📚 Year 10 SQA Chemistry: Case Study Practical Drills | SQA 化学：案例分析实战演练
Case studies are a powerful way to connect theoretical chemistry to real-world applications. In SQA National 5 Chemistry, being able to analyse scenarios, interpret data and apply calculations is essential for success. This article presents eight practical case studies covering redox, acids and bases, equilibrium, stoichiometry, polymers and energy. Each one is worked through in the same way you would answer it in an exam: read the context, extract the data, choose the chemistry, then calculate and explain.
案例分析是把理论化学与现实应用联系起来的有效方式。在 SQA National 5 化学中，能够分析场景、解读数据并运用计算是取得好成绩的关键。本文呈现八个实战案例，涵盖氧化还原、酸碱、化学平衡、化学计量、聚合物和能量。每一个案例都按照考试答题的思路完整演练：读背景、提取数据、选定知识点，然后计算并解释。
1. How to Attack an SQA Case Study | 如何攻克 SQA 案例题
A case study question wraps familiar chemistry inside an unfamiliar context - a kitchen, a factory, a hospital or a recycling plant. The chemistry itself always comes from the course; only the wrapping is new. Your job is to unwrap it quickly and get to the marks.
案例题只是把熟悉的化学知识包裹在不熟悉的背景中——厨房、工厂、医院或回收站。化学本身永远来自课程内容，只是外包装是新的。你要做的就是快速拆开包装，拿到分数。
Read the stem twice. On the first pass, ignore the numbers and work out what the story is about. On the second pass, underline the command words and the data you will need, then decide which topic from the course is being tested.
把题干读两遍。第一遍忽略数字，弄清故事讲的是什么；第二遍划出指令词和你会用到的数据，再判断考查的是课程中的哪个主题。
- Command words: state, describe, explain, calculate, suggest, compare.
指令词：陈述、描述、解释、计算、提出建议、比较。 - Data extraction: pull numbers into a small table before you calculate.
数据提取：计算前先把数字整理成一个小表格。 - Show every step: equation, moles, ratio, answer with units.
展示每一步：方程式、物质的量、比例、带单位的答案。 - Answer in the language of the question, and never leave a calculation blank - a correct method often earns partial marks.
用题目的语言作答，计算题不要留空——正确的方法通常也能得分。
2. Case Study 1: The Rusting Experiment (Redox) | 案例一：生锈实验（氧化还原）
Rusting is a redox reaction in which iron loses electrons to form iron(II) ions, which are then further oxidised to hydrated iron(III) oxide. In an experiment, a student placed iron nails in four test tubes: (A) dry air only, (B) boiled water covered with oil, (C) tap water exposed to air, and (D) saltwater exposed to air.
生锈是一种氧化还原反应：铁失去电子形成铁(II)离子，随后进一步被氧化为水合铁(III)氧化物。在一次实验中，学生把铁钉放入四支试管：(A) 只有干燥空气，(B) 煮沸过的水并用油覆盖，(C) 暴露在空气中的自来水，(D) 暴露在空气中的盐水。
| Tube 试管 | Conditions 条件 | Result after one week 一周后结果 | Reason 原因 |
| A | Dry air only 只有干燥空气 | No rust 无锈 | No water 没有水 |
| B | Boiled water + oil 煮沸水加油封 | No rust 无锈 | No oxygen 没有氧气 |
| C | Tap water + air 自来水加空气 | Rust 生锈 | Both present 两者都有 |
| D | Saltwater + air 盐水加空气 | Fastest rust 生锈最快 | Electrolyte speeds it up 电解质加速 |
After one week, only the nails in C and D showed significant rust, with D rusting fastest. Analysis: the nail in A had no water and the nail in B had no oxygen, so neither could rust. The salt in D acts as an electrolyte, allowing ions to move freely and speeding up electron transfer.
一周后，只有 C 和 D 中的铁钉明显生锈，且 D 生锈最快。分析：A 中的铁钉没有水，B 中的铁钉没有氧气，因此都无法生锈。D 中的盐充当电解质，使离子可以自由移动，从而加快电子转移。
Therefore, both oxygen and water are necessary for rusting, and ionic compounds such as salt accelerate the process. This is why cars rust faster near the coast and why salty winter roads damage vehicles.
因此，生锈既需要氧气也需要水，而且盐等离子化合物会加速这一过程。这就是海边汽车更容易生锈、冬季撒盐的道路会腐蚀车辆的原因。
Fe(s) → Fe²⁺(aq) + 2e⁻
O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
The first half-equation shows oxidation (loss of electrons) and the second shows reduction (gain of electrons). Adding them after balancing the electrons gives the overall redox equation for rusting.
