$1 \times 10^{-3} \mathrm{~mol}$ of $\mathrm{Cd}(\mathrm{CN})_{2}$ is placed in $1 \mathrm{~L}$ of buffer solution at $\mathrm{pH}=8.5$. It is known that the cumulative stability constants of $\mathrm{CN}^{-}$ to $\mathrm{Cd}^{2+}$ are $10^{5.48}, 10^{10.60}, 10^{15.23}, 10^{18.78}$, respectively, and the solubility product constant of $\mathrm{Cd}(\mathrm{OH})_{2}$ is $K_{\mathrm{sp}}=4 \times 10^{-13}$, and the dissociation constant of $\mathrm{HCN}$ is $K_{\mathrm{a}}=6.2 \times 10^{-10}$. Coordination of cadmium species and $\mathrm{OH^-}$ is ignored.
The following statements are made:

1. The species conservation of cyanide ions in the solution is $\mathrm{[CN^-]+[HCN]+[Cd(CN)^+]+2[Cd(CN)_2]+3[Cd(CN)_3^-]+4[Cd(CN)_4^{2-}]=2\times10^{-3}~mol/L}$

2. The charge conservation in the solution is $\mathrm{2[Cd^{2+}]+[Cd(CN)^+]+[H^+]=[Cd(CN)_3^-]+2[Cd(CN)_4^{2-}]+[CN^-]+[OH^-]}$

3. Let $\mathrm{[CN^-]}=a.b\times10^{-c}\mathrm{~mol/L}$ (retain two significant figures), then $a+b+c=10$

4. The system will produce $\mathrm{Cd(OH)}_2$ precipitate

What is the sum of the serial numbers of all correct statements?
将 $1 \times 10^{-3} \mathrm{~mol}$ 的 $\mathrm{Cd}(\mathrm{CN})_{2}$ 放入 $1 \mathrm{~L} ~\mathrm{pH}=8.5$ 的缓冲溶液中。已知 $\mathrm{CN}^{-}$对 $\mathrm{Cd}^{2+}$ 的累计稳定常数分别为 $10^{5.48}, 10^{10.60}, 10^{15.23}, 10^{18.78}$，$ \mathrm{Cd}(\mathrm{OH})_{2}$的溶度积常数为 $K_{\mathrm{sp}}=4 \times 10^{-13}$，$\mathrm{HCN}$ 的解离常数为 $K_{\mathrm{a}}=6.2 \times 10^{-10}$ 。忽略含镉物种和 $\mathrm{OH^-}$ 的配位。
有如下说法：

1.溶液中氰根离子的物种守恒有 $\mathrm{[CN^-]+[HCN]+[Cd(CN)^+]+2[Cd(CN)_2]+3[Cd(CN)_3^-]+4[Cd(CN)_4^{2-}]=2\times10^{-3}~mol/L}$

2.溶液中的电荷守恒有 $\mathrm{2[Cd^{2+}]+[Cd(CN)^+]+[H^+]=[Cd(CN)_3^-]+2[Cd(CN)_4^{2-}]+[CN^-]+[OH^-]}$

3.设$\mathrm{[CN^-]}=a.b\times10^{-c}\mathrm{~mol/L}$（保留两位有效数字），有$a+b+c=10$

4.体系会产生 $\mathrm{Cd(OH)}_2$ 沉淀

所有正确说法的序号和为多少？
1
3
4
6
7
8
All of the above options are incorrect.
1
3
4
6
7
8
以上选项均不对
This question cannot be solved using charge conservation because the buffer ion concentration changes before and after the addition of $\mathrm{Cd(CN)_2}$, its relative buffer ion concentration is small, but the relative concentrations of cadmium-containing and cyanide-containing species are large and unknown. Statement 1 regarding material balance is correct, and statement 2 regarding charge conservation is incorrect because buffer ions are not considered.

<Checkpoint>[Statement 1 is correct](1)</Checkpoint>

<Checkpoint>[Statement 2 is incorrect](2)</Checkpoint>

According to material balance $\mathrm{[CN^-]+[HCN]+[Cd(CN)^+]+2[Cd(CN)_2]+3[Cd(CN)_3^-]+4[Cd(CN)_4^{2-}]=2\times10^{-3}~mol/L}$, express the concentrations of species other than $\mathrm{CN^-}$ in terms of $\mathrm{[CN^-]}$.

$\mathrm{[HCN]=\frac{[CN^-][H^+]}{\mathit{K}_a}}$

<Checkpoint>[$\mathrm{[HCN]=\frac{[CN^-][H^+]}{\mathit{K}_a}}$](0.5)</Checkpoint>

According to the distribution coefficient, we have:

$\mathrm{[Cd(CN)^+]=\mathit{C}_{Cd}\cdot\frac{\beta_1[CN^-]}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$

$\mathrm{[Cd(CN)_2]=\mathit{C}_{Cd}\cdot\frac{\beta_2[CN^-]^2}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$

$\mathrm{[Cd(CN)_3^-]=\mathit{C}_{Cd}\cdot\frac{\beta_3[CN^-]^3}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$

$\mathrm{[Cd(CN)_4^{2-}]=\mathit{C}_{Cd}\cdot\frac{\beta_4[CN^-]^4}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$

Where $C_\mathrm{Cd}=1\times10^{-3}\mathrm{~mol/L}$ .

