A certain organic solvent can extract salicylic acid from water, given its distribution coefficient $\rm{K_D=100}$, and the dimerization equilibrium constant of salicylic acid $\rm{K=50}$. It is known that the $\rm{pK_{a1}=2.98, pK_{a2}=13.66}$ of salicylic acid.

The following questions are asked:

a. Calculate the distribution ratio $\rm{D}$ of salicylic acid extracted by an equal volume of the solvent in water when the total concentration of salicylic acid in the original water is $\rm{0.1 mol/L}$ at $\rm{pH=1.0}$. (The answer should be taken to three significant figures)

b. Calculate the distribution ratio $\rm{D}$ of salicylic acid extracted by an equal volume of the solvent in water when the total concentration of salicylic acid in the original water is $\rm{0.1 mol/L}$ under human physiological conditions. (The answer should be taken to two significant figures)

c. Calculate the $\rm{pH}$ of the aqueous system when the distribution ratio $\rm{D=1}$. (The answer should be taken to one significant figure)

Which of the following statements is correct (calculation error within 2% can be considered correct):
某有机溶剂可以从水中萃取水杨酸，设其分配系数$\rm{K_D=100}$，水杨酸二聚平衡常数$\rm{K=50}$。已知水杨酸的$\rm{pK_{a1}=2.98, pK_{a2}=13.66}$。

有以下问题：

a. 计算$\rm{pH=1.0}$时，原水中水杨酸总浓度为$\rm{0.1 mol/L}$时等体积的该溶剂在水中萃取水杨酸的分配比$\rm{D}$。（答案取三位有效数字）

b. 计算人体生理条件下，原水中水杨酸总浓度为$\rm{0.1 mol/L}$时等体积的该溶剂在水中萃取水杨酸的分配比$\rm{D}$。（答案取两位有效数字）

c. 计算分配比$\rm{D=1}$时水体系的$\rm{pH}$。（答案取一位有效数字）

下列说法正确的是（计算误差在2%内可视为正确）：
All other options are incorrect
The answer to problem a is greater than 400
The answer to question a is less than 360.
The answer to problem b is 0.0021
The answer to problem b is 0.0051
The answer to problem c is 5.2.
The answer to problem c is 5.6
The answer to problem c is 5.4.
其他选项均不正确
问题a的答案大于400
问题a的答案小于360
问题b的答案为0.0021
问题b的答案为0.0051
问题c的答案为5.2
问题c的答案为5.6
问题c的答案为5.4
According to the second-level ionization constant of salicylic acid, the second-level ionization of salicylic acid is extremely weak and can be ignored. Therefore, it is denoted as $\rm{HA}$.

<Checkpoint>[The second-level ionization of salicylic acid is extremely weak and can be ignored](1)</Checkpoint>

The suffix $\rm{(w)}$ represents species in water, and $\rm{(o)}$ represents species in organic solvent.

Problem a:

According to the equilibrium equation:

$\rm{HA(w) = H^+ + A^-(w) \quad K_a=[H^+][A^-]/[HA(w)]}$

$\rm{HA(w) = HA(o) \quad K_{D}=[HA(o)]/[HA(w)]}$

$\rm{2HA(o) = (HA)_2(o) \quad K=[(HA)_2(o)] /[HA(o)]^2}$

Equal volume extraction, so the material balance can be written as:

$\rm{2[(HA)_2(o)]+[HA(o)]+[HA(w)]+[A^- (w)]=0.1 ~mol \cdot L^{-1}}$

<Checkpoint>[Material balance $\rm{2[(HA)_2(o)]+[HA(o)]+[HA(w)]+[A^- (w)]=0.1 ~mol \cdot L^{-1}}$](1)</Checkpoint>

When $\rm{pH=1}$, we have $\rm{\delta_{HA} \approx 1}$, ignoring $\rm{[A^-(w)]}$

<Checkpoint>[When $\rm{pH=1}$, we have $\rm{\delta_{HA} \approx 1}$, ignoring $\rm{[A^-(w)]}$](1)</Checkpoint>

Rearranging and substituting gives:

$$
\rm{2K[HA(o)]^2+(K_D+1)[HA(o)]/K_D-0.1=0}
$$

Solving for

$\rm{[HA(o)]=0.0270 ~mol \cdot L^{-1}}$

$\rm{[HA(w)]=0.000270 ~mol \cdot L^{-1}}$

$\rm{[(HA)_2(o)]=0.0364 ~mol \cdot L^{-1}}$

$\rm{D=(2[(HA)_2(o)]+[HA(o)])/[HA(w)]=370}$

<Checkpoint>[$\rm{D=(2[(HA)_2(o)]+[HA(o)])/[HA(w)]=370}$ ](1)</Checkpoint>

Problem b:

The $\rm{pH}$ of the human physiological environment is $\rm{pH = 7.4}$. At this time, $\rm{\delta_{A^-} \approx 1}$, ignoring $\rm{[HA(w)]}$.

