Bright green **A** (ω<sub>Cu</sub> = 57.5%) and blue **B** (ω<sub>Cu</sub> = 55.3%) are widely found in nature, and these two minerals often coexist. Heating a mixture of copper sulfate solution and sodium carbonate in a sealed tube to 150–170 °C yields **A** (**Reaction 1**), while a lower reaction temperature produces **B** (**Reaction 2**). (ω% represents mass fraction.)

Heating **A** or **B** within a certain temperature range decomposes them to produce copper oxide. Take \( 7.96\,\text{mg} \) of \(\text{CuO}\) in a beaker, add \( 100\,\text{mL} \) of deionized water, and stir continuously while adding solid \((\text{NH}_4)_2\text{CO}_3\). Assuming the solution volume remains constant at \( 100\,\text{mL} \), calculate the minimum mass of \((\text{NH}_4)_2\text{CO}_3\) required to fully dissolve the solid (assuming no other basic salt precipitates form and no \(\text{CO}_2\) escapes the system). Also, determine the chemical equations for **Reaction 1** and **Reaction 2**.

Given conditions:

**Dissolution equilibrium**:  
   \[
   \text{CuO} + \text{H}_2\text{O} \rightleftharpoons \text{Cu}^{2+} + 2\,\text{OH}^-
   \]  
   Equilibrium constant:  
   \[
   K_s = 2.5 \times 10^{-21}
   \]

 **Solubility product constants**:  
   - \(\text{Cu(OH)}_2\):  
     \[
     K_{\text{sp}}(\text{Cu(OH)}_2) = 2.2 \times 10^{-20}
     \]  
   - \(\text{CuCO}_3\):  
     \[
     K_{\text{sp}}(\text{CuCO}_3) = 1.4 \times 10^{-10}
     \]

**Ionization constants of carbonic acid**:  
   - First ionization:  
     \[
     K_{a1}(\text{H}_2\text{CO}_3) = 4.2 \times 10^{-7}
     \]  
   - Second ionization:  
     \[
     K_{a2}(\text{H}_2\text{CO}_3) = 5.6 \times 10^{-11}
     \]

 **Ionic product of water**:  
   \[
   K_w = 1.0 \times 10^{-14}
   \]

 **Acid dissociation constant of ammonium ion**:  
   \[
   \text{p}K_a(\text{NH}_4^+) = 9.26
   \]

 **Cumulative stability constants for \(\text{Cu}^{2+}-\text{NH}_3\)**:  
   \[
   \begin{aligned}
   \lg\beta_1 &= 4.31, \\
   \lg\beta_2 &= 7.98, \\
   \lg\beta_3 &= 11.02, \\
   \lg\beta_4 &= 13.32, \\
   \lg\beta_5 &= 12.86.
   \end{aligned}
   \]
亮绿色的 **A**（ω<sub>Cu</sub> = 57.5%）与蓝色的 **B**（ω<sub>Cu</sub> = 55.3%）广泛存在于自然界中，这两种矿物在自然界中常共生。将硫酸铜溶液与碳酸钠的混合物在密闭管中加热至 150–170 °C 可制得 **A**（**反应1**），若反应温度较低则制得 **B**（**反应2**）。（ω%代表质量分数）

在一定温度范围内加热 **A** 或 **B**，会分解产生氧化铜；取 \( 7.96\,\text{mg} \) \(\text{CuO}\) 于烧杯中，加入 \( 100\,\text{mL} \) 去离子水，在不断搅拌下加入 \((\text{NH}_4)_2\text{CO}_3\) 固体。假设溶液体积保持 \( 100\,\text{mL} \) 不变，通过计算求出使固体完全溶解需加入的最小 \((\text{NH}_4)_2\text{CO}_3\) 质量（假设不生成其他碱式盐沉淀，且无 \(\text{CO}_2\) 溢出体系）。并确定**反应1**和**反应2**的化学方程式。


