The molecular structure of the drug in a certain tablet is shown in the following figure:

<MultiModal>![O=C(C1=C(OC(C)=O)C=CC=C1)O](/media/uploads/1acb97b5-d5b1-4128-a0b6-8b760f6567a3.jpg)</MultiModal>

The method for determining the drug content is as follows: Weigh $1.03$ g of tablet powder, add $15 \mathrm {mL}$ of water and shake well, then extract four times with $50 \mathrm { mL }$ of solvent A, wash and evaporate to dryness, dissolve the residue with $25 \mathrm {mL}$ of solvent B, add 3 drops of indicator, cool the conical flask to $T ^ { \circ } C$, and titrate to the endpoint with NaOH solution with a concentration of $0.05034 \mathrm { m o l } { \cdot } { \tt L } ^ { - 1 }$, consuming $24.78\mathrm {mL}$. Based on this, the following statement that is correct is:
某药片中的药物分子结构如下图：

<MultiModal>![O=C(C1=C(OC(C)=O)C=CC=C1)O](/media/uploads/1acb97b5-d5b1-4128-a0b6-8b760f6567a3.jpg)</MultiModal>

测定药物含量的方法如下：称取 $1.03$克药片粉末，加入 $15\mathrm{mL}$ 水后摇匀，再用 $50\mathrm{mL}$ 溶剂A分四次提取，洗涤后蒸干，用$25\mathrm{mL}$ 溶剂B溶解残渣，加入指示剂3滴，冷却锥形瓶至$T^{\circ}C$ ，用浓度为 $0.05034\mathrm{mol}{\cdot}{\tt L}^{-1}$ 的NaOH溶液滴定至终点，消耗 $24.78\mathrm {mL}$，据此判断下列说法中正确的是：
All other options are incorrect
Due to the ester group hydrolyzing under alkaline conditions, consuming 3 equivalents of NaOH, the mass fraction of the drug in the tablet is $7.27{\%}$
Titration endpoint has a relatively strong alkalinity, thymolphthalein can be selected as an indicator.
The drug molecule is highly polar, and highly polar DMF can be selected as solvent A for extraction.
Molecules of drugs possess intermolecular hydrogen bonds; protic solvent methanol can be selected as solvent B during titration.
To make the titration more accurate, the temperature T can be maintained at around 45 degrees Celsius to promote rapid reaction.
其他选项均不正确
由于酯基会在碱性条件下会水解，消耗3当量NaOH，因此药片中药品的质量分数为$7.27{\%}$
滴定终点碱性较强，可选用百里酚酞作指示剂
药物分子极性较大，可选用高极性的DMF作为萃取用的溶剂A
药物分子存在分子间氢键，可选用质子溶剂甲醇作为滴定时的溶剂B
为了使滴定更加准确，温度T可以维持在45摄氏度左右以促进反应快速发生
This question examines acid-base titration related content. First, calculate the amount of substance of NaOH consumed in the titration: $0.05034{\times}24.78{\times}10^{-3}=1.247{\times}10^{-3}\mathrm{mol}$. Considering the reaction ratio of acetylsalicylic acid and NaOH, first, the carboxyl group in the molecule will react with 1 NaOH. At the same time, the system maintains an acidic to weakly alkaline environment, the alkalinity of this system is not so strong, and the time of titration is normally short, so the ester group will not be completely hydrolyzed.

<Checkpoint>[Ester group will not be completely hydrolyzed](1)</Checkpoint>

Therefore, acetylsalicylic acid and NaOH react in a $1:1$ ratio, so option B is incorrect.

<Checkpoint>[Acetylsalicylic acid and NaOH react in a $1:1$ ratio](1)</Checkpoint>

Therefore, the amount of substance of acetylsalicylic acid in the sample is $1.247{\times}10^{-3}\mathrm{mol}$, the molecular weight of acetylsalicylic acid is $180.159\mathrm {g/mol}$, and the mass of acetylsalicylic acid is $180.159{\times}1.247{\times}10^{-3}=0.2247\mathrm{g}$

The correct mass fraction should be $0.2247{\div}1.03{\times}100{\%}=21.8{\%}$

<Checkpoint>[Mass fraction is $21.8{\%}$](1)</Checkpoint>

The pKa of aspirin is approximately 3.5, and it can be calculated that the pH at the endpoint is approximately 8, which is weakly alkaline.

<Checkpoint>[Endpoint pH is approximately 8](1)</Checkpoint>

The color change range of thymolphthalein is 9.3-10.5

<Checkpoint>[Color change range of thymolphthalein is 9.3-10.5](1)</Checkpoint>

Therefore, thymolphthalein cannot be selected as an indicator.