<Checkpoint>[The $\rm{pH}$ of the human physiological environment is $\rm{pH = 7.4}$. At this time, $\rm{\delta_{A^-} \approx 1}$, ignoring $\rm{[HA(w)]}$.](1)</Checkpoint>

Therefore, we have:

$\rm{[HA(o)]/[A^-(w)]=K_D[H^+]/K_{a1}=0.00380}$

<Checkpoint>[$\rm{[HA(o)]/[A^-(w)]=K_D[H^+]/K_{a1}=0.00380}$](1)</Checkpoint>

Substituting into the material balance and rearranging gives

$\rm{100[HA(o)]^2 + 264[HA(o)] - 0.1 = 0}$

Solving for
$\rm{[HA(o)]=0.000379 ~mol \cdot L^{-1}}$

$\rm{[A^-(w)]=0.10 ~mol \cdot L^{-1}}$

$\rm{[(HA)_2(o)]=0.0000144 ~mol \cdot L^{-1}}$

$\rm{D=(2[(HA)_2(o)]+[HA(o)])/[A^-(w)]=0.0041}$

<Checkpoint>[$\rm{D=(2[(HA)_2(o)]+[HA(o)])/[A^-(w)]=0.0041}$ ](1)</Checkpoint>

Problem c:

$\rm{D=1}$, obviously $\rm{pH}$ is between 1 and 7.4, ignoring the second-level ionization.

Material balance: $\rm{2[(HA)_{2}(o)]+[HA(o)]=0.05 {~mol} \cdot L^{-1} ;[HA(w)]+[{A}^{-}(w)]=0.05 {~mol} \cdot L^{-1}}$

<Checkpoint>[Material balance: $\rm{2[(HA)_{2}(o)]+[HA(o)]=0.05 {~mol} \cdot L^{-1} ;[HA(w)]+[{A}^{-}(w)]=0.05 {~mol} \cdot L^{-1}}$](1)</Checkpoint>

Substituting the equilibrium constant and rearranging gives:
$\rm{100[HA(o)]^{2}+[HA(o)]-0.05=0}$

$\rm{[HA(o)]=0.0179 {~mol} \cdot L^{-1}}$

$\rm{[(HA)_{2}(o)]=0.016 {~mol} \cdot L^{-1}}$

$\rm{[HA(w)]=0.000179 {~mol} \cdot L^{-1}}$

$\rm{[{A}^{-}(w)]=0.05 {~mol} \cdot L^{-1}}$

$\rm{[{H}^{+}]=K_{{a}}[HA(w)] /[{A}^{-}(w)]=3.75 \times 10^{-6} {~mol} \cdot L^{-1}}$

$\rm{{pH}=5.42}$

<Checkpoint>[$\rm{{pH}=5.42}$](1)</Checkpoint>

Since the number of decimal places is calculated based on the significant figures of pH, retain one significant digit, which is 5.4.

Therefore, option H is correct.
根据水杨酸的二级电离常数可知，水杨酸二级电离极其微弱，可忽略。因此将其记作$\rm{HA}$。

<Checkpoint>[水杨酸二级电离极其微弱，可忽略](1)</Checkpoint>

后缀$\rm{(w)}$表示水中物种，$\rm{(o)}$表示有机溶剂中物种。

问题a:

根据平衡式有：

$\rm{HA(w) = H^+ + A^-(w) \quad K_a=[H^+][A^-]/[HA(w)]}$

$\rm{HA(w) = HA(o) \quad K_{D}=[HA(o)]/[HA(w)]}$

$\rm{2HA(o) = (HA)_2(o) \quad K=[(HA)_2(o)] /[HA(o)]^2}$

等体积萃取, 故物料守恒可写作: 