已知条件：

**溶解平衡**：  
   \[
   \text{CuO} + \text{H}_2\text{O} \rightleftharpoons \text{Cu}^{2+} + 2\,\text{OH}^-
   \]  
   平衡常数：  
   \[
   K_s = 2.5 \times 10^{-21}
   \]

 **溶度积常数**：  
   - \(\text{Cu(OH)}_2\)：  
     \[
     K_{\text{sp}}(\text{Cu(OH)}_2) = 2.2 \times 10^{-20}
     \]  
   - \(\text{CuCO}_3\)：  
     \[
     K_{\text{sp}}(\text{CuCO}_3) = 1.4 \times 10^{-10}
     \]

**碳酸的电离常数**：  
   - 一级电离：  
     \[
     K_{a1}(\text{H}_2\text{CO}_3) = 4.2 \times 10^{-7}
     \]  
   - 二级电离：  
     \[
     K_{a2}(\text{H}_2\text{CO}_3) = 5.6 \times 10^{-11}
     \]

 **水的离子积**：  
   \[
   K_w = 1.0 \times 10^{-14}
   \]

 **铵根离子的酸解离常数**：  
   \[
   \text{p}K_a(\text{NH}_4^+) = 9.26
   \]

 **\(\text{Cu}^{2+}-\text{NH}_3\) 累积稳定常数**：  
   \[
   \begin{aligned}
   \lg\beta_1 &= 4.31, \\
   \lg\beta_2 &= 7.98, \\
   \lg\beta_3 &= 11.02, \\
   \lg\beta_4 &= 13.32, \\
   \lg\beta_5 &= 12.86.
   \end{aligned}
   \]
In the chemical equation for **Reaction 1**, the sum of the coefficients of the products is 1 greater than the sum of the coefficients of the reactants.
In the chemical equation for **Reaction 2**, the sum of the coefficients of the products is 1 greater than the sum of the coefficients of the reactants.
After the solid is completely dissolved, the pH of the solution is approximately 8.20.
After the solid has just completely dissolved, the amount of ammonia bound to $\text{Cu}$ accounts for approximately 0.38 of the total ammonia content (including $\text{NH}_3$ bound to $\text{Cu}$, free $\text{NH}_3$, and free $\text{NH}_4^{+}$).
To dissolve the solid, the minimum mass of \((\text{NH}_4)_2\text{CO}_3\) to be added is approximately 0.27 g.
To dissolve the solid, the minimum mass of \((\text{NH}_4)_2\text{CO}_3\) to be added is approximately 0.49 g.
To dissolve the solid, the minimum mass of \((\text{NH}_4)_2\text{CO}_3\) to be added is approximately 0.61 g.
To dissolve the solid, the minimum mass of \((\text{NH}_4)_2\text{CO}_3\) to be added is approximately 0.85 g.
**反应1**的化学方程式中生成物系数之和比反应物系数之和多1
**反应2**的化学方程式中生成物系数之和比反应物系数之和多1
固体刚好完全溶解后，溶液的pH大约为8.20
固体刚好完全溶解后，与$\text{Cu}$结合的氨量约占总氨量（包括与$\text{Cu}$结合的$\text{NH}_3$、自由$\text{NH}_3$和自由$\text{NH}_4^{+}$）的0.38
欲使固体溶解，加入的\((\text{NH}_4)_2\text{CO}_3\) 最小质量约为0.27 g
欲使固体溶解，加入的\((\text{NH}_4)_2\text{CO}_3\) 最小质量约为0.49 g
欲使固体溶解，加入的\((\text{NH}_4)_2\text{CO}_3\) 最小质量约为0.61 g
欲使固体溶解，加入的\((\text{NH}_4)_2\text{CO}_3\) 最小质量约为0.85 g
From the problem description, it is known that **A** and **B** are both basic copper carbonates. Based on the mass fraction of $\text{Cu}$, the chemical formulas of **A** and **B** can be deduced as $\text{Cu}_2\text{(OH)}_2\text{CO}_3$ and $\text{Cu}_3\text{(OH)}_2\text{(CO}_3)_2$, respectively.

