A-Level Chemistry Unit 3 June 2022: Mastering the Titration Experiment | A-Level 化学 Unit 3 2022年6月：掌握滴定实验

📚 A-Level Chemistry Unit 3 June 2022: Mastering the Titration Experiment | A-Level 化学 Unit 3 2022年6月：掌握滴定实验

The June 2022 Edexcel A-Level Chemistry Unit 3 paper remains one of the most talked‑about assessments of practical skills. It challenged students with a classic two‑stage titration task: standardising hydrochloric acid against a primary standard solution of anhydrous sodium carbonate, then using that standardised acid to determine the concentration of a sodium hydroxide solution. This article unpacks every critical step of that experiment, from making up the standard solution to calculating uncertainties, so you can master the practical techniques and numerical reasoning required.

2022年6月爱德思A‑Level化学Unit 3试卷至今仍是师生讨论最多的实验技能考核之一。它以一道经典的双段式滴定题考查学生：先用无水碳酸钠基准溶液标定盐酸，再用标定后的酸测定氢氧化钠溶液的浓度。本文拆解该实验的每一个关键环节——从配制标准溶液到计算不确定度——帮助你彻底掌握所需的技术操作与数据推理。

1. The Context of the June 2022 Unit 3 Paper | Jun22 Unit 3 试卷背景

Edexcel’s Unit 3 (Practical Skills in Chemistry I) in June 2022 was structured around three main questions, with Question 1 worth the largest share of marks. It required candidates to describe and evaluate the preparation of a standard solution of Na₂CO₃, perform a series of titrations with HCl, and then use their acid titre to find the concentration of NaOH. The question also probed understanding of indicator choice, concordant results, percentage uncertainty, and common procedural errors.

爱德思Unit 3（化学实验技能I）2022年6月试卷围绕三道大题展开，其中第1题占分最重。题目要求学生描述并评价Na₂CO₃标准溶液的配制，完成一系列与HCl的滴定，再运用酸滴定数据求算NaOH的浓度。此外，还深入考查了指示剂选择、平行结果的一致性、百分数不确定度以及常见的操作错误。

2. Overview of the Core Titration Chemistry | 核心滴定化学概述

The titration sequence hinges on two reactions. First, the standardisation of HCl:

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

滴定序列依赖于两个反应。首先是盐酸的标定：

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Second, the assay of the unknown NaOH solution:

HCl + NaOH → NaCl + H₂O

其次是对未知NaOH溶液的测定：

HCl + NaOH → NaCl + H₂O

Both are neutralisation reactions, but the first involves a carbonate that reacts with acid in a 1:2 molar ratio. Methyl orange is the indicator of choice for the Na₂CO₃/HCl titration because the end‑point occurs at a pH close to 4, avoiding the carbonic acid/bicarbonate buffer region. The NaOH/HCl titration can use either phenolphthalein or methyl orange.

两者都是中和反应，但第一个涉及碳酸盐，其与酸的摩尔比为1:2。Na₂CO₃/HCl滴定选用甲基橙作指示剂，因为终点出现在pH≈4的区域，避开了碳酸/碳酸氢盐缓冲区。NaOH/HCl滴定则可使用酚酞或甲基橙。

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3. Preparing the Standard Solution of Na₂CO₃ | 制备碳酸钠标准溶液

1. Accurately weigh about 1.30 g of anhydrous sodium carbonate using a balance reading to 0.001 g. Record the exact mass (e.g., 1.325 g).

1. 用精度为0.001 g的天平准确称取约1.30 g无水碳酸钠，并记录确切质量（如1.325 g）。

2. Transfer the solid to a 100 cm³ beaker, add about 50 cm³ of deionised water, and stir until fully dissolved.

2. 将固体转移至100 cm³烧杯中，加入约50 cm³去离子水，搅拌至完全溶解。

3. Using a funnel, pour the solution into a 250 cm³ volumetric flask. Rinse the beaker and funnel several times with deionised water and add the washings to the flask.

3. 借助漏斗将溶液转入250 cm³容量瓶。用去离子水反复洗涤烧杯和漏斗，洗涤液一并转入容量瓶。

4. Add deionised water until the bottom of the meniscus sits exactly on the graduation mark. Stopper the flask and invert it 10–15 times to homogenise the solution.

