZnS)}{K_{a_1}K_{a_2}}=1.7\times10^{-2}$
<Checkpoint>[$K=1.7\times 10^{-2}$](1)</Checkpoint>

When $\mathrm{ZnS}$ is completely dissolved, $0.01\ \mathrm{mol\cdot L}\ Zn^{2+}$ and $0.01\ \mathrm{mol\cdot L}\ H_2S$ are generated. The equilibrium $[H^{+}]$ can be calculated:

$[H^+]_{eq}=\sqrt{\frac{0.01\times 0.01}{K}}=0.077\ \mathrm{mol/L}$

<Checkpoint>[$[H^+]_{eq}=0.077\ \mathrm{mol/L}$](1)</Checkpoint>

Since dissolution occurs, $[H^+]_{total}=0.02+[H^+]_{eq}=0.097\ \mathrm{mol/L}$

<Checkpoint>[$[H^+]_{total}=0.097\ \mathrm{mol/L}$](1)</Checkpoint>
这是一道非常经典的物理化学题目。发生反应的方程式为$\mathrm{ZnS+2H^+=Zn^{2+}+H_2S}$。可以计算，该反应的平衡常数为：

<Checkpoint>[$\mathrm{ZnS+2H^+=Zn^{2+}+H_2S}$](1)</Checkpoint>

$K=\frac{[Zn^{2+}] [H_2S]}{[H^+] ^2}=\frac{[Zn^{2+}] [H_2S] [S^{2-}]}{[H^+] ^2[S^{2-}]}=K_{sp}(ZnS)\frac{[H_2S]}{[H^+] ^2[S^{2-}]}=\frac{K_{sp}(ZnS)}{K_{a_1}K_{a_2}}=1.7\times10^{-2}$

<Checkpoint>[$K=1.7\times 10^{-2}$](1)</Checkpoint>

当$\mathrm{ZnS}$完全溶解后，生成$0.01\ \mathrm{mol\cdot L}\ Zn^{2+}$和$0.01\ \mathrm{mol\cdot L}\ H_2S$。由于$K_{a_1}<10^{-5}$，可以忽略$H_2S$的解离，从而$c_{H_2S}=[H_2S]_{eq}$。

<Checkpoint>[由于$K_{a_1}<10^{-5}$，可以忽略$H_2S$的解离，从而$c_{H_2S}=[H_2S]_{eq}$](1)</Checkpoint>

可以计算出平衡时$[H^{+}]$：

$[H^+]_{eq}=\sqrt{\frac{0.01\times 0.01}{K}}=0.077\ \mathrm{mol/L}$

<Checkpoint>[$[H^+]_{eq}=0.077\ \mathrm{mol/L}$](1)</Checkpoint>

由于发生溶解过程，$[H^+]_{total}=0.02+[H^+]_{eq}=0.097\ \mathrm{mol/L}$

<Checkpoint>[$[H^+]_{total}=0.097\ \mathrm{mol/L}$](1)</Checkpoint>