📚 Year 12 CIE Chemistry: Case Study Practical Exercises | CIE 化学：案例分析实战演练
Case studies in A-Level Chemistry bridge the gap between textbook theory and real laboratory decision-making. This article walks you through a worked practical scenario: determining the purity of a chalk sample via two independent methods. You will see how to handle raw data, perform multi‑step calculations, evaluate reliability, and link findings to core topics such as stoichiometry, acid‑base titrations, and gas laws.
A-Level 化学中的案例分析能将课本理论与实验室的实际决策联系起来。本文将通过一个完整的实践情景——用两种独立方法测定粉笔样品的纯度——带你处理原始数据、进行多步计算、评估可靠性，并将结果与计量学、酸碱滴定和气体定律等核心主题挂钩。
1. The Scenario – Analysing Chalk Purity | 情景设定 – 分析粉笔纯度
A student is given a stick of white chalk and asked to determine its percentage by mass of calcium carbonate, CaCO₃. The teacher suggests two possible approaches: a back titration with standard hydrochloric acid and sodium hydroxide, and a gas collection experiment in which the carbon dioxide evolved is measured. The student obtains the following raw materials: approximately 0.5 g samples of the ground chalk, 0.500 mol dm⁻³ HCl, 0.100 mol dm⁻³ NaOH, standard laboratory glassware and a gas syringe.
一位学生拿到一支白色粉笔，需要测定其中碳酸钙 (CaCO₃) 的质量百分含量。老师建议了两种可行方法：用标准盐酸和氢氧化钠进行返滴定，以及通过收集所释放的二氧化碳气体进行测量。学生得到以下原料：约 0.5 g 磨细的粉笔样品、0.500 mol dm⁻³ HCl、0.100 mol dm⁻³ NaOH、标准实验室玻璃器皿和一支气体注射器。
2. The Chemistry Behind the Methods | 方法背后的化学原理
Chalk reacts with hydrochloric acid according to the equation:
粉笔与盐酸的反应方程式如下：
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
In a back titration, an excess of HCl is added to the sample and the unreacted acid is determined by titration with standard NaOH. The difference between the initial and remaining amount of HCl gives the quantity that actually reacted with CaCO₃. In the gas collection method, the CO₂ released is captured in a gas syringe; using the molar volume of a gas at room temperature and pressure (RTP, 24.0 dm³ mol⁻¹) the amount of carbonate can be calculated directly.
返滴定中，先向样品加入过量 HCl，再用标准 NaOH 滴定未反应的酸。初始 HCl 量与剩余量之差即为与 CaCO₃ 实际反应的量。气体收集法里，释放的 CO₂ 被收集在气体注射器中；利用室温常压 (RTP) 下气体摩尔体积 (24.0 dm³ mol⁻¹) 可直接算出碳酸盐的物质的量。
3. Method 1 – Back Titration in Detail | 方法一 – 详细返滴定
The student accurately weighs 0.500 g of the powdered chalk and transfers it into a conical flask. 50.0 cm³ of 0.500 mol dm⁻³ HCl is added from a volumetric pipette. After the effervescence ceases, the mixture is carefully transferred to a 250 cm³ volumetric flask and made up to the mark with deionised water. A 25.0 cm³ aliquot is then titrated against 0.100 mol dm⁻³ NaOH using phenolphthalein indicator. The procedure is repeated until three concordant titres are obtained; the mean titre is 17.20 cm³.