4. 加去离子水至弯月面底部恰好与刻度线相切。塞好瓶塞，翻转摇匀10–15次。

5. Calculate the concentration of the standard Na₂CO₃ solution. For 1.325 g in 250 cm³ (0.250 dm³), using M(Na₂CO₃) = 105.99 g mol⁻¹:

c(Na₂CO₃) = (1.325 g ÷ 105.99 g mol⁻¹) ÷ 0.250 dm³ = 0.0500 mol·dm⁻³

5. 计算Na₂CO₃标准溶液的浓度。以1.325 g溶于250 cm³（0.250 dm³）为例，Na₂CO₃摩尔质量为105.99 g mol⁻¹：

c(Na₂CO₃) = (1.325 g ÷ 105.99 g mol⁻¹) ÷ 0.250 dm³ = 0.0500 mol·dm⁻³

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4. Setting Up the Burette and Pipette | 安装滴定管和移液管

Rinse a 50 cm³ burette first with deionised water, then with a small portion of the HCl solution. Fill the burette above the zero mark, remove the funnel, and run some liquid through the tip to expel air bubbles. Record the initial burette reading to the nearest 0.05 cm³.

先用去离子水、再用少量盐酸溶液润洗50 cm³滴定管。将滴定管装液至零刻度以上，取出漏斗，放出一部分溶液以排出尖嘴中的气泡。记录初始读数，估计至0.05 cm³。

Rinse a 25 cm³ pipette with deionised water and then with the Na₂CO₃ standard solution. Pipette exactly 25.0 cm³ of the standard solution into a clean conical flask. Add 3–4 drops of methyl orange indicator.

用去离子水及Na₂CO₃标准溶液先后润洗25 cm³移液管。准确移取25.0 cm³标准溶液至洁净锥形瓶中，加入3–4滴甲基橙指示剂。

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5. Performing the Rough Titration | 进行粗滴定

Place the conical flask on a white tile under the burette tip. Swirl the flask while running the acid from the burette. When the colour starts to change from yellow to orange, slow down. The rough end‑point is the first permanent orange‑pink colour. Record the final burette reading. This rough titre gives an estimate of the volume needed; it is not used in the calculation of the mean.

将锥形瓶置于滴定管尖嘴下方的白瓷砖上。一边摇晃锥形瓶，一边从滴定管放出酸液。当颜色开始由黄变橙时减慢流速。粗滴定的终点是第一次出现的持久橙粉色。记录终点读数。粗滴定体积仅用于预估所需体积，不参与平均值的计算。

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6. Accurate Titrations and Achieving Concordancy | 精确滴定与平行的一致性

Refill the burette and record a new initial reading. Repeat the titration, but as you approach the expected end‑point (about 2 cm³ before the rough titre), add acid drop by drop. For the final drops, allow a fraction of a drop to hang from the tip and wash it into the flask with deionised water. The end‑point is taken when the solution just turns orange and does not revert upon swirling.

重新装液，记录新初始读数。重复滴定，但在接近预期终点（约低于粗滴定值2 cm³时）逐滴加入。最后几滴可让半滴悬挂在尖嘴处，用去离子水冲入瓶内。当溶液刚好变为橙色且摇匀后不褪色，即为终点。

Continue performing titrations until three titres agree within 0.10 cm³ of each other. For the June 2022 paper, candidates then selected the concordant titres (e.g., 24.80, 24.75, 24.85 cm³) and calculated the mean (24.80 cm³).

继续进行滴定，直至三次滴定读数彼此相差在0.10 cm³以内。2022年6月试卷中，考生需选出平行一致的滴定值（如24.80、24.75、24.85 cm³）并计算平均值（24.80 cm³）。

Titration	初始读数/cm³	终点读数/cm³	体积/cm³
Rough	0.00	25.10	25.10
1	0.15	24.95	24.80
2	0.10	24.85	24.75
3	0.20	25.05	24.85

Concordant titres: 24.80, 24.75, 24.85 cm³ → Mean titre = 24.80 cm³

平行滴定值：24.80、24.75、24.85 cm³ → 平均体积 = 24.80 cm³

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7. Calculating Concentrations from Titration Data | 从滴定数据计算浓度

Step A – Concentration of HCl:

步骤A – 盐酸浓度：

n(Na₂CO₃) in 25.0 cm³ = 0.0500 mol·dm⁻³ × (25.0/1000) dm³ = 0.00125 mol

n(Na₂CO₃) = 0.0500 × 0.0250 = 0.00125 mol

From the equation, 1 mol Na₂CO₃ reacts with 2 mol HCl, so n(HCl) = 2 × 0.00125 mol = 0.00250 mol.

由方程式，1 mol Na₂CO₃与2 mol HCl反应，故 n(HCl) = 2 × 0.00125 mol = 0.00250 mol。

c(HCl) = n(HCl) / mean titre in dm³ = 0.00250 mol / 0.02480 dm³ = 0.101 mol·dm⁻³ (to 3 significant figures).

c(HCl) = 0.00250 ÷ 0.02480 = 0.101 mol·dm⁻³

Step B – Concentration of NaOH:

步骤B – NaOH浓度：

25.0 cm³ of the unknown NaOH solution was titrated with the 0.101 mol·dm⁻³ HCl. Suppose concordant titres gave a mean volume of 23.10 cm³.

n(HCl) used = 0.101 mol·dm⁻³ × (23.10/1000) dm³ = 0.00233 mol

n(HCl) = 0.101 × 0.02310 = 0.00233 mol

The 1:1 ratio gives n(NaOH) = 0.00233 mol in 25.0 cm³. Hence c(NaOH) = 0.00233 mol / 0.0250 dm³ = 0.0933 mol·dm⁻³ (3 s.f.).

