Calcium carbonate is also one of the important components of scale. A large amount of scale may cause uneven heat transfer, and even cause explosions in boilers. Therefore, efficient water softening is one of the important topics.

Given the following thermodynamic data: (Air $p(\text{CO}_2) = 0.034 \text{ kPa}$)

| Substance | $\Delta_f H_m^{\ominus} / \text{kJ} \cdot \text{mol}^{-1}$ | $S_m^{\ominus} / \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$ |
|---|---|---|
| $\text{CaCO}_3 (\text{s})$ | -1207.6 | 91.7 |
| $\text{Mg(OH)}_2 (\text{s})$ | -924.7 | 63.24 |
| $\text{H}_2\text{O} (\text{l})$ | -285.8 | 69.95 |
| $\text{CO}_2 (\text{g})$ | -393.5 | 213.8 |
| $\text{Ca}^{2+} (\text{aq.})$ | -543.0 | -56.2 |
| $\text{Mg}^{2+} (\text{aq.})$ | -467.0 | -137.4 |
| $\text{OH}^- (\text{aq.})$ | -230.0 | -10.90 |
| $\text{CO}_3^{2-} (\text{aq.})$ | -675.2 | -50.0 |
| $\text{HCO}_3^-(\text{aq.})$ | -689.9 | 98.4 |

Hard water often leaves scale in kettles. If boiling ( $T=373 \text{ K}$) water with $\text{pH}=7.35$ at a certain normal temperature ($T=298 \text{ K}$) (the water only contains carbonate buffer pairs) does not produce $\text{CaCO}_3$ or $\text{Mg(OH)}_2$ scale, find the maximum hardness of the water (here it refers to the sum of the concentrations of calcium and magnesium ions in the water, which needs to be converted into the mass of calcium carbonate, in units of $\text{ppm}$, i.e., $\rm mg/L$).
碳酸钙也是水垢的重要成分之一，大量水垢可能引起传热不均，在锅炉内甚至引发爆炸，故而水的高效软化是重要课题之一。

已知下列热力学数据：(空气中 $p(\text{CO}_2) = 0.034 \text{ kPa}$)

| 物质 | $\Delta_f H_m^{\ominus} / \text{kJ} \cdot \text{mol}^{-1}$ | $S_m^{\ominus} / \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$ |
|---|---|---|
| $\text{CaCO}_3 (\text{s})$ | -1207.6 | 91.7 |
| $\text{Mg(OH)}_2 (\text{s})$ | -924.7 | 63.24 |
| $\text{H}_2\text{O} (\text{l})$ | -285.8 | 69.95 |
| $\text{CO}_2 (\text{g})$ | -393.5 | 213.8 |
| $\text{Ca}^{2+} (\text{aq.})$ | -543.0 | -56.2 |
| $\text{Mg}^{2+} (\text{aq.})$ | -467.0 | -137.4 |
| $\text{OH}^- (\text{aq.})$ | -230.0 | -10.90 |
| $\text{CO}_3^{2-} (\text{aq.})$ | -675.2 | -50.0 |
| $\text{HCO}_3^-(\text{aq.})$ | -689.9 | 98.4 |

 硬水常在水壶中残余水垢，若煮沸($T=373 \text{ K}$)某常温($T=298 \text{ K}$)下 $\text{pH}=7.35$ 的水(水中只含有碳酸盐的缓冲对)后没有产生 $\text{CaCO}_3$ 或 $\text{Mg(OH)}_2$ 水垢，求水的最大硬度(在这里是指水中钙、镁离子的浓度之和，需要折算为碳酸钙的质量，单位为 $\text{ppm}$，即$\rm mg/L$)。
$730 \text{ ppm}$
$430 \text{ ppm}$
$7.3 \text{ ppm}$
$300 \text{ ppm}$
$350 \text{ ppm}$
$600 \text{ ppm}$
$3.5 \text{ ppm}$
$730 \text{ ppm}$
$430 \text{ ppm}$
$7.3 \text{ ppm}$
$300 \text{ ppm}$
$350 \text{ ppm}$
$600 \text{ ppm}$
$3.5 \text{ ppm}$
First, calculate the equilibrium constant of the reaction: $CO_2(g) + OH^-(a.q.) = HCO_3^-(a.q.)$

Reaction $\Delta_r H_m^{\ominus} = -66.4 \text{ kJ} \cdot \text{mol}^{-1}$，$\Delta_r S_m^{\ominus} = -104.5 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$