学生准确称量 0.500 g 磨细的粉笔并移入锥形瓶。用移液管加入 50.0 cm³ 0.500 mol dm⁻³ HCl。待泡腾停止后，将混合物小心转移至 250 cm³ 容量瓶中，用去离子水定容。然后取 25.0 cm³ 等分试样，以酚酞为指示剂，用 0.100 mol dm⁻³ NaOH 滴定。重复操作至获得三次吻合滴定值，平均滴定体积为 17.20 cm³。
4. Data and Calculations for Back Titration | 返滴定数据与计算
First, calculate the amount of HCl that remained in the 25.0 cm³ aliquot:
首先，计算 25.0 cm³ 等分试样中剩余的 HCl 量：
n(HCl) in aliquot = c(NaOH) × V(NaOH) = 0.100 mol dm⁻³ × 0.01720 dm³ = 0.00172 mol
This value is then scaled up to the full 250 cm³ volumetric flask:
然后将此值换算对至整个 250 cm³ 容量瓶：
n(HCl) total remaining = 0.00172 mol × (250 cm³ / 25.0 cm³) = 0.0172 mol
The amount of HCl originally added to the chalk was:
最初加入粉笔的 HCl 量为：
n(HCl) initial = 0.500 mol dm⁻³ × 0.0500 dm³ = 0.0250 mol
The amount that reacted with CaCO₃ is therefore:
因此，与 CaCO₃ 反应的 HCl 量为：
n(HCl) reacted = 0.0250 mol − 0.0172 mol = 0.00780 mol
From the stoichiometry, 2 moles of HCl react with 1 mole of CaCO₃, so:
根据计量关系，2 mol HCl 与 1 mol CaCO₃ 反应，故：
n(CaCO₃) = 0.00780 mol ÷ 2 = 0.00390 mol
Using the molar mass of CaCO₃ (100.1 g mol⁻¹), the mass of pure carbonate is:
利用 CaCO₃ 的摩尔质量 (100.1 g mol⁻¹)，纯碳酸盐的质量为：
mass(CaCO₃) = 0.00390 mol × 100.1 g mol⁻¹ = 0.390 g
The percentage purity by mass is therefore (0.390 g / 0.500 g) × 100% = 78.1%. The remainder of the sample consists of inert binders and impurities.
由此质量百分纯度为 (0.390 g / 0.500 g) × 100% = 78.1%。样品其余部分为惰性粘结剂和杂质。
5. Method 2 – Gas Collection Experiment | 方法二 – 气体收集实验
The student uses a second 0.500 g portion of the same chalk powder. It is placed in a side‑arm flask connected to a gas syringe. Excess 2 mol dm⁻³ HCl is added through a dropping funnel, and the plunger movement is recorded until no further gas is evolved. After allowing the apparatus to cool back to room temperature and adjusting for atmospheric pressure, the final volume of CO₂ collected is 76.5 cm³.
学生取同一粉笔粉末的另一份 0.500 g 样品，置于连接气体注射器的支管烧瓶中。通过滴液漏斗加入过量 2 mol dm⁻³ HCl，记录活塞移动直至不再产生气体。待装置冷却至室温并校正大气压强后，最终收集到 CO₂ 体积为 76.5 cm³。
6. Calculations from Gas Volume Data | 气体体积数据计算
Under RTP conditions, 1 mole of any gas occupies 24.0 dm³. Convert the volume to dm³ and calculate the amount of CO₂:
在 RTP 条件下，1 mol 任何气体的体积为 24.0 dm³。将体积转换为 dm³ 并计算 CO₂ 的量：
n(CO₂) = 0.0765 dm³ ÷ 24.0 dm³ mol⁻¹ = 0.00319 mol
The reaction equation shows a 1:1 ratio between CaCO₃ and CO₂, so n(CaCO₃) is also 0.00319 mol. The corresponding mass is:
反应方程式显示 CaCO₃ 与 CO₂ 的化学计量比为 1:1，因此 n(CaCO₃) 亦为 0.00319 mol。对应的质量为：
mass(CaCO₃) = 0.00319 mol × 100.1 g mol⁻¹ = 0.319 g
This gives a purity of (0.319 g / 0.500 g) × 100% = 63.8%, substantially lower than the value obtained by back titration.
由此纯度为 (0.319 g / 0.500 g) × 100% = 63.8%，明显低于返滴定所得数值。
7. Comparing the Two Purity Results | 两种纯度结果比较
The back titration suggests the chalk is about 78% CaCO₃, while the gas collection method indicates only 64%. The large discrepancy immediately prompts a critical evaluation. Back titrations tend to be more reliable for carbonate samples because the technique avoids loss of the measured species: the excess acid is retained in solution. Gas collection, on the other hand, is vulnerable to CO₂ dissolving in water, leaking from connections, or incomplete flushing of the gas syringe. The lower purity from the gas method can be rationalised by such losses, which lead to a smaller volume being recorded and hence a reduced calculated mass of carbonate.
