The following reactions control the stepwise hydrolysis and sulfur substitution sequence of molybdate and thiomolybdate in dilute aqueous solution (the concentration of water is already included in the equilibrium constant):

$\mathrm{MoS_4^{2-}} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{MoOS_3^{2-}} + \mathrm{H_2S(aq.)} \quad K_1 = 1.3 \times 10^{-5}$

$\mathrm{MoOS_3^{2-}} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{MoO_2S_2^{2-}} + \mathrm{H_2S(aq.)} \quad K_2 = 1.0 \times 10^{-5}$

$\mathrm{MoO_2S_2^{2-}} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{MoO_3S^{2-}} + \mathrm{H_2S(aq.)} \quad K_3 = 1.6 \times 10^{-5}$

$\mathrm{MoO_3S^{2-}} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{MoO_4^{2-}} + \mathrm{H_2S(aq.)} \quad K_4 = 6.5 \times 10^{-6}$

$\mathrm{MoS_4^{2-}}$ with an initial concentration of $2.00\times10^{-7}\ \mathrm{mol\;L^{-1}}$ undergoes hydrolysis in a buffer solution at $\text{pH}=5.00$. Considering the ionization and volatilization of $\mathrm{H_2S(aq.)}$, it is known that:

- The buffer capacity is large enough that the $\text{pH}$ does not change, and the buffer species do not interfere with the given reactions.
- $\mathrm{p}K_{a1} = 6.88,\; \mathrm{p}K_{a2} = 12.90$ for $\mathrm{H_2S}$
- The solution volume is $V_{\rm aq}=1.00\ \mathrm{L}$, and the gas phase volume above is $V_{\rm g}=5.00\ \mathrm{L}$ (the whole system is closed)
- Henry's constant for $\mathrm{H_2S}$ is $H = 9.741\times10^{-3}\ \mathrm{atm}/(\mathrm{mol}\cdot\mathrm{m^{-3}})$
- Take $T=298.15\ \mathrm{K}$, and the ideal gas constant is $R=0.082057\ \mathrm{L\;atm\;mol^{-1}\;K^{-1}}$

Based on the above information, calculate the concentration of each species, and based on the concentration of each species, calculate the ratio of the product of the concentrations/partial pressures of all molybdenum-containing species to the product of the concentrations of all sulfur-containing, molybdenum-free species (normalization is required, the concentration is in $\dfrac{c}{c^{\ominus}}$, and the partial pressure is in $\dfrac{p}{p^{\ominus}}$). The final result should retain the appropriate number of significant figures, and choose the option within $10\%$ of the calculated result; otherwise, choose option A: All other options are incorrect.
以下反应控制了稀薄水溶液中钼酸根与硫代钼酸根的逐步水解与硫取代序列（水的浓度已包含在平衡常数中）：

$\mathrm{MoS_4^{2-}} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{MoOS_3^{2-}} + \mathrm{H_2S(aq.)} \quad K_1 = 1.3 \times 10^{-5}$

$\mathrm{MoOS_3^{2-}} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{MoO_2S_2^{2-}} + \mathrm{H_2S(aq.)} \quad K_2 = 1.0 \times 10^{-5}$

$\mathrm{MoO_2S_2^{2-}} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{MoO_3S^{2-}} + \mathrm{H_2S(aq.)} \quad K_3 = 1.6 \times 10^{-5}$

$\mathrm{MoO_3S^{2-}} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{MoO_4^{2-}} + \mathrm{H_2S(aq.)} \quad K_4 = 6.5 \times 10^{-6}$

初始浓度为 $2.00\times10^{-7}\ \mathrm{mol\;L^{-1}}$ 的 $\mathrm{MoS_4^{2-}}$ 在 $\text{pH}=5.00$ 的缓冲溶液中发生水解。考虑到 $\mathrm{H_2S(aq.)}$ 的电离与挥发，已知：

