The laboratory uses the following method to analyze the content of a bottle of commercially available resorcinol:

(1) Accurately weigh $\mathrm {1 . 8928 \ \  g }$ of potassium ferricyanide, add $4 0\  \mathrm { m L }$ of water to dissolve, add $2 \ \mathrm { g } \mathrm { K I }$ , $2 \  \mathrm { mL }$ of hydrochloric acid, let stand for 1 min, then add $1 0 \ \mathrm { m L } \ 3 0 0 \ \mathrm { g { \cdot } L ^ { - 1 } }\  \mathrm { Z n S O _ { 4 } }$ , titrate with $\mathrm { 0 . 2 \ m o l { \cdot } L ^ { - 1 } \ N a _ { 2 } S _ { 2 } O _ { 3 } }$ , add $5\  \mathrm { m L } \ \ 0 . 5\%$ starch after the color becomes lighter, titrate until the blue color disappears, and consume $\mathrm { N a _ { 2 } S _ { 2 } O _ { 3 } }$ standard titration solution $2 6 . 9 2 \ \mathrm { m L }$

(2) Weigh $0 . 096 6 \ \mathrm { g }$ of resorcinol, dissolve in $2 0 ~ \mathrm { m L }$ of water, add $5 0 \ \mathrm { m L }\  \mathrm { p H } = 5$ of HAc-NaAc solution, add $5 0 . 00 \ \mathrm { m L \ \ 0 . 1 \ \ m o l { \cdot } L ^ { - 1 } }$ iodine solution, react for 5 minutes, titrate with the above $\mathrm { N a _ { 2 } S _ { 2 } O _ { 3 } }$ standard solution, and consume $\mathrm { N a _ { 2 } S _ { 2 } O _ { 3 } }$ standard solution $2 2 . 4 4\  \mathrm { m L }$. At the same time, perform a blank experiment, consuming $\mathrm { N a _ { 2 } S _ { 2 } O _ { 3 } }$ standard solution $4 5 . 6 8 \ \mathrm { m L }$.

The following statements exist:

1. Phenol can also be titrated using a similar method.
2. Phloroglucinol can also be titrated using a similar method.
3. The sum of the coefficients of the chemical equation in the first step of step (2) is 8.
4. $\mathrm{ZnSO_4}$ is a masking agent to mask the color of $\mathrm{Fe(CN)_6^{3-}}$.
5. The content of resorcinol in this sample is 94.26%.

Which of the above statements are correct (options involving calculations are correct if the error is less than 1%)
实验室用如下方法分析一瓶市售间苯二酚的含量：

（1）准确称取 $\mathrm {1 . 8 9 2 8 \ \  g }$ 铁氰化钾，加 $4 0\  \mathrm { m L }$ 水溶解，加 $2 \ \mathrm { g } \mathrm { K I }$ ， $2 \  \mathrm { mL }$ 盐酸，静置1min 后加 $1 0 \ \mathrm { m L } \ 3 0 0 \ \mathrm { g { \cdot } L ^ { - 1 } }\  \mathrm { Z n S O _ { 4 } }$ ，用 $\mathrm { 0 . 2 \ m o l { \cdot } L ^ { - 1 } \ N a _ { 2 } S _ { 2 } O _ { 3 } }$ 滴定，待颜色变浅后加入 $5\  \mathrm { m L } \ \ 0 . 5\%$淀 粉，滴定至蓝色消失，消耗 $\mathrm { N a _ { 2 } S _ { 2 } O _ { 3 } }$ 标准滴定液 $2 6 . 9 2 \ \mathrm { m L }$

（2）称取 $0 . 0 9 6 6 \ \mathrm { g }$间苯二酚，用 $2 0 ~ \mathrm { m L }$ 水溶解，加 $5 0 \ \mathrm { m L }\  \mathrm { p H } = 5$ 的HAc-NaAc溶液，加入 $5 0 . 0 0 \ \mathrm { m L \ \ 0 . 1 \ \ m o l { \cdot } L ^ { - 1 } }$ 碘液，反应5分钟，用上述 $\mathrm { N a _ { 2 } S _ { 2 } O _ { 3 } }$ 标准溶液滴定，消耗 $\mathrm { N a _ { 2 } S _ { 2 } O _ { 3 } }$ 标准溶液 $2 2 . 4 4\  \mathrm { m L }$。同时进行空白实验，消耗 $\mathrm { N a _ { 2 } S _ { 2 } O _ { 3 } }$ 标准溶液 $4 5 . 6 8 \ \mathrm { m L }$。

存在以下说法：

1. 苯酚也可以使用类似方法进行滴定。
2. 均苯三酚也可以使用类似方法进行滴定。
3. 步骤(2)中第一步化学方程式的系数和为8。
4. $\mathrm{ZnSO_4}$为掩蔽剂，为了掩蔽$\mathrm{Fe(CN)_6^{3-}}$的颜色。
5. 该样品中间苯二酚的含量为94.26%。

