📚 Year 9 Edexcel Chemistry: Case Study Practical Drills | 案例分析实战演练
In this article we walk through a series of real-world investigation scenarios that test your understanding of key Edexcel Year 9 Chemistry topics. You will collect evidence, analyse data, write balanced equations, and perform simple quantitative calculations. Each case builds your ability to think like a scientist and answer the type of extended questions that feature in the Edexcel exam. Feel confident to pause and try each step on your own before reading the solution.
本文通过一系列现实世界的调查场景，帮助你巩固爱德思九年级化学的核心知识点。你将学习收集证据、分析数据、书写配平方程式并进行简单的定量计算。每个案例都培养你像科学家一样思考的能力，让你更有把握应对爱德思考试中的拓展题。请在阅读解答之前，先自己动手尝试每一步。
1. Case Study: Acidification of Crystal Lake | 案例研究：水晶湖的酸化问题
Local residents have noticed that fish in Crystal Lake are dying and the water smells slightly sour. The lake is located downwind of a coal-fired power station. Scientists suspect that oxides of sulfur and nitrogen released into the air are forming acid rain. Your task is to investigate water samples and suggest a chemical treatment to reverse the acidification. This case connects the topics of acid deposition, pH testing, neutralisation and stoichiometry.
当地居民发现水晶湖中的鱼正在死亡，湖水带有轻微的酸味。该湖泊位于一家燃煤发电厂的下风向。科学家怀疑排放到空气中的硫氧化物和氮氧化物正在形成酸雨。你的任务是分析水样并提出一种化学治理方案来逆转酸化。这个案例连接了酸沉降、pH 测试、中和反应和化学计量等知识点。
2. Collecting and Testing Water Samples | 收集与水样测试
Three 1‑litre water samples were collected from different points: A (near the power plant), B (middle of the lake) and C (a nearby unaffected stream). Each sample was filtered and its pH measured with a calibrated pH probe at 25 °C. The table below shows the results.
我们从三个地点各采集了 1 升水样：A（靠近发电厂）、B（湖中心）和 C（附近未受影响的溪流）。每个样品经过滤后在 25 °C 下用校准后的 pH 探头测量 pH 值。下表展示了测量结果。
| Sample | pH | Observation |
| A – near plant | 3.2 | faint yellow tint |
| B – lake centre | 4.5 | slight cloudiness |
| C – stream | 6.8 | clear |
The pH probe gave precise readings; acid rain typically has a pH below 5.6. Sample A is strongly acidic, confirming the power station's impact. The higher pH in sample C shows that limestone rocks in the stream bed may be naturally neutralising the rain water.
pH探头提供了精确读数；酸雨的 pH 通常低于 5.6。样品 A 呈强酸性，证实了发电厂的影响。样品 C 的 pH 较高，表明溪流河床的石灰岩可能正在自然中和雨水中的酸。
3. Using Indicators and pH Probes | 使用指示剂和pH探头
Before the pH probe was used, a technician tested samples with universal indicator paper. Sample A turned red/orange, sample B turned orange/yellow, and sample C turned green. Universal indicator is a mixture of dyes that change colour across the pH scale. A red colour indicates pH 1–3, green is neutral around pH 7. The probe results confirmed the approximate colour chart readings but were far more accurate. When you answer exam questions about indicators, always describe both the initial colour and the final colour after addition.
在使用 pH 探头之前，技术员用广谱试纸测试了样品。样品 A 变为红/橙色，样品 B 变为橙/黄色，样品 C 变为绿色。广谱指示剂是一种在不同 pH 范围内变色的混合染料。红色指示 pH 1–3，绿色接近中性 pH 7。探头的测量结果证实了色卡的大致读数，但精度要高得多。在回答有关指示剂的试题时，务必同时描述加入指示剂前后的颜色变化。
4. Neutralisation Calculations with Limestone | 用石灰石进行中和反应计算
ise the acid. Laboratory analysis showed that the main acid present is sulfuric acid, H₂SO₄, with an estimated total amount of 0.050 mol of H⁺ ions in the 1‑ the same acidity, calculate the mass of limestone needed. The neutralisation reaction is:
市政厅计划向湖中撒入石灰石粉末 CaCO₃ 来中和酸性。实验室分析表明样品中主要的酸是硫酸 H₂SO₄，在 1 升样品 A 中估计含有 0.050 mol 的 H⁺ 离子。假设整个湖泊（5.0 × 10⁷ L）具有相同的酸度，请计算所需的石灰石质量。中和反应如下：
CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g)
d = (2.5 × 10⁶) ÷ 2 = 1.25 × 10⁶ mol. Molar mass of CaCO₃ = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹. Mass = 1.25 × 10⁶ mol × 100.1 g mol⁻¹ = 1.25 × 10⁸ g, which is 125 tonnes. This calculation shows how a simple laboratory titration can be scaled up to solve a real environmental problem.