- 缓冲容量足够大，$\text{pH}$ 不会发生变化，且缓冲物种不会干扰给出的反应
- $\mathrm{H_2S}$ 的 $ \mathrm{p}K_{a1} = 6.88,\; \mathrm{p}K_{a2} = 12.90$
- 溶液体积 $V_{\rm aq}=1.00\ \mathrm{L}$，上方气相体积 $V_{\rm g}=5.00\ \mathrm{L}$（整体为密闭系统）
- $\mathrm{H_2S}$ 的亨利常数 $H = 9.741\times10^{-3}\ \mathrm{atm}/(\mathrm{mol}\cdot\mathrm{m^{-3}})$
- 取 $T=298.15\ \mathrm{K}$，理想气体常数 $R=0.082057\ \mathrm{L\;atm\;mol^{-1}\;K^{-1}}$

根据以上信息，计算各物种的浓度，并根据各物种的浓度，计算所有含钼物种与所有含硫不含钼的物种的浓度/气压乘积的浓度乘积的比值（要求归一化，浓度以 $\dfrac{c}{c^{\ominus}}$ 计，气压以 $\dfrac{p}{p^{\ominus}}$ 计）。最终结果保留合适的有效数字位数，选择与计算结果相差 $10\%$ 以内的选项，否则选择选项 A：其他选项均不正确。
All other options are incorrect
$3.1\times 10^{-5}$
$7.7\times 10^{-8}$
$2.2\times 10^{-10}$
$3.9\times 10^{-12}$
$4.3\times 10^{-13}$
$9.1\times 10^{-14}$
$3.1\times 10^{-16}$
$5.9\times 10^{-18}$
$9.3\times 10^{-20}$
$1.7\times 10^{-25}$
$2.8\times 10^{-28}$
其他选项均不正确
$3.1\times 10^{-5}$
$7.7\times 10^{-8}$
$2.2\times 10^{-10}$
$3.9\times 10^{-12}$
$4.3\times 10^{-13}$
$9.1\times 10^{-14}$
$3.1\times 10^{-16}$
$5.9\times 10^{-18}$
$9.3\times 10^{-20}$
$1.7\times 10^{-25}$
$2.8\times 10^{-28}$
Known
- Initial: $[{\rm MoS_4^{2-}}]_0=C_0=2.00\times10^{-7}\ \mathrm{mol\;L^{-1}}$
- Equilibrium constants for stepwise hydrolysis of $\rm MoS_4^{2-}$: $K_1=1.3\times10^{-5},\;K_2=1.0\times10^{-5},\;K_3=1.6\times10^{-5},\;K_4=6.5\times10^{-6}$
- The $\text{pH}$ of the solution is fixed at $5.00$
- Ionization of $\mathrm{H_2S}$: $\mathrm{p}K_{a1}=6.88,\;\mathrm{p}K_{a2}=12.90$.
- System volume: solution $V_{\rm aq}=1.00\ \mathrm{L}$, gas phase $V_g=5.00\ \mathrm{L}$
- Henry's constant (original units): $H=9.741\times10^{-3}\ \mathrm{atm}/(\mathrm{mol\;m^{-3}})$. Converted to the form with $\mathrm{mol\;L^{-1}}$ yields $H' = H\times1000 = 9.741\ \mathrm{atm\;L\;mol^{-1}}$
- Take $T=298.15\ \mathrm{K},\;R=0.082057\ \mathrm{L\;atm\;mol^{-1}\;K^{-1}}$

Let the concentration of molecular $\mathrm{H_2S}$ in the solution at equilibrium be $x=[\mathrm{H_2S(aq)}]$.

<Checkpoint>[Let the concentration of molecular $\mathrm{H_2S}$ in the solution at equilibrium be $x=[\mathrm{H_2S(aq)}]$](0.5)</Checkpoint>

Define the concentrations of Mo species:
$[A_0]=[\mathrm{MoS_4^{2-}}],\;[A_1]=[\mathrm{MoOS_3^{2-}}],\;[A_2]=[\mathrm{MoO_2S_2^{2-}}],\;[A_3]=[\mathrm{MoO_3S^{2-}}],\;[A_4]=[\mathrm{MoO_4^{2-}}]$.



