📚 Year 10 OCR Chemistry: Case Study Practical Exercises | Year 10 OCR 化学：案例分析实战演练
Case studies bring together your knowledge of chemical principles, practical techniques, and data analysis – exactly the skills tested in OCR GCSE Chemistry. By working through real-world experiments with unexpected results, you learn to spot errors, suggest improvements, and calculate precisely.
案例分析将化学原理、实验技术和数据分析融为一体——这正是 OCR 化学考试所考查的技能。通过分析出现意外结果的真实实验，你将学会发现错误、提出改进并进行精确计算。
1. Why Case Studies Matter | 案例分析的重要性
OCR exam papers assess AO3 (Analysis and Evaluation) by asking you to comment on experimental data, identify variables, and explain sources of error. Regular case-study practice transforms a set of numbers into a scientific argument.
OCR 试卷通过让你评论实验数据、识别变量和解释误差来源来考核 AO3（分析与评估）。经常练习案例分析能将一组数据转化为科学论证。
Every case study reinforces quantitative skills – converting units, using mole ratios, and calculating percentage yields. It also trains you to think about the reliability of methods and how to control risks.
每个案例都强化定量技能——单位换算、运用摩尔比和计算产率。它还能训练你思考方法的可靠性以及如何控制风险。
2. Case Study 1: Water Purity Investigation | 案例一：水纯度调查
62 g. After adding 50.0 cm³ of sample and heating to dryness, the dish + residue weighed 45.74 g. The expected dissolved-solids content for safe water is 0.5 g/dm³.
一名学生检测河水样品中的溶解固体。她记录：空蒸发皿质量 = 45.62 g。加入 50.0 cm³ 样品并加热蒸干后，蒸发皿 + 残渣质量为 45.74 g。安全用水的预期溶解固体含量为 0.5 g/dm³。
Mass of solid = 45.74 – ntal concentration = 0.12 g / 0.050 dm³ = 2.4 g/dm³, almost five times the expected value.
固体质量 = 45.74 – 45.62 = 0.12 g。所用体积 = 0.050 dm³。实验浓度 = 0.12 g / 0.050 dm³ = 2.4 g/dm³，几乎是预期值的五倍。
Likely errors: the dish was not heated to constant mass – some water remained; solid splashed out during vigorous boiling; or the dish was not cleaned, adding extra mass.
可能的误差：蒸发皿未加热至恒重——仍有水分残留；猛烈沸腾时固体溅出；或蒸发皿未清洁干净，增加了额外质量。
.01 g. Use a steam bath instead of a direct flame. Run a blank trial with distilled water.
改进：反复加热、冷却、称量，直到连续两次质量偏差在 ±0.01 g 以内。使用水浴代替直接火焰。用蒸馏水做空白试验。
Calculating percentage error: (2.4 – 0.5) / 0.5 × 100% = 380%. This large error highlights a systematic flaw, not just random scatter.
计算百分误差：(2.4 – 0.5) / 0.5 × 100% = 380%。这么大的误差表明存在系统缺陷，而不仅仅是随机波动。
3. Case Study 2: Acid-Base Titration | 案例二：酸碱滴定
lein. Burette readings: initial 0.50 cm³, final 24.80 cm³. Volume of HCl used = 24.30 cm³.
一名学生用 0.100 mol/dm³ 盐酸滴定 25.0 cm³ 氢氧化钠溶液，以酚酞为指示剂。滴定管读数：初 0.50 cm³，终 24.80 cm³。盐酸用量 = 24.30 cm³。
Moles of HCl = 0.100 × (24.30/1000) = 0.00243 mol
NaOH + HCl → NaCl + H₂O, so moles NaOH = 0.00243 mol
Concentration of NaOH = 0.00243 / 0.0250 = 0.0972 mol/dm³
The result, 0.0972 mol/dm³, is lower than the expected 0.100 mol/dm³. One error could be that the pipette used for NaOH was wet with water, diluting the base before titration.
结果为 0.0972 mol/dm³，低于预期的 0.100 mol/dm³。一个可能的误差是用于量取 NaOH 的移液管沾有水，在滴定前稀释了碱液。
Another error: the student may have stopped the titration too early, before the pale pink colour persisted for 30 seconds. This would reduce the titre volume and give a lower calculated base concentration.
另一个误差：学生可能过早结束滴定，未能让淡粉红色持续 30 秒。这会减少滴定管读数，使计算所得的碱浓度偏低。
Improvements: rinse the pipette with the NaOH solution before use. Place a white tile under the flask to judge colour change accurately. Repeat titres until two readings agree within 0.10 cm³.
改进：在使用前用 NaOH 溶液润洗移液管。在锥形瓶下放白瓷砖以便准确判断颜色变化。重复滴定，直到两次读数之差在 0.10 cm³ 以内。
4. Case Study 3: Electrolysis of Copper(II) Sulfate | 案例三：硫酸铜电解
5.55 g (gain 0.35 g). Initial mass of anode = 45.80 g, final mass = 45.44 g (loss 0.36 g).
在一次电解实验中，将铜电极放入 CuSO₄ 溶液。学生记录：阴极初始质量 =
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