<Checkpoint>[The molecular formula of **A** is $\text{Cu}_2\text{(OH)}_2\text{CO}_3$](1)</Checkpoint>

<Checkpoint>[The molecular formula of **B** is $\text{Cu}_3\text{(OH)}_2\text{(CO}_3)_2$](1)</Checkpoint>

Thus, the reactants and products can be determined, and the chemical equations for **Reaction 1** and **Reaction 2** are as follows:

**Reaction 1**:  
 \[
   2\,\text{CuSO}_4 + 2\,\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow \text{Cu}_2(\text{OH})_2\text{CO}_3 + 2\,\text{Na}_2\text{SO}_4 + \text{CO}_2 
   \]

**Reaction 2**:  
   \[
   3\,\text{CuSO}_4 + 3\,\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow \text{Cu}_3(\text{OH})_2(\text{CO}_3)_2 + 3\,\text{Na}_2\text{SO}_4 + \text{CO}_2 
   \]

<Checkpoint>[The chemical equation for **Reaction 1** is \[    2\,\text{CuSO}_4 + 2\,\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow \text{Cu}_2(\text{OH})_2\text{CO}_3 + 2\,\text{Na}_2\text{SO}_4 + \text{CO}_2     \]](1)</Checkpoint>

<Checkpoint>[The chemical equation for **Reaction 2** is \[    3\,\text{CuSO}_4 + 3\,\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow \text{Cu}_3(\text{OH})_2(\text{CO}_3)_2 + 3\,\text{Na}_2\text{SO}_4 + \text{CO}_2     \]](1)</Checkpoint>

Since \(\text{CuO}\) is highly insoluble, \((\text{NH}_4)_2\text{CO}_3\) is present in large excess compared to \(\text{CuO}\). Therefore, it can be assumed that the pH of the solution after the solid dissolves is approximately equal to the pH of a pure \((\text{NH}_4)_2\text{CO}_3\) solution. For a pure \((\text{NH}_4)_2\text{CO}_3\) solution, the charge balance equation is:  
$[\mathrm{NH}_4^{+}] + [\mathrm{H}^{+}] = [\mathrm{OH}^{-}] + [\mathrm{HCO}_3^{-}] + 2[\mathrm{CO}_3^{2-}]$

Upon analysis, the analytical concentration of \((\text{NH}_4)_2\text{CO}_3\) is greater than $1.0 \times 10^{-3}\ \mathrm{mol/L}$, which is significantly higher than $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$. Thus, $[\mathrm{H}^{+}]$ and $[\mathrm{OH}^{-}]$ in the above equation can be neglected, simplifying the charge balance equation to:  
$[\mathrm{NH}_4^{+}] = [\mathrm{HCO}_3^{-}] + 2[\mathrm{CO}_3^{2-}]$

Let the analytical concentration of \((\text{NH}_4)_2\text{CO}_3\) be $c$. According to the distribution coefficient formulas:

$[\mathrm{NH}_4^{+}] = 2c · \frac{[\mathrm{H}^{+}]}{[\mathrm{H}^{+}] + K_a(\mathrm {NH}_4^+)}$

$[\mathrm{HCO}_3^{-}] = c · \frac{K_{a1}[\mathrm{H}^{+}]}{[\mathrm{H}^{+}]^2 + K_{a1}[\mathrm{H}^{+}] + K_{a1}K_{a2}}$

$[\mathrm{CO}_3^{2-}] = c · \frac{ K_{a1}K_{a2}}{[\mathrm{H}^{+}]^2 + K_{a1}[\mathrm{H}^{+}] + K_{a1}K_{a2}}$

By solving these equations, the concentration of $\mathrm{H}^{+}$ after the solid dissolves is calculated as $[\mathrm{H}^{+}] = 6.4 \times 10^{-10}\ \mathrm{mol/L}$, and $[\mathrm{OH}^{-}] = 1.6 \times 10^{-5}\ \mathrm{mol/L}$.

