The organic extractant $\rm{HA}$ can extract $\mathrm{Ce^{3+}}$ from an aqueous solution to the organic phase. $\rm{HA}$ is almost insoluble in water, but exists in the organic phase only as a dimer of $\mathrm{H_2A_2}$. To determine the specific form of this cerium complex in the organic phase, the following experiment was performed:

A tracer amount of ${ }^{144} \mathrm{Ce(NO_3)_3}$ was dissolved in water and acidified with a small amount of nitric acid to prepare an aqueous phase containing ${ }^{144} \mathrm{Ce}^{3+}$. $\rm{2.50 ~mL}$ of the above aqueous phase and $\rm{2.50 ~mL}$ of the organic phase containing $\rm{HA}$ were taken and mixed thoroughly for extraction at room temperature. After equilibrium was reached, $\rm{1.00 ~mL}$ each of the aqueous phase and the organic phase were taken, and the ratio of their radioactivity intensities was measured. Another portion of the equilibrated aqueous phase was taken and its $\mathrm{pH}$ value was measured. When the initial concentration of $\mathrm{H_2A_2}$ is $\rm{c_0(H_2A_2)=0.240 ~mol/L}$, the ratio of radioactivity intensity of the organic phase to the aqueous phase at equilibrium was measured to be 2929:416, and the aqueous phase $\rm{[\mathrm{H}^{+}]=3.39\times10^{-2}~mol/L}$. When $\rm{c_0(H_2A_2)=0.120 ~mol/L}$, the ratio of radioactivity intensity of the organic phase to the aqueous phase at equilibrium was measured to be 1542:1682, and the aqueous phase $\rm{[\mathrm{H}^{+}]=3.42\times10^{-2}~mol/L}$. It is known that the organic phase does not contain nitrate ions.

Please determine the equilibrium constant of the extraction reaction through calculation and reasoning, and select the correct equilibrium constant from the following options:

(Hydrolysis of $\mathrm{Ce^{3+}}$ and complexation of $\mathrm{Ce^{3+}}$ with $\mathrm{NO_3^-}$ are not considered. A calculation error within 5% is considered correct, and take two significant figures).
有机萃取剂$\rm{HA}$能将水溶液中的$\mathrm{Ce^{3+}}$萃取至有机相。$\rm{HA}$ 在水中几乎不溶解, 而在有机相中仅以 $\mathrm{H_2A_2}$ 的二聚体形式存在。为测定该铈配合物在有机相的具体形式, 进行如下实验: 

将示踪量${ }^{144} \mathrm{Ce(NO_3)_3}$溶于水, 并用少量硝酸酸化, 制成含${ }^{144} \mathrm{Ce}^{3+}$的水相; 移取上述水相及含$\rm{HA}$的有机相各$\rm{2.50 ~mL}$于室温下充分混合萃取, 待平衡后, 取水相和有机相各$\rm{1.00 ~mL}$, 测定其放射性强度之比; 另取一份平衡后的水相, 测定其 $\mathrm{pH}$ 值。当$\mathrm{H_2A_2}$的初始浓度$\rm{c_0(H_2A_2)=0.240 ~mol/L}$时, 测得平衡时有机相与水相放射性强度比为2929:416, 水相$\rm{[\mathrm{H}^{+}]=3.39\times10^{-2}~mol/L}$; 当$\rm{c_0(H_2A_2)=0.120 ~mol/L}$时, 测得平衡时有机相与水相放射性强度比为 1542:1682, 水相$\rm{[\mathrm{H}^{+}]=3.42\times10^{-2}~mol/L}$。已知有机相不含有硝酸根离子。

请通过计算和推理确定萃取反应的平衡常数，并选择下列选项中的平衡常数正确的一个：

(不考虑$\mathrm{Ce^{3+}}$的水解及$\mathrm{Ce}^{3+}$与$\mathrm{NO_3^-}$的络合，计算误差在5%以内视为正确，取两位有效数字)。
All other options are incorrect
0.010
0.014
0.025
0.30
1.5
0.52
0.030
0.0022
其他选项均不正确
0.010
0.014
0.025
0.30
1.5
0.52
0.030
0.0022
The extracted substance is electrically neutral because there are no nitrate radicals, and $\rm{Ce^{3+}}$ is coordinated with at least three $\rm{A^-}$ ions.

<Checkpoint>[The extracted substance is electrically neutral because there are no nitrate radicals, and $\rm{Ce^{3+}}$ is coordinated with at least three $\rm{A^-}$ ions.](1)</Checkpoint>

Consider the possible extraction reaction equation:

$\rm{Ce^{3+} + 1.5H_2A_2 = Ce(A)_3 + 3H^+}$

From the two extraction experiments, the ratio of radiation intensities is the ratio of $\rm{Ce(III)}$ concentrations, so the equilibrium constant can be listed:

$\rm{K=\frac{[H^+]^3[Ce(A)_3]}{[H_2A_2]^{1.5}[Ce^{3+}]}}$

<Checkpoint>[$\rm{K=\frac{[H^+]^3[Ce(A)_3]}{[H_2A_2]^{1.5}[Ce^{3+}]}}$](1)</Checkpoint>

Since $\rm{Ce(III)}$ is a tracer amount, $\rm{c_0(H_2A_2)\approx [H_2A_2]}$.

