📚 KS3 CIE Chemistry: Case Study Practice | KS3 CIE 化学：案例分析实战演练
Case studies bring chemistry to life by showing how concepts work in real scenarios. In KS3 CIE Chemistry, you are often asked to apply your knowledge to practical situations – whether it is separating a mixture, analysing experimental data or explaining an environmental problem. This article walks you through a series of carefully chosen case studies covering core topics: separation techniques, solubility, acids and bases, reactivity of metals, gases, rusting, extraction, energy changes and water treatment. Each case includes a clear problem, data where needed, step‑by‑step analysis and the chemical reasoning behind the answers. By working through these examples, you will sharpen your investigative skills and learn how to approach unfamiliar contexts with confidence.
案例分析让化学变得生动，展示概念在真实情境中如何应用。在 KS3 CIE 化学中，你常需要将知识应用于实际场景——无论是分离混合物、分析实验数据还是解释环境问题。本文将通过一系列精选案例，带你演练分离技术、溶解度、酸碱、金属反应性、气体、生锈、提取、能量变化和水处理等核心主题。每个案例都包含明确的问题、必要的数据、逐步分析以及答案背后的化学原理。完成这些练习后，你的探究能力将得到强化，学会如何自信地应对陌生情境。
1. Understanding Case Studies in Chemistry | 理解化学中的案例分析
A case study in KS3 chemistry is a detailed description of a real‑life problem or experiment. It may include observations, numbers in a table or a diagram of apparatus. Your job is to identify the relevant science, interpret data, make predictions and suggest improvements. The key is not just to recall facts but to use them in a logical sequence. Start by reading the scenario carefully, highlight keywords (such as 'separate', 'acid', 'temperature', 'gas produced') and think about which topic from your syllabus connects to it.
在 KS3 化学中，案例研究是对一个真实问题或实验的详细描述。它可能包含观察记录、表格中的数字或装置图。你的任务是找出相关的科学原理、解释数据、作出预测并提出改进。关键在于不仅要回忆事实，还要用逻辑顺序运用它们。先仔细阅读情境，标出关键词（如“分离”“酸”“温度”“产生的气体”），然后思考它与你大纲中的哪个主题相关。
Most case study questions follow a pattern: the scenario presents a problem, some data or observations are given, and then you are asked to explain, calculate or design an experiment. Always link your answers to scientific ideas – for example, cite the particle model for changes of state, or the reactivity series for displacement reactions. If you are asked to suggest an improvement, think about accuracy, reliability and safety.
大多数案例分析题有固定模式：情境提出一个问题，给出一些数据或观察结果，然后要求你解释、计算或设计实验。回答时总要联系科学概念——例如，用粒子模型解释状态变化，或用反应性顺序解释置换反应。如果要求提出改进，考虑准确性、可靠性和安全性。
2. Case Study 1: Purifying Rock Salt | 案例分析 1：提纯岩盐
Scenario: A student collected some rock salt (a mixture of white sodium chloride and brown sand) from a beach. She wants to obtain pure, dry salt crystals. Describe a step‑by‑step method she can use, explaining the purpose of each step.
情境：一位同学从海滩收集了一些岩盐（白色氯化钠和棕色沙子的混合物）。她想得到纯净干燥的食盐晶体。请描述她可以采用的逐步方法，并解释每一步的目的。
Analysis: Rock salt is a mixture of soluble salt and insoluble sand. The correct sequence is: grinding, dissolving, filtering and evaporating. First, crush the rock salt with a mortar and pestle to increase the surface area so the salt dissolves faster. Then, add the crushed mixture to a beaker of warm water and stir well. Sodium chloride dissolves, but sand does not. Next, filter the mixture through filter paper in a funnel. The sandy residue stays on the paper, while the filtrate is salt solution (brine). Finally, heat the filtrate gently in an evaporating dish to evaporate the water. Slow evaporation gives larger crystals. To get dry crystals, pat them with filter paper or leave them in a warm place.
分析：岩盐是可溶性盐和不溶性沙子的混合物。正确顺序是：研磨、溶解、过滤、蒸发。首先用研钵研杵研碎岩盐，增大表面积使盐溶解更快。然后将研碎的混合物加入装有温水的烧杯中充分搅拌。氯化钠溶解，沙子不溶。接着用漏斗和滤纸过滤混合物。沙质残渣留在滤纸上，滤液为盐水。最后将滤液在蒸发皿中缓缓加热，蒸发水分。缓慢蒸发可得到较大晶体。要获得干燥晶体，可用滤纸吸干或放在温暖处晾干。
This case tests your knowledge of separation techniques based on solubility. The key idea is that filtration separates an insoluble solid from a liquid, while evaporation (or crystallisation) isolates a dissolved solid from a solution.
