$\rm HA$ is a monoprotic weak acid. The solubility $s(\mathrm{mol} / \mathrm{L})$ of the sparingly soluble salt $\rm MA$ in buffer solutions of different acidity is experimentally studied to determine the solubility product constant of $\rm MA$ and the ionization constant of $\rm HA$. The experimental data are shown in the table below:

|pH|$s(10^{-4}~\mathrm{mol/L})$|
|---|---|
|4.00|2.7|
|3.70|3.2|
|3.52|3.5|
|3.40|3.9|
|3.22|4.5|

Please find the variables with relevant relationships through the equilibrium and conservation relationships in the solution, and obtain the equation by fitting the experimental data, and use the obtained results to find $K_{\mathrm{sp}}$ and $K_{\mathrm{a}}$.
$\rm HA$ 是一元弱酸。现通过实验研究难溶盐 $\rm MA$ 在不同酸度的缓冲溶液中的溶解度 $s(\mathrm{mol} / \mathrm{L})$, 来求出 $\rm MA$的溶度积常数以及 $\rm HA$ 的电离常数。实验数据如下表格所示:

|pH|$s(10^{-4}~\mathrm{mol/L})$|
|---|---|
|4.00|2.7|
|3.70|3.2|
|3.52|3.5|
|3.40|3.9|
|3.22|4.5|

请通过溶液中的平衡和守恒关系,找到具有相关关系的变量,并由实验数据作拟合得到方程,并用所得结果求出 $K_{\mathrm{sp}}$ 和 $K_{\mathrm{a}}$ 。(涉及计算的选项, 需要保证有效数字最后一位正确)
pH has a linear relationship with $s$
$\rm [H^+]$ and $s$ have a linear relationship.
$K_{\mathrm{sp}}=4.9\times10^{-8}$
$K_{\mathrm{a}}=1.9\times10^{-4}$
$K_{\mathrm{sp}}=4.8\times10^{-6}$
$K_{\mathrm{a}}=2.\times10^{-4}$
All other options are incorrect.
Apart from G and H, there are more than one correct option.
pH与$s$具有线性关系
$\rm [H^+]$与$s$具有线性关系
$K_{\mathrm{sp}}=4.9\times10^{-8}$
$K_{\mathrm{a}}=1.9\times10^{-4}$
$K_{\mathrm{sp}}=4.8\times10^{-6}$
$K_{\mathrm{a}}=2.3\times10^{-4}$
除G, H外其他选项均不正确
除G, H外有多于一个选项正确
According to the definition of $K _ { \mathrm { s p } } $, the following derivation can be made:

$K _ { \mathrm { s p } } = [ \mathrm { M ^ { + } } ] [ \mathrm { A ^ { - } } ] = [ \mathrm { M ^ { + } } ] c ( \mathrm { H A } ) \delta ( \mathrm { A ^ { - } } ) = \frac{[ \mathrm { M ^ { + } } ] ^ { 2 } K _ { \mathrm { a } }}{[ \mathrm { H ^ { + } } ] + K _ { \mathrm { a } }}  = \frac{s^ { 2 } K _ { \mathrm { a } }}{[ \mathrm { H ^ { + } } ] + K _ { \mathrm { a } }}$

<Checkpoint>[Derivation yields $K _ { \mathrm { s p } } =  \frac{s^ { 2 } K _ { \mathrm { a } }}{[ \mathrm { H ^ { + } } ] + K _ { \mathrm { a } }}$ or equivalent form](1)</Checkpoint>

After rearrangement, the linear equation is obtained
$$
s^2=\frac{K _ { \mathrm { s p } } }{K _ { \mathrm { a } }}[ \mathrm { H ^ { + } } ]+K _ { \mathrm { s p } } 
$$

<Checkpoint>[The linear equation is $s^2=\frac{K _ { \mathrm { s p } } }{K _ { \mathrm { a } }}[ \mathrm { H ^ { + } } ]+K _ { \mathrm { s p } } $](1)</Checkpoint>

Therefore, experimental data can be substituted for linear regression, yielding $s^2=2.6\times10^{-4}\mathrm{[H^+]}+4.8\times10^{-8}$

<Checkpoint>[The slope of the linear equation is $2.6\times10^{-4}$, and the intercept is $4.8\times10^{-8}$](1)</Checkpoint>

That is, $\frac{K _ { \mathrm { s p } } }{K _ { \mathrm { a } }}=2.6\times10^{-4}$, $K _ { \mathrm { s p } }=4.8\times10^{-8}$, therefore $K_{\mathrm{a}}=1.9\times10^{-4}$

<Checkpoint>[$K _ { \mathrm { s p } }=4.8\times10^{-8}$, $K_{\mathrm{a}}=1.9\times10^{-4}$](1)</Checkpoint>
根据$K _ { \mathrm { s p } } $的定义可以做以下推导：

$K _ { \mathrm { s p } } = [ \mathrm { M ^ { + } } ] [ \mathrm { A ^ { - } } ] = [ \mathrm { M ^ { + } } ] c ( \mathrm { H A } ) \delta ( \mathrm { A ^ { - } } ) = \frac{[ \mathrm { M ^ { + } } ] ^ { 2 } K _ { \mathrm { a } }}{[ \mathrm { H ^ { + } } ] + K _ { \mathrm { a } }}  = \frac{s^ { 2 } K _ { \mathrm { a } }}{[ \mathrm { H ^ { + } } ] + K _ { \mathrm { a } }}$

<Checkpoint>[推导得出$K _ { \mathrm { s p } } =  \frac{s^ { 2 } K _ { \mathrm { a } }}{[ \mathrm { H ^ { + } } ] + K _ { \mathrm { a } }}$或等价形式](1)</Checkpoint>

整理后得到线性方程
$$
s^2=\frac{K _ { \mathrm { s p } } }{K _ { \mathrm { a } }}[ \mathrm { H ^ { + } } ]+K _ { \mathrm { s p } } 
$$

<Checkpoint>[线性方程为$s^2=\frac{K _ { \mathrm { s p } } }{K _ { \mathrm { a } }}[ \mathrm { H ^ { + } } ]+K _ { \mathrm { s p } } $](1)</Checkpoint>

因此可以代入实验数据进行线性回归，得到$s^2=2.6\times10^{-4}\mathrm{[H^+]}+4.8\times10^{-8}$

<Checkpoint>[线性方程的斜率为$2.6\times10^{-4}$, 截距为$4.8\times10^{-8}$](1)</Checkpoint>

即$\frac{K _ { \mathrm { s p } } }{K _ { \mathrm { a } }}=2.6\times10^{-4}$, $K _ { \mathrm { s p } }=4.8\times10^{-8}$, 因此$K_{\mathrm{a}}=1.9\times10^{-4}$

<Checkpoint>[$K _ { \mathrm { s p } }=4.8\times10^{-8}$, $K_{\mathrm{a}}=1.9\times10^{-4}$](1)</Checkpoint>