For aqueous solutions containing iodide ions (excluding other ions that may affect the determination), the following method can be used to determine the content of iodide ions:

1. Under neutral or weakly acidic conditions, add excess ${\rm Br}_2$ for oxidation. After removing the remaining ${\rm Br}_2$, add excess ${\rm KI}$;
2. Extract with ${\rm CCl}_4$ (approximately considered complete extraction). After separating the aqueous phase, back-extract with an aqueous solution of ${\rm N}_2{\rm H}_4$;
3. Under neutral or weakly acidic conditions, oxidize with excess ${\rm Br}_2$. After removing the remaining ${\rm Br}_2$, add excess ${\rm KI}$;
4. Appropriately acidify the solution, use **X** as an indicator, and titrate with **Y** standard solution.

In a certain determination, $25.00\textup{ mL}$ of the test solution containing ${\rm KI}$ was taken. After the above operations, $20.06\textup{ mL}$ of $0.1000\textup{ mol/L}$ **Y** standard solution was consumed.

Which of the following options is correct?
对于含有碘离子（不考虑可能影响测定的其它离子）的水溶液，可使用如下方法测定其中碘离子的含量：

1. 在中性或弱酸性条件下，加入过量的${\rm Br}_2$进行氧化，除去剩余的${\rm Br}_2$后，再加入过量的${\rm KI}$；
2. 用${\rm CCl}_4$进行萃取（近似认为完全萃取），分去水相后，用${\rm N}_2{\rm H}_4$的水溶液进行反萃取；
3. 在中性或弱酸性条件下，用过量的${\rm Br}_2$氧化，除去剩余的${\rm Br}_2$后，再加入过量的${\rm KI}$；
4. 适当酸化溶液，以**X**为指示剂，用**Y**标准溶液进行滴定。

在某次测定中，取含${\rm KI}$的试液$25.00\textup{ mL}$，经过上述操作后最终消耗$0.1000\textup{ mol/L}$的**Y**标准溶液$20.06\textup{ mL}$。

下列选项正确的是：
All other options are incorrect
**X** should be sodium diphenylamine sulfonate, and **Y** should be sodium thiosulfate.
Under neutral or weakly acidic conditions, the purpose of adding excess ${\rm Br}_2$ is to oxidize ${\rm I}^-$ to ${\rm I}_2$
The target substance for back extraction using an aqueous solution of ${\rm N}_2{\rm H}_4$ is ${\rm I}^-$ in the organic phase
The concentration of ${\rm I}^-$ in the original solution is $13.37\textup{ mmol/L}$.
The concentration of ${\rm I}^-$ in the original solution is $2.229\textup{ mmol/L}$.
The concentration of ${\rm I}^-$ in the original solution is $80.24\textup{ mmol/L}$.
其他选项均不正确
**X**应为二苯胺磺酸钠，**Y**应为硫代硫酸钠
在中性或弱酸性条件下，加入过量的${\rm Br}_2$的目的是将${\rm I}^-$氧化为${\rm I}_2$
用${\rm N}_2{\rm H}_4$的水溶液进行反萃取的目标物为有机相中的${\rm I}^-$
原始溶液中的${\rm I}^-$的浓度为$13.37\textup{ mmol/L}$
原始溶液中的${\rm I}^-$的浓度为$2.229\textup{ mmol/L}$
原始溶液中的${\rm I}^-$的浓度为$80.24\textup{ mmol/L}$
Under neutral or weakly acidic conditions, the addition of excess ${\rm Br}_2$ can oxidize ${\rm I}^-$ to ${\rm IO}_3^-$:
$$
{\rm I}^-+3{\rm Br}_2+3{\rm H}_2{\rm O}\to{\rm IO}_3^-+6{\rm Br}^-+6{\rm H}^+
$$

<Checkpoint>[${\rm Br}_2$ can oxidize ${\rm I}^-$ to ${\rm IO}_3^-$](1)</Checkpoint>

<Checkpoint>[$ {\rm I}^-+3{\rm Br}_2+3{\rm H}_2{\rm O}\to{\rm IO}_3^-+6{\rm Br}^-+6{\rm H}^+ $](1)</Checkpoint>

After removing the remaining ${\rm Br}_2$, adding excess ${\rm KI}$ can reduce ${\rm IO}_3^-$ to ${\rm I}_2$:
$$
{\rm IO}_3^-+5{\rm I}^-+6{\rm H}^+\to3{\rm I}_2+3{\rm H}_2{\rm O}
$$

<Checkpoint>[$ {\rm IO}_3^-+5{\rm I}^-+6{\rm H}^+\to3{\rm I}_2+3{\rm H}_2{\rm O} $](1)</Checkpoint>

After extracting ${\rm I}_2$ into the organic phase with ${\rm CCl}_4$, back extraction is performed with an aqueous solution of ${\rm N}_2{\rm H}_4$ to return ${\rm I}_2$ to the aqueous phase in the form of ${\rm I}^-$:

<Checkpoint>[Back extraction is performed with an aqueous solution of ${\rm N}_2{\rm H}_4$ to return ${\rm I}_2$ to the aqueous phase in the form of ${\rm I}^-$](1)</Checkpoint>

${\rm N}_2{\rm H}_4+2{\rm I}_2\to{\rm N}_2+4{\rm H}^++4{\rm I}^-$

<Checkpoint>[${\rm N}_2{\rm H}_4+2{\rm I}_2\to{\rm N}_2+4{\rm H}^++4{\rm I}^-$](1)</Checkpoint>

The oxidation extraction process in steps 1 and 2 is equivalent to amplifying the content of ${\rm I}^-$ by 6 times:

$$
{\rm I}^-\sim{\rm IO}_3^-\sim3{\rm I}_2\sim6{\rm I}^-
$$

<Checkpoint>[The oxidation extraction process in steps 1 and 2 is equivalent to amplifying the content of ${\rm I}^-$ by 6 times](1)</Checkpoint>

The repeated operations in steps 3 and 4 once again amplify the concentration of ${\rm I}^-$ by 6 times:

$$
6{\rm I}^-\sim6{\rm IO}_3^-\sim18{\rm I}_2(\sim36{\rm I}^-)
$$

Therefore, this method totally amplifies the initial iodide ion concentration by 36 times.