<Checkpoint>[Thymolphthalein cannot be selected as an indicator](1)</Checkpoint>

The requirement for solvent A is that it has good solubility for aspirin and is easy to separate by rotary evaporation, that is, it has a lower boiling point.

<Checkpoint>[Boiling point of solvent A needs to be low](2)</Checkpoint>

Solvent A cannot be DMF, which has a high boiling point and is miscible with water.

<Checkpoint>[Solvent A cannot be DMF](1)</Checkpoint>

Solvent B is required to be usable for acid-base titration, that is, it must be a protic solvent.

<Checkpoint>[Solvent B is a protic solvent](1)</Checkpoint>

And in order to prevent partial hydrolysis of the ester group from producing acidic carboxyl groups, which would affect the titration results, solvent B cannot be water.

<Checkpoint>[Solvent B cannot be water](1)</Checkpoint>

Therefore, solvent B can be methanol, and even if transesterification occurs, it will not affect the titration results. Option E is reasonable.

<Checkpoint>[Solvent B can be methanol](1)</Checkpoint>

Acid-base neutralization reactions are very fast and generally do not require heating to promote them.

<Checkpoint>[Neutralization reaction is fast, no heating required](1)</Checkpoint>

Heating will instead promote the reaction between the water formed during the titration process and the ester group to produce acidic carboxyl groups, which will affect the titration results.

<Checkpoint>[Heating will promote ester hydrolysis, affecting titration](1)</Checkpoint>

Therefore, the entire titration process must be kept at a low temperature and cannot be carried out at 45 degrees Celsius. Option F is incorrect.

<Checkpoint>[Titration cannot be carried out at 45 degrees Celsius](1)</Checkpoint>
本题目考察酸碱滴定相关内容，首先要计算出滴定消耗的NaOH的物质的量为$0.05034{\times}24.78{\times}10^{-3}=1.247{\times}10^{-3}\mathrm{mol}$，考虑乙酰水杨酸和NaOH的反应比例，首先分子中的羧基会和1个NaOH反应，同时体系一直维持酸性至弱碱性环境，体系碱性不强，滴定时间也较短，酯基不会完全水解。

<Checkpoint>[酯基不会完全水解](1)</Checkpoint>

所以乙酰水杨酸和NaOH以$1:1$的比例反应，B选项错误。

<Checkpoint>[乙酰水杨酸和NaOH$1:1$反应](1)</Checkpoint>

所以样品中的乙酰水杨酸的物质的量为$1.247{\times}10^{-3}\mathrm{mol}$，乙酰水杨酸的分子量为$180.159\mathrm {g/mol}$，乙酰水杨酸的质量为$180.159{\times}1.247{\times}10^{-3}=0.2247\mathrm{g}$

正确的质量分数应为$0.2247{\div}1.03{\times}100{\%}=21.8{\%}$

<Checkpoint>[质量分数为$21.8{\%}$](1)</Checkpoint>

阿司匹林的pKa约为3.5，可算出终点的pH约为8，属于弱碱性.

<Checkpoint>[终点pH约为8](1)</Checkpoint>

百里酚酞的变色范围为9.3-10.5.

<Checkpoint>[百里酚酞的变色范围为9.3-10.5](1)</Checkpoint>

因此指示剂不可选择百里酚酞.

<Checkpoint>[指示剂不可选百里酚酞](1)</Checkpoint>

溶剂A的要求为对阿司匹林溶解性好且易于旋蒸分离，即沸点较低

<Checkpoint>[溶剂A沸点需较低](2)</Checkpoint>

溶剂A不可选择高沸点且与水混溶的DMF

<Checkpoint>[溶剂A不可选择DMF](1)</Checkpoint>

溶剂B要求为可用于酸碱滴定，即须为质子溶剂。

<Checkpoint>[溶剂B为质子溶剂](1)</Checkpoint>

且为了防止酯基部分水解产生酸性羧基，影响滴定结果，溶剂B不能是水。

<Checkpoint>[溶剂B不可为水](1)</Checkpoint>

所以溶剂B可以选择甲醇，即使发生酯交换也不影响滴定结果，E选项合理.

<Checkpoint>[溶剂B可以是甲醇](1)</Checkpoint>

酸碱中和反应非常快，一般不需要加热促进。

<Checkpoint>[中和反应快，无需加热](1)</Checkpoint>

加热反而会促进滴定过程中形成的水和酯基反应产生酸性羧基，影响滴定结果。

<Checkpoint>[加热会促进酯基水解，影响滴定](1)</Checkpoint>

因此滴定全程需保持较低温度，不可在45摄氏度进行。F选项错误

<Checkpoint>[滴定不可在45摄氏度进行](1)</Checkpoint>