Mining minerals generates a large amount of wastewater. The steps for the highly selective iodometric determination of cobalt content are as follows:

$\textcircled{1}$ Weigh $\mathrm{0.2517}$ g of arsenic trioxide standard, dissolve in $\mathrm {10\%NaOH}$, and dilute to $\mathrm {250.0mL}$.

$\textcircled{2}$ Take $\mathrm {500.0mL}$ of wastewater, evaporate and concentrate it, add a certain amount of nitric acid-hydrochloric acid, evaporate to dryness, dissolve soluble salts with an appropriate amount of water, add nitric acid and potassium chlorate again, boil, filter the precipitate, and wash the precipitate.

$\textcircled{3}$ Concentrate the solution volume to $\mathrm{30-40 mL}$, add $\mathrm{25}$ g of ammonium nitrate, add saturated ammonium citrate solution and an appropriate amount of ammonia water to bring the solution to $\mathrm {pH=9- 9.5}$, transfer to $\mathrm {50.0mL}$ of iodine standard solution, shake and let stand for a period of time.

$\textcircled{4}$ Titrate the excess iodine with sodium arsenite solution. Add starch near the end point. Take the reading after the blue color disappears. The blank experiment consumed $\mathrm {30.46mL}$ of sodium arsenite, and the titration experiment consumed $\mathrm {14.65mL}$ of sodium arsenite.

Based on the above content, select the correct answer.
开采矿物会产生大量废水，具有较高选择性的碘量法测定钴含量步骤如下：

$\textcircled{1}$ 称取$\mathrm{0.2517}$克三氧化二砷基准物，溶于 $\mathrm {10\%NaOH}$ ，定容至 $\mathrm {250.0mL}$。

$\textcircled{2}$ 取废水 $\mathrm {500.0mL}$ ，蒸发浓缩后加入一定量硝酸-盐酸，蒸发至干，用适量水溶解可溶性盐，再次加入硝酸和氯酸钾，煮沸，过滤沉淀，洗涤沉淀。

$\textcircled{3}$ 将溶液体积浓缩至 $\mathrm{30-40 mL}$ ，加入$\mathrm{25}$克硝酸铵，加入柠檬酸铵饱和溶液和适量氨水，使溶液 $\mathrm {pH=9- 9.5}$ ，移入 $\mathrm {50.0mL}$ 碘标准溶液，震摇后静置一段时间。

$\textcircled{4}$ 用亚砷酸钠溶液滴定过量的碘，近终点时加入淀粉，待蓝色褪去后读数。空白实验消耗 亚砷酸钠 $\mathrm {30.46mL}$ ，滴定实验消耗亚砷酸钠 $\mathrm {14.65mL}$。

根据以上内容，选择正确答案。
All other options are incorrect
The chemical equation for the titration of iodine with arsenite solution in the system is as follows: $\mathrm{AsO_3^{3-} + I_2 + 2NH_3 + H_2O \rightarrow AsO_4^{3-} + 2I^- + 2NH_4^+}$
It is possible to accurately weigh sublimed and purified iodine solid to prepare a standard iodine solution, and then use a $ \mathrm { Na_ 2S _ 2O_3 }$ solution to replace the sodium arsenite solution for titration.
The concentration of Co element in the sample is $\mathrm{18.96 mg/L}$
The concentration of Co element in the sample is $\mathrm{9.48 mg/L}$
The concentration of Co element in the sample is $\mathrm{37.93 mg/L}$
其他选项均不正确
体系中亚砷酸溶液滴定碘的化学方程式如下：$\mathrm{AsO_3^{3-} + I_2 + 2NH_3 + H_2O \rightarrow AsO_4^{3-} + 2I^- + 2NH_4^+}$
可以准确称取升华提纯的碘固体配制碘标准溶液，再用 $ \mathrm { Na_ 2S _ 2O_3 }$ 溶液代替亚砷酸钠溶液进行滴定。
试样中Co元素浓度为$\mathrm{18.96 mg/L}$
试样中Co元素浓度为$\mathrm{9.48 mg/L}$
试样中Co元素浓度为$\mathrm{37.93 mg/L}$
In the chemical reaction equation, the $\mathrm{pKa}$ of the third ionization step of arsenic acid is $\mathrm{11.50}$, and the $\mathrm{pKa}$ of the second ionization step is $\mathrm{6.96}$. The $\mathrm{pKa}$ of the ammonium ion is $\mathrm{9.25}$.

<Checkpoint>[The $\mathrm{pKa}$ of the third ionization step of arsenic acid is $\mathrm{11.50}$, and the $\mathrm{pKa}$ of the second ionization step is $\mathrm{6.96}$. The $\mathrm{pKa}$ of the ammonium ion is $\mathrm{9.25}$](1)</Checkpoint>

Therefore, the existing form of arsenic acid molecules in the reaction equation should be $\mathrm{HAsO_4^{2-}}$.