<Checkpoint>[After the solid dissolves, $[\mathrm{H}^{+}] = 6.4 \times 10^{-10}\ \mathrm{mol/L}$](2)</Checkpoint>

Since $\mathrm{CuO}$ is less soluble than $\mathrm{Cu(OH)}_2$, when the solid just dissolves, the following condition holds:  
$[\mathrm{Cu}^{2+}][\mathrm{OH}^{-}]^2 = K_s$.  
Thus, the concentration of $\mathrm{Cu}^{2+}$ when the solid just dissolves is calculated as $[\mathrm{Cu}^{2+}] = 1.0 \times 10^{-11}\ \mathrm{mol/L}$.

<Checkpoint>[When the solid just dissolves, $[\mathrm{Cu}^{2+}] = 1.0 \times 10^{-11}\ \mathrm{mol/L}$](1)</Checkpoint>

Given that $c(\mathrm{Cu}^{2+}) = 1.0 \times 10^{-3}\ \mathrm{mol/L}$, the side reaction coefficient for $\mathrm{Cu}^{2+}$ is:  
$α = \frac{ c(\mathrm{Cu}^{2+})}{[\mathrm{Cu}^{2+}]} = 1.0 \times 10^{8}$.

According to the side reaction coefficient formula:  
$α = 1 + β_1[\mathrm{NH}_3] + β_2[\mathrm{NH}_3]^2 + β_3[\mathrm{NH}_3]^3 + β_4[\mathrm{NH}_3]^4 + β_5[\mathrm{NH}_3]^5 = 1.0 \times 10^{8}$,

the free $[\mathrm{NH}_3]$ is determined to be $[\mathrm{NH}_3] = 4.5 \times 10^{-2}\ \mathrm{mol/L}$.

<Checkpoint>[Free $[\mathrm{NH}_3] = 4.5 \times 10^{-2}\ \mathrm{mol/L}$](1)</Checkpoint>

From the distribution coefficient:  
$[\mathrm{NH}_3] = c(\mathrm{NH}_4^{+}) · \frac{K_a(\mathrm {NH}_4^+)}{[\mathrm{H}^{+}] + K_a(\mathrm {NH}_4^+)}$,  
the concentration $c(\mathrm{NH}_4^{+})$ is solved as $c(\mathrm{NH}_4^{+}) = 9.8 \times 10^{-2}\ \mathrm{mol/L}$.

Additionally, the amount of ammonia bound to $\text{Cu}$ can be calculated using the complex distribution coefficient:  
$c_{Cu}(\mathrm{NH}_3) = [\mathrm{Cu}^{2+}] · (β_1[\mathrm{NH}_3] + 2β_2[\mathrm{NH}_3]^2 + 3β_3[\mathrm{NH}_3]^3 + 4β_4[\mathrm{NH}_3]^4 + 5β_5[\mathrm{NH}_3]^5) = 3.91 \times 10^{-3}\ \mathrm{mol/L}$.

Thus, the total $\text{NH}_3$ concentration is $0.102\ \mathrm{mol/L}$.

<Checkpoint>[The total $\text{NH}_3$ concentration is $0.102\ \mathrm{mol/L}$](2)</Checkpoint>

From this, the amount of ammonia bound to $\text{Cu}$ accounts for approximately 0.038 of the total ammonia, which can be neglected in terms of its consumption during solid dissolution.

<Checkpoint>[The amount of ammonia bound to $\text{Cu}$ accounts for approximately 0.038 of the total ammonia](1)</Checkpoint>

In summary, the analytical concentration of ammonium carbonate is half of the total $\text{NH}_3$ concentration. Thus, the analytical concentration of ammonium carbonate and the minimum mass of ammonium carbonate to be added are calculated as $0.051\ \mathrm{mol/L}$ and $0.49\ \mathrm{g}$, respectively.