<Checkpoint>[$\Delta_r H_m^{\ominus} = -66.4 \text{ kJ} \cdot \text{mol}^{-1}$，$\Delta_r S_m^{\ominus} = -104.5 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$](1)</Checkpoint>

At $298 \text{ K}$, the reaction has $\Delta_r G_m^{\ominus} = -35.3 \text{ kJ} \cdot \text{mol}^{-1}$，$K_1^{\ominus} = 1.5 \times 10^6$

<Checkpoint>[At $298 \text{ K}$, $\Delta_r G_m^{\ominus} = -35.3 \text{ kJ} \cdot \text{mol}^{-1}$，$K_1^{\ominus} = 1.5 \times 10^6$](1)</Checkpoint>

At $373 \text{ K}$, $\Delta_r G_m^{\ominus} = -27.4 \text{ kJ} \cdot \text{mol}^{-1}$，$K_2^{\ominus} = 6.9 \times 10^3$

<Checkpoint>[At $373 \text{ K}$, $\Delta_r G_m^{\ominus} = -27.4 \text{ kJ} \cdot \text{mol}^{-1}$，$K_2^{\ominus} = 6.9 \times 10^3$](1)</Checkpoint>

It is easy to calculate that the solution at room temperature contains $[HCO_3^-]_1 = 1.14 \times 10^{-4} \text{ mol} \cdot \text{L}^{-1}$

<Checkpoint>[The solution at room temperature contains $[HCO_3^-]_1 = 1.14 \times 10^{-4} \text{ mol} \cdot \text{L}^{-1}$](1)</Checkpoint>

After boiling, the equilibrium shifts, $ [HCO_3^-]_2 + [OH^-]_2 = [HCO_3^-]_1 + [OH^-]$

Solving for $[HCO_3^-]_2 = 8.0 \times 10^{-5} \text{ mol} \cdot \text{L}^{-1}$ $[OH^-]_2 = 3.4 \times 10^{-5} \text{ mol} \cdot \text{L}^{-1}$

<Checkpoint>[After boiling, $[HCO_3^-]_2 = 8.0 \times 10^{-5} \text{ mol} \cdot \text{L}^{-1}$ $[OH^-]_2 = 3.4 \times 10^{-5} \text{ mol} \cdot \text{L}^{-1}$](1)</Checkpoint>

Consider the reaction $HCO_3^-(a.q.) + OH^-(a.q.) = CO_3^{2-}(a.q.) + H_2O(l)$

The equilibrium constant of the reaction at $373 \text{ K}$ is $K^{\ominus} = 1.7 \times 10^2$, at this time $[CO_3^{2-}] = 4.6 \times 10^{-7} \text{ mol} \cdot \text{L}^{-1}$

<Checkpoint>[At $373 \text{ K}$, $K^{\ominus} = 1.7 \times 10^2$, at this time $[CO_3^{2-}] = 4.6 \times 10^{-7} \text{ mol} \cdot \text{L}^{-1}$](1)</Checkpoint>

Then consider $Ca^{2+}(a.q.) + CO_3^{2-}(a.q.) = CaCO_3(s)$

The equilibrium constant of the reaction at $373 \text{ K}$ is $K^{\ominus} = 7.1 \times 10^8$, under critical conditions $[Ca^{2+}]_{\text{max}} = 3.0 \times 10^{-3} \text{ mol} \cdot \text{L}^{-1}$

Consider the reaction $Mg^{2+}(a.q.) + 2OH^-(a.q.) = Mg(OH)_2(s)$

The equilibrium constant of the reaction at $373 \text{ K}$ is $K^{\ominus} = 2.0 \times 10^{11}$. Under critical conditions, $[Mg^{2+}]_{\text{max}} = 4.3 \times 10^{-3} \text{ mol} \cdot \text{L}^{-1}$

<Checkpoint>[Under critical conditions, $[Mg^{2+}]_{\text{max}} = 4.3 \times 10^{-3} \text{ mol} \cdot \text{L}^{-1}$](1)</Checkpoint>
Adding the calculated critical calcium and magnesium ion concentrations, the maximum hardness of water is $= 730 \text{ ppm}$

<Checkpoint>[The maximum hardness of water $= 730 \text{ ppm}$](1)</Checkpoint>
首先计算反应的平衡常数：$CO_2(g) + OH^-(a.q.) = HCO_3^-(a.q.)$

