Known: For the diprotic acid $H_2CO_3$, $K_{a1} = 4.2 * 10^{-7}$, $K_{a2} = 5.6 * 10^{-11}$, and for $CaCO_3$, $K_{sp} = 2.9 * 10^{-9}$. Calculate the solubility of $CaCO_3$ in pure water and in 1L of 0.01 mol/L $NaHCO_3$ solution (in mol/L), respectively. The following options provide possible answers for the two solubilities, with the units already given in the question. Choose the correct option (an error within 15% is considered correct):
已知： 二元酸$H_2CO_3$的$K_{a1} =4.2* 10^{-7}$ ,$K_{a2} =5.6* 10^{-11}$，$CaCO_3$的$K_{sp}=2.9*10^{-9}$。试分别计算$CaCO_3$在纯水和1L 0.01 mol/L 的$NaHCO_3$溶液中的溶解度（单位均为mol/L），下列选项依次给出了两个溶解度的可能答案，单位已经在问题中给出，请选择正确的一项（误差在15%以内视为正确）：
$5*10^{-5},1.8*10^{-5}$
$5*10^{-5},2.5*10^{-5}$
$5*10^{-5},3.3*10^{-5}$
$7*10^{-5},1.8*10^{-5}$
$7*10^{-5},2.5*10^{-5}$
$7*10^{-5},3.3*10^{-5}$
$9*10^{-5},1.8*10^{-5}$
$9*10^{-5},2.5*10^{-5}$
$9*10^{-5},3.3*10^{-5}$
All other options are incorrect.
$5*10^{-5},1.8*10^{-5}$
$5*10^{-5},2.5*10^{-5}$
$5*10^{-5},3.3*10^{-5}$
$7*10^{-5},1.8*10^{-5}$
$7*10^{-5},2.5*10^{-5}$
$7*10^{-5},3.3*10^{-5}$
$9*10^{-5},1.8*10^{-5}$
$9*10^{-5},2.5*10^{-5}$
$9*10^{-5},3.3*10^{-5}$
其他选项均不正确
### In pure water:
$CaCO_3\to Ca^{2+}+CO_3^{2-}$，Let the solubility of $CaCO_3$ be x mol/L, then the total concentration of $Ca^{2+}$ and $H_2CO_3$ in the solution are both x mol/L. The concentration of $CO_3^{2-}$ can be expressed using the distribution coefficient. Therefore, according to the precipitation-dissolution equilibrium:

$ [Ca^{2+}][CO_3^{2-}]=K_{sp}$

<Checkpoint>[Precipitation-dissolution equilibrium: $ [Ca^{2+}][CO_3^{2-}]=K_{sp}$](1)</Checkpoint>

$x^2\cdot \frac{K_{a1}K_{a2}}{[H^+]^2+K_{a1}[H^+]+K_{a1}K_{a2}}=K_{sp}$

In addition, according to charge conservation, we can list:

$[H^+]+2[Ca^{2+}]=2[CO_3^{2-}]+[HCO_3^-]+[OH^-] $

<Checkpoint>[Charge conservation: $[H^+]+2[Ca^{2+}]=2[CO_3^{2-}]+[HCO_3^-]+[OH^-] $](1)</Checkpoint>

$[H^+]+2x=x\cdot \frac{2K_{a1}K_{a2}+K_{a1}[H^+]}{[H^+]^2+K_{a1}[H^+]+K_{a1}K_{a2}}+[OH^-]$

Solving the simultaneous equations yields $x=1.0*10^{-4}mol/L$

<Checkpoint>[The solubility of $CaCO_3$ in pure water is $1.0*10^{-4}mol/L$](1)</Checkpoint>

### In $NaHCO_3$ solution:
$CaCO_3\to Ca^{2+}+CO_3^{2-}$，Let the solubility of $CaCO_3$ be x mol/L, then the total concentration of $Ca^{2+}$ in the solution is x mol/L, and the total concentration of $H_2CO_3$ is (x+0.01) mol/L. The concentration of $CO_3^{2-}$ can be expressed using the distribution coefficient. Therefore, according to the precipitation-dissolution equilibrium:

