📚 Case Study Practice Drills | 案例分析实战演练
Case studies are a powerful way to apply your chemistry knowledge to real-world scenarios. In Year 9 CAIE Chemistry, you are expected to analyse data, make predictions, and justify conclusions based on your understanding of core principles. This article presents eight practical case studies covering topics from particle theory to experimental design, each with step-by-step reasoning to help you think like a scientist.
案例分析是将化学知识应用于真实场景的有效方法。在九年级 CAIE 化学课程中，你需要分析数据、作出预测并根据核心原理证明结论。本文提供八个实践案例，涵盖从粒子理论到实验设计等主题，每个案例都配有逐步推理过程，帮助你像科学家一样思考。
1. Identifying an Unknown Substance | 鉴定未知物质
A student finds a white crystalline solid in the lab. It dissolves in water to form a colourless solution. When tested with a pH meter, the solution gives a pH of 7. On heating, the solid melts at 801 °C and conducts electricity only when molten or dissolved in water.
一名学生在实验室发现了一种白色晶体固体。它溶于水形成无色溶液。用 pH 计测试时，该溶液的 pH 值为 7。加热时，该固体在 801 °C 熔化，并且只有在熔融或溶于水时才能导电。
What type of bonding is present in the solid? Justify your answer using three pieces of evidence from the observations.
该固体中存在哪种类型的化学键？用观测到的三个证据证明你的答案。
The solid shows giant ionic bonding. First, the high melting point of 801 °C indicates strong electrostatic forces between ions, requiring large amounts of energy to overcome. Second, electrical conductivity only in the liquid or aqueous state confirms ions are free to move in those states, but fixed in the solid lattice. Third, a neutral pH of 7 when dissolved suggests it is a salt of a strong acid and strong base, consistent with an ionic compound like sodium chloride.
该固体表现出巨型离子键。首先，801 °C 的高熔点表明离子间存在强大的静电引力，需要大量能量才能克服。其次，只有在液态或水溶液中才能导电，证实离子在这两种状态下能自由移动，而在固态晶格中被束缚。第三，溶解后 pH 为 7，表明它是强酸和强碱形成的盐，与氯化钠等离子化合物一致。
Additional evidence includes the brittle nature of most ionic solids and their cleavage along crystal planes, though not mentioned here. The student could further confirm by flame test showing characteristic yellow if sodium is present.
进一步的证据包括大多数离子固体的脆性及其沿晶面解理，尽管此处未提及。如果钠存在，学生还可以通过火焰测试显示特征黄色来进一步确认。
2. Conservation of Mass in Reactions | 化学反应的质量守恒
A teacher demonstrates a reaction in a sealed conical flask. She places 5.0 g of calcium carbonate chips and 20.0 g of dilute hydrochloric acid in the flask, quickly inserts a stopper, and places the flask on a balance. The total mass before reaction is 125.0 g. Bubbles of gas form, and the mass reading stays at 125.0 g throughout. After the reaction, the flask is opened, and the mass drops to 123.2 g.
一名教师在一个密封锥形瓶中演示化学反应。她将 5.0 克碳酸钙块和 20.0 克稀盐酸放入瓶中，迅速塞上瓶塞，并将锥形瓶放在天平上。反应前总质量为 125.0 克。有气泡产生，质量读数始终保持在 125.0 克。反应结束后，打开瓶塞，质量降至 123.2 克。
Explain why the mass did not change while the flask was sealed, and account for the mass loss when opened.
解释为什么在密封时质量没有变化，并说明打开瓶塞后质量损失的原因。
While sealed, the mass remained constant because the law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. The carbon dioxide gas produced was trapped inside the flask, so the total mass of the system (reactants and products) stayed the same. When the stopper was removed, the denser carbon dioxide gas escaped and diffused into the air, causing the measured mass to decrease by 1.8 g, which corresponds to the mass of CO₂ released.