Express each $\mathrm{Mo}$ species using the equilibrium constant, from the reaction equilibrium (using concentrations instead of activities):

$$
[A_1]=\frac{K_1}{x}[A_0]
$$
$$
[A_2]=\frac{K_1K_2}{x^2}[A_0]
$$
$$
[A_3]=\frac{K_1K_2K_3}{x^3}[A_0]
$$
$$
[A_4]=\frac{K_1K_2K_3K_4}{x^4}[A_0]
$$

Let:

$$
D(x)=1+\frac{K_1}{x}+\frac{K_1K_2}{x^2}+\frac{K_1K_2K_3}{x^3}+\frac{K_1K_2K_3K_4}{x^4}
$$

Then the molybdenum conservation gives:

$$
[A_0]=\frac{C_0}{D(x)}
$$

<Checkpoint>[[$\mathrm{MoS_4^{2-}}$] and the relationship between the total $\mathrm{Mo}$ concentration is $[A_0]=\dfrac{C_0}{D(x)}$ ](1)</Checkpoint>

Let:

$$
S(x)=\frac{K_1}{x}+\frac{2K_1K_2}{x^2}+\frac{3K_1K_2K_3}{x^3}+\frac{4K_1K_2K_3K_4}{x^4},
$$

Then the "total equivalents of $\mathrm{H_2S}$ produced" can be written from the equivalents of $\mathrm{H_2S}$ generated in each step (calculated per liter of solution):

$$
\text{produced }\mathrm{H_2S} = \frac{C_0}{D(x)}\,S(x)
$$

<Checkpoint>[The relationship for the total equivalents of $\mathrm{H_2S}$ produced is $\text{produced }\mathrm{H_2S} = \dfrac{C_0}{D(x)}\,S(x)$](1)</Checkpoint>

Further, consider the partitioning relationship of $\mathrm{H_2S}$ between the solution and the gas phase (under the condition of fixed $\text{pH}=5$)

In the solution, the total dissolved sulfur equivalents (in $\mathrm{H_2S}$ equivalents) is:

$$
c_{\rm aq} = x\Big(1+\alpha_1+\alpha_2\Big)
$$

Where:

$$
\alpha_1=\frac{[\mathrm{HS^-}]}{[\mathrm{H_2S}]}=10^{\mathrm{pH}-\mathrm{p}K_{a1}}
$$
$$
\alpha_2=\frac{[\mathrm{S^{2-}}]}{[\mathrm{H_2S}]}=10^{2\mathrm{pH}-\mathrm{p}K_{a1}-\mathrm{p}K_{a2}}
$$

<Checkpoint>[According to $\mathrm{p}K_a$, the relationship for the total dissolved sulfur equivalents is $ c_{\rm aq} = x\Big(1+\alpha_1+\alpha_2\Big)$](1)</Checkpoint>

The partial pressure of $\mathrm{H_2S}$ in the gas phase is given by Henry's law:

$$
p = H'\,x
$$

The number of moles in the gas phase (calculated per liter of solution):

$$
n_{\rm g\;per\;L}=\frac{p\,V_g}{R T}=\frac{H' x\,V_g}{R T}
$$

Overall mass conservation ($\mathrm{H_2S}$ equivalents):

$$
\frac{C_0}{D(x)}S(x) \;=\; x\Big(1+\alpha_1+\alpha_2\Big) \;+\; \frac{H' x\,V_g}{R T}
$$

<Checkpoint>[The overall mass conservation relationship for $\mathrm{H_2S}$ is $ \dfrac{C_0}{D(x)}S(x) \;=\; x\Big(1+\alpha_1+\alpha_2\Big) \;+\; \dfrac{H' x\,V_g}{R T} $](1)</Checkpoint>

Simplified, we obtain the equation (with unknown $x$):

$$
\,x \;=\; \dfrac{C_0\,S(x)}{D(x)\;F}\,
$$

Where:
$$
F \equiv \Big(1+\alpha_1+\alpha_2\Big)+\dfrac{H'V_g}{R T}.
$$

<Checkpoint>[Obtain the equation with unknown $x$, $x \;=\; \dfrac{C_0\,S(x)}{D(x)\;F}\,,\quad F \equiv \Big(1+\alpha_1+\alpha_2\Big)+\dfrac{H'V_g}{R T}$](1)</Checkpoint>

Further, use numerical root-finding methods such as the bisection method or Newton's method to solve this equation, $f(x)=\dfrac{C_0 S(x)}{D(x)} - x\cdot F=0$.