<Checkpoint>[This method totally amplifies the initial iodide ion concentration by 36 times](1)</Checkpoint>

The content of ${\rm I}_2$ should be titrated using starch as an indicator (**X**) and ${\rm Na}_2{\rm S}_2{\rm O}_3$ as a titrant (**Y**), with the titration relationship being:
$$
{\rm I}_2+2{\rm S}_2{\rm O}_3^{2-}\to2{\rm I}^-+{\rm S}_4{\rm O}_6^{2-}
$$

<Checkpoint>[**X** is starch and **Y** is ${\rm Na}_2{\rm S}_2{\rm O}_3$](1)</Checkpoint>

Therefore, the titration consumption of ${\rm Na}_2{\rm S}_2{\rm O}_3$ is:
$$
0.1000\times20.06=2.006\textup{ mmol}
$$
Corresponding to ${\rm I}^-$ in the original solution:
$$
\frac{1}{36}\times2.006=5.572\times10^{-2}\textup{ mmol}
$$
Then the concentration of ${\rm I}^-$ in the original solution is:
$$
5.572\times10^{-2}/\frac{25.00}{1000}=2.229\textup{ mmol/L}
$$

<Checkpoint>[The concentration of ${\rm I}^-$ in the original solution is $2.229\textup{ mmol/L}$](2)</Checkpoint>
在中性或弱酸性条件下，加入过量的${\rm Br}_2$可以将${\rm I}^-$氧化为${\rm IO}_3^-$：
$$
{\rm I}^-+3{\rm Br}_2+3{\rm H}_2{\rm O}\to{\rm IO}_3^-+6{\rm Br}^-+6{\rm H}^+
$$

<Checkpoint>[${\rm Br}_2$可以将${\rm I}^-$氧化为${\rm IO}_3^-$](1)</Checkpoint>

<Checkpoint>[$ {\rm I}^-+3{\rm Br}_2+3{\rm H}_2{\rm O}\to{\rm IO}_3^-+6{\rm Br}^-+6{\rm H}^+ $](1)</Checkpoint>

除去剩余的${\rm Br}_2$后，再加入过量的${\rm KI}$，可以将${\rm IO}_3^-$还原为${\rm I}_2$：
$$
{\rm IO}_3^-+5{\rm I}^-+6{\rm H}^+\to3{\rm I}_2+3{\rm H}_2{\rm O}
$$

<Checkpoint>[$ {\rm IO}_3^-+5{\rm I}^-+6{\rm H}^+\to3{\rm I}_2+3{\rm H}_2{\rm O} $](1)</Checkpoint>

用${\rm CCl}_4$萃取${\rm I}_2$到有机相后，又用${\rm N}_2{\rm H}_4$的水溶液进行反萃取，使${\rm I}_2$以${\rm I}^-$的形式回到水相：

<Checkpoint>[${\rm N}_2{\rm H}_4$的水溶液进行反萃取，使${\rm I}_2$以${\rm I}^-$的形式回到水相](1)</Checkpoint>

${\rm N}_2{\rm H}_4+2{\rm I}_2\to{\rm N}_2+4{\rm H}^++4{\rm I}^-$

<Checkpoint>[${\rm N}_2{\rm H}_4+2{\rm I}_2\to{\rm N}_2+4{\rm H}^++4{\rm I}^-$](1)</Checkpoint>

第1，2步的氧化萃取过程相当于将${\rm I}^-$的含量放大了6倍：

$$
{\rm I}^-\sim{\rm IO}_3^-\sim3{\rm I}_2\sim6{\rm I}^-
$$

<Checkpoint>[第1，2步的氧化萃取过程相当于将${\rm I}^-$的含量放大了6倍](1)</Checkpoint>

之后3，4两步重复的操作，再一次将${\rm I}^-$的浓度放大了6倍：

$$
6{\rm I}^-\sim6{\rm IO}_3^-\sim18{\rm I}_2(\sim36{\rm I}^-)
$$

因此该方法总计将初始碘离子浓度放大了36倍。

<Checkpoint>[该方法总计将初始碘离子浓度放大了36倍](1)</Checkpoint>

滴定${\rm I}_2$的含量应当以淀粉为指示剂（**X**）、${\rm Na}_2{\rm S}_2{\rm O}_3$为滴定剂（**Y**），滴定关系为：
$$
{\rm I}_2+2{\rm S}_2{\rm O}_3^{2-}\to2{\rm I}^-+{\rm S}_4{\rm O}_6^{2-}
$$

<Checkpoint>[**X**为淀粉，**Y**为${\rm Na}_2{\rm S}_2{\rm O}_3$](1)</Checkpoint>

故滴定消耗量了${\rm Na}_2{\rm S}_2{\rm O}_3$：
$$
0.1000\times20.06=2.006\textup{ mmol}
$$
对应于原始溶液中的${\rm I}^-$：
$$
\frac{1}{36}\times2.006=5.572\times10^{-2}\textup{ mmol}
$$
则原始溶液中的${\rm I}^-$的浓度为：
$$
5.572\times10^{-2}/\frac{25.00}{1000}=2.229\textup{ mmol/L}
$$

<Checkpoint>[原始溶液中的${\rm I}^-$的浓度为$2.229\textup{ mmol/L}$](2)</Checkpoint>