📚 Case Study: Determining Ethanoic Acid in Vinegar by Titration | 案例分析：滴定法测定食醋中乙酸含量
This case study focuses on a classic Year 12 AQA Chemistry practical: determining the concentration of ethanoic acid in household vinegar using acid-base titration. You will follow a complete workflow from preparing solutions, performing the titration, recording data, calculating the acid content, and critically evaluating errors and uncertainties. This hands-on exercise mirrors the style of application and analysis questions found in AQA examinations, strengthening your skills in quantitative chemistry, practical techniques, and scientific reasoning.
本案例研究聚焦于 AQA 化学 Year 12 的经典实验：利用酸碱滴定法测定家用食醋中乙酸的浓度。你将体验完整的实验流程，包括溶液配制、滴定操作、数据记录、含量计算以及误差和不确定度的批判性分析。这种动手实践的方式模拟了 AQA 考试中常见的应用与分析题型，帮助你巩固定量化学、实验技术和科学推理能力。
1. Background and Theory | 背景与理论
de (NaOH), a neutralisation reaction occurs in a 1:1 molar ratio: CH₃COOH + NaOH → CH₃COONa + H₂O. The salt formed, sodium ethanoate, partially hydrolyses in water, making the equivalence point slightly alkaline (pH > 7). Therefore, an indicator that changes colour in the basic range is required. Phenolphthalein, which turns from colourless to pale pink between pH 8.2 and 10.0, is an ideal choice for this weak acid-strong base titration.
食醋是含有弱一元酸——乙酸（CH₃COOH）的水溶液。用强碱如氢氧化钠（NaOH）滴定时，发生 1:1 摩尔比的中和反应：CH₃COOH + NaOH → CH₃COONa + H₂O。生成的盐醋酸钠在水中部分水解，使等当点呈弱碱性（pH > 7）。因此，需要选用在碱性范围内变色的指示剂。酚酞在 pH 8.2 至 10.0 之间由无色变为淡粉红色，是这种弱酸-强碱滴定的理想选择。
The concentration of ethanoic acid in vinegar is typically about 5% mass/volume (around 0.8 mol dm⁻³), which is too concentrated for direct titration with a standard 0.10 mol dm⁻³ NaOH solution. A precise dilution must be performed before the titration, usually by a factor of 10, to bring the acid concentration into a suitable range where the titre volume is large enough to minimise relative measurement uncertainty.
食醋中乙酸的浓度通常约为 5% 质量/体积（约 0.8 mol dm⁻³），浓度偏高，不适合直接用 0.10 mol dm⁻³ 的标准 NaOH 溶液滴定。滴定前必须准确稀释，通常稀释 10 倍，将酸浓度调节到合适范围，使滴定体积足够大，从而减小相对测量不确定度。
2. Experimental Procedure | 实验步骤
First, prepare the standard solution: a 0.100 mol dm⁻³ sodium hydroxide solution is provided, already standardised against a primary standard such as potassium hydrogenphthalate. Record the exact concentration from the label. Using a 25.0 cm³ volumetric pipette, transfer exactly 25.0 cm³ of the vinegar into a 250 cm³ volumetric flask. Make up to the mark with deionised water, stopper and invert several times to mix thoroughly. This gives a ten-fold diluted vinegar sample. Rinse the pipette with the diluted solution, then transfer 25.0 cm³ aliquots of this dilution into a clean 250 cm³ conical flask. Add 2–3 drops of phenolphthalein indicator. Fill a 50 cm³ burette with the standard NaOH solution, ensuring the tip is filled and no air bubbles are present. Record the initial burette reading to the nearest 0.05 cm³. Titrate the diluted vinegar solution with swirling, adding NaOH dropwise near the endpoint until a faint permanent pink colour appears. Record the final reading. Perform a rough trial first to estimate the endpoint volume, then carry out three accurate titrations, refilling the burette and recording all data in a suitable table.
首先准备标准溶液：提供已用邻苯二甲酸氢钾等基准物质标定好的 0.100 mol dm⁻³ 氢氧化钠溶液，记录标签上的准确浓度。用 25.0 cm³ 移液管准确移取 25.0 cm³ 食醋至 250 cm³ 容量瓶中，加去离子水定容至刻度，塞紧瓶塞并反复倒转混匀，得到稀释 10 倍的醋样。用该稀释液润洗移液管后，准确移取 25.0 cm³ 等分试样至洁净的 250 cm³ 锥形瓶中，加入 2–3 滴酚酞指示剂。向 50 cm³ 滴定管中加入标准 NaOH 溶液，确保管尖充满液体且无气泡。记录初始读数，精确到 0.05 cm³。边摇动边用稀释醋液滴定，接近终点时逐滴加入，直至出现微弱的永久性粉红色。记录终读数。先进行一次粗滴定估计终点体积，然后重新装液，进行三次精确滴定，将所有数据记录在合适的表格中。
3. Data and Observations | 数据与观察
An example data set from a Year 12 student is presented in the table below. The initial and final burette readings and the calculated titre volumes are recorded. Note that the rough titration is used only as a guide; the three subsequent accurate runs are used for averaging.