First, calculate the obtained intermediate constant values:

$$
\alpha_1 = 10^{5.00-6.88} = 1.3183\times10^{-2}
$$
$$
\alpha_2 = 10^{10.00-19.78} = 1.6596\times10^{-10}
$$

$$
H' = 9.741\ \mathrm{atm\cdot L\;mol^{-1}}
$$
$$
\frac{H'V_g}{R T} \approx 1.9908
$$

$$
F \approx (1+\alpha_1+\alpha_2) + 1.9908 \approx 3.0040
$$

<Checkpoint>[$ F \approx (1+\alpha_1+\alpha_2) + 1.9908 \approx 3.0040$ ](1)</Checkpoint>

Finally, solving yields:

$$
\,x=[\mathrm{H_2S(aq)}] \;=\; 2.6363\times10^{-7}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[The concentration of $\mathrm{H_2S}$ is $x=2.6363\times10^{-7}\ \mathrm{mol\;L^{-1}}$](3)</Checkpoint>

Next, use $x$ to calculate the concentrations of other species, obtaining:

$$
D(x)\approx 2.914291\times10^{6}
$$

$$
[A_0]=[\mathrm{MoS_4^{2-}}]=\frac{C_0}{D(x)}\approx 6.8627\times10^{-14}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[\mathrm{MoS_4^{2-}}]=6.8627\times10^{-14}\ \mathrm{mol\;L^{-1}}$](0.5)</Checkpoint>

$$
[A_1]=[{\rm MoOS_3^{2-}}]=\frac{K_1}{x}[A_0]\approx 3.3841\times10^{-12}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[{\rm MoOS_3^{2-}}]=3.3841\times10^{-12}\ \mathrm{mol\;L^{-1}}$](0.5)</Checkpoint>

$$
[A_2]=[{\rm MoO_2S_2^{2-}}]\approx 1.2836\times10^{-10}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[{\rm MoO_2S_2^{2-}}]=1.2836\times10^{-10}\ \mathrm{mol\;L^{-1}}$](0.5)</Checkpoint>

$$
[A_3]=[{\rm MoO_3S^{2-}}]\approx 7.7905\times10^{-9}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[{\rm MoO_3S^{2-}}]=7.7905\times10^{-9}\ \mathrm{mol\;L^{-1}}$](0.5)</Checkpoint>

$$
[A_4]=[{\rm MoO_4^{2-}}]\approx 1.9208\times10^{-7}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[{\rm MoO_4^{2-}}]=1.9208\times10^{-7}\ \mathrm{mol\;L^{-1}}$](0.5)</Checkpoint>

Ionization products in solution:

$$
[\mathrm{HS^-}]=\alpha_1\,x \approx 3.4754\times10^{-9}\ \mathrm{mol\;L^{-1}}
$$
$$
[\mathrm{S^{2-}}]=\alpha_2\,x \approx 4.3752\times10^{-17}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[\mathrm{HS^-}]=\alpha_1\,x \approx 3.4754\times10^{-9}\ \mathrm{mol\;L^{-1}}$](1)</Checkpoint>


<Checkpoint>[$ [\mathrm{S^{2-}}]=\alpha_2\,x \approx 4.3752\times10^{-17}\ \mathrm{mol\;L^{-1}}$](1)</Checkpoint>

Gas phase species:

$$
p=H'x\approx 2.5680\times10^{-6}\ \mathrm{atm}
$$

<Checkpoint>[Partial pressure of $\mathrm{H_2S}$ gas $p= 2.5680\times10^{-6}\ \mathrm{atm}$](0.5)</Checkpoint>