$\rm{2[(HA)_2(o)]+[HA(o)]+[HA(w)]+[A^- (w)]=0.1 ~mol \cdot L^{-1}}$

<Checkpoint>[物料守恒$\rm{2[(HA)_2(o)]+[HA(o)]+[HA(w)]+[A^- (w)]=0.1 ~mol \cdot L^{-1}}$](1)</Checkpoint>

$\rm{pH=1}$ 时,有 $\rm{\delta_{HA} \approx 1}$, 忽略 $\rm{[A^-(w)]}$

<Checkpoint>[$\rm{pH=1}$ 时,有 $\rm{\delta_{HA} \approx 1}$, 忽略 $\rm{[A^-(w)]}$](1)</Checkpoint>

整理代入得:

$$
\rm{2K[HA(o)]^2+(K_D+1)[HA(o)]/K_D-0.1=0}
$$

解得

$\rm{[HA(o)]=0.0270 ~mol \cdot L^{-1}}$

$\rm{[HA(w)]=0.000270 ~mol \cdot L^{-1}}$

$\rm{[(HA)_2(o)]=0.0364 ~mol \cdot L^{-1}}$

$\rm{D=(2[(HA)_2(o)]+[HA(o)])/[HA(w)]=370}$ 

<Checkpoint>[$\rm{D=(2[(HA)_2(o)]+[HA(o)])/[HA(w)]=370}$ ](1)</Checkpoint>

问题b:

人体生理环境的$\rm{pH = 7.4}$，此时有 $\rm{\delta_{A^-} \approx 1}$, 忽略 $\rm{[HA(w)]}$。

<Checkpoint>[人体生理环境的$\rm{pH = 7.4}$，此时有 $\rm{\delta_{A^-} \approx 1}$, 忽略 $\rm{[HA(w)]}$。](1)</Checkpoint>

因此有：

$\rm{[HA(o)]/[A^-(w)]=K_D[H^+]/K_{a1}=0.00380}$

<Checkpoint>[$\rm{[HA(o)]/[A^-(w)]=K_D[H^+]/K_{a1}=0.00380}$](1)</Checkpoint>

带入物料守恒整理可得

$\rm{100[HA(o)]^2 + 264[HA(o)] - 0.1 = 0}$

解得
$\rm{[HA(o)]=0.000379 ~mol \cdot L^{-1}}$

$\rm{[A^-(w)]=0.10 ~mol \cdot L^{-1}}$

$\rm{[(HA)_2(o)]=0.0000144 ~mol \cdot L^{-1}}$

$\rm{D=(2[(HA)_2(o)]+[HA(o)])/[A^-(w)]=0.0041}$ 

<Checkpoint>[$\rm{D=(2[(HA)_2(o)]+[HA(o)])/[A^-(w)]=0.0041}$ ](1)</Checkpoint>

问题c:

$\rm{D=1}$, 显然$\rm{pH}$在1与7.4之间, 忽略二级电离。

物料守恒：$\rm{2[(HA)_{2}(o)]+[HA(o)]=0.05 {~mol} \cdot L^{-1} ;[HA(w)]+[{A}^{-}(w)]=0.05 {~mol} \cdot L^{-1}}$

<Checkpoint>[物料守恒：$\rm{2[(HA)_{2}(o)]+[HA(o)]=0.05 {~mol} \cdot L^{-1} ;[HA(w)]+[{A}^{-}(w)]=0.05 {~mol} \cdot L^{-1}}$](1)</Checkpoint>

平衡常数代入整理得:
$\rm{100[HA(o)]^{2}+[HA(o)]-0.05=0}$

$\rm{[HA(o)]=0.0179 {~mol} \cdot L^{-1}}$

$\rm{[(HA)_{2}(o)]=0.016 {~mol} \cdot L^{-1}}$

$\rm{[HA(w)]=0.000179 {~mol} \cdot L^{-1}}$

$\rm{[{A}^{-}(w)]=0.05 {~mol} \cdot L^{-1}}$

$\rm{[{H}^{+}]=K_{{a}}[HA(w)] /[{A}^{-}(w)]=3.75 \times 10^{-6} {~mol} \cdot L^{-1}}$

$\rm{{pH}=5.42}$

<Checkpoint>[$\rm{{pH}=5.42}$](1)</Checkpoint>

由于pH有效数字计算其小数点后的位数，故保留一位有效数字为$5.4$。

因此选项H正确。