在密封状态下，质量保持恒定，因为质量守恒定律指出，化学反应中物质不能被创造或消灭。产生的二氧化碳气体被截留在瓶内，因此系统（反应物和生成物）的总质量保持不变。当拔掉瓶塞时，密度较大的二氧化碳气体逸散到空气中，导致测得的质量减少了 1.8 克，这与释放出的 CO₂ 质量一致。
3. Using Indicators to Classify Acids and Bases | 使用指示剂分类酸与碱
A student tests three colourless solutions X, Y, and Z with universal indicator and litmus paper. Solution X turns blue litmus red and gives a universal indicator colour of red. Solution Y turns red litmus blue and gives a purple colour with universal indicator. Solution Z shows no change to litmus and gives green with universal indicator.
一名学生用通用指示剂和石蕊试纸测试三种无色溶液 X、Y 和 Z。溶液 X 使蓝色石蕊变红，通用指示剂显示红色。溶液 Y 使红色石蕊变蓝，通用指示剂显示紫色。溶液 Z 对石蕊无变化，通用指示剂显示绿色。
Classify each solution as acidic, alkaline, or neutral. Suggest the approximate pH range for each, and name a common laboratory substance that matches each description.
将每种溶液分类为酸性、碱性或中性。推测每种溶液的近似 pH 范围，并说出一个符合每种描述的常见实验室物质。
Solution X is acidic. The red colour with universal indicator corresponds to a pH of about 1–2, indicating a strong acid such as hydrochloric acid. Solution Y is alkaline; the purple colour indicates a pH of about 13–14, typical of strong bases like sodium hydroxide solution. Solution Z is neutral with a pH around 7; pure water or sodium chloride solution would match this behaviour. Litmus paper confirms these classifications: no change for neutral, red for acid, blue for alkali.
溶液 X 是酸性。通用指示剂的红色对应 pH 值大约 1–2，表明是强酸，如盐酸。溶液 Y 是碱性；紫色表明 pH 值约 13–14，典型的强碱如氢氧化钠溶液。溶液 Z 呈中性，pH 值约 7；纯水或氯化钠溶液符合此表现。石蕊试纸验证了这些分类：中性无变化，酸变红，碱变蓝。
4. Factors Affecting Reaction Rate | 影响反应速率的因素
An experiment investigates the rate of reaction between marble chips and excess dilute hydrochloric acid. The volume of carbon dioxide gas collected is recorded every 20 seconds under three conditions: A – large chips at 25 °C; B – powdered marble at 25 °C; C – large chips at 40 °C. The curves of gas volume against time are plotted. Curve B is steepest and reaches maximum volume fastest. Curve C is steeper than A, but both A and C eventually reach the same final volume as B.
一项实验研究大理石碎片与过量稀盐酸的反应速率。在三种条件下每 20 秒记录收集到的二氧化碳体积：A – 25 °C 下的大块碎片；B – 25 °C 下的粉末状大理石；C – 40 °C 下的大块碎片。绘制出气体体积随时间变化的曲线。曲线 B 最陡，最快达到最大体积。曲线 C 比 A 陡，但 A 和 C 最终达到与 B 相同的最终体积。
Explain why curve B is steepest, and why increasing temperature (C) only affects the rate but not the final volume.
解释为什么曲线 B 最陡，以及为什么升高温度（C）只影响速率而不影响最终体积。
Curve B is steepest because it uses powdered marble, which has a much larger surface area than large chips. A larger surface area allows more frequent collisions between reactant particles per unit time, increasing the rate of reaction. The same mass of marble produces the same total volume of CO₂, so all conditions reach the same final volume as the excess acid ensures complete reaction.
曲线 B 最陡，因为使用了粉末状大理石，其表面积远大于大块碎片。更大的表面积使单位时间内反应物粒子之间的碰撞更加频繁，提高了反应速率。相同质量的大理石产生相同总量的二氧化碳，因此所有条件都达到相同的最终体积，因为过量酸保证了反应完全。
Increasing temperature to 40 °C (C) gives reactant particles more kinetic energy, causing more frequent and more energetic successful collisions per second, so the rate rises and the curve is steeper than A. However, temperature does not change the amount of marble, so the maximum yield of gas remains unchanged.