Verify the correctness of the results by verifying the material balance of $\mathrm{Mo}$:

$$
\sum_{i=0}^4 [A_i] = 2.0000\times10^{-7}\ \mathrm{M}= C_0
$$

<Checkpoint>[Verify $ \sum_{i=0}^4 [A_i] = C_0 $](0.5)</Checkpoint>

Finally, according to the requirements of the question, calculate:

$$
P_{\rm S}=\frac{[\mathrm{H_2S(aq)}]}{c^\ominus}\cdot\frac{[\mathrm{HS^-}]}{c^\ominus}\cdot\frac{[\mathrm{S^{2-}}]}{c^\ominus}\cdot\frac{p(\mathrm{H_2S})}{p^\ominus}\approx 1.02944\times10^{-37}
$$
$$
P_{\rm Mo}=\prod_{i=0}^{4}\frac{[A_i]}{c^\ominus}
=\frac{[{\rm MoS_4^{2-}}]\,[{\rm MoOS_3^{2-}}]\,[{\rm MoO_2S_2^{2-}}]\,[{\rm MoO_3S^{2-}}]\,[{\rm MoO_4^{2-}}]}{(c^\ominus)^5}\approx 4.46088\times10^{-50}
$$

According to the question, retain two significant figures, and the final ratio is:

$$
\dfrac{P_{\rm Mo}}{P_{\rm S}}=4.3333\times 10^{-13}\approx4.3\times 10^{-13}
$$

<Checkpoint>[The final ratio is $4.3\times 10^{-13}$](2)</Checkpoint>
已知
- 初始：$[{\rm MoS_4^{2-}}]_0=C_0=2.00\times10^{-7}\ \mathrm{mol\;L^{-1}}$
- $\rm MoS_4^{2-}$ 逐级水解的平衡常数：$K_1=1.3\times10^{-5},\;K_2=1.0\times10^{-5},\;K_3=1.6\times10^{-5},\;K_4=6.5\times10^{-6}$
- 溶液的 $\text{pH}$ 固定为 $5.00$
- $\mathrm{H_2S}$ 的电离：$\mathrm{p}K_{a1}=6.88,\;\mathrm{p}K_{a2}=12.90$。
- 体系体积：溶液 $V_{\rm aq}=1.00\ \mathrm{L}$，气相 $V_g=5.00\ \mathrm{L}$
- 亨利常数（原单位）：$H=9.741\times10^{-3}\ \mathrm{atm}/(\mathrm{mol\;m^{-3}})$。换算为与 $\mathrm{mol\;L^{-1}}$ 的形式得 $H' = H\times1000 = 9.741\ \mathrm{atm\;L\;mol^{-1}}$
- 取 $T=298.15\ \mathrm{K},\;R=0.082057\ \mathrm{L\;atm\;mol^{-1}\;K^{-1}}$

设平衡时溶液中分子型 $\mathrm{H_2S}$ 浓度为 $x=[\mathrm{H_2S(aq)}]$。

<Checkpoint>[设平衡时溶液中分子型 $\mathrm{H_2S}$ 浓度 $x=[\mathrm{H_2S(aq)}]$](0.5)</Checkpoint>

定义 Mo 物种浓度：
$[A_0]=[\mathrm{MoS_4^{2-}}],\;[A_1]=[\mathrm{MoOS_3^{2-}}],\;[A_2]=[\mathrm{MoO_2S_2^{2-}}],\;[A_3]=[\mathrm{MoO_3S^{2-}}],\;[A_4]=[\mathrm{MoO_4^{2-}}]$。



用平衡常数表示各 $\mathrm{Mo}$ 物种，由反应平衡（以浓度代替活度）：

$$
[A_1]=\frac{K_1}{x}[A_0]
$$
$$
[A_2]=\frac{K_1K_2}{x^2}[A_0]
$$
$$
[A_3]=\frac{K_1K_2K_3}{x^3}[A_0]
$$
$$
[A_4]=\frac{K_1K_2K_3K_4}{x^4}[A_0]
$$

令：

$$
D(x)=1+\frac{K_1}{x}+\frac{K_1K_2}{x^2}+\frac{K_1K_2K_3}{x^3}+\frac{K_1K_2K_3K_4}{x^4}
$$

则钼守恒给出：

$$
[A_0]=\frac{C_0}{D(x)}
$$

<Checkpoint>[$[\mathrm{MoS_4^{2-}}]$ 与 $\mathrm{Mo}$ 总浓度的关系为 $[A_0]=\dfrac{C_0}{D(x)}$ ](1)</Checkpoint>

令:

$$
S(x)=\frac{K_1}{x}+\frac{2K_1K_2}{x^2}+\frac{3K_1K_2K_3}{x^3}+\frac{4K_1K_2K_3K_4}{x^4},
$$

