Quantitative Analysis of Strong Oxidants

$\mathrm{NaBiO_3}$ and $\mathrm{PbO_2}$ are strong oxidants commonly used in industry. Their redox properties can be utilized to conduct volumetric analysis studies.

$\mathrm{NaBiO_3}$ is a light yellow powder that is insoluble in cold water. Accurately weigh $0.8671\mathrm{~g}$ of sodium bismuthate sample, add an appropriate amount of dilute sulfuric acid and $\mathrm{MnSO_4}$ solution to completely convert it to $\mathrm{Bi^{3+}}$, and then titrate with a $\mathrm{H_2C_2O_4}$ standard solution with a concentration of $0.1080~\mathrm{mol \cdot L^{-1}}$. When the purple-red color of the solution just disappears, $24.89\mathrm{mL}$ of $\mathrm{H_2C_2O_4}$ solution is consumed.

$\mathrm{PbO_2}$ is easily decomposed by heat. Heating a certain amount of $\mathrm{PbO_2}$ yields some solid $(x\mathrm{PbO_2} \cdot y\mathrm{PbO})$. To investigate the composition of this solid, a certain mass of the solid is taken and added to $30\mathrm{~mL~CH_3COOH-CH_3COONa}$ buffer solution, and then a sufficient amount of $\mathrm{KI}$ solid is added. The conical flask is shaken to completely dissolve the solid, resulting in a brown solution. Using starch as an indicator, the solution is titrated with $0.1020~\mathrm{mol \cdot L^{-1}}$ of $\mathrm{Na_2S_2O_3}$ standard solution until the blue color of the solution just disappears, consuming $24.70~\mathrm{mL}$ of $\mathrm{Na_2S_2O_3}$ standard solution. The same mass of the above solid is mixed with dilute nitric acid to obtain a brownish-black powder. The filtrate obtained by filtration is added with a sufficient amount of $\mathrm{K_2CrO_4}$ solution to convert it to $\mathrm{PbCrO_4}$ precipitate. The precipitate is washed and dried, and the mass of the solid obtained is $0.611\mathrm{~g}$.

Determine the correctness of the following statements
1. The purity of the sodium bismuthate sample (calculated as the mass fraction of $\mathrm{NaBiO_3}$) is $87.69\%$
2. Determine that $x=2, y=3$ through calculation
3. The most accurate chemical equation for the reaction of $\mathrm{PbO_2}$ dissolved in $\mathrm{CH_3COOH-CH_3COONa}$ buffer solution with $\mathrm{KI}$ is
$\mathrm{PbO_2} + 5\mathrm{KI} + 4\mathrm{CH_3COOH} \rightarrow \mathrm{PbI_2} + \mathrm{KI_3} + 2\mathrm{H_2O} + 4\mathrm{CH_3COOK}$
强氧化剂的定量分析

$\mathrm{NaBiO_3}$ 和 $\mathrm{PbO_2}$ 都是工业中常用的强氧化剂，利用其氧化还原特性可以展开容量分析研究。

$\mathrm{NaBiO_3}$ 是一种浅黄色粉末，不溶于冷水。准确称取 $0.8671\mathrm{~g}$ 铋酸钠样品，加入适量稀硫酸和 $\mathrm{MnSO_4}$ 溶液使其完全转化为 $\mathrm{Bi^{3+}}$，再用浓度为 $0.1080~\mathrm{mol \cdot L^{-1}}$ 的 $\mathrm{H_2C_2O_4}$ 标准溶液滴定，当溶液紫红色恰好褪去时，消耗 $\mathrm{H_2C_2O_4}$ 溶液 $24.89\mathrm{mL}$。

$\mathrm{PbO_2}$ 受热易分解，将一定量的 $\mathrm{PbO_2}$ 加热得到了一些固体 $(x\mathrm{PbO_2} \cdot y\mathrm{PbO})$。为探究该固体的组成，取一定质量该固体加入到 $30\mathrm{~mL~CH_3COOH-CH_3COONa}$ 缓冲溶液中，再加入足量 $\mathrm{KI}$ 固体，振荡锥形瓶使固体完全溶解，得到棕色溶液。以淀粉为指示剂，用 $0.1020~\mathrm{mol \cdot L^{-1}}$ 的 $\mathrm{Na_2S_2O_3}$ 标准溶液滴定至溶液蓝色恰好消失，消耗 $\mathrm{Na_2S_2O_3}$ 标准溶液 $24.70~\mathrm{mL}$。取相同质量的上述固体与稀硝酸混合，得到棕黑色粉末，过滤出的滤液加入足量的 $\mathrm{K_2CrO_4}$ 溶液将其转化为 $\mathrm{PbCrO_4}$ 沉淀，洗涤烘干沉淀，得固体质量为 $0.611\mathrm{~g}$。

判断以下说法的正确性  
1. 铋酸钠样品的纯度（以$\mathrm{NaBiO_3}$的质量分数计）为$87.69\%$  
2. 通过计算确定$x=2, y=3$  
3. $\mathrm{PbO_2}$ 溶于 $\mathrm{CH_3COOH-CH_3COONa}$缓冲溶液与$\mathrm{KI}$反应的最准确的化学方程式为  
$\mathrm{PbO_2} + 5\mathrm{KI} + 4\mathrm{CH_3COOH} \rightarrow \mathrm{PbI_2} + \mathrm{KI_3} + 2\mathrm{H_2O} + 4\mathrm{CH_3COOK}$
1 2 3 are all incorrect.
1 2 3 are all correct
Only 1 is correct.
Only 2 is correct.
Only 3 is correct.
Only 1 error
Only 2 errors
Only 3 errors
1 2 3均不正确
1 2 3均正确
仅1正确
仅2正确
仅3正确
仅1错误
仅2错误
仅3错误
Stoichiometric ratio: $\rm 5NaBiO_{3} \sim 2MnO_{4}^{-} \sim 5H_{2}C_{2}O_{4}$ 

