Partial electrode potentials are shown below:

| Half-reaction                                      | Standard Electrode Potential |
|---------------------------------------------|--------------|
| $4\mathbf{A}O_2(aq) + 4H^++ 6e^- \to \mathbf{A}_4O_6^{2-} + 2H_2O$       | 0.51V        |
| $2\mathbf{A}O_2(aq) + 2H^+ + 4e^- \to \mathbf{A}_2O_3^{2-} + H_2O$         | 0.40V        |
| $*2\mathbf{A}O_3^{2-} + 3H_2O + 4e^- \to \mathbf{A}_2O_3^{2-} + 6OH^-$        | -0.58V       |
| $*\mathbf{A}O_4^{2-} + 4H_2O + 6e^- \to \mathbf{A} + 8OH^-$              | -0.66V       |
|$ \mathbf{B}_2 + 2e^- \to 2\mathbf{B}^- $                             | 0.535V       |

Note: The standard electrode potentials of reactions marked with "$*$" are values at $\mathrm { p H } { = } 1 4$; in this problem, the temperature is assumed to be 298K, and the two-stage ionization constants of $H_2\mathbf{A}O_3$ are $1.3\times10^{-2}$ and $6.3\times10^{-8}$, respectively.

Then, the equilibrium constant $K_1$ for the titration of $\mathbf{B}_2$ by $\mathbf{A}_2O_3^{2-}$ (one of the products is $\mathbf{A}_4O_6^{2-}$) and the equilibrium constant $K_2$ for the reaction $\mathbf{A}O_2+H_2O\to H_2\mathbf{A}O_3$ are respectively (retain two significant figures, an error within 5% is considered correct):
部分电极电势如下所示：

| 半反应                                      | 标准电极电势 |
|---------------------------------------------|--------------|
| $4\mathbf{A}O_2(aq) + 4H^++ 6e^- \to \mathbf{A}_4O_6^{2-} + 2H_2O$       | 0.51V        |
| $2\mathbf{A}O_2(aq) + 2H^+ + 4e^- \to \mathbf{A}_2O_3^{2-} + H_2O$         | 0.40V        |
| $*2\mathbf{A}O_3^{2-} + 3H_2O + 4e^- \to \mathbf{A}_2O_3^{2-} + 6OH^-$        | -0.58V       |
| $*\mathbf{A}O_4^{2-} + 4H_2O + 6e^- \to \mathbf{A} + 8OH^-$              | -0.66V       |
|$ \mathbf{B}_2 + 2e^- \to 2\mathbf{B}^- $                             | 0.535V       |

注：带"$*$"的反应标准电极电势为 $\mathrm { p H } { = } 1 4$ 时的值；本题中温度默认为298K，$H_2\mathbf{A}O_3$的两级电离常数分别为$1.3\times10^{-2}$和$6.3\times10^{-8}$。

则$\mathbf{A}_2O_3^{2-}$滴定$\mathbf{B}_2$(产物之一为$\mathbf{A}_4O_6^{2-}$)的平衡常数$K_1$和反应$\mathbf{A}O_2+H_2O\to H_2\mathbf{A}O_3$的平衡常数$K_2$分别是（保留两位有效数字，误差在5%以内为正确）：
All other options are incorrect
$5.3\times10^{13}$, 1.7
$5.3\times10^{13}$, 1.5
$5.3\times10^{14}$, 1.7
$5.3\times10^{14}$, 1.5
$5.3\times10^{15}$, 1.7
$5.3\times10^{15}$，1.5
$3.5\times10^{13}$, 1.7
$3.5\times10^{13}$, 1.5
$3.5\times10^{14}$, 1.7
$3.5\times10^{14}$, 1.5
$3.5\times10^{15}$, 1.7
$3.5\times10^{15}$, 1.5
$5.3\times10^{13}$，1.9
$5.3\times10^{14}$，1.9
$5.3\times10^{15}$，1.9
$3.5\times10^{13}$，1.9
$3.5\times10^{14}$，1.9
$3.5\times10^{15}$，1.9
其他选项均不正确
$5.3\times10^{13}$，1.7
$5.3\times10^{13}$，1.5
$5.3\times10^{14}$，1.7
$5.3\times10^{14}$，1.5
$5.3\times10^{15}$，1.7
$5.3\times10^{15}$，1.5
$3.5\times10^{13}$，1.7
$3.5\times10^{13}$，1.5
$3.5\times10^{14}$，1.7
$3.5\times10^{14}$，1.5
$3.5\times10^{15}$，1.7
$3.5\times10^{15}$，1.5
$5.3\times10^{13}$，1.9
$5.3\times10^{14}$，1.9
$5.3\times10^{15}$，1.9
$3.5\times10^{13}$，1.9
$3.5\times10^{14}$，1.9
$3.5\times10^{15}$，1.9
### Calculate $K_1$

Connect the battery in series with a standard hydrogen electrode, and calculate the change in Gibbs free energy of the reaction (hydrogen electrode changes are omitted here)