则由每步生成 $\mathrm{H_2S}$ 的当量可写出“产生的  $\mathrm{H_2S}$ 总当量”（按每升溶液计）：

$$
\text{produced }\mathrm{H_2S} = \frac{C_0}{D(x)}\,S(x)
$$

<Checkpoint>[产生的 $\mathrm{H_2S}$ 总当量的关系式为 $\text{produced }\mathrm{H_2S} = \dfrac{C_0}{D(x)}\,S(x)$](1)</Checkpoint>

进一步，考虑溶液与气相中 $\mathrm{H_2S}$ 的分配关系（在 $\text{pH}=5$  固定的条件下）

在溶液中，总溶解硫当量（以 $\mathrm{H_2S}$ 当量计）为：

$$
c_{\rm aq} = x\Big(1+\alpha_1+\alpha_2\Big)
$$

其中：

$$
\alpha_1=\frac{[\mathrm{HS^-}]}{[\mathrm{H_2S}]}=10^{\mathrm{pH}-\mathrm{p}K_{a1}}
$$
$$
\alpha_2=\frac{[\mathrm{S^{2-}}]}{[\mathrm{H_2S}]}=10^{2\mathrm{pH}-\mathrm{p}K_{a1}-\mathrm{p}K_{a2}}
$$

<Checkpoint>[根据 $\mathrm{p}K_a$，总溶解硫当量的关系式为 $ c_{\rm aq} = x\Big(1+\alpha_1+\alpha_2\Big)$](1)</Checkpoint>

气相中 $\mathrm{H_2S}$ 分压由亨利定律给出：

$$
p = H'\,x
$$

气相摩尔数（按每升溶液计）：

$$
n_{\rm g\;per\;L}=\frac{p\,V_g}{R T}=\frac{H' x\,V_g}{R T}
$$

总体质量守恒（$\mathrm{H_2S}$ 当量）：

$$
\frac{C_0}{D(x)}S(x) \;=\; x\Big(1+\alpha_1+\alpha_2\Big) \;+\; \frac{H' x\,V_g}{R T}
$$

<Checkpoint>[$\mathrm{H_2S}$ 的总体质量守恒关系为 $ \dfrac{C_0}{D(x)}S(x) \;=\; x\Big(1+\alpha_1+\alpha_2\Big) \;+\; \dfrac{H' x\,V_g}{R T} $](1)</Checkpoint>

化简得方程（未知量为 $x$）：

$$
\,x \;=\; \dfrac{C_0\,S(x)}{D(x)\;F}\,
$$

其中：
$$
F \equiv \Big(1+\alpha_1+\alpha_2\Big)+\dfrac{H'V_g}{R T}.
$$

<Checkpoint>[得未知量为 $x$的方程 $x \;=\; \dfrac{C_0\,S(x)}{D(x)\;F}\,,\quad F \equiv \Big(1+\alpha_1+\alpha_2\Big)+\dfrac{H'V_g}{R T}$](1)</Checkpoint>

进一步，使用二分法或牛顿法等数值求根的方法，求解这个方程，$f(x)=\dfrac{C_0 S(x)}{D(x)} - x\cdot F=0$。

首先，计算得到的中间常量值：

$$
\alpha_1 = 10^{5.00-6.88} = 1.3183\times10^{-2}
$$
$$
\alpha_2 = 10^{10.00-19.78} = 1.6596\times10^{-10}
$$

$$
H' = 9.741\ \mathrm{atm\cdot L\;mol^{-1}}
$$
$$
\frac{H'V_g}{R T} \approx 1.9908
$$

$$
F \approx (1+\alpha_1+\alpha_2) + 1.9908 \approx 3.0040
$$

<Checkpoint>[$ F \approx (1+\alpha_1+\alpha_2) + 1.9908 \approx 3.0040$ ](1)</Checkpoint>

最终，进行求解得到：

$$
\,x=[\mathrm{H_2S(aq)}] \;=\; 2.6363\times10^{-7}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$\mathrm{H_2S}$ 的浓度为 $x=2.6363\times10^{-7}\ \mathrm{mol\;L^{-1}}$](3)</Checkpoint>