<Checkpoint>[The stoichiometric ratio of $\rm NaBiO_{3}$ to $\rm H_{2}C_{2}O_{4}$ is 5:5, i.e., 1:1](1)</Checkpoint>

$n ( \mathrm { N a B i O _ { 3 } } ) = n ( \mathrm { H _ { 2 } C _ { 2 } O _ { 4 } } ) = 0 . 1 0 8 0 \times 2 4 . 8 9 \times 1 0 ^ { - 3 } = 2 . 6 8 8 \times 1 0 ^ { - 3 } ~ ( \mathrm { m o l } )$

$w\left(\mathrm{NaBiO}_{3}\right)=2.688 \times 10^{-3} \times 279.97 / 0.8671=86.79 \% $. Therefore, description 1 is incorrect.

<Checkpoint>[$w\left(\mathrm{NaBiO}_{3}\right)=86.79 \% $, description 1 is incorrect](1)</Checkpoint>

Stoichiometric ratio: $\rm PbO_{2} \sim I_{2} \sim 2S_{2}O_{3}^{2-}$ 

$n\left(\mathrm{PbO}_{2}\right)=1 / 2 \times 0.1020 \times 24.70 \times 10^{-3}=1.260 \times 10^{-3}(\mathrm{~mol})  $ 

<Checkpoint>[Obtained by redox titration: $n\left(\mathrm{PbO}_{2}\right)=1.260 \times 10^{-3}(\mathrm{~mol})  $ ](1)</Checkpoint>

Since $\rm PbO_2$ is insoluble in dilute nitric acid, the gravimetric method corresponds to $\rm PbO$.

<Checkpoint>[$\rm PbO_2$ is insoluble in dilute nitric acid](1)</Checkpoint>

$n(\mathrm{PbO})=0.611 / 323.2=1.89 \times 10^{-3}(\mathrm{~mol}) $

<Checkpoint>[$n(\mathrm{PbO})=1.89 \times 10^{-3}(\mathrm{~mol}) $](0.5)</Checkpoint>

$x / y=n\left(\mathrm{PbO}_{2}\right) / n(\mathrm{PbO})=1.260 \times 10^{-3} / 1.89 \times 10^{-3}=0.667\approx2 / 3$

Therefore, $x=2, y=3$, description 2 is correct. 

<Checkpoint>[$x / y\approx2 / 3$, $x=2, y=3$, description 2 is correct](1)</Checkpoint>

Adding sufficient solid $\rm KI$ can complex and dissolve $\rm PbI_2$, so it is not reasonable for the product to contain precipitate $\rm PbI_2$. Description 3 is incorrect.

<Checkpoint>[It is not reasonable for the product to contain precipitate $\rm PbI_2$. Description 3 is incorrect](1)</Checkpoint>
化学计量比:$\rm 5NaBiO_{3} \sim 2MnO_{4}^{-} \sim 5H_{2}C_{2}O_{4}$ 

<Checkpoint>[$\rm NaBiO_{3}$与$\rm H_{2}C_{2}O_{4}$的计量比为5:5, 即1:1](1)</Checkpoint>

$n ( \mathrm { NaBiO _ { 3 } } ) = n ( \mathrm { H _ { 2 } C _ { 2 } O _ { 4 } } ) = 0 . 1 0 8 0 \times 2 4 . 8 9 \times 1 0 ^ { - 3 } = 2 . 6 8 8 \times 1 0 ^ { - 3 } ~ ( \mathrm { m o l } )$

$w\left(\mathrm{NaBiO}_{3}\right)=2.688 \times 10^{-3} \times 279.97 / 0.8671=86.79 \% $. 故描述1错误

<Checkpoint>[$w\left(\mathrm{NaBiO}_{3}\right)=86.79 \% $, 描述1错误](1)</Checkpoint>

计量比:$\rm PbO_{2} \sim I_{2} \sim 2S_{2}O_{3}^{2-}$ 

$n\left(\mathrm{PbO}_{2}\right)=1 / 2 \times 0.1020 \times 24.70 \times 10^{-3}=1.260 \times 10^{-3}(\mathrm{~mol})  $ 

<Checkpoint>[通过氧化还原滴定得出$n\left(\mathrm{PbO}_{2}\right)=1.260 \times 10^{-3}(\mathrm{~mol})  $ ](1)</Checkpoint>

由于$\rm PbO_2$不溶于稀硝酸, 重量法对应于$\rm PbO$.

<Checkpoint>[$\rm PbO_2$不溶于稀硝酸](1)</Checkpoint>

$n(\mathrm{PbO})=0.611 / 323.2=1.89 \times 10^{-3}(\mathrm{~mol}) $

<Checkpoint>[$n(\mathrm{PbO})=1.89 \times 10^{-3}(\mathrm{~mol}) $](0.5)</Checkpoint>

$x / y=n\left(\mathrm{PbO}_{2}\right) / n(\mathrm{PbO})=1.260 \times 10^{-3} / 1.89 \times 10^{-3}=0.667\approx2 / 3$

故 $x=2, y=3$, 描述2正确 。 

<Checkpoint>[$x / y\approx2 / 3$, $x=2, y=3$, 描述2正确](1)</Checkpoint>

加入足量$\rm KI$固体可以使$\rm PbI_2$络合溶解, 因此生成物中出现沉淀$\rm PbI_2$不够合理. 描述3错误

<Checkpoint>[生成物中出现沉淀$\rm PbI_2$不够合理. 描述3错误](1)</Checkpoint>