下面展示了一组来自 Year 12 学生的示例数据，表中记录了滴定管的初始读数、末读数以及计算出的滴定体积。请注意，粗滴定仅供预估终点体积，随后的三次精确滴定用于取平均值。
| Titration | Initial reading / cm³ | Final reading / cm³ | Volume of NaOH used / cm³ |
|---|---|---|---|
| Rough | 0.00 | 24.30 | 24.30 |
| 1 (accurate) | 0.00 | 24.00 | 24.00 |
| 2 (accurate) | 0.10 | 24.15 | 24.05 |
| 3 (accurate) | 0.05 | 24.10 | 24.05 |
. The average volume of NaOH used is therefore (24.00 + 24.05 + 24.05) / 3 = 24.033 cm³, which is rounded to 24.03 cm³ for calculations. Observations during the titration: the diluted vinegar was initially colourless; near the endpoint, the pink colour from the phenolphthalein appeared momentarily and then disappeared on swirling; finally, one drop of NaOH produced a faint permanent pink colour that persisted for at least 30 seconds.
三次精确滴定体积非常接近，分别为 24.00 cm³、24.05 cm³ 和 24.05 cm³。粗滴定值（24.30 cm³）不参与平均值计算。因此，所用 NaOH 的平均体积为 (24.00 + 24.05 + 24.05) / 3 = 24.033 cm³，计算时取 24.03 cm³。滴定过程中的观察：稀释后的醋液最初为无色；临近终点时，酚酞的粉红色在摇动中瞬间出现又消失；最终，一滴 NaOH 使溶液呈现可维持至少 30 秒的淡永久性粉红色。
4. Calculations | 计算
The calculation sequence follows the standard mole method. First, calculate the amount (in moles) of sodium hydroxide delivered from the burette.
计算步骤按标准摩尔法进行。首先，计算从滴定管放出的氢氧化钠的物质的量。
n(NaOH) = c(NaOH) × V(NaOH)
OH) = 0.100 mol dm⁻³, we obtain n(NaOH) = 0.100 mol dm⁻³ × 0.02403 dm³ = 2.403 × 10⁻³ mol.
使用准确滴定的平均体积，单位为 dm³：V = 24.03 cm³ = 0.02403 dm³。已知 c(NaOH) = 0.100 mol dm⁻³，可得 n(NaOH) = 0.100 mol dm⁻³ × 0.02403 dm³ = 2.403 × 10⁻³ mol。
n(CH₃COOH) = n(NaOH)
centration of the diluted vinegar is then found by dividing the amount by the aliquot volume in dm³.
由于反应计量比为 1:1，因此所取 25.0 cm³ 稀释醋样中的乙酸物质的量也为 2.403 × 10⁻³ mol。稀释醋液的浓度可通过物质的量除以等分试样的体积（dm³）求得。
c(diluted) = n / V(aliquot) = 2.403 × 10⁻³ mol / 0.0250 dm³ = 0.09612 mol dm⁻³
The original undiluted vinegar was prepared by a ten-fold dilution, so its concentration is 10 times greater.
原始未稀释的食醋经 10 倍稀释而成，因此其浓度为稀释液的 10 倍。
c(original) = 0.09612 mol dm⁻³ × 10 = 0.9612 mol dm⁻³
tration to mass concentration (g dm⁻³) using the molar mass of CH₃COOH, M = 60.05 g mol⁻¹.
为了以更贴近消费者的形式表示乙酸含量，可利用摩尔质量 M(CH₃COOH) = 60.05 g mol⁻¹，将摩尔浓度转换为质量浓度（g dm⁻³）。
Mass concentration = c × M = 0.9612 mol dm⁻³ × 60.05 g mol⁻¹ = 57.73 g dm⁻³
Assuming the density of the vinegar is very close to that of water (1 g cm⁻³), this corresponds to an ethanoic acid content of approximately 5.77% w/v, which is within the expected range for commercial vinegar (typically 4–8%).
假设食醋的密度与水相近（1 g cm⁻³），则该结果相当于乙酸的体积质量百分含量约为 5.77% w/v，处于市售食醋的预期范围（通常为 4–8%）之内。
5. Uncertainty Analysis | 不确定度分析
For the 25.0 cm³ volumetric pipette used to transfer the diluted vinegar, the manufacturer' erance is typically ±0.06 cm³. The percentage uncertainty is therefore (0.06 / 25.0) × 100% = 0.24%.
每一次测量都带有不确定度，AQA 要求考生能够识别并合成仪器不确定度。用于移取稀释醋液的 25.0 cm³ 移液管，制造商标称允差通常为 ±0.06 cm³，因此百分比不确定度为 (0.06 / 25.0) × 100% = 0.24%。
reading. Since a titre volume is obtained from the difference between two readings (final – titre of 24.03 cm³, the percentage uncertainty in the titre volume is (0.10 / 24.03) × 100% = 0.416% ette contributions (both random and systematic effects are considered additive here): 0.24% + 0.416% ≈ 0.66% e used to estimate the absolute uncertainty in the final ethanoic acid concentration: 0.66% ⁻³ is about 0.0063 mol dm⁻³. Hence, the concentration can be reported as 0.961 ± 0.006 mol
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