接下来，用 $x$ 计算其他物种浓度，计算得到：

$$
D(x)\approx 2.914291\times10^{6}
$$

$$
[A_0]=[\mathrm{MoS_4^{2-}}]=\frac{C_0}{D(x)}\approx 6.8627\times10^{-14}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[\mathrm{MoS_4^{2-}}]=6.8627\times10^{-14}\ \mathrm{mol\;L^{-1}}$](0.5)</Checkpoint>

$$
[A_1]=[{\rm MoOS_3^{2-}}]=\frac{K_1}{x}[A_0]\approx 3.3841\times10^{-12}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[{\rm MoOS_3^{2-}}]=3.3841\times10^{-12}\ \mathrm{mol\;L^{-1}}$](0.5)</Checkpoint>

$$
[A_2]=[{\rm MoO_2S_2^{2-}}]\approx 1.2836\times10^{-10}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[{\rm MoO_2S_2^{2-}}]=1.2836\times10^{-10}\ \mathrm{mol\;L^{-1}}$](0.5)</Checkpoint>

$$
[A_3]=[{\rm MoO_3S^{2-}}]\approx 7.7905\times10^{-9}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[{\rm MoO_3S^{2-}}]=7.7905\times10^{-9}\ \mathrm{mol\;L^{-1}}$](0.5)</Checkpoint>

$$
[A_4]=[{\rm MoO_4^{2-}}]\approx 1.9208\times10^{-7}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[{\rm MoO_4^{2-}}]=1.9208\times10^{-7}\ \mathrm{mol\;L^{-1}}$](0.5)</Checkpoint>

溶液中电离产物：

$$
[\mathrm{HS^-}]=\alpha_1\,x \approx 3.4754\times10^{-9}\ \mathrm{mol\;L^{-1}}
$$
$$
[\mathrm{S^{2-}}]=\alpha_2\,x \approx 4.3752\times10^{-17}\ \mathrm{mol\;L^{-1}}
$$

<Checkpoint>[$[\mathrm{HS^-}]=\alpha_1\,x \approx 3.4754\times10^{-9}\ \mathrm{mol\;L^{-1}}$](1)</Checkpoint>


<Checkpoint>[$ [\mathrm{S^{2-}}]=\alpha_2\,x \approx 4.3752\times10^{-17}\ \mathrm{mol\;L^{-1}}$](1)</Checkpoint>

气相物种：

$$
p=H'x\approx 2.5680\times10^{-6}\ \mathrm{atm}
$$

<Checkpoint>[$\mathrm{H_2S}$ 气体分压 $p= 2.5680\times10^{-6}\ \mathrm{atm}$](0.5)</Checkpoint>

验证结果的正确性，根据 $\mathrm{Mo}$ 的物料守恒进行验证：

$$
\sum_{i=0}^4 [A_i] = 2.0000\times10^{-7}\ \mathrm{M}= C_0
$$

<Checkpoint>[验证 $ \sum_{i=0}^4 [A_i] = C_0 $](0.5)</Checkpoint>

最终，按照题目要求，计算：

$$
P_{\rm S}=\frac{[\mathrm{H_2S(aq)}]}{c^\ominus}\cdot\frac{[\mathrm{HS^-}]}{c^\ominus}\cdot\frac{[\mathrm{S^{2-}}]}{c^\ominus}\cdot\frac{p(\mathrm{H_2S})}{p^\ominus}\approx 1.02944\times10^{-37}
$$
$$
P_{\rm Mo}=\prod_{i=0}^{4}\frac{[A_i]}{c^\ominus}
=\frac{[{\rm MoS_4^{2-}}]\,[{\rm MoOS_3^{2-}}]\,[{\rm MoO_2S_2^{2-}}]\,[{\rm MoO_3S^{2-}}]\,[{\rm MoO_4^{2-}}]}{(c^\ominus)^5}\approx 4.46088\times10^{-50}
$$

根据题目，最终保留两位有效数字，最终比值为：

$$
\dfrac{P_{\rm Mo}}{P_{\rm S}}=4.3333\times 10^{-13}\approx4.3\times 10^{-13}
$$

<Checkpoint>[最终比值为 $4.3\times 10^{-13}$](2)</Checkpoint>