上述说法中正确的是（涉及计算的选项，误差小于1%即正确）
1,2
1,3
1,4
1,5
2,3
2,4
2.5
3,4
3,5
4,5
1, 2, 3
1, 2, 4
1, 2, 5
1, 3, 4
1,3,5
1,4,5
2,3,4
2, 3, 5
2,4,5
3, 4, 5
1, 2, 3, 4
1, 2, 3, 5
1, 2, 4, 5
1,3,4,5
2, 3, 4, 5
1, 2, 3, 4, 5
1,2
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4,5
1,2,3
1,2,4
1,2,5
1,3,4
1,3,5
1,4,5
2,3,4
2,3,5
2,4,5
3,4,5
1,2,3,4
1,2,3,5
1,2,4,5
1,3,4,5
2,3,4,5
1,2,3,4,5
This is a relatively basic iodometric titration question. Among them, step (1) is the titration of $\mathrm{Na_2S_2O_3}$ standard solution; step (2) is the titration of the unknown sample of resorcinol.

The analysis is as follows:

For statements 1 and 2, the principle of resorcinol titration in step (2) is that the phenolic hydroxyl group, as an electron-donating group, can undergo aromatic electrophilic substitution reaction, replacing all three H atoms at the ortho- and para- positions with I, thereby performing iodometric back titration.

<Checkpoint>[Phenolic hydroxyl group is an ortho-para directing group and will undergo electrophilic substitution](1)</Checkpoint>

However, phenol has a low electron-donating ability, making it difficult for aromatic electrophilic substitution to occur at all three positions, so it cannot be used for quantitative analysis; phloroglucinol has a stronger electron-donating ability, so a similar method can be used for titration. Statement 1 is incorrect, statement 2 is correct.

<Checkpoint>[Phenol has a weak electron-donating ability and cannot be used for quantitative analysis; phloroglucinol has a stronger electron-donating ability and can be analyzed](1)</Checkpoint>

<Checkpoint>[Statement 1 is incorrect, statement 2 is correct](1)</Checkpoint>

For statement 3, according to the above analysis, the chemical equation that occurs is $\mathrm{C_6H_6O_2+3I_2=C_6H_3O_2I_3+3HI}$, and the sum of the coefficients of the chemical equation is 8, so statement 3 is correct.

<Checkpoint>[$\mathrm{C_6H_6O_2+3I_2=C_6H_3O_2I_3+3HI}$](1)</Checkpoint>

<Checkpoint>[The sum of the coefficients of the chemical equation is 8, so statement 3 is correct](0.5)</Checkpoint>

For statement 4, $\mathrm{ZnSO_4}$ can react with $\mathrm{Fe(CN)_6^{4-}}$ to form a precipitate, thereby removing the influence of $\mathrm{Fe(CN)_6^{4-}}$ coloration on the iodometric titration endpoint. However, although the statement points out that $\mathrm{ZnSO_4}$ is a masking agent, the masked ion is incorrect, so statement 4 is incorrect.

<Checkpoint>[$\mathrm{ZnSO_4}$ can remove the influence of $\mathrm{Fe(CN)_6^{4-}}$ coloration on the iodometric titration endpoint](1)</Checkpoint>

<Checkpoint>[Masked ion is incorrect, so statement 4 is incorrect](0.5)</Checkpoint>

For statement 5, calculate according to the above analysis. The molecular weight of $\mathrm{K_3Fe(CN)_6}$ is 329.27 $\mathrm{g/mol}$, then $\mathrm{n_{K_3Fe(CN)_6}=\frac{1.8928}{329.27}=0.005748 \ \ mol}$. According to charge conservation, 1 molecule of $\mathrm{K_3Fe(CN)_6}$ corresponds to 1 molecule of $\mathrm{Na_2S_2O_3}$.

<Checkpoint>[1 molecule of $\mathrm{K_3Fe(CN)_6}$ corresponds to 1 molecule of $\mathrm{Na_2S_2O_3}$](0.5)</Checkpoint>

Therefore, the amount of $\mathrm{Na_2S_2O_3}$ consumed by titration can be calculated as 0.005748 mol, and the concentration of $\mathrm{Na_2S_2O_3}$ is $\mathrm{c_{Na_2S_2O_3}=\frac{0.005748}{26.92\times 10^{-3}}=0.2135 \ \ mol/L}$.

From the back titration, it can be calculated that the amount of $\mathrm{I_2}$ consumed in the reaction process is $\mathrm{n_{I_2}=\frac{1}{2}\times(45.68-22.44)\times 10^{-3}\times{0.2135}=2.481×10^{-3} \ \ mol}$. Since 1 molecule of resorcinol corresponds to the consumption of 3 molecules of $\mathrm{I_2}$, the amount of resorcinol can be calculated as $\mathrm{8.27×10^{-4} \ \ mol}$. The molecular weight of resorcinol is $\mathrm{110.11 \ \ g/mol}$, and the mass of resorcinol can be calculated as $\mathrm{0.09106 \ \ g}$. Thus, the mass fraction of resorcinol is calculated as $\omega=\frac{0.09106}{0.0966}\times 100\%=94.26\%$, so statement 5 is correct.