温度升高到 40 °C（C）使反应物粒子获得更多动能，导致每秒成功碰撞更频繁且能量更高，因此速率加快，曲线比 A 更陡。然而，温度不改变大理石的质量，因此气体的最大产量保持不变。
5. Separating Mixtures: A Practical Challenge | 分离混合物：实践挑战
A mixture contains sand, iron filings, sodium chloride, and water. Design a sequence of separation techniques to obtain each component in a pure, dry state. Justify your choice of technique at each step.
一种混合物包含沙子、铁屑、氯化钠和水。设计一系列分离技术以获得每种成分的纯净、干燥状态。在每一步中证明你选择的技术的合理性。
First, remove iron filings using a magnet. Iron is magnetic, so it can be easily attracted and separated from the non-magnetic components. Next, filter the remaining mixture. Sand is insoluble and will remain as residue on the filter paper, while salt solution passes through as filtrate. Wash and dry the sand. Finally, crystallise the sodium chloride by heating the salt solution to evaporate most of the water, then allowing it to cool for crystals to form. Dry the crystals on filter paper. Distillation could also be used to recover pure water, but simple evaporation achieves salt recovery adequately.
首先，用磁铁吸除铁屑。铁具有磁性，因此可轻易被吸引并从非磁性成分中分离出来。接着，过滤剩余混合物。沙子不溶于水，会留在滤纸上作为残渣，而盐溶液则作为滤液通过。洗涤并干燥沙子。最后，通过加热盐溶液蒸发大部分水，然后冷却使其结晶，从而得到氯化钠。在滤纸上干燥晶体。也可以使用蒸馏来回收纯水，但简单的蒸发足以回收盐。
6. Atomic Structure and the Periodic Table | 原子结构与元素周期表
Element Q has atomic number 17 and mass number 35. It forms a diatomic molecule and reacts vigorously with sodium, a metal from Group 1, to produce a white solid.
元素 Q 的原子序数为 17，质量数为 35。它形成双原子分子，并与第 1 族的金属钠剧烈反应，生成白色固体。
Identify element Q. Draw its electronic structure using shell notation, and explain why it is placed in Group 17 and Period 3 of the Periodic Table.
鉴定元素 Q。用电子层符号画出其电子结构，并解释为什么它位于周期表的第 17 族和第 3 周期。
Element Q is chlorine. Its atomic number is 17, so it has 17 protons and 17 electrons. The electron configuration is 2,8,7. It is in Group 17 because it has 7 electrons in its outer shell, making it a halogen that gains 1 electron to achieve a stable octet. It is in Period 3 because its electrons occupy three shells. The diatomic molecule Cl₂ and reaction with sodium to form NaCl (white solid) confirm its identity.
元素 Q 是氯。原子序数为 17，因此它有 17 个质子和 17 个电子。电子排布为 2,8,7。它位于第 17 族，因为它的最外层有 7 个电子，是卤素，通过获得 1 个电子达到稳定八隅体。它位于第 3 周期，因为它的电子占据三个电子层。双原子分子 Cl₂ 以及与钠反应生成 NaCl（白色固体）证实了它的身份。
7. Writing and Balancing Chemical Equations | 书写与配平化学方程式
Propane (C₃H₈) burns completely in oxygen to produce carbon dioxide and water. A student writes the equation: C₃H₈ + O₂ → CO₂ + H₂O. She counts 3 carbon, 8 hydrogen, and 2 oxygen atoms on the reactant side, but only 1 carbon, 2 hydrogen, and 3 oxygen on the product side.
丙烷（C₃H₈）在氧气中完全燃烧，生成二氧化碳和水。一名学生写下方程式：C₃H₈ + O₂ → CO₂ + H₂O。她计算出反应物一侧有 3 个碳原子、8 个氢原子和 2 个氧原子，但生成物一侧仅有 1 个碳原子、2 个氢原子和 3 个氧原子。
Balance the equation and explain the meaning of the coefficients in terms of moles of molecules. Show your working.
