Organic extractant $\mathrm{H}\bf{A}$ can extract $\mathrm{Ce^{3+}}$ from aqueous solution into the organic phase in the form of $\mathrm{Ce}(\bf{A})_x(\mathrm{H}\bf{A})_y$. $\mathrm{H}\bf{A}$ is almost insoluble in water, but exists in the organic phase only as a dimer of $\mathrm{{H_2}\bf{A}\mathrm{_2}}$. In order to determine the specific form of this cerium complex in the organic phase, the following experiment was carried out: A tracer amount of $^{144}\mathrm{Ce}^{3+}$ was dissolved in water and acidified with a small amount of nitric acid to prepare an aqueous phase containing $^{144}\mathrm{Ce}^{3+}$; $\mathrm{2.50~mL}$ each of the above aqueous phase and the organic phase containing $\mathrm{H}\bf{A}$ were transferred and thoroughly mixed and extracted at room temperature. After equilibrium, $\mathrm{1.00~mL}$ 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 to measure its $\mathrm{pH}$ value. When the initial concentration of $\mathrm{{H_2}\bf{A}\mathrm{_2}}$ is $c_{0}(\mathrm{{H_2}\bf{A}\mathrm{_2}}) = 0.240~\mathrm{mol/L}$, the ratio of the radioactivity intensities of the organic phase to the aqueous phase at equilibrium is measured to be $\mathrm2929:416$, and the aqueous phase $\mathrm{[H^+]} = 3.39 \times 10^{-2}~\mathrm{mol/L}$; when $c_{0}(\mathrm{{H_2}\bf{A}\mathrm{_2}}) = 0.120~\mathrm{mol/L}$, the ratio of the radioactivity intensities of the organic phase to the aqueous phase at equilibrium is measured to be $\mathrm1542:1682$, and the aqueous phase $\mathrm{[H^+]} = 3.42 \times 10^{-2}~\mathrm{mol/L}$.
The following statements are made:

1. The sum of $x$ and $y$ is an even number (treat $x$ and $y$ as integers).
2. The sum of the coefficients on both sides of the simplified and balanced extraction reaction equation are not equal.
3. The equilibrium constant of the extraction reaction is smaller than $\mathrm0.023$.
4. The equilibrium constant of the extraction reaction is smaller than $\mathrm0.015$.
5. The equilibrium constant should be taken to $\mathrm2$ significant figures.

The hydrolysis of $\mathrm{Ce^{3+}}$ and the complexation of $\mathrm{Ce^{3+}}$ with $\mathrm{NO_3^-}$ are not considered in the calculation of the equilibrium constant. What is the sum of the correct serial numbers?
有机萃取剂 $\mathrm{H}\bf{A}$ 能将水溶液中的 $\mathrm{Ce^{3+}}$ 以  $\mathrm{Ce}(\bf{A})_x(\mathrm{H}\bf{A})_y$  的形式萃取至有机相。$\mathrm{H}\bf{A}$ 在水中几乎不溶解，而在有机相中仅以 $\mathrm{{H_2}\bf{A}\mathrm{_2}}$ 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 $^{144}\mathrm{Ce}^{3+}$ 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 $\mathrm{H}\bf{A}$ 的有机相各 $\mathrm{2.50~mL}$ 于室温下充分混合萃取，待平衡后，取水相和有机相各 $\mathrm{1.00~mL}$ ，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 $\mathrm{{H_2}\bf{A}\mathrm{_2}}$  的初始浓度 $c_{0}(\mathrm{{H_2}\bf{A}\mathrm{_2}}) = 0.240~\mathrm{mol/L}$ 时，测得平衡时有机相与水相放射性强度比为$\mathrm2929:416$，水相 $\mathrm{[H^+]} = 3.39 \times 10^{-2}~\mathrm{mol/L}$ ；当 $c_{0}(\mathrm{{H_2}\bf{A}\mathrm{_2}}) = 0.120~\mathrm{mol/L}$ 时，测得平衡时有机相与水相放射性强度比为$\mathrm1542:1682$，水相 $\mathrm{[H^+]} = 3.42 \times 10^{-2}~\mathrm{mol/L}$ 。
有以下说法：

1. $x$ 和 $y$ 相加是偶数（将 $x$ 和 $y$ 处理为整数）。
2. 化简配平后的萃取反应方程式两边系数和不相等。
3. 萃取反应的平衡常数比 $\mathrm0.023$ 小。
4. 萃取反应的平衡常数比 $\mathrm0.015$ 小。
5. 平衡常数应该取 $\mathrm2$ 位有效数字。