<Checkpoint>[The existing form of arsenic acid molecules should be $\mathrm{HAsO_4^{2-}}$](1)</Checkpoint>

The reaction of $\mathrm{I_2}$ with $\mathrm{S_2O_3^{2-}}$ under alkaline conditions is non-quantitative and may produce high-valent sulfur products, so it cannot be used for titration. Therefore, option C is incorrect.

<Checkpoint>[The reaction of $\mathrm{I_2}$ with $\mathrm{S_2O_3^{2-}}$ under alkaline conditions is non-quantitative](1)</Checkpoint>

$\mathrm{As(III)}$ will be oxidized to $\mathrm{As(V)}$, $\mathrm{I_2(0)}$ will be reduced to negative one valence, and $\mathrm{Co(II)}$ will be oxidized to $\mathrm{Co(III)}$. Therefore, the ratio of the reaction is as follows: $ \mathrm{As_2O_3 \sim 2I_2 \sim 4Co^{2+}}$, when the reaction ratio is judged incorrectly, option D or option E will be obtained.

<Checkpoint>[Reaction ratio $ \mathrm{As_2O_3 \sim 2I_2 \sim 4Co^{2+}}$](1)</Checkpoint>

$\mathrm{c(As_2O_3) = \frac{m(As_2O_3)}{{V} \cdot M(As_2O_3)} = 0.005089mol \cdot L^{-1}}$

<Checkpoint>[$\mathrm{c(As_2O_3) =0.005089mol \cdot L^{-1}}$](1)</Checkpoint>

The difference between the blank experiment and the titration experiment consuming $\mathrm{As(III)}$ corresponds to the content of $\mathrm{Co}$: $\mathrm{{n}(Co) = c(As_2O_3) \cdot (V_1-V_2) \cdot 4 = 0.3218 mmol}$

<Checkpoint>[$\mathrm{{n}(Co) =0.3218 mmol}$](1)</Checkpoint>

$\mathrm{\omega(Co) = \frac{n(Co) \cdot M(Co)}{V} = 37.93 mg/L}$ , so the correct option is F.

<Checkpoint>[$\mathrm{\omega(Co) =37.93 mg/L}$  ](1)</Checkpoint>
化学反应方程式中，砷酸分子第三步电离的$\mathrm{pKa}$为$\mathrm{11.50}$，第二步电离的$\mathrm{pKa}$为$\mathrm{6.96}$。铵根离子的$\mathrm{pKa}$为$\mathrm{9.25}$。

<Checkpoint>[砷酸分子第三步电离的$\mathrm{pKa}$为$\mathrm{11.50}$，第二步电离的$\mathrm{pKa}$为$\mathrm{6.96}$。铵根离子的$\mathrm{pKa}$为$\mathrm{9.25}$](1)</Checkpoint>

因此，反应方程式中砷酸分子的存在形式应为$\mathrm{HAsO_4^{2-}}$，所以选项B错误。

<Checkpoint>[砷酸分子的存在形式应为$\mathrm{HAsO_4^{2-}}$。](1)</Checkpoint>

碱性条件下$\mathrm{I_2}$与$\mathrm{S_2O_3^{2-}}$的反应不定量，可能生成高价硫产物，无法用作滴定。所以选项C错误。

<Checkpoint>[碱性条件下$\mathrm{I_2}$与$\mathrm{S_2O_3^{2-}}$的反应不定量](1)</Checkpoint>

$\mathrm{As(III)}$会被氧化为$\mathrm{As(V)}$，$\mathrm{I_2(0)}$会被还原为负一价，$\mathrm{Co(II)}$被氧化为$\mathrm{Co(III)}$，因此反应的比例如下： $ \mathrm{As_2O_3 \sim 2I_2 \sim 4Co^{2+}}$，当反应比例判断出错，会得到选项D或选项E。

<Checkpoint>[反应比例$ \mathrm{As_2O_3 \sim 2I_2 \sim 4Co^{2+}}$](1)</Checkpoint>

$\mathrm{c(As_2O_3) = \frac{m(As_2O_3)}{V} \cdot M(As_2O_3) = 0.005089mol \cdot L^{-1}}$

<Checkpoint>[$\mathrm{c(As_2O_3) =0.005089mol \cdot L^{-1}}$](1)</Checkpoint>

空白实验与滴定实验消耗$\mathrm{As(III)}$的差，对应着$\mathrm{Co}$的含量：$\mathrm{{n}(Co) = c(As_2O_3) \cdot (V_1-V_2) \cdot 4 = 0.3218 mmol}$

<Checkpoint>[$\mathrm{{n}(Co) =0.3218 mmol}$](1)</Checkpoint>

$\mathrm{\omega(Co) = \frac{n(Co) \cdot M(Co)}{V} = 37.93 mg/L}$ ，因此正确选项为F，故而选项A错误。

<Checkpoint>[$\mathrm{\omega(Co) =37.93 mg/L}$  ](1)</Checkpoint>