该案例考察基于溶解性的分离技术知识。核心思想是：过滤将不溶性固体从液体中分离，而蒸发（或结晶）则从溶液中析出溶解的固体。
3. Case Study 2: Investigating Solubility of Potassium Nitrate | 案例分析 2：探究硝酸钾的溶解度
Scenario: A laboratory investigation measured the mass of potassium nitrate (KNO₃) that dissolves in 100 g of water at different temperatures. The results are shown in the table:
情境：一项实验室研究测量了不同温度下能溶解在 100 g 水中的硝酸钾 (KNO₃) 质量。结果如下表：
| Temperature (°C) | Solubility (g/100 g water) |
|---|---|
| 0 | 13 |
| 20 | 32 |
| 40 | 64 |
| 60 | 110 |
| 80 | 170 |
Questions: (a) Plot a solubility curve. (b) Describe the trend. (c) Predict the solubility at 50 °C. (d) What mass of KNO₃ would crystallise if a saturated solution at 80 °C in 50 g water is cooled to 20 °C?
问题：(a) 绘制溶解度曲线。(b) 描述趋势。(c) 预测 50 °C 时的溶解度。(d) 若将 80 °C 下 50 g 水配制的饱和溶液冷却至 20 °C，会有多少克 KNO₃ 结晶析出？
rature. (b) As temperature increases, solubility increases; the relationship is not linear – e dissolved mass is 170 × (50/100) = 85 g. At 20 °C, 32 g in 100 g water gives 16 g in 50 g water. Mass crystallised = 85 – move faster, so more solute particles can be separated and surrounded by water molecules.
分析：(a) 曲线快速上升，表明 KNO₃ 溶解度随温度升高而增大。(b) 温度升高，溶解度增大；关系非直线——高温段曲线更陡。(c) 由曲线读得约 85 g/100 g 水。(d) 80 °C 时 100 g 水溶解 170 g，则 50 g 水溶解 170×(50/100)=85 g。20 °C 时 100 g 水溶解 32 g，故 50 g 水含 16 g。结晶质量 = 85 – 16 = 69 g。此题用到粒子模型：温度更高时粒子运动更快，更多溶质粒子可被水分子分离并包围。
4. Case Study 3: Neutralising Acidic Soil | 案例分析 3：中和酸性土壤
Scenario: A farmer finds that her soil has a pH of 4.5, which is too acidic for wheat to grow well. She plans to add a common substance to raise the pH to about 6.5. Identify a suitable substance, write a word and symbol equation for the reaction, and explain how the neutralisation works using the ion concept.
情境：一位农民测得土壤 pH 为 4.5，太酸不适宜小麦生长。她打算添加一种常见物质，使 pH 升至约 6.5。请推荐一种合适的物质，写出文字及符号方程式，并用离子概念解释中和过程。
rd equation: calcium carbonate + nitric acid → calcium nitrate + carbon dioxide + water. Symbol equation: CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l). In the soil, acids release H⁺ ions. Carbonate ions (CO₃²⁻) or hydroxide ions (OH⁻) from the added base react with H⁺ to form water and carbon dioxide. This removes excess H⁺, raising the pH. Limestone is cheap and slow‑acting, ideal for long‑term pH adjustment.
分析：农民可用石灰石粉（碳酸钙）或熟石灰（氢氧化钙）。碳酸钙与酸反应。文字方程式：碳酸钙 + 硝酸 → 硝酸钙 + 二氧化碳 + 水。符号方程式：CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l)。在土壤中，酸释放 H⁺ 离子。来自添加碱的碳酸根离子 (CO₃²⁻) 或氢氧根离子 (OH⁻) 与 H⁺ 反应生成水和二氧化碳。这样清除了过量的 H⁺，pH 上升。石灰石廉价且作用缓慢，适合长期调节土壤酸度。
This case links neutralisation, ionic equations and real‑world agriculture. It reinforces the idea that neutralisation is the reaction between H⁺ and OH⁻/carbonate ions.
该案例将中和反应、离子方程式与实际农业联系，巩固了中和反应是 H⁺ 与 OH⁻ 或碳酸根离子反应的概念。
5. Case Study 4: Identifying Unknown Gases | 案例分析 4：鉴别未知气体
Scenario: In the lab, three gas jars contain colourless gases collected from reactions. Jar A gas relights a glowing splint. Jar B gas burns with a squeaky pop. Jar C gas turns limewater milky. Name the gases and describe the chemical test that confirms each one.