<Checkpoint>[The analytical concentration of ammonium carbonate is $0.051\ \mathrm{mol/L}$](1)</Checkpoint>

<Checkpoint>[The minimum mass of ammonium carbonate to be added is $0.49\ \mathrm{g}$](2)</Checkpoint>
由题目描述，可知 **A** 和  **B** 都是$\text{Cu}$的碱式碳酸盐。根据$\text{Cu}$的质量分数，可推出 **A** 和  **B** 的化学式分别是$\text{Cu}_2\text{(OH)}_2\text{CO}_3$和$\text{Cu}_3\text{(OH)}_2\text{(CO}_3)_2$。

<Checkpoint>[**A** 的分子式为$\text{Cu}_2\text{(OH)}_2\text{CO}_3$](1)</Checkpoint>

<Checkpoint>[**B** 的分子式为$\text{Cu}_3\text{(OH)}_2\text{(CO}_3)_2$](1)</Checkpoint>

由此可确定反应物和生成物，从而确定**反应1**和**反应2**化学方程式如下：

**反应 1**：  
 \[
   2\,\text{CuSO}_4 + 2\,\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow \text{Cu}_2(\text{OH})_2\text{CO}_3 + 2\,\text{Na}_2\text{SO}_4 + \text{CO}_2 
   \]

**反应 2**：  
   \[
   3\,\text{CuSO}_4 + 3\,\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow \text{Cu}_3(\text{OH})_2(\text{CO}_3)_2 + 3\,\text{Na}_2\text{SO}_4 + \text{CO}_2 
   \]

<Checkpoint>[**反应1**的化学方程式为 \[    2\,\text{CuSO}_4 + 2\,\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow \text{Cu}_2(\text{OH})_2\text{CO}_3 + 2\,\text{Na}_2\text{SO}_4 + \text{CO}_2     \]](1)</Checkpoint>

<Checkpoint>[**反应2**的化学方程式为 \[    3\,\text{CuSO}_4 + 3\,\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow \text{Cu}_3(\text{OH})_2(\text{CO}_3)_2 + 3\,\text{Na}_2\text{SO}_4 + \text{CO}_2     \]](1)</Checkpoint>

由于\(\text{CuO}\)很难溶，故\((\text{NH}_4)_2\text{CO}_3\)相较于\(\text{CuO}\)是大过量的，故可认为固体溶解后体系的pH约等于纯\((\text{NH}_4)_2\text{CO}_3\)溶液的pH。对于纯\((\text{NH}_4)_2\text{CO}_3\)溶液，有电荷守恒式：$[\mathrm{NH}_4^{+}] + [\mathrm{H}^{+}] = [\mathrm{OH}^{-}] + [\mathrm{HCO}_3^{-}] + 2[\mathrm{CO}_3^{2-}]$

而经分析，\((\text{NH}_4)_2\text{CO}_3\)的分析浓度大于$1.0 \times 10^{-3}\ \mathrm{mol/L}$，远高于$\mathrm{H}^{+}$和$\mathrm{OH}^{-}$，因此上式中$[\mathrm{H}^{+}]$和$[\mathrm{OH}^{-}]$可忽略，故电荷守恒式变成：$[\mathrm{NH}_4^{+}] = [\mathrm{HCO}_3^{-}] + 2[\mathrm{CO}_3^{2-}]$

设\((\text{NH}_4)_2\text{CO}_3\)的分析浓度为$c$，根据分布系数公式：

$[\mathrm{NH}_4^{+}] = 2c · \frac{[\mathrm{H}^{+}]}{[\mathrm{H}^{+}] + K_a(\mathrm {NH}_4^+)}$

$[\mathrm{HCO}_3^{-}] = c · \frac{K_{a1}[\mathrm{H}^{+}]}{[\mathrm{H}^{+}]^2 + K_{a1}[\mathrm{H}^{+}] + K_{a1}K_{a2}}$

$[\mathrm{CO}_3^{2-}] = c · \frac{ K_{a1}K_{a2}}{[\mathrm{H}^{+}]^2 + K_{a1}[\mathrm{H}^{+}] + K_{a1}K_{a2}}$