1:1摩尔比下，n(NaOH) = 0.00233 mol（25.0 cm³中），因此 c(NaOH) = 0.00233 ÷ 0.0250 = 0.0933 mol·dm⁻³（三位有效数字）。

c(NaOH) = 0.0933 mol·dm⁻³

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8. Propagation of Errors and Uncertainty | 误差传递与不确定度

Instrument uncertainties for typical school apparatus:

• Balance: ±0.001 g (for a reading of 1.325 g, percentage uncertainty = (0.001/1.325)×100 = 0.0755%)
• 250 cm³ volumetric flask: ±0.15 cm³ → (0.15/250)×100 = 0.060%
• 25 cm³ pipette: ±0.06 cm³ → (0.06/25.0)×100 = 0.24%
• Burette: each reading ±0.05 cm³, so a titre difference involves two readings → absolute uncertainty ±0.10 cm³. For a mean titre of 24.80 cm³, percentage uncertainty = (0.10/24.80)×100 = 0.403%

学校常用仪器的仪器误差：

• 天平：±0.001 g（对于1.325 g读数，百分数不确定度 = (0.001/1.325)×100 = 0.0755%）
• 250 cm³容量瓶：±0.15 cm³ → 0.060%
• 25 cm³移液管：±0.06 cm³ → 0.24%
• 滴定管：每读数±0.05 cm³，一次滴定体积差包含两次读数，绝对不确定度±0.10 cm³。平均体积24.80 cm³时，百分数不确定度 = (0.10/24.80)×100 = 0.403%

总百分比不确定度（对HCl浓度）≈ √(0.0755² + 0.060² + 0.24² + 0.403²) ≈ 0.48%。该不确定度传递至NaOH浓度时还会叠加第二次滴定的0.43%（若NaOH滴定均值23.10 cm³，不确定度 ±0.10 cm³，百分数为0.433%），因此最终c(NaOH)的总不确定度约为 √(0.48² + 0.433²) ≈ 0.65% 。这为评价实验精度提供了定量依据。

Total percentage uncertainty for c(HCl) ≈ √(0.0755² + 0.060² + 0.24² + 0.403²) ≈ 0.48%. When propagated to the NaOH determination, the second titration adds a further 0.43% (assuming a titre around 23.10 cm³ with ±0.10 cm³), giving a combined uncertainty of about 0.65%. This quantitative justification is often required in the Unit 3 exam to discuss reliability.

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9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

1. Using a wet pipette without rinsing with the standard solution – this dilutes the aliquot and lowers the titre, leading to an overestimation of HCl concentration. Always rinse the pipette with the solution it will deliver.

1. 移液管未用标准溶液润洗——这会稀释分取液，使滴定体积减小，导致HCl浓度被高估。务必用待移取的溶液润洗移液管。

2. Adding several millilitres of indicator – methyl orange is a weak acid; too much indicator consumes titrant and shifts the end‑point. Use exactly 3‑4 drops.

2. 加入数毫升指示剂——甲基橙是弱酸，过量指示剂会消耗滴定剂并偏移终点。严格使用3–4滴。

3. Recording the titre before the colour becomes permanent – the end‑point must persist after swirling. In CO₂‑producing titrations, the colour may fade slowly due to dissolved CO₂; gently boiling the solution briefly before the final approach can help.

3. 在颜色尚未持久时记录滴定体积——终点必须在摇匀后不褪色。产生CO₂的滴定中，因溶解的CO₂可能导致颜色缓慢消退；可在接近终点时短暂微沸溶液以驱除CO₂。

4. Failing to note initial and final burette readings to 2 decimal places – always record to the nearest 0.05 cm³, e.g., 0.00, 24.80.

4. 未将滴定管初、终读数记录至小数点后两位——始终估读到0.05 cm³，如0.00、24.80。

5. Using the rough titre in the mean – the rough titre is only a guide. Mark schemes reject means that include the rough trial.

5. 将粗滴定体积用于平均值计算——粗滴定仅为参考，评分标准不接受包含粗滴定的平均值。

6. Distilled water in the conical flask – adding water does not change the number of moles of the aliquot, so it does not affect the titre, yet many students wrongly think it contaminates the sample. Clarifying this misconception is essential.

6. 锥形瓶中的蒸馏水——加水不会改变分取液中的物质的量，因此不影响滴定体积，但许多学生误认为会污染样品。澄清

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