📚 Titration Case Study: Determining Acetic Acid in Vinegar | 滴定案例分析：测定食醋中乙酸含量
This article walks through a real titration scenario in a Year 11 CAIE Chemistry laboratory: using a standard sodium hydroxide solution to determine the acetic acid content in a sample of commercial vinegar. You will follow the steps of the experiment, perform calculations step by step, analyse the results, and evaluate possible errors and improvements. The case is designed to strengthen both your practical skills and your ability to answer exam-style questions on acid–base titrations.
本文将通过一个真实的滴定案例——用标准氢氧化钠溶液测定市售食醋样品中乙酸的含量——带你演练Year 11 CAIE化学实验的全过程。你将逐步跟随实验步骤、进行定量计算、分析结果，并评估误差与改进方法。本案例旨在巩固你的实验技能以及解答酸碱滴定考题的能力。
1. The Scenario | 案例情景
A student investigates the acetic acid content of a bottle of clear vinegar bought from a supermarket. The label states that the vinegar contains 5% (w/v) acetic acid. The student plans to use a standard 0.100 mol dm⁻³ sodium hydroxide solution and phenolphthalein indicator to check the manufacturer’s claim.
一名学生要检测从超市购买的一瓶清澈食醋中乙酸的含量。标签注明该食醋含5%（质量/体积）乙酸。学生计划用0.100 mol dm⁻³的标准氢氧化钠溶液和酚酞指示剂来验证厂商的标注。
2. Principle of the Titration | 滴定原理
Acetic acid (CH₃COOH) is a weak monoprotic acid. It reacts with sodium hydroxide (NaOH), a strong base, in a 1 : 1 mole ratio. The reaction produces sodium acetate and water. Phenolphthalein changes from colourless to a permanent pale pink at the endpoint (pH ~8.2–10). The volume of NaOH required to neutralise a known volume of diluted vinegar allows the amount of acetic acid to be calculated.
乙酸（CH₃COOH）是一种弱的一元酸，它与强碱氢氧化钠（NaOH）以1 : 1的摩尔比反应，生成乙酸钠和水。终点时（pH约8.2–10）酚酞由无色变为持久的淡粉色。根据中和一定体积稀释食醋所消耗的NaOH体积，可计算出乙酸含量。
3. Procedure Outline | 操作步骤概述
The student first dilutes the original vinegar. A 10.0 cm³ pipette is used to transfer exactly 10.0 cm³ of the vinegar into a 100 cm³ volumetric flask. Deionised water is added to the graduation mark, and the flask is stoppered and shaken to mix thoroughly. A 25.0 cm³ pipette is then used to transfer 25.0 cm³ of the diluted vinegar solution into a clean conical flask, and 3 drops of phenolphthalein indicator are added. The solution is titrated against the 0.100 mol dm⁻³ NaOH solution until the first permanent pink colour appears. The titration is repeated until three consistent titres are obtained.
学生首先稀释原食醋。用10.0 cm³移液管精确移取10.0 cm³食醋至100 cm³容量瓶中，加去离子水至刻度线，塞上瓶塞摇匀混合。然后用25.0 cm³移液管移取25.0 cm³稀释后的醋溶液至洁净的锥形瓶中，加入3滴酚酞指示剂。用0.100 mol dm⁻³ NaOH溶液滴定，直至溶液出现持久的浅粉色。重复滴定，得到三次一致的滴定体积。
4. Data Table | 数据记录表
The student records the following burette readings (initial and final) for three successful runs. All volumes are measured to the nearest 0.05 cm³.