反应 $\Delta_r H_m^{\ominus} = -66.4 \text{ kJ} \cdot \text{mol}^{-1}$，$\Delta_r S_m^{\ominus} = -104.5 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$

<Checkpoint>[$\Delta_r H_m^{\ominus} = -66.4 \text{ kJ} \cdot \text{mol}^{-1}$，$\Delta_r S_m^{\ominus} = -104.5 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$](1)</Checkpoint>

反应在 $298 \text{ K}$ 下 $\Delta_r G_m^{\ominus} = -35.3 \text{ kJ} \cdot \text{mol}^{-1}$，$K_1^{\ominus} = 1.5 \times 10^6$

<Checkpoint>[$298 \text{ K}$ 下 $\Delta_r G_m^{\ominus} = -35.3 \text{ kJ} \cdot \text{mol}^{-1}$，$K_1^{\ominus} = 1.5 \times 10^6$](1)</Checkpoint>

在 $373 \text{ K}$ 下 $\Delta_r G_m^{\ominus} = -27.4 \text{ kJ} \cdot \text{mol}^{-1}$，$K_2^{\ominus} = 6.9 \times 10^3$

<Checkpoint>[$373 \text{ K}$ 下 $\Delta_r G_m^{\ominus} = -27.4 \text{ kJ} \cdot \text{mol}^{-1}$，$K_2^{\ominus} = 6.9 \times 10^3$](1)</Checkpoint>

不难计算得出，常温下溶液中含 $[HCO_3^-]_1 = 1.14 \times 10^{-4} \text{ mol} \cdot \text{L}^{-1}$

<Checkpoint>[常温下溶液中含 $[HCO_3^-]_1 = 1.14 \times 10^{-4} \text{ mol} \cdot \text{L}^{-1}$](1)</Checkpoint>

煮沸后平衡发生移动，$ [HCO_3^-]_2 + [OH^-]_2 = [HCO_3^-]_1 + [OH^-]$

解得 $[HCO_3^-]_2 = 8.0 \times 10^{-5} \text{ mol} \cdot \text{L}^{-1}$ $[OH^-]_2 = 3.4 \times 10^{-5} \text{ mol} \cdot \text{L}^{-1}$

<Checkpoint>[煮沸后$[HCO_3^-]_2 = 8.0 \times 10^{-5} \text{ mol} \cdot \text{L}^{-1}$ $[OH^-]_2 = 3.4 \times 10^{-5} \text{ mol} \cdot \text{L}^{-1}$](1)</Checkpoint>

考虑反应 $HCO_3^-(a.q.) + OH^-(a.q.) = CO_3^{2-}(a.q.) + H_2O(l)$

反应在 $373 \text{ K}$ 下的平衡常数 $K^{\ominus} = 1.7 \times 10^2$，此时$[CO_3^{2-}] = 4.6 \times 10^{-7} \text{ mol} \cdot \text{L}^{-1}$

<Checkpoint>[$373 \text{ K}$ 下 $K^{\ominus} = 1.7 \times 10^2$，此时$[CO_3^{2-}] = 4.6 \times 10^{-7} \text{ mol} \cdot \text{L}^{-1}$](1)</Checkpoint>

再考虑 $Ca^{2+}(a.q.) + CO_3^{2-}(a.q.) = CaCO_3(s)$

反应在 $373 \text{ K}$ 下的平衡常数 $K^{\ominus} = 7.1 \times 10^8$，临界情形下 $[Ca^{2+}]_{\text{max}} = 3.0 \times 10^{-3} \text{ mol} \cdot \text{L}^{-1}$

考虑反应 $Mg^{2+}(a.q.) + 2OH^-(a.q.) = Mg(OH)_2(s)$

反应在 $373 \text{ K}$ 平衡常数 $K^{\ominus} = 2.0 \times 10^{11}$ 临界情形下 $[Mg^{2+}]_{\text{max}} = 4.3 \times 10^{-3} \text{ mol} \cdot \text{L}^{-1}$

<Checkpoint>[临界情形下 $[Mg^{2+}]_{\text{max}} = 4.3 \times 10^{-3} \text{ mol} \cdot \text{L}^{-1}$](1)</Checkpoint>
将计算得到的临界钙、镁离子浓度相加，得到水的最大硬度为 $= 730 \text{ ppm}$

<Checkpoint>[水的最大硬度 $= 730 \text{ ppm}$](1)</Checkpoint>