The test solution with volume \( V_{1} \), containing a certain concentration of peroxodisulfate ions, is acidified with sulfuric acid, and then transferred into an excess volume \( V_{2} \) of a standard sodium arsenite solution with concentration \( c_{1} \). The mixture is boiled. After adding potassium bromide solution and letting it stand for a while, the solution is titrated with a standard potassium bromate solution of concentration \( c_{2} \) until the solution turns yellow, consuming a volume \( V_{3} \) of the bromate solution. Subsequently, the same standard sodium arsenite solution is used to titrate the solution until it becomes colorless, consuming a volume \( V_{4} \). Please select the correct option. Note that when balancing the equations, the coefficients should be in the simplest integer ratio.
将体积为 $V _ { 1 }$ 的含有一定浓度过硫酸根离子的待测液用硫酸酸化，移入浓度 $c _ { 1 }$ 的亚砷酸钠标准溶液 $V _ { 2 }$ (过量)，煮沸。加入溴化钾溶液，放置片刻，用浓度 $c _ { 2 }$ 的溴酸钾标准溶液滴定至溶液呈黄色，消耗溴酸钾体积 $ V _ { 3 } $，再用同样的亚砷酸钠标准溶液滴定至无色时为止，消耗 $V _ { 4 }$ 。请选择正确的选项，需要注意的是，有关配平方程式的选项时，系数应该为最简整数比。
During the titration process, $\mathrm{S_2O_8^{2-}}$ can react with $\mathrm{AsO_2^-}$ under acidic conditions to produce $\mathrm{H_3AsO_4}$. This equation requires 3 water molecules on the left side to be balanced.
During the titration process, $\mathrm{H_3AsO_3}$ reacts with $\mathrm{BrO_3^-}$, and the sum of the coefficients on the right side of the balanced equation is 1 more than the sum of the coefficients on the left side.
During the titration process, $\mathrm {BrO_3^-}$ and $\mathrm {Br^-}$ react under acidic conditions, with the sum of coefficients on the left side of the equation being 11 and the sum on the right side being 6.
During the titration process, after adding the sodium arsenite standard solution again, $\mathrm{AsO_2^-}$ reacts with an elemental substance produced in the reaction of $\mathrm{BrO_3^-}$ added in the previous step. The sum of the coefficients on both sides of the balanced reaction equation is equal.
The expression for the sulfate ion concentration is $$ c=[c_1(V_2+V_4)-2c_2V_3]/V_1$$
Acidification with hydrochloric acid can also be performed before titration.
滴定过程中，$\mathrm {S_2O_8^{2-}}$ 在酸性条件下可以和 $\mathrm { AsO_2^- }$ 反应得到 $\mathrm {H_3AsO_4}$，该方程式配平左边需要3个水分子。
滴定过程中，$\mathrm {H_3AsO_3}$ 和 $\mathrm {BrO_3^-}$ 反应，方程式配平后右边系数和比左边系数和多1。
滴定过程中，$\mathrm {BrO_3^-}$ 和 $\mathrm {Br^-}$ 在酸性条件下反应，方程式左边系数和为11，右边系数和为6。
滴定过程中，再次加入亚砷酸钠标准溶液后，$\mathrm {AsO_2^-}$ 与上一步加入的 $\mathrm {BrO_3^-}$ 的反应过程中产生的一种单质发生反应，反应方程式配平后左右两边系数和相等。
硫酸根离子浓度的表达式为$$ c=[c_1(V_2+V_4)-2c_2V_3]/V_1$$
滴定前也可以使用盐酸进行酸化
Under acidic conditions, persulfate ions and arsenite ions react, with the reaction equation as follows:  
$$\mathrm {H^+ + S_2O_8^{2-} + AsO_2^- +2 H_2O = 2HSO_4^− + H_3AsO_4}$$  

<Checkpoint>[Equation 1$$\mathrm {H^+ + S_2O_8^{2-} + AsO_2^- +2 H_2O = 2HSO_4^− + H_3AsO_4}$$](1)</Checkpoint>  

Upon adding bromate ions, arsenous acid and bromate ions undergo a redox reaction, described by the equation:  
$$\mathrm {3 H_3AsO_3+BrO_3^-=3 H_3AsO_4+Br^-}$$  

<Checkpoint>[Equation 2$$\mathrm {3 H_3AsO_3+BrO_3^-=3 H_3AsO_4+Br^-}$$](1)</Checkpoint>  