学生记录了三次成功滴定的滴定管读数（初始和终末），所有体积均读取至0.05 cm³。
| Titration | Initial reading / cm³ | Final reading / cm³ | Volume used / cm³ | Best titres |
|---|---|---|---|---|
| Trial 1 | 0.00 | 21.10 | 21.10 | |
| Trial 2 | 0.00 | 20.90 | 20.90 | ✓ |
| Trial 3 | 0.00 | 20.80 | 20.80 | ✓ |
The student discards Trial 1 because it is a rough titration. The two concordant titres are 20.90 cm³ and 20.80 cm³, giving an average titre of 20.85 cm³. (For three concordant readings, the average would be used; here two values are close enough, so the mean is accepted.)
学生剔除了第一次作为粗略滴定。两次一致量为20.90 cm³和20.80 cm³，平均用量为20.85 cm³。（若三次均一致可使用平均值；此例两次接近，故取均值。）
5. Balanced Chemical Equation | 配平化学方程式
CH₃COOH + NaOH → CH₃COONa + H₂O
The equation confirms a 1 : 1 molar ratio between acetic acid and sodium hydroxide.
该方程式表明乙酸与氢氧化钠以1 : 1摩尔比反应。
6. Calculating Moles of NaOH Used | 计算所用NaOH的摩尔数
× volume (dm³). Convert the average titre to dm³: 20.85 cm³ = 20.85 ÷ 1000 = 0.02085 dm³.
NaOH的摩尔数 = 浓度（mol dm⁻³）× 体积（dm³）。将平均用量换算为dm³：20.85 cm³ = 20.85 ÷ 1000 = 0.02085 dm³。
n(NaOH) = 0.100 mol dm⁻³ × 0.02085 dm³ = 0.002085 mol
7. Moles and Concentration of Acetic Acid | 乙酸的摩尔数与浓度
ar (10.0 cm³) was diluted to 100 cm³, and a 25.0 cm³ aliquot was taken for titration. Therefore, the volume of original vinegar represented by the 25.0 cm³ aliquot is (10.0 cm³) × (25.0 / 100) = 2.5 cm³. So the number of moles of acetic acid in 2.5 cm³ of the original vinegar is 0.002085 mol.
由反应式可知，25.0 cm³稀释醋液中CH₃COOH的摩尔数 = 0.002085 mol。需要考虑稀释倍数。原食醋（10.0 cm³）被稀释至100 cm³，移取其中25.0 cm³进行滴定。因此，所取25.0 cm³稀释液对应的原食醋体积为 10.0 cm³ × (25.0/100) = 2.5 cm³。所以2.5 cm³原食醋中乙酸的摩尔数为0.002085 mol。
Concentration of original vinegar = n / V = 0.002085 mol / 0.0025 dm³ = 0.834 mol dm⁻³
8. Conversion to Percentage Mass/Volume | 转换为质量体积百分比
oncentration × molar mass = 0.834 mol dm⁻³ × 60 g mol⁻¹ = 50.04 g dm⁻³. To express this as % 1 dm³ = 1000 cm³, so the mass per 100 cm³ is 5.004 g. The vinegar therefore contains 5.00% (w/v) acetic acid (to three significant figures).
乙酸（CH₃COOH）的摩尔质量为60 g mol⁻¹。质量浓度 = 物质的量浓度 × 摩尔质量 = 0.834 mol dm⁻³ × 60 g mol⁻¹ = 50.04 g dm⁻³。表示为%(w/v)：1 dm³ = 1000 cm³，因此每100 cm³的质量为5.004 g。该食醋中乙酸含量为5.00% (w/v)（保留三位有效数字）。
9. Comparison with the Label Claim | 与标签声称值对比
The experimentally determined value, 5.00% (w/v), matches the label claim of 5% text, you might be asked to calculate the percentage error relative to the stated value: (|5.00 – 5.00| / 5.00) × 100% = 0% in this idealised case, though in real measurements a small discrepancy is expected.
实验测定的5.00% (w/v)与标签标注的5% (w/v)在误差范围内一致。学生可得出结论：厂商的标注得到证实。在考试中，可能需要计算相对误差：(|5.00 – 5.00| / 5.00) × 100% = 0%（本理想化案例），实际测量中会存在微小偏差。
10. Sources of Error | 误差来源
The following common errors could affect the accuracy and precision of the titration result.
