Ammonium sulfide is a widely used qualitative analytical reagent. To prepare this reagent, hydrogen sulfide gas is introduced into 4~5 mol/L aqueous ammonia, followed by the addition of water. Pure solutions cannot be obtained by this method: when absorbing gas below or above the stoichiometric amount, the solution will contain excess ammonia or ammonium hydrosulfide.

Dilute 10.00 mL of the ammonium sulfide test solution to 1.000 L, then transfer 10.00 mL of the resulting stock solution to a distillation flask and add approximately 40 mL of water. Next, add 25.00 mL of 0.1 mol/L cadmium nitrate solution to the collection flask (where the distilled components will condense), and introduce 20.00 mL of 0.02498 mol/L sulfuric acid into the distillation flask.

Distill approximately half of the solution from the distillation flask into the collection flask. (A yellow precipitate can be observed in the collection flask.)

Thoroughly wash the remaining contents in the distillation flask into a conical flask, add a few drops of methyl red indicator, and titrate with 0.05002 mol/L NaOH solution, consuming 10.97 mL of NaOH at the endpoint.

Add bromine water to the solution in the collection flask (the precipitate dissolves), heat and boil the solution for 15 minutes to remove excess bromine. The bromine oxidizes all sulfur-containing ions to sulfate ions. Titrate the solution in the collection flask with 0.1012 mol/L NaOH, consuming 15.01 mL.

Calculate the pH of this reagent solution and select the closest option.

$\mathrm{NH_3: K_b = 1.76×10^{-5}; H_2S: K_{a1} = 1.1×10^{-7}, K_{a2} = 1.3×10^{-13}}$
硫化铵是广泛使用的定性分析化学试剂，配制此试剂时，将硫化氢气体导入4~5 mol/L的氨水中，之后再加水。用这个方法无法配制得到纯的溶液：当吸收了过低或过高于化学计量的气体时，溶液中会含有过量的氨或硫氢化铵。

将 10.00 mL 的硫化铵试液稀释到1.000 L，取 10.00 mL 所得到的储存溶液移至一个蒸馏瓶并且加入大约 40 mL 的水。然后，将 25.00 mL 的 0.1 mol/L 硝酸镉溶液加到收集瓶(蒸馏出的成分将凝结于此)。并且将 20.00 mL 的 0.02498 mol/L 硫酸溶液加入蒸馏瓶。

将蒸馏瓶中大约一半的溶液蒸馏至收集瓶。(收集瓶中可以看到黄色的沉淀物。）

将留在蒸馏瓶中的物质完全洗涤至锥形瓶，加入数滴甲基红指示剂之后以0.05002 mol/L NaOH溶液滴定，终点时消耗NaOH溶液10.97 mL。

将溴水加入收集瓶中的溶液(沉淀物溶解)，加热煮沸溶液15分钟除去多余的溴。溴将所有含硫离子氧化为硫酸根离子。收集瓶中的溶液用 0.1012 mol/L NaOH 滴定，消耗 15.01 mL 。

计算此试剂溶液的pH，选择最接近的选项。

$\mathrm{NH_3: K_b = 1.76×10^{-5}; H_2S: K_{a1} = 1.1×10^{-7}, K_{a2} = 1.3×10^{-13}}$
3.72
4.13
4.89
5.02
6.27
7.33
7.92
8.54
9.54
10.06
10.54
11.46
12.09
12.68
13.38
3.72
4.13
4.89
5.02
6.27
7.33
7.92
8.54
9.54
10.06
10.54
11.46
12.09
12.68
13.38
Ions remaining in the distillation flask are $\mathrm{NH_4^+}$ and $\mathrm{H^+}$, and only $\mathrm{H^+}$ is consumed when methyl red is used as an indicator.