$ [Ca^{2+}][CO_3^{2-}]=K_{sp}$

$x(x+0.01)\cdot \frac{K_{a1}K_{a2}}{[H^+]^2+K_{a1}[H^+]+K_{a1}K_{a2}}=K_{sp}$

In addition, according to charge conservation, we can list another equation:

$[H^+]+2[Ca^{2+}]+[Na^+]=2[CO_3^{2-}]+[HCO_3^-]+[OH^-] $

<Checkpoint>[Charge conservation: $[H^+]+2[Ca^{2+}]+[Na^+]=2[CO_3^{2-}]+[HCO_3^-]+[OH^-] $](1)</Checkpoint>

$[H^+]+2x+0.01=(x+0.01)\cdot \frac{2K_{a1}K_{a2}+K_{a1}[H^+]}{[H^+]^2+K_{a1}[H^+]+K_{a1}K_{a2}}+[OH^-]$

Solving the simultaneous equations yields $x=2.3*10^{-5}mol/L$

<Checkpoint>[The solubility of $CaCO_3$ in 1L 0.01 mol/L $NaHCO_3$ solution is $2.3*10^{-5}mol/L$](1)</Checkpoint>

Therefore, choose option H
### 纯水中：
$CaCO_3\to Ca^{2+}+CO_3^{2-}$，设$CaCO_3$溶解度为x mol/L，则溶液中$Ca^{2+}$和$H_2CO_3$的总浓度均为x mol/L，$CO_3^{2-}$的浓度使用分布系数可以表示出来，因此根据沉淀溶解平衡有：

$ [Ca^{2+}][CO_3^{2-}]=K_{sp}$

<Checkpoint>[沉淀溶解平衡：$ [Ca^{2+}][CO_3^{2-}]=K_{sp}$](1)</Checkpoint>

$x^2\cdot \frac{K_{a1}K_{a2}}{[H^+]^2+K_{a1}[H^+]+K_{a1}K_{a2}}=K_{sp}$

除此之外，再根据电荷守恒可以列出：

$[H^+]+2[Ca^{2+}]=2[CO_3^{2-}]+[HCO_3^-]+[OH^-] $

<Checkpoint>[电荷守恒：$[H^+]+2[Ca^{2+}]=2[CO_3^{2-}]+[HCO_3^-]+[OH^-] $](1)</Checkpoint>

$[H^+]+2x=x\cdot \frac{2K_{a1}K_{a2}+K_{a1}[H^+]}{[H^+]^2+K_{a1}[H^+]+K_{a1}K_{a2}}+[OH^-]$

联立方程解得$x=1.0*10^{-4}mol/L$

<Checkpoint>[$CaCO_3$在纯水中溶解度为$1.0*10^{-4}mol/L$](1)</Checkpoint>

### $NaHCO_3$溶液中：
$CaCO_3\to Ca^{2+}+CO_3^{2-}$，设$CaCO_3$溶解度为x mol/L，则溶液中$Ca^{2+}$的总浓度为x mol/L，$H_2CO_3$的总浓度为(x+0.01) mol/L，$CO_3^{2-}$的浓度使用分布系数可以表示出来，因此根据沉淀溶解平衡有：

$ [Ca^{2+}][CO_3^{2-}]=K_{sp}$

$x(x+0.01)\cdot \frac{K_{a1}K_{a2}}{[H^+]^2+K_{a1}[H^+]+K_{a1}K_{a2}}=K_{sp}$

除此之外，再根据电荷守恒可以列出另一个方程：

$[H^+]+2[Ca^{2+}]+[Na^+]=2[CO_3^{2-}]+[HCO_3^-]+[OH^-] $

<Checkpoint>[电荷守恒：$[H^+]+2[Ca^{2+}]+[Na^+]=2[CO_3^{2-}]+[HCO_3^-]+[OH^-] $](1)</Checkpoint>

$[H^+]+2x+0.01=(x+0.01)\cdot \frac{2K_{a1}K_{a2}+K_{a1}[H^+]}{[H^+]^2+K_{a1}[H^+]+K_{a1}K_{a2}}+[OH^-]$

联立方程解得$x=2.3*10^{-5}mol/L$

<Checkpoint>[$CaCO_3$在1L 0.01 mol/L 的$NaHCO_3$溶液中溶解度为$2.3*10^{-5}mol/L$](1)</Checkpoint>

因此选择H选项