<Checkpoint>[Since $\rm{Ce(III)}$ is a tracer amount, $\rm{c_0(H_2A_2)\approx [H_2A_2]}$](1)</Checkpoint>

Therefore, the following equation can be listed according to the two experiments:
$\rm{K=\frac{0.0339^3\times2929}{416\times0.24^{1.5}}=\frac{0.0342^3\times1542}{1682\times0.12^{1.5}}}$

Obviously, this equation does not hold. Therefore, some $\rm{HA}$ is also coordinated with $\rm{Ce(III)}$ in the organic phase;

<Checkpoint>[Therefore, some $\rm{HA}$ is also coordinated with $\rm{Ce(III)}$ in the organic phase](1)</Checkpoint>

At this time, the extraction reaction can be written as

$\rm{Ce^{3+} + xH_2A_2 = Ce(A)_3(HA)_{2x-3} + 3H^+}$

<Checkpoint>[$\rm{Ce^{3+} + xH_2A_2 = Ce(A)_3(HA)_{2x-3} + 3H^+}$](1)</Checkpoint>

List the equation according to the equilibrium constant:

$\rm{K=\frac{0.0339^3\times2929}{416\times0.24^{x}}=\frac{0.0342^3\times1542}{1682\times0.12^{x}}}$

Solving for $\rm{x=2.9\approx 3}$. That is, the existence form of $\rm{Ce(III)}$ in the organic phase is $\rm{Ce(A)_3(HA)_3}$.

<Checkpoint>[The existence form of $\rm{Ce(III)}$ in the organic phase is $\rm{Ce(A)_3(HA)_3}$](1)</Checkpoint>

Substituting, we get the equilibrium constant $\rm{K = 0.019}$.

<Checkpoint>[Equilibrium constant $\rm{K = 0.019}$](1)</Checkpoint>

Therefore, all options are incorrect. Option A is correct.
萃取后的物质为电中性，因为无硝酸根，$\rm{Ce^{3+}}$至少与三个$\rm{A^-}$离子配合。

<Checkpoint>[萃取后的物质为电中性，因为无硝酸根，$\rm{Ce^{3+}}$至少与三个$\rm{A^-}$离子配合](1)</Checkpoint>

考虑可能的萃取反应方程式：

$\rm{Ce^{3+} + 1.5H_2A_2 = Ce(A)_3 + 3H^+}$

由两次萃取实验可知，放射强度比即为$\rm{Ce(III)}$的浓度之比，因此可以列出平衡常数：

$\rm{K=\frac{[H^+]^3[Ce(A)_3]}{[H_2A_2]^{1.5}[Ce^{3+}]}}$

<Checkpoint>[$\rm{K=\frac{[H^+]^3[Ce(A)_3]}{[H_2A_2]^{1.5}[Ce^{3+}]}}$](1)</Checkpoint>

由于$\rm{Ce(III)}$为示踪量，有$\rm{c_0(H_2A_2)\approx [H_2A_2]}$。

<Checkpoint>[由于$\rm{Ce(III)}$为示踪量，有$\rm{c_0(H_2A_2)\approx [H_2A_2]}$](1)</Checkpoint>

因此根据两次实验可列出方程：
$\rm{K=\frac{0.0339^3\times2929}{416\times0.24^{1.5}}=\frac{0.0342^3\times1542}{1682\times0.12^{1.5}}}$

很显然该方程并不成立。因此还有部分$\rm{HA}$与有机相的$\rm{Ce(III)}$发生配合；

<Checkpoint>[还有部分$\rm{HA}$与有机相的$\rm{Ce(III)}$发生配合](1)</Checkpoint>

此时萃取反应可写为

$\rm{Ce^{3+} + xH_2A_2 = Ce(A)_3(HA)_{2x-3} + 3H^+}$

<Checkpoint>[$\rm{Ce^{3+} + xH_2A_2 = Ce(A)_3(HA)_{2x-3} + 3H^+}$](1)</Checkpoint>

根据平衡常数列出方程：

$\rm{K=\frac{0.0339^3\times2929}{416\times0.24^{x}}=\frac{0.0342^3\times1542}{1682\times0.12^{x}}}$

解得$\rm{x=2.9\approx 3}$。即$\rm{Ce(III)}$在有机相的存在形式为$\rm{Ce(A)_3(HA)_3}$。

<Checkpoint>[$\rm{Ce(III)}$在有机相的存在形式为$\rm{Ce(A)_3(HA)_3}$](1)</Checkpoint>

带入可得平衡常数$\rm{K = 0.019}$.

<Checkpoint>[平衡常数$\rm{K = 0.019}$](1)</Checkpoint>

因此所有选项均不正确。选项A正确。