Excess bromate ions and bromide ions undergo a comproportionation reaction to form $\mathrm Br_2$, turning the solution yellow, as shown in the equation:  
$$\mathrm {BrO_3^-+5Br^-+6H^+=3Br_2 +3H_2O}$$  

<Checkpoint>[Equation 3$$\mathrm {BrO_3^-+5Br^-+6H^+=3Br_2 +3H_2O}$$](1)</Checkpoint>  

After adding sodium arsenite standard solution, arsenite ions react with $\mathrm Br_2$, causing the solution to become colorless, as represented by the equation:  
$$\mathrm {Br_2 + AsO_2^- + 2H_2O =2Br^- +H_3AsO_4 +H^+ }$$  

<Checkpoint>[Equation 4$$\mathrm {Br_2 + AsO_2^- + 2H_2O =2Br^- +H_3AsO_4 +H^+ }$$](1)</Checkpoint>  

Arsenous acid reacts with persulfate and bromate ions. The stoichiometric ratio of arsenous acid to persulfate is 1:1,  

<Checkpoint>[The stoichiometric ratio of arsenous acid to persulfate is 1:1](1)</Checkpoint>  

and the stoichiometric ratio of arsenous acid to bromate is 3:1.  

<Checkpoint>[The stoichiometric ratio of arsenous acid to bromate is 3:1](1)</Checkpoint>  

The total amount of arsenous acid involved in the reaction minus the amount reacting with bromate yields the remaining amount reacting with persulfate. Knowing the volume of the persulfate-containing solution allows determination of the persulfate ion concentration:  
$$ c=[c_1(V_2+V_4)-3c_2V_3]/V_1$$  

<Checkpoint>[Persulfate ion concentration $c=[c_1(V_2+V_4)-3c_2V_3]/V_1$](2)</Checkpoint>
酸性条件下过硫酸根离子和亚砷酸根离子进行反应，反应方程式为$$\mathrm {H^+ + S_2O_8^{2-} + AsO_2^- +2 H_2O = 2HSO_4^− + H_3AsO_4}$$

<Checkpoint>[方程式1$$\mathrm {H^+ + S_2O_8^{2-} + AsO_2^- +2 H_2O = 2HSO_4^− + H_3AsO_4}$$](1)</Checkpoint>

加入溴酸根离子后，亚砷酸和溴酸根离子发生氧化还原反应，方程式为
$$\mathrm {3 H_3AsO_3+BrO_3^-=3 H_3AsO_4+Br^-}$$

<Checkpoint>[方程式2$$\mathrm {3 H_3AsO_3+BrO_3^-=3 H_3AsO_4+Br^-}$$](1)</Checkpoint>

过量的溴酸根离子和溴离子发生归中反应生成$\mathrm Br_2$, 溶液变黄，方程式为
$$\mathrm {BrO_3^-+5Br^-+6H^+=3Br_2 +3H_2O}$$

<Checkpoint>[方程式3$$\mathrm {BrO_3^-+5Br^-+6H^+=3Br_2 +3H_2O}$$](1)</Checkpoint>

加入亚砷酸钠标准溶液后，亚砷酸根离子和$\mathrm Br_2$发生反应，溶液变为无色，方程式为
$$\mathrm {Br_2 + AsO_2^- + 2H_2O =2Br^- +H_3AsO_4 +H^+ }$$

<Checkpoint>[方程式4$$\mathrm {Br_2 + AsO_2^- + 2H_2O =2Br^- +H_3AsO_4 +H^+ }$$](1)</Checkpoint>

亚砷酸和过硫酸根以及溴酸根发生反应，亚砷酸和过硫酸根反应的系数比为1：1，

<Checkpoint>[亚砷酸和过硫酸根反应的系数比为1：1](1)</Checkpoint>

亚砷酸和溴酸根反应的系数比为3:1，

<Checkpoint>[亚砷酸和溴酸根反应的系数比为3：1](1)</Checkpoint>

反应的所有亚砷酸的物质的量减去其中亚砷酸和溴酸根反应的物质的量，之后剩下的为亚砷酸和过硫酸根反应的物质的量，知道含有过硫酸根离子溶液的体积即可得知过硫酸根离子的浓度
$$ c=[c_1(V_2+V_4)-3c_2V_3]/V_1$$

<Checkpoint>[过硫酸根离子的浓度 $c=[c_1(V_2+V_4)-3c_2V_3]/V_1$](2)</Checkpoint>