<Checkpoint>[$\mathrm{[Cd(CN)^+]=\mathit{C}_{Cd}\cdot\frac{\beta_1[CN^-]}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$](0.5)</Checkpoint>

<Checkpoint>[$\mathrm{[Cd(CN)_2]=\mathit{C}_{Cd}\cdot\frac{\beta_2[CN^-]^2}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$](0.5)</Checkpoint>

<Checkpoint>[$\mathrm{[Cd(CN)_3^-]=\mathit{C}_{Cd}\cdot\frac{\beta_3[CN^-]^3}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$](0.5)</Checkpoint>

<Checkpoint>[$\mathrm{[Cd(CN)_4^{2-}]=\mathit{C}_{Cd}\cdot\frac{\beta_4[CN^-]^4}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$](0.5)</Checkpoint>

Substituting into the material balance equation, we obtain $\mathrm{[CN^-]=1.4\times10^{-5}~mol/L}$. According to the notation in statement 3, $a+b+c=10$, so statement 3 is correct.

<Checkpoint>[$\mathrm{[CN^-]=1.4\times10^{-5}~mol/L}$](0.5)</Checkpoint>

<Checkpoint>[$a+b+c=10$](0.5)</Checkpoint>

According to the distribution coefficient, $\mathrm{[Cd^{2+}]=\mathit{C}_{Cd}\cdot\frac{1}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}=5.9\times10^{-5}~mol/L}$

$Q=\mathrm{[Cd^{2+}][OH^-]^2/(1~mol/L)^3=5.9\times10^{-16}}<K_\mathrm{sp}$

<Checkpoint>[$Q=5.9\times10^{-16}<K_\mathrm{sp}$](1)</Checkpoint>

$\mathrm{Cd(OH)}_2$ precipitate will not be produced, so statement 4 is incorrect.

In summary, the correct statements are 1 and 3, so choose option C.
该题不能用电荷守恒解答，因为加入 $\mathrm{Cd(CN)_2}$ 前后缓冲离子浓度有变化，其相对缓冲离子浓度较小，但相对含镉和含氰物种浓度较大，且未知。说法1的物种守恒正确，说法2的电荷守恒错误，因为未考虑缓冲离子。

<Checkpoint>[说法1正确](1)</Checkpoint>

<Checkpoint>[说法2错误](2)</Checkpoint>

依据物种守恒 $\mathrm{[CN^-]+[HCN]+[Cd(CN)^+]+2[Cd(CN)_2]+3[Cd(CN)_3^-]+4[Cd(CN)_4^{2-}]=2\times10^{-3}~mol/L}$，将 $\mathrm{CN^-}$ 外的物种浓度用 $\mathrm{[CN^-]}$表示。

$\mathrm{[HCN]=\frac{[CN^-][H^+]}{\mathit{K}_a}}$

<Checkpoint>[$\mathrm{[HCN]=\frac{[CN^-][H^+]}{\mathit{K}_a}}$](0.5)</Checkpoint>

根据分布系数，有：

$\mathrm{[Cd(CN)^+]=\mathit{C}_{Cd}\cdot\frac{\beta_1[CN^-]}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$

$\mathrm{[Cd(CN)_2]=\mathit{C}_{Cd}\cdot\frac{\beta_2[CN^-]^2}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$

$\mathrm{[Cd(CN)_3^-]=\mathit{C}_{Cd}\cdot\frac{\beta_3[CN^-]^3}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$

$\mathrm{[Cd(CN)_4^{2-}]=\mathit{C}_{Cd}\cdot\frac{\beta_4[CN^-]^4}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$

其中 $C_\mathrm{Cd}=1\times10^{-3}\mathrm{~mol/L}$ 。

<Checkpoint>[$\mathrm{[Cd(CN)^+]=\mathit{C}_{Cd}\cdot\frac{\beta_1[CN^-]}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$](0.5)</Checkpoint>

<Checkpoint>[$\mathrm{[Cd(CN)_2]=\mathit{C}_{Cd}\cdot\frac{\beta_2[CN^-]^2}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$](0.5)</Checkpoint>

<Checkpoint>[$\mathrm{[Cd(CN)_3^-]=\mathit{C}_{Cd}\cdot\frac{\beta_3[CN^-]^3}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$](0.5)</Checkpoint>

<Checkpoint>[$\mathrm{[Cd(CN)_4^{2-}]=\mathit{C}_{Cd}\cdot\frac{\beta_4[CN^-]^4}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}}$](0.5)</Checkpoint>

代入物料守恒方程，解得 $\mathrm{[CN^-]=1.4\times10^{-5}~mol/L}$，按说法3的记法，$a+b+c=10$，说法3正确。

<Checkpoint>[$\mathrm{[CN^-]=1.4\times10^{-5}~mol/L}$](0.5)</Checkpoint>

<Checkpoint>[$a+b+c=10$](0.5)</Checkpoint>

根据分布系数，$\mathrm{[Cd^{2+}]=\mathit{C}_{Cd}\cdot\frac{1}{1+\beta_1[CN^-]+\beta_2[CN^-]^2+\beta_3[CN^-]^3+\beta_4[CN^-]^4}=5.9\times10^{-5}~mol/L}$

$Q=\mathrm{[Cd^{2+}][OH^-]^2/(1~mol/L)^3=5.9\times10^{-16}}<K_\mathrm{sp}$

<Checkpoint>[$Q=5.9\times10^{-16}<K_\mathrm{sp}$](1)</Checkpoint>

不会产生 $\mathrm{Cd(OH)}_2$ 沉淀，说法4错误。

综上所述，正确的说法为1和3，选C选项。