以下常见误差可能影响滴定结果的准确度和精密度。
- Parallax error in reading the burette – reading the meniscus from an angle can introduce systematic error. / 滴定管读数视差——从侧面读取弯月面会引入系统误差。
- Not rinsing the burette with the NaOH solution – water left in the burette dilutes the alkali, making the titre larger. / 未用NaOH溶液润洗滴定管——管中残留的水会稀释碱液，增大滴定体积。
- Not rinsing the pipette with the diluted vinegar – water in the pipette dilutes the acid solution, resulting in a lower titre. / 移液管未用稀释醋液润洗——管内残留水会稀释酸液，导致滴定体积偏小。
- Not filling the tip of the burette – an air bubble initially present may be displaced during titration, causing an overestimation of the volume delivered. / 滴定管尖端未充满——起始气泡在滴定过程中被排出，使记录用量偏高。
- Overshooting the endpoint – adding too much NaOH beyond the colour change leads to a titre larger than the true value. / 过量滴定——添加NaOH超出变色点，导致体积偏大。
- Using too much indicator – more than 3 drops of phenolphthalein can affect the pH at which the colour change is observed. / 指示剂用量过多——超过3滴的酚酞会影响颜色变化的pH，干扰判断。
11. Suggestions for Improvement | 改进建议
To improve accuracy, the student should use a white tile under the conical flask to observe the phenolphthalein colour change more sharply. The burette tap should be operated slowly near the endpoint, adding NaOH dropwise with constant swirling. Repeating the titration several more times would increase confidence in the mean titre. Using a pH meter instead of an indicator could identify the true equivalence point more precisely, especially since acetic acid is weak and the endpoint is at a slightly basic pH.
为提高准确性，锥形瓶下方应放置白色瓷砖，以便更清晰地观察酚酞变色。临近终点时应缓慢转动活塞，逐滴加入NaOH并持续摇动。增加滴定次数可获得更可靠的平均值。使用pH计代替指示剂可更精确地找到等当点，尤其因为乙酸为弱酸，终点在微碱性区域。
12. Exam-Style Questions Derived from This Case | 由此案例衍生的常见考题
Analysing this practical scenario helps you tackle typical CAIE questions. Consider the following examples:
分析这一实际情境有助于应对CAIE常见考题。思考以下问题：
- Why is phenolphthalein a suitable indicator? It changes colour sharply in the pH range 8.2–10, which matches the steep rise near the equivalence point of a strong base–weak acid titration. / 为什么酚酞是合适的指示剂？ 它在pH 8.2–10范围内发生突跃变色，与强碱–弱酸滴定等当点附近的陡升段匹配。
- Why was the vinegar diluted before titration? Dilution brings the acetic acid concentration into a range that gives a convenient titration volume with the standard NaOH solution, reducing the relative error when reading the burette. / 为什么要先稀释食醋？ 稀释可使乙酸浓度降低到与标准NaOH溶液滴定产生适当体积的范围内，从而减小滴定管读数带来的相对误差。
- How would the result be affected if the volumetric flask was not inverted thoroughly after dilution? The solution would be unevenly mixed, so the 25.0 cm³ aliquot might not contain a representative amount of acid, leading to inaccurate results. / 如果稀释后容量瓶未充分摇匀，对结果有何影响？ 溶液不均匀，25.0 cm³等分试样不能代表整体，导致结果不准确。
- Calculate the mass of NaHCO₃ needed to prepare 250 cm³ of 0.100 mol dm⁻³ Na₂CO₃ free solution. This extension tests understanding of molar mass and solution preparation. / 计算配制250 cm³ 0.100 mol dm⁻³ Na₂CO₃溶液所需NaHCO₃的质量。 这类延伸题检验摩尔质量与溶液配制的掌握。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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