联立方程，可以计算出固体溶解后$[\mathrm{H}^{+}] = 6.4 \times 10^{-10}\ \mathrm{mol/L}$，$[\mathrm{OH}^{-}] = 1.6 \times 10^{-5}\ \mathrm{mol/L}$

<Checkpoint>[固体溶解后，$[\mathrm{H}^{+}] = 6.4 \times 10^{-10}\ \mathrm{mol/L}$](2)</Checkpoint>

由于$\mathrm{CuO}$比$\mathrm{Cu(OH)}_2$更难溶，则固体恰好溶解时，有：$[\mathrm{Cu}^{2+}][\mathrm{OH}^{-}]^2 = K_s$，故可算出固体刚好溶解时$[\mathrm{Cu}^{2+}] = 1.0 \times 10^{-11}\ \mathrm{mol/L}$

<Checkpoint>[固体刚好溶解时$[\mathrm{Cu}^{2+}] = 1.0 \times 10^{-11}\ \mathrm{mol/L}$](1)</Checkpoint>

而$c(\mathrm{Cu}^{2+}) = 1.0 \times 10^{-3}\ \mathrm{mol/L}$，故$\mathrm{Cu}^{2+}$的副反应系数为：$α = \frac{ c(\mathrm{Cu}^{2+})}{[\mathrm{Cu}^{2+}]} = 1.0 \times 10^{8}$

根据副反应系数公式:

$α = 1 + β_1[\mathrm{NH}_3] + β_2[\mathrm{NH}_3]^2 + β_3[\mathrm{NH}_3]^3 + β_4[\mathrm{NH}_3]^4 + β_5[\mathrm{NH}_3]^5 = 1.0 \times 10^{8}$

可以得到自由$[\mathrm{NH}_3] = 4.5 \times 10^{-2}\ \mathrm{mol/L}$

<Checkpoint>[自由$[\mathrm{NH}_3] = 4.5 \times 10^{-2}\ \mathrm{mol/L}$](1)</Checkpoint>

由分布系数：

$[\mathrm{NH}_3] = c(\mathrm{NH}_4^{+}) · \frac{K_a(\mathrm {NH}_4^+)}{[\mathrm{H}^{+}] + K_a(\mathrm {NH}_4^+)}$

解得$c(\mathrm{NH}_4^{+}) = 9.8 \times 10^{-2}\ \mathrm{mol/L}$

又可根据配合物分布系数算出与$\text{Cu}$结合的氨量：

$c_{Cu}(\mathrm{NH}_3) = [\mathrm{Cu}^{2+}] · (β_1[\mathrm{NH}_3] + 2β_2[\mathrm{NH}_3]^2 + 3β_3[\mathrm{NH}_3]^3 + 4β_4[\mathrm{NH}_3]^4 + 5β_5[\mathrm{NH}_3]^5) = 3.91 \times 10^{-3}\ \mathrm{mol/L}$

故总$\text{NH}_3$浓度为$0.102\ \mathrm{mol/L}$.

<Checkpoint>[总$\text{NH}_3$浓度为$0.102\ \mathrm{mol/L}$](2)</Checkpoint>

由此算出与$\text{Cu}$结合的氨量约占总氨量的0.038，故可以忽略溶解固体对氨的消耗。

<Checkpoint>[与$\text{Cu}$结合的氨量约占总氨量的0.038](1)</Checkpoint>

综上，碳酸铵分析浓度为总$\text{NH}_3$浓度的二分之一，由此可算出碳酸铵分析浓度和加入碳酸铵的最小质量分别为$0.051\ \mathrm{mol/L}$和$0.49\ \mathrm{g}$

<Checkpoint>[碳酸铵分析浓度为$0.051\ \mathrm{mol/L}$](1)</Checkpoint>

<Checkpoint>[加入碳酸铵的最小质量为$0.49\ \mathrm{g}$](2)</Checkpoint>