情境：实验室中，三个集气瓶装有由反应制得的无色气体。瓶 A 气体使带火星木条复燃；瓶 B 气体燃烧发出尖锐爆鸣声；瓶 C 气体使石灰水变浑浊。说出这些气体的名称，并描述确认每种气体的化学检验方法。
The test: hold a lit splint at the mouth of the jar; hydrogen burns rapidly with a squeaky pop, forming water. Jar C is carbon dioxide. Test: bubble the gas through limewater; CO₂ reacts with calcium hydroxide to form insoluble calcium carbonate, turning the water milky. Equation: Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l).
分析：瓶 A 为氧气。检验：将带火星木条伸入气体中，木条复燃，因氧气助燃。瓶 B 是氢气。检验：在瓶口用点燃的木条测试，氢气急速燃烧发出爆鸣声，生成水。瓶 C 是二氧化碳。检验：将气体通入石灰水中，CO₂ 与氢氧化钙反应生成不溶的碳酸钙，使溶液变为浑浊。方程式：Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l)。
imewater turns cloudy → CO₂. Also, chlorine (which turns damp litmus paper red then white) and ammonia (turns red litmus blue) may appear.
此案例要求回忆标准气体检验方法。测验是特定的：带火星木条复燃 → O₂；燃着木条发出爆鸣声 → H₂；石灰水变浑浊 → CO₂。此外，氯气（使湿润红色石蕊试纸先变红后变白）和氨气（使红色石蕊试纸变蓝）也可能出现。
6. Case Study 5: Rusting of Iron – Investigation and Prevention | 案例分析 5：铁的生锈——探究与防护
Scenario: A student set up four test tubes. Tube 1: iron nail in dry air (with anhydrous calcium chloride). Tube 2: iron nail in boiled water (oil layer on top). Tube 3: iron nail half‑submerged in water (open to air). Tube 4: iron nail wrapped in copper foil, half‑submerged in water. After two weeks, only nails in tubes 3 and 4 showed rust. Explain the results and suggest why copper increased rusting.
情境：某同学准备了四支试管。试管 1：铁钉置于干燥空气中（放有无水氯化钙）。试管 2：铁钉放在煮沸过的水中（上盖油层）。试管 3：铁钉一半浸入水中（敞口）。试管 4：铁钉包裹铜箔，一半浸入水中。两周后，仅试管 3 和 4 的铁钉生锈。解释结果，并说明为何铜加快了生锈。
l prevents redissolution) → no rust. Tube 3 has water and oxygen → rust. Tube 4 has water, oxygen, and copper in contact with iron; copper is less reactive and acts as a cathode, causing iron to lose electrons more readily (sacrificial corrosion). Thus, iron rusts faster. Prevention methods include painting, oiling, galvanising (zinc coating) or using sacrificial protection.
分析：生锈需要氧气和水同时存在。试管 1 仅有干燥空气（无水）→ 不生锈。试管 2 有水但无氧气（煮沸赶走了空气，油层防止氧气重新溶解）→ 不生锈。试管 3 既有水又有氧气 → 生锈。试管 4 有水、氧气，且铜与铁接触；铜较不活泼，作阴极，使铁更易失去电子（牺牲腐蚀），因此铁锈得更快。防护方法包括涂漆、抹油、镀锌（锌保护层）或利用牺牲阳极保护。
This investigation highlights the conditions necessary for rusting and the principles of electrochemical corrosion. Students can design their own experiments to test protective methods.
此探究突出了锈蚀所需的条件和电化学腐蚀原理。学生可自行设计实验来测试各种防护方法。
7. Case Study 6: Extracting Copper from Malachite | 案例分析 6：从孔雀石中提取铜
Scenario: Malachite is a green mineral containing copper carbonate (CuCO₃). A chemist heats malachite strongly, producing black copper oxide and carbon dioxide. The black solid is then heated with charcoal powder to yield reddish‑brown copper metal. (a) Write word and symbol equations for both reactions. (b) Name the type of each reaction. (c) Suggest why charcoal is used instead of just heating copper oxide alone.
情境：孔雀石是一种含碳酸铜 (CuCO₃) 的绿色矿物。化学家强热孔雀石，产生黑色氧化铜和二氧化碳。然后黑色固体与木炭粉混合加热，得到红棕色金属铜。(a) 写出两步反应的文字及符号方程式。(b) 指出每个反应的类型。(c) 说明为何要用木炭，而不能仅靠加热氧化铜得到铜。
Analysis: Step 1: thermal decomposition – CuCO₃(s) → CuO(s) + CO₂(g). Step 2: reduction – ent: it removes oxygen from copper oxide. Heating copper oxide alone does not break the Cu– tive element (carbon) is needed to reduce the metal. This relates to the reactivity series – carbon is above copper, so it can displace copper from its oxide.