平衡常数的计算不考虑 $\mathrm{Ce^{3+}}$ 的水解及 $\mathrm{Ce}^{3+}$ 与 $\mathrm{NO_3^-}$ 的络合。请问正确的序号相加结果为：
1
2
3
4
5
7
9
10
1
2
3
4
5
7
9
10
Since the extracted substance $\mathrm{Ce(\bf{A})}_x\mathrm{(H}\bf{A)}_y$ is electrically neutral, $x = 3$;

<Checkpoint>[$x = 3$](1)</Checkpoint>

The extraction reaction equation is:

$\mathrm{Ce^{3+}(w)} + [(3+y)/2]~\mathrm{{H_2\bf{A}}_2(o)} = \mathrm{Ce(A)}_x\mathrm{{(H}\bf{A)}}_y{(o)}+3~\mathrm{H^+(w)}$

<Checkpoint>[The extraction reaction equation is: $\mathrm{Ce^{3+}(w)} + [(3+y)/2]~\mathrm{H_2\bf{A}_2(o)} = \mathrm{Ce(\bf{A})}_x\mathrm{(H\bf{A})}_y{(o)}+3~\mathrm{H^+(w)}$](2)</Checkpoint>

$K =\frac{2929\times[3.39 \times 10^{-2}]^{3}}{416\times 0.240^{(3+y)/2}} = \frac{1542\times[3.42 \times 10^{-2}]^{3}}{1682\times 0.120^{(3+y)/2}}$

<Checkpoint>[Equilibrium constant calculation equation: $K =\frac{2929\times[3.39 \times 10^{-2}]^{3}}{416\times 0.240^{(3+y)/2}} = \frac{1542\times[3.42 \times 10^{-2}]^{3}}{1682\times 0.120^{(3+y)/2}}$](2)</Checkpoint>

Solving the above equation, we obtain: $y = 2.9 \approx 3$; the value of $K$ is between $1.50×10^{-2}\sim2.30×10^{-2}$, retaining $3$ significant figures.

<Checkpoint>[$y = 2.9 \approx 3$](1)</Checkpoint>

<Checkpoint>[The value of $K$ is between $1.50×10^{-2}\sim2.30×10^{-2}$](1)</Checkpoint>

$x=3,y=3$, so the sum is equal to $6$, which is an even number. Statement 1 is correct.

The equilibrium constant of the extraction reaction is smaller than 0.023 and larger than 0.015, statement 3 is correct, and 4 is wrong.

Therefore, statements 1 and 3 are correct, choose option D.
由于萃取后的物质 $\mathrm{Ce(\bf{A})}_x\mathrm{(H}\bf{A)}\mathrm{_y}$ 为电中性，故 $x = 3$ ；

<Checkpoint>[$x = 3$](1)</Checkpoint>

萃取反应方程式为：

$\mathrm{Ce^{3+}(w)} + [(3+y)/2]~\mathrm{{H_2\bf{A}}_2(o)} = \mathrm{Ce(A)}_x\mathrm{{(H}\bf{A)}}_y{(o)}+3~\mathrm{H^+(w)}$

<Checkpoint>[萃取反应方程式为 ：$\mathrm{Ce^{3+}(w)} + [(3+y)/2]~\mathrm{H_2\bf{A}_2(o)} = \mathrm{Ce(\bf{A})}_x\mathrm{(H\bf{A})}_y{(o)}+3~\mathrm{H^+(w)}$](2)</Checkpoint>

$K =\frac{2929\times[3.39 \times 10^{-2}]^{3}}{416\times 0.240^{(3+y)/2}} = \frac{1542\times[3.42 \times 10^{-2}]^{3}}{1682\times 0.120^{(3+y)/2}}$

<Checkpoint>[平衡常数计算方程式：$K =\frac{2929\times[3.39 \times 10^{-2}]^{3}}{416\times 0.240^{(3+y)/2}} = \frac{1542\times[3.42 \times 10^{-2}]^{3}}{1682\times 0.120^{(3+y)/2}}$](2)</Checkpoint>

解上述方程，求得：$y = 2.9 \approx 3$；$K$值在 $1.50×10^{-2}\sim2.30×10^{-2}$ 之间，保留 $3$ 位有效数字。

<Checkpoint>[$y = 2.9 \approx 3$](1)</Checkpoint>

<Checkpoint>[$K$值在 $1.50×10^{-2}\sim2.30×10^{-2}$ 之间](1)</Checkpoint>

$x=3,y=3$,所以相加等于$6$,为偶数。说法1正确。

萃取反应的平衡常数比 
0.023小，比0.015大，说法3正确，4错误。

因此说法1, 3正确，选择选项D。