<Checkpoint>[Only $\mathrm{H^+}$ is titrated when methyl red is used as an indicator](1)</Checkpoint>

The remaining $\mathrm{NH_4^+}$ is produced by the protonation of $\mathrm{NH_4^+}$ by $\mathrm{H_2SO_4}$, therefore:
$\mathrm{c(NH_3) = (0.02000×0.02498×2-0.05002×0.01097)×}\frac{100}{0.01000}\mathrm{=4.505 (M)}$

<Checkpoint>[Ammonium sulfide reagent solution contains 4.505 M $\mathrm{NH_3}$](1)</Checkpoint>

Two reactions occurred in the receiving flask:

$\mathrm{Cd^2+ + H_2S = CdS + 2H^+}$
$\mathrm{CdS+ 4Br_2 + 4H_2O = CdSO_4 + 8HBr}$

Therefore, one equivalent of $\mathrm{H_2S}$ produces ten equivalents of $\mathrm{H^+}$.

<Checkpoint>[$\mathrm{H_2S-10H^+}$](1)</Checkpoint>

Thus: $\mathrm{c(H_2S) = 0.1012×0.01501×0.1×}\frac{100}{0.01000}\mathrm{=1.519 (M)}$

That is, $\mathrm{c(total S) = 1.519 M}$ in the reagent.

<Checkpoint>[$\mathrm{c(total S) = 1.519 M}$](1)</Checkpoint>

$\mathrm{c(NH_4HS) = 4.505-1.519 = 2.986 M}$

<Checkpoint>[$\mathrm{c(NH_3) = 4.505-1.519 = 2.986 M}$](1)</Checkpoint>

Charge balance:
$\mathrm{[NH_4^+] + [H^+] = 2[S^{2-}] + [HS^-] + [OH^-]}$

<Checkpoint>[$\mathrm{[NH_4^+] + [H^+] = 2[S^{2-}] + [HS^-] + [OH^-]}$](1)</Checkpoint>

Substituting into the distribution fraction expression, we get $\mathrm{[H^+] = 2.88×10^{-10} M}$

<Checkpoint>[$\mathrm{[H^+] = 2.88×10^{-10} M}$](1)</Checkpoint>

pH = 9.54

<Checkpoint>[pH = 9.54](1)</Checkpoint>
留在蒸馏瓶中的离子有$\mathrm{NH_4^+}$和$\mathrm{H^+}$，以甲基红为指示剂只会消耗掉$\mathrm{H^+}$

<Checkpoint>[以甲基红为指示剂只会滴定$\mathrm{H^+}$](1)</Checkpoint>

剩下的$\mathrm{NH_4^+}$都是$\mathrm{H_2SO_4}$质子化$\mathrm{NH_4^+}$产生的，因此：
$\mathrm{c(NH_3) = (0.02000×0.02498×2-0.05002×0.01097)×}\frac{100}{0.01000}\mathrm{=4.505 (M)}$

<Checkpoint>[硫化铵试剂溶液含4.505 M的$\mathrm{NH_3}$](1)</Checkpoint>

接收瓶发生了两个反应：

$\mathrm{Cd^2+ + H_2S = CdS + 2H^+}$
$\mathrm{CdS+ 4Br_2 + 4H_2O = CdSO_4 + 8HBr}$

因此一当量$\mathrm{H_2S}$产生十当量$\mathrm{H^+}$。

<Checkpoint>[$\mathrm{H_2S-10H^+}$](1)</Checkpoint>

故:$\mathrm{c(H_2S) = 0.1012×0.01501×0.1×}\frac{100}{0.01000}\mathrm{=1.519 (M)}$

即试剂中$\mathrm{c(总S) = 1.519 M}$

<Checkpoint>[$\mathrm{c(总S) = 1.519 M}$](1)</Checkpoint>

$\mathrm{c(NH_4HS) = 4.505-1.519 = 2.986 M}$

<Checkpoint>[$\mathrm{c(NH_3) = 4.505-1.519 = 2.986 M}$](1)</Checkpoint>

电荷守恒：
$\mathrm{[NH_4^+] + [H^+] = 2[S^{2-}] + [HS^-] + [OH^-]}$

<Checkpoint>[$\mathrm{[NH_4^+] + [H^+] = 2[S^{2-}] + [HS^-] + [OH^-]}$](1)</Checkpoint>

代入分布分数表达式，解得$\mathrm{[H^+] = 2.88×10^{-10} M}$

<Checkpoint>[$\mathrm{[H^+] = 2.88×10^{-10} M}$](1)</Checkpoint>

pH = 9.54

<Checkpoint>[pH = 9.54](1)</Checkpoint>