分析：第一步：热分解反应 – CuCO₃(s) → CuO(s) + CO₂(g)。第二步：还原反应 – 2CuO(s) + C(s) → 2Cu(s) + CO₂(g)。木炭（碳）作为还原剂，从氧化铜中夺氧。单独加热氧化铜不能打断 Cu–O 键；需要用更活泼的元素（碳）来还原金属。这与反应性顺序有关——碳在铜之上，故可从其氧化物中置换出铜。
This case shows the industrial principle of extracting metals. Low‑reactivity metals like copper can be produced by reduction with carbon, while highly reactive metals need electrolysis.
此案例展示了金属提取的工业原理。低活性金属如铜可用碳还原获得，而高活性金属则需要电解法制取。
8. Case Study 7: Energy Changes in Chemical Reactions | 案例分析 7：化学反应中的能量变化
Scenario: A student mixes 25 cm³ of hydrochloric acid (1 mol/dm³) with 25 cm³ of sodium hydroxide (1 mol/dm³) in a polystyrene cup. The temperature rises from 21 °C to 29 °C. A similar experiment with ammonium nitrate and water causes a temperature drop. (a) Classify each reaction as exothermic or endothermic. (b) Calculate the heat energy released (assume 4.2 J/g°C for the solution and total mass 50 g). (c) Draw a simple energy level diagram for neutralisation.
情境：某同学在聚苯乙烯杯中混合 25 cm³ 盐酸 (1 mol/dm³) 和 25 cm³ 氢氧化钠溶液 (1 mol/dm³)。温度从 21 °C 升至 29 °C。另一实验将硝酸铵溶于水，温度下降。(a) 判断每个反应是放热还是吸热。(b) 计算释放的热量（假设溶液比热容 4.2 J/g°C，总质量 50 g）。(c) 绘制中和反应的简单能级图。
Analysis: (a) Acid‑base neutralisation is exothermic (gives out heat) – temperature increases. Dissolving ammonium nitrate is endothermic (takes in heat) – temperature decreases. (b) Temperature change ΔT = 29 – i) at a higher energy level; products (salt + water) at a lower energy level; an arrow downwards showing energy released to the surroundings. Label activation energy and ΔH negative.
分析：(a) 酸碱中和是放热反应（释放热量）——温度升高。硝酸铵溶于水为吸热反应（吸收热量）——温度降低。(b) 温度变化 ΔT = 29 – 21 = 8 °C。释放能量 = 质量 × 比热容 × ΔT = 50 g × 4.2 J/g°C × 8 °C = 1680 J (1.68 kJ)。(c) 能级图：反应物（酸 + 碱）在较高能级；生成物（盐 + 水）在较低能级；向下的箭头表示向环境释放能量。标出活化能与负值 ΔH。
This case practises data handling and the link between observable temperature changes and bond breaking/forming.
该案例练习了数据处理，并将可观察的温度变化与化学键断裂和形成联系起来。
9. Case Study 8: Water Treatment Plant | 案例分析 8：水处理工厂
Scenario: Describe the steps used in a municipal water treatment plant to turn river water into safe drinking water, and explain the chemistry behind each step: screening, sedimentation, filtration, carbon filtration, and chlorination.
情境：描述市政水处理厂将河水净化为安全饮用水的步骤，并解释每步的化学原理：粗筛、沉降、砂滤、活性炭过滤和氯化消毒。
Analysis: (1) Screening: a metal grid removes large objects such as sticks and leaves. (2) Sedimentation: water flows into large tanks where aluminium sulfate (or iron sulfate) is added; tiny particles clump together (coagulation) and settle as sludge. (3) Filtration: water passes through beds of gravel and sand, removing finer suspended solids. (4) Carbon filtration: activated charcoal adsorbs organic compounds and chlorine, improving taste and removing odours. (5) Chlorination: a small amount of chlorine gas (or sodium hypochlorite) is added to kill bacteria and other pathogens. The chemistry: chlorine reacts with water to produce hypochlorous acid (HOCl), a powerful disinfectant. The addition of fluoride may also be considered for dental health. This case tests knowledge of separation and chemical treatment processes.
分析：(1) 粗筛：金属栅格拦截树枝树叶等大件杂物。(2) 沉降：水流进入大池，加入硫酸铝（或硫酸铁），细小颗粒凝聚增大并沉淀为污泥。(3) 过滤：水流经砾石和砂层，去除较细的悬浮物。(4) 活性炭过滤：活性炭吸附有机物和余氯，改善味道并除臭。(5) 氯化消毒：加入少量氯气（或次氯酸钠）杀死细菌等病原体。化学原理：氯与水反应生成次氯酸 (HOCl)，是一种强消毒剂。有些水厂还加氟化物以保护牙齿。该案例检验分离与化学处理工艺的知识。
10. Key Tips for Case Study Questions | 案例分析题的关键技巧
Always read the question twice: first to grasp the scenario, second to underline data and command words
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