首先，求湖中 H⁺ 的物质的量：0.050 mol L⁻¹ × 5.0 × 10⁷ L = 2.5 × 10⁶ mol。从方程式可知，1 mol CaCO₃ 与 2 mol H⁺ 反应，因此所需 CaCO₃ 的物质的量 = (2.5 × 10⁶) ÷ 2 = 1.25 × 10⁶ mol。CaCO₃ 的摩尔质量 = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹。质量 = 1.25 × 10⁶ mol × 100.1 g mol⁻¹ = 1.25 × 10⁸ g，即 125 吨。这个计算展示了如何将实验室滴定结果放大，用于解决实际环境问题。
5. Identifying Ions in the Water | 鉴定水中的离子
To identify the pollutants, students carried out flame tests and anion tests on sample A. A flame test gave a faint brick‑red colour, suggesting the presence of Ca²⁺. Adding barium chloride solution followed by dilute hydrochloric acid produced a white precipitate that did not dissolve, confirming sulfate ions, SO₄²⁻. The sequence of tests is crucial; barium chloride is used because barium sulfate is insoluble, whereas adding acid removes any carbonate or sulfite precipitates that might interfere.
为了鉴定污染物，学生们对样品 A 进行了焰色反应和阴离子测试。焰色反应呈现微弱的砖红色，提示存在 Ca²⁺。加入氯化钡溶液后再加入稀盐酸，产生了不溶解的白色沉淀，确认了硫酸根离子 SO₄²⁻。测试顺序很关键；使用氯化钡是因为硫酸钡不溶于水，而加入酸可以消除可能干扰的碳酸盐或亚硫酸盐沉淀。
Another portion of the water was tested with silver nitrate solution acidified with nitric acid – no precipitate formed, ruling out chloride ions. The ion tests remind us that qualitative analysis is about observing changes and linking them to specific ions.
另取部分水样用硝酸酸化的硝酸银溶液测试，没有沉淀生成，排除了氯离子的存在。这些离子测试提醒我们，定性分析就是观察变化并将其与特定离子联系起来。
6. Precipitation Reactions to Remove Pollutants | 通过沉淀反应去除污染物
In addition to neutralisation, some metal ions can be removed by precipitation. The lake water also contains a small amount of lead(II) ions, Pb²⁺, from industrial runoff. Adding sodium carbonate solution forms a white precipitate of lead carbonate:
除了中和反应，某些金属离子也可以通过沉淀去除。湖水中还含有少量来自工业径流的铅(II)离子 Pb²⁺。加入碳酸钠溶液会生成白色碳酸铅沉淀：
Pb²⁺(aq) + CO₃²⁻(aq) → PbCO₃(s) ↓
d 0.1 mol L⁻¹ sodium carbonate, then filter and dry the precipitate. The balanced ionic equation gives a 1:1 mole ratio, so if you use 25.0 cm³ of each solution, you would expect 25.0 cm³ × 0.1 mol dm⁻³ = 0.0025 mol of PbCO₃. Its mass = 0.0025 mol × (207.2 + 12.0 + 48.0) g mol⁻¹ = 0.0025 × 267.2 = 0.668 g. This quantitative treatment of precipitation is excellent practice for Edexcel GCSE foundations.
沉淀物可以通过过滤除去，使水体更清洁。在课堂实验中，你可以混合等体积的 0.1 mol L⁻¹ 硝酸铅和 0.1 mol L⁻¹ 碳酸钠溶液，然后过滤并干燥沉淀物。配平的离子方程式给出 1:1 的摩尔比，因此若使用每种溶液 25.0 cm³，预计可得 0.0025 mol 的 PbCO₃。其质量 = 0.0025 mol × (207.2 + 12.0 + 48.0) g mol⁻¹ = 0.0025 × 267.2 = 0.668 g。这种对沉淀反应的量化处理是爱德思 GCSE 基础的绝佳练习。
7. Combustion Analysis of a Fuel Sample | 燃料样品的燃烧分析
A second case shifts to the power station itself. A solid fuel sample was analysed to determine its empirical formula. 1.00 g of the fuel was burned completely in excess oxygen. The products were passed through anhydrous calcium chloride to absorb water and then through soda lime to absorb carbon dioxide. The mass increases recorded were: water 1.26 g, carbon dioxide 3.08 g.