配平该方程式，并根据分子摩尔数解释系数的含义。写出你的推导过程。
To balance, we ensure equal numbers of each atom type on both sides. Start with carbon: C₃H₈ → 3CO₂ gives 3 carbon atoms on right. Then hydrogen: 8 H atoms on left require 4H₂O on right, because 4 × 2 = 8. Now count oxygen in products: 3CO₂ contributes 6 O atoms, and 4H₂O contributes 4 O atoms, total 10 O atoms. Thus we need 5 O₂ molecules on left (5 × 2 = 10). The balanced equation is: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. This means 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide and 4 moles of water.
配平时，我们要确保每种原子在两边的数量相等。从碳开始：C₃H₈ → 3CO₂ 使右边有 3 个碳原子。然后是氢：左边有 8 个氢原子需要右边有 4H₂O，因为 4 × 2 = 8。现在计算产物中的氧：3CO₂ 提供 6 个氧原子，4H₂O 提供 4 个氧原子，共 10 个氧原子。因此左边需要 5 个 O₂ 分子（5 × 2 = 10）。配平后的方程式为：C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。这表示 1 摩尔丙烷与 5 摩尔氧反应，生成 3 摩尔二氧化碳和 4 摩尔水。
8. Experimental Design and Error Analysis | 实验设计与误差分析
A student wants to determine the percentage purity of a sample of potassium carbonate by reacting it with dilute nitric acid and measuring the volume of carbon dioxide produced. She uses the apparatus: a conical flask with a delivery tube leading to an inverted measuring cylinder in a trough of water. She weighs 1.00 g of the sample and excess acid, but the volume of gas collected is always less than the theoretical value. She notices bubbles escaping around the stopper and that the acid was added without quickly sealing the flask.
一名学生想通过将碳酸钾样品与稀硝酸反应并测量产生的二氧化碳体积来确定样品的纯度。她使用的装置：一个带导气管的锥形瓶，导气管连接到水槽中倒置的量筒。她称取 1.00 克样品和过量酸，但收集到的气体体积始终低于理论值。她注意到瓶塞周围有气泡逸出，并且加酸时未立即密封锥形瓶。
Identify two sources of error and suggest improvements. Write the balanced equation for the reaction and explain how she should calculate percentage purity from the correct gas volume.
指出两个误差来源并提出改进建议。写出该反应的配平方程式，并说明应如何根据正确的气体体积计算纯度百分比。
Error 1: Gas loss before sealing the flask. Acid should be placed in a small test tube inside the flask, and the reaction initiated by shaking the sealed system to mix acid and solid. Error 2: Leaks around the stopper; a well-fitted rubber bung and possibly a side-arm flask with a secure delivery tube would prevent gas escape. The balanced equation is: K₂CO₃ + 2HNO₃ → 2KNO₃ + CO₂ + H₂O. Use the collected CO₂ volume at room temperature and pressure to calculate moles of CO₂ (using molar volume 24 dm³ mol⁻¹ at r.t.p.). The moles of K₂CO₃ equal moles of CO₂. Multiply by the molar mass of K₂CO₃ (138 g mol⁻¹) to find the mass of pure K₂CO₃. Percentage purity = (mass of pure K₂CO₃ / mass of sample) × 100%.
误差 1：密封前气体损失。应将酸放在锥形瓶内的小试管中，通过摇动密封系统使酸与固体混合来启动反应。误差 2：瓶塞周围的泄漏；使用紧密的橡胶塞，可能的话用带支管的锥形瓶和安全的导气管防止气体逸出。配平方 程式为：K₂CO₃ + 2HNO₃ → 2KNO₃ + CO₂ + H₂O。使用在室温和常压下收集到的 CO₂ 体积，计算 CO₂ 的摩尔数（在室温下，摩尔体积为 24 dm³ mol⁻¹）。K₂CO₃ 的摩尔数等于 CO₂ 的摩尔数。乘以 K₂CO₃ 的摩尔质量（138 g mol⁻¹）得到纯 K₂CO₃ 的质量。纯度百分比 =（纯 K₂CO₃ 质量 / 样品质量）× 100%。
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