$4\mathbf{A}O_2(aq)+4H^++8e^-\to 2\mathbf{A}_2O_3^{2-}+2H_2O$

$\Delta _rG_m^{\circ}=-nEF=-308.752kJ/mol$

$4\mathbf{A}O_2(aq)+4H^++6e^-\to \mathbf{A}_4O_6^{2-}+2H_2O$

$\Delta _rG_m^{\circ}=-nEF=-295.244kJ/mol$

$\mathbf{A}_4O_6^{2-}+2e^-\to 2\mathbf{A}_2O_3^{2-}$

$\Delta _rG_m^{\circ}=-13.508kJ/mol$

Therefore $E=0.070V$

<Checkpoint>[The standard electrode potential of $\mathbf{A}_4O_6^{2-}+2e^-\to 2\mathbf{A}_2O_3^{2-}$ is 0.070V](2)</Checkpoint>

Therefore, the electromotive force of $\mathbf{A}_2O_3^{2-}$ titrating $\mathbf{B}_2$ is 0.465V,

$K_1=e^{\frac{nEF}{RT}}=5.3\times10^{15}$

<Checkpoint>[The equilibrium constant $K_1$ is $5.3\times10^{15}$](1)</Checkpoint>

### Calculate $K_2$
Convert the data under alkaline conditions:

$2\mathbf{A}O_3^{2-}+6H^++4e^-\to \mathbf{A}_2O_3^{2-}+3H_2O$

$E^{'}=0.6623V$

$2\mathbf{A}O_3^{2-}+2H^++2H_2\to \mathbf{A}_2O_3^{2-}+3H_2O$

$K=6.051\times10^{44}$

$2\mathbf{A}O_2+2H_2\to \mathbf{A}_2O_3^{2-}+H_2O+2H^+$

$K=1.114\times10^{27}$

Subtract to get

$\mathbf{A}O_2+H_2O\to \mathbf{A}O_3^{2-}+2H^+$

$K=1.357\times10^{-9}$

<Checkpoint>[The equilibrium constant of $\mathbf{A}O_2+H_2O\to \mathbf{A}O_3^{2-}+2H^+$ is $K=1.357\times10^{-9}$](1)</Checkpoint>

Thus we obtain

$\mathbf{A}O_2+H_2O\to H_2\mathbf{A}O_3$

$K=1.656$

<Checkpoint>[The equilibrium constant of $\mathbf{A}O_2+H_2O\to H_2\mathbf{A}O_3$ is $K_2=1.656$](2)</Checkpoint>
### 计算$K_1$

将电池与标准氢电极串联，计算反应的Gibbs自由能变（此处省略氢电极变化）

$4\mathbf{A}O_2(aq)+4H^++8e^-\to 2\mathbf{A}_2O_3^{2-}+2H_2O$

$\Delta _rG_m^{\circ}=-nEF=-308.752kJ/mol$

$4\mathbf{A}O_2(aq)+4H^++6e^-\to \mathbf{A}_4O_6^{2-}+2H_2O$

$\Delta _rG_m^{\circ}=-nEF=-295.244kJ/mol$

$\mathbf{A}_4O_6^{2-}+2e^-\to 2\mathbf{A}_2O_3^{2-}$

$\Delta _rG_m^{\circ}=-13.508kJ/mol$

从而$E=0.070V$

<Checkpoint>[$\mathbf{A}_4O_6^{2-}+2e^-\to 2\mathbf{A}_2O_3^{2-}$的标准电极电势为0.070V](2)</Checkpoint>

从而$\mathbf{A}_2O_3^{2-}$滴定$\mathbf{B}_2$的电动势为0.465V,

$K_1=e^{\frac{nEF}{RT}}=5.3\times10^{15}$

<Checkpoint>[平衡常数$K_1$为$5.3\times10^{15}$](1)</Checkpoint>

### 计算$K_2$
将碱性条件的数据转化：

$2\mathbf{A}O_3^{2-}+6H^++4e^-\to \mathbf{A}_2O_3^{2-}+3H_2O$

$E^{'}=0.6623V$

$2\mathbf{A}O_3^{2-}+2H^++2H_2\to \mathbf{A}_2O_3^{2-}+3H_2O$

$K=6.051\times10^{44}$

$2\mathbf{A}O_2+2H_2\to \mathbf{A}_2O_3^{2-}+H_2O+2H^+$

$K=1.114\times10^{27}$

做差得到

$\mathbf{A}O_2+H_2O\to \mathbf{A}O_3^{2-}+2H^+$

$K=1.357\times10^{-9}$

<Checkpoint>[$\mathbf{A}O_2+H_2O\to \mathbf{A}O_3^{2-}+2H^+$的平衡常数$K=1.357\times10^{-9}$](1)</Checkpoint>

从而得到

$\mathbf{A}O_2+H_2O\to H_2\mathbf{A}O_3$

$K=1.656$

<Checkpoint>[$\mathbf{A}O_2+H_2O\to H_2\mathbf{A}O_3$的平衡常数$K_2=1.656$](2)</Checkpoint>