第二个案例转向发电厂本身。为了确定一种固体燃料的经验式，对其进行了分析。取 1.00 g 燃料在过量的氧气中完全燃烧。生成产物先通过无水氯化钙吸收水分，再通过碱石灰吸收二氧化碳。记录的质量增量为：水 1.26 g，二氧化碳 3.08 g。
ounted for = 0.840 + 0.140 = 0.980 g. Since only 1.00 g of fuel was burned, the remaining 0.02 g is likely due to experimental error or trace elements; we assume the fuel is a hydrocarbon. Step 2: moles of C = 0.840/12.0 = 0.0700 mol; moles of H = 0.140/1.0 = 0.140 mol. Divide by the smallest number: C 0.0700/0.0700 = 1, H 0.140/0.0700 = 2. So the empirical formula is CH₂. This technique is often tested in Edexcel's calculation questions; always show your working step by step.
求经验式。第一步：碳的质量 = (12.0/44.0) × 3.08 g = 0.840 g。氢的质量 = (2.0/18.0) × 1.26 g = 0.140 g。被计算的质量总和 = 0.840 + 0.140 = 0.980 g。由于只燃烧了 1.00 g 燃料，剩余的 0.02 g 很可能是实验误差或微量元素；我们假设燃料是碳氢化合物。第二步：C 的物质的量 = 0.840/12.0 = 0.0700 mol；H 的物质的量 = 0.140/1.0 = 0.140 mol。除以最小数值：C 0.0700/0.0700 = 1，H 0.140/0.0700 = 2。因此经验式为 CH₂。这种技巧在爱德思的计算题中经常出现；务必逐步展示你的计算过程。
8. Linking the Case to the Reactivity Series | 将案例联系到活动性顺序
Why does limestone work for neutralisation but not for removing lead ions? That depends on the reactivity series. Calcium is more reactive than hydrogen, so calcium carbonate reacts with acids, releasing CO₂. However, calcium is more reactive than lead, so adding CaCO₃ does not displace lead from solution. A precipitation method using carbonate ions is needed instead. Understanding the reactivity series helps you predict whether a metal will react with an acid or displace another metal. In the context of the lake, several treatments work together: neutralisation with limestone, precipitation with sodium carbonate, and aeration to oxidise any remaining Fe²⁺ to insoluble Fe(OH)₃.
为什么石灰石能用于中和反应，但不能去除铅离子？这取决于活动性顺序。钙比氢活泼，所以碳酸钙与酸反应，释放出 CO₂。然而，钙比铅活泼，因此加入 CaCO₃ 并不能从溶液中置换出铅。反而需要利用碳酸根离子进行沉淀。理解活动性顺序可以帮助你预测金属是否能与酸反应，或是否能置换另一种金属。在湖泊治理的背景下，多种处理方法协同作用：用石灰石中和、用碳酸钠沉淀，以及通过曝气将残留的 Fe²⁺ 氧化为不溶的 Fe(OH)₃。
9. Writing Balanced Equations for Each Reaction | 为每个反应书写配平方程式
A vital skill tested by Edexcel is writing balanced chemical equations, including state symbols. For the neutralisation: CaCO₃(s) + H₂SO₄(aq) → CaSO₄(aq) + H₂O(l) + CO₂(g). For the precipitation: Pb(NO₃)₂(aq) + Na₂CO₃(aq) → PbCO₃(s) + 2NaNO₃(aq). Always check that the number of atoms of each element is the same on both sides. State symbols: (s) – solid, (l) – liquid, (g) – gas, (aq) – aqueous (dissolved in water). Getting state symbols correct can earn you additional marks in the exam.
爱德思考试中一项重要的技能是书写配平的化学方程式，并标注状态符号。中和反应：CaCO₃(s) + H₂SO₄(aq) → CaSO₄(aq) + H₂O(l) + CO₂(g)。沉淀反应：Pb(NO₃)₂(aq) + Na₂CO₃(aq) → PbCO₃(s) + 2NaNO₃(aq)。务必检查每种元素的原子数在两边是否相等。状态符号：(s) – 固体，(l) – 液体，(g) – 气体，(aq) – 水溶液。正确使用状态符号可以为你在考试中赢得额外分数。
10. Evaluating the Experimental Methods | 实验方法的评估
No experiment is perfect. In the lake investigation, the pH probe may have been contaminated if not rinsed between measurements, leading to inaccurate readings. The mass of CaCO₃ calculated assumes complete reaction, but in reality some limestone may settle to the bottom without reacting fully. In the combustion analysis, if the apparatus was not swept with dry air after burning, some CO₂ or H₂O might remain in the tube, lowering the recorded mass. Good scientific practice includes identifying these sources of error and suggesting improvements, such as repeating titrations or using a more precise balance.