返滴定结果显示粉笔中 CaCO₃ 含量约 78%，而气体收集法仅 64%。这种巨大的差异立即引发批判性评估。返滴定对碳酸盐样品通常更可靠，因为该技术避免了被测物种的损失：过量的酸始终留存于溶液中。另一方面，气体收集法容易因 CO₂ 溶于水、连接处漏气或气体注射器未完全排空而产生误差。气体法纯度偏低可由这些损失解释，它们导致记录体积偏小，进而使算出的碳酸盐质量偏低。
8. Identifying Key Sources of Error | 识别主要误差来源
For the back titration, the main sources of uncertainty are the precision of the volumetric glassware, the judgement of the indicator end‑point, and potential loss of acid during the transfer to the volumetric flask. However, through careful technique and repetition, these can be minimised. The gas collection method has more fundamental flaws: carbon dioxide is fairly soluble in water, so some gas remains dissolved in the reaction mixture; the bung and tubing may not be completely airtight; and the gas syringe may stick or need correction for temperature differences between the reaction and the room. All of these factors systematically reduce the recorded volume.
返滴定中，不确定度的主要来源是容量玻璃器皿的精密度、指示剂终点的判断以及转移至容量瓶过程中酸的潜在损失。但通过细致操作和重复，这些影响可降到最低。气体收集法存在的缺陷则更为根本：二氧化碳在水中溶解度较高，部分气体会留在反应混合液中；橡皮塞和管道可能不完全气密；气体注射器可能卡滞或需校正反应温度与室温的差异。所有这些因素都会系统地降低记录体积。
9. Practical Improvements and Refinements | 实际改进与优化
To improve the back titration, a student could rinse the conical flask several times during the transfer to ensure all dissolved HCl enters the volumetric flask. Using a more dilute NaOH solution for the final titration would require a larger titre, reducing the relative uncertainty in reading the burette. For the gas collection, one effective modification is to saturate the acid with carbon dioxide before the experiment so that less product gas dissolves during the reaction. A gas burette or inverted measuring cylinder over water (with a delivery tube) can also be used, but the water must be pre‑saturated with CO₂. Repeating the measurement with a cold‑finger condenser to return water vapour to the flask would also improve accuracy.
改进返滴定时，学生可在转移过程中多次冲洗锥形瓶，确保所有溶解的 HCl 都进入容量瓶。最后滴定时使用更稀的 NaOH 溶液会使滴定体积增大，从而降低读取滴定管的相对不确定度。对于气体收集，一个有效的改进是实验前用二氧化碳将盐酸预饱和，以减少反应中产物的溶解损失。也可以使用气量管或将倒置量筒通过导管排水集气，但所用的水须预通 CO₂ 饱和。加用冷指冷凝器使水蒸气回流至烧瓶内重复测量也可提高准确度。
10. Extending the Case – Exam-Style Questions | 拓展案例 – 考试题型
This case study can be extended with typical CIE examination tasks. Consider the following: (a) Explain why it is necessary to dilute the reaction mixture before performing the aliquot titration. (b) A student suggests that the gas syringe result is more accurate because it avoids the use of an indicator. Comment on this claim using the data above. (c) If the chalk were 100% CaCO₃, what volume of CO₂ would be expected from 0.500 g of sample? (d) The teacher proposes a third method: heating the chalk strongly and measuring the mass loss due to decomposition of CaCO₃. Write the equation for the thermal decomposition and state one advantage of this approach. Thinking through such questions reinforces your understanding of experimental design, quantitative reasoning, and the evaluation of uncertainties.
此案例可拓展出典型的 CIE 考试题。思考以下问题：(a) 解释为什么在取等分滴定前必须稀释反应混合物。(b) 有学生认为气体注射器结果更准确，因为它避免使用指示剂。请结合上述数据评述这一说法。(c) 若粉笔为 100% CaCO₃，0.500 g 样品预期能收集到多少体积的 CO₂？(d) 老师提出第三种方法：将粉笔强热并测量因 CaCO₃ 分解造成的质量损失。写出热分解方程式并说明该方法的一个优点。思考这些问题能强化你对实验设计、量化推理和不确定度评估的理解。
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