<Checkpoint>[Mass fraction of resorcinol is $\omega=94.26\%$](2)</Checkpoint>

<Checkpoint>[Statement 5 is correct](0.5)</Checkpoint>

Therefore, the correct statements are 2, 3, and 5, and option R is correct.

<Checkpoint>[The correct statements are 2, 3, and 5, and option R is correct.](1)</Checkpoint>
这是一道较为基础的碘量法滴定题目。其中，步骤(1)为对于$\mathrm{Na_2S_2O_3}$标准溶液的滴定；步骤(2)为对未知样中间二苯酚的滴定。



分析说法如下：



对于说法1和说法2，步骤(2)对间二苯酚滴定的原理在于，酚羟基作为给电子基团，可以发生芳香亲电取代反应，将邻对位的三个H均取代为I，从而进行碘量法的返滴定。

<Checkpoint>[酚羟基是邻对位定位基，将发生亲电取代](1)</Checkpoint>

但是，苯酚由于给电子能力较低，很难发生三个位置均进行芳香亲电取代，所以无法进行定量分析；均苯三酚给电子能力更强，所以可以使用类似方法滴定。说法1错误，说法2正确。

<Checkpoint>[苯酚给电子能力较弱，不能用于定量分析；均苯三酚给电子能力更强，可以分析](1)</Checkpoint>

<Checkpoint>[说法1错误，说法2正确](1)</Checkpoint>


对于说法3，根据上述分析，发生的化学方程式为$\mathrm{C_6H_6O_2+3I_2=C_6H_3O_2I_3+3HI}$，化学方程式系数和为8，说法3正确。

<Checkpoint>[$\mathrm{C_6H_6O_2+3I_2=C_6H_3O_2I_3+3HI}$](1)</Checkpoint>

<Checkpoint>[化学方程式系数和为8，说法3正确](0.5)</Checkpoint>

对于说法4，$\mathrm{ZnSO_4}$可以和$\mathrm{Fe(CN)_6^{4-}}$发生作用，形成沉淀，从而除去$\mathrm{Fe(CN)_6^{4-}}$显色对于碘量法滴定终点的影响。但是，说法中虽然指出了$\mathrm{ZnSO_4}$为掩蔽剂，但掩蔽离子错误，故说法4错误。

<Checkpoint>[$\mathrm{ZnSO_4}$可以除去$\mathrm{Fe(CN)_6^{4-}}$显色对于碘量法滴定终点的影响](1)</Checkpoint>

<Checkpoint>[掩蔽离子错误，故说法4错误](0.5)</Checkpoint>

对于说法5，根据上述分析进行计算。$\mathrm{K_3Fe(CN)_6}$的分子量为329.27 $\mathrm{g/mol}$，则$\mathrm{n_{K_3Fe(CN)_6}=\frac{1.8928}{329.27}=0.005748 \ \ mol}$，由电荷守恒，1分子$\mathrm{K_3Fe(CN)_6}$对应1分子$\mathrm{Na_2S_2O_3}$。

<Checkpoint>[1分子$\mathrm{K_3Fe(CN)_6}$对应1分子$\mathrm{Na_2S_2O_3}$](0.5)</Checkpoint>

因此，可以计算滴定消耗$\mathrm{Na_2S_2O_3}$的量为0.005748 mol，$\mathrm{Na_2S_2O_3}$的浓度为$\mathrm{c_{Na_2S_2O_3}=\frac{0.005748}{26.92\times 10^{-3}}=0.2135 \ \ mol/L}$。

由返滴定可以计算，反应过程消耗$\mathrm{I_2}$的量$\mathrm{n_{I_2}=\frac{1}{2}\times(45.68-22.44)\times 10^{-3}\times{0.2135}=2.481×10^{-3} \ \ mol}$。由于1分子间苯二酚对应消耗3分子$\mathrm{I_2}$，可以计算出间苯二酚的量为$\mathrm{8.27×10^{-4} \ \ mol}$。间苯二酚的分子量为$\mathrm{110.11 \ \ g/mol}$，可以计算出间苯二酚的质量为$\mathrm{0.09106 \ \ g}$。从而算出间苯二酚质量分数为$\omega=\frac{0.09106}{0.0966}\times 100\%=94.26\%$，说法5正确。

<Checkpoint>[间苯二酚质量分数为$\omega=94.26\%$](2)</Checkpoint>

<Checkpoint>[说法5正确](0.5)</Checkpoint>

因此，正确的说法为2，3，5，选项R正确。

<Checkpoint>[正确的说法为2，3，5，选项R正确。](1)</Checkpoint>