没有任何实验是完美的。在湖泊调查中，如果 pH 探头在测量之间没有冲洗干净，可能会被污染，导致读数不准确。计算得出的 CaCO₃ 质量假设了反应完全进行，但实际上部分石灰石可能会沉积到湖底而不完全反应。在燃烧分析中，如果燃烧后没有用干燥的空气吹扫装置，部分 CO₂ 或 H₂O 可能会残留在管内，降低记录到的质量增量。良好的科学实践包括识别这些误差来源并提出改进建议，例如重复滴定或使用更精密的天平。
% error = (accepted value − experimental value) / accepted value × 100%
. If a student collected only 0.610 g, the percentage error is (0.668 − 0.610)/0.668 × 100% = 8.7% ete drying or loss during filtration. A lower yield suggests the procedure needs refining.
对于碳酸铅沉淀，预期质量为 0.668 g。如果学生只收集到 0.610 g，则百分比误差为 (0.668 − 0.610)/0.668 × 100% = 8.7%。这可能是由于干燥不完全或过滤过程中有损失。产率偏低说明实验步骤需要优化。
11. Connecting to the Edexcel Exam: Command Words | 联系爱德思考试：指令词
Edexcel questions often use command words such as 'describe', 'explain', 'calculate' and 'evaluate'. In this case study, 'describe' the colour change when universal indicator is added to acid rain (turns red/orange). 'Explain' why limestone neutralises the acid using the concept of H⁺ ions. 'Calculate' the mass of limestone needed using mole ratios. 'Evaluate' the chosen method by discussing its advantages and limitations. Practising these command words with real data builds confidence for the written paper.
爱德思试题经常使用诸如“描述”、“解释”、“计算”和“评价”等指令词。在本案例研究中，“描述”将广谱指示剂加入酸雨后的颜色变化（变为红/橙色）。“解释”为什么石灰石能中和酸性，运用 H⁺ 离子的概念。“计算”运用摩尔比求出所需石灰石的质量。“评价”所选方法，讨论其优点和局限性。用真实数据练习这些指令词，可以增强你对笔试的信心。
12. Revision Summary and Exam-Style Questions | 复习总结和考试风格问题
erforming empirical formula calculations, linking to the reactivity series, and evaluating methods. All these skills are firmly rooted in the Edexcel Year 9 Chemistry specification. To consolidate, try these questions: (1) Explain why lake water near a power station may have a lower pH than a stream 10 km away. (2) Describe a chemical test, other than pH measurement, to show that the lake water contains H₂SO₄. (3) Calculate the mass of CaCO₃ required to neutralise 0.10 mol of H₂SO₄. (4) A 2.00 g sample of a different fuel produces 5.86 g CO₂ and 2.40 g H₂O; determine its empirical formula. Answers: (1) Acidic gases from the power station dissolve in rain. The stream flows over limestone which neutralises acid. (2) Add barium chloride solution and dilute HCl; a white precipitate of BaSO₄ forms. (3) Reaction shows 1:1 mole ratio, mass = 0.10 mol × 100.1 g mol⁻¹ = 10.01 g. (4) Mass C = (12.0/44.0)×5.86 = 1.60 g, H = (2.0/18.0)×2.40 = 0.267 g, remainder O = 2.00 − (1.60+0.267) = 0.133 g; moles: C=0.1333, H=0.267, O=0.00831; divide by 0.00831 gives C₁₆H₃₂O? Wait, that doesn't simplify nicely. Let' C₁₆H₃₂O. This is a plausible longer-chained compound; the principle of the working is what's important.)
我们涉及了：采集水样并测量 pH、使用指示剂和探头、书写中和与沉淀方程式、进行经验式计算、联系活动性顺序以及评估实验方法。所有这些技能都牢牢扎根于爱德思九年级化学大纲。为巩固所学，请尝试以下问题：(1) 解释为什么发电厂附近的湖水 pH 比 10 公里以外的溪流低。(2) 描述一种除 pH 测量外的化学检验方法，证明湖水中含有 H₂SO₄。(3) 计算中和 0.10 mol H₂SO₄ 所需 CaCO₃ 的质量。(4) 另取 2.00 g 燃料样品充分燃烧，得到 5.86 g CO₂ 和 2.40 g H₂O，求其经验式。参考答案：(1) 发电厂排放的酸性气体溶解在雨水中；溪流流经石灰岩，中和了酸性。(2) 滴加氯化钡溶液和稀盐酸，生成白色硫酸钡沉淀。(3) 反应摩尔比 1:1，质量 = 0.10 mol × 100.1 g mol⁻¹ = 10.01 g。(4) 碳质量 = (12.0/44.0)×5.86 = 1.60 g，氢质量 = (2.0/18.0)×2.40 = 0.267 g，余量为氧 = 2.00 − (1.60+0.267) = 0.133 g；摩尔数：C=0.1333, H=0.267, O=0.00831；除以 0.00831 得 C₁₆H₃₂O。虽然较为复杂，但解题思路更为重要。）
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