第一个半反应式表示氧化（失去电子），第二个表示还原（得到电子）。把电子数配平后将两式相加，即可得到生锈的总氧化还原方程式。
3. Case Study 2: Preventing Corrosion | 案例二：防止腐蚀
Prevention methods include coating with zinc (galvanising), painting, greasing or using sacrificial anodes. A paint layer simply blocks water and oxygen from reaching the iron, so it works only as long as the barrier is unbroken.
防锈方法包括镀锌（镀锌处理）、涂漆、涂油脂或使用牺牲阳极。油漆层只是阻止水和氧气接触铁，因此只有在涂层完好时才会起作用。
Zinc is more reactive than iron, so it oxidises preferentially and protects the iron even if the coating is scratched. This is called sacrificial protection, or cathodic protection when a block of metal is bolted to a pipeline or a ship's hull.
锌比铁更活泼，因此会优先被氧化，即使涂层被刮伤也能保护铁。这称为牺牲保护；当把金属块固定在管道或船体上时，则称为阴极保护。
Zn(s) → Zn²⁺(aq) + 2e⁻
A useful examination trap: magnesium, aluminium and zinc all sit above iron in the reactivity series and can act as sacrificial anodes, but copper and silver cannot, because they are less reactive than iron.
考试中常见的陷阱：镁、铝和锌在金属活动性顺序中都排在铁前面，可以作为牺牲阳极，但铜和银不行，因为它们比铁不活泼。
Both cases illustrate the reactivity series and the link between electron transfer and everyday corrosion control. In a written answer, always state the comparison explicitly: "zinc is more reactive than iron, therefore zinc loses electrons in preference to iron, therefore the iron does not rust."
两个案例都体现了金属活动性顺序以及电子转移与日常防腐之间的联系。在书面作答时，一定要明确写出比较关系："锌比铁活泼，因此锌优先失去电子，所以铁不会生锈。"
4. Case Study 3: Acid-Alkali Titration | 案例三：酸碱滴定
A student titrates 25.0 cm³ of 0.100 mol l⁻¹ sodium hydroxide solution with hydrochloric acid of unknown concentration. Using phenolphthalein, the solution changes from pink to colourless at the end point. The concordant titres average 22.4 cm³. Calculate the concentration of the acid.
一名学生用未知浓度的盐酸滴定 25.0 cm³、浓度为 0.100 mol l⁻¹ 的氢氧化钠溶液。使用酚酞指示剂，终点时溶液由粉红色变为无色。三次吻合滴定的平均值为 22.4 cm³。求盐酸的浓度。
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
mol. Step 2: the equation shows a 1:1 ratio, so the moles of HCl are also 2.50 × 10⁻³ mol.
第一步：氢氧化钠的物质的量。n = c × V，V 以升为单位，所以 n = 0.100 × 0.0250 = 2.50 × 10⁻³ mol。第二步：方程式中比例为 1:1，所以 HCl 的物质的量也是 2.50 × 10⁻³ mol。
. Step 4: c = n ÷ V = 2.50 × 10⁻³ ÷ 0.0224 = 0.112 mol l⁻¹ (to three significant figures).
第三步：把滴定体积除以 1000 换算为升，得到 0.0224 l。第四步：c = n ÷ V = 2.50 × 10⁻³ ÷ 0.0224 = 0.112 mol l⁻¹（保留三位有效数字）。
Practical points that carry marks: rinse the burette with the acid, not water; read the bottom of the meniscus at eye level; repeat until two or three concordant titres agree within 0.2 cm³; and remember that the end point is where the indicator changes colour, which should be as close as possible to the equivalence point.
实验操作的得分点：用待测酸润洗滴定管而不是用水；视线与凹液面最低处齐平读数；重复滴定直到两到三次结果吻合（相差不超过 0.2 cm³）；并记住终点是指示剂变色的时刻，应尽量接近等当点。
5. Case Study 4: pH, Neutralisation and Everyday Acids | 案例四：pH、中和与生活中的酸
A class tests household substances with Universal Indicator and a pH probe: lemon juice (pH 2), vinegar (pH 3), milk (pH 6.5), pure water (pH 7), baking soda solution (pH 9) and bleach (pH 13). The results show the pH scale runs from below 7 for acids to above 7 for alkalis.
某班级用通用指示剂和 pH 探头检测家庭常见物质：柠檬汁（pH 2）、食醋（pH 3）、牛奶（pH 6.5）、纯水（pH 7）、小苏打溶液（pH 9）和漂白剂（pH 13）。结果显示 pH 标度从低于 7 的酸性到高于 7 的碱性。
The same data can be used to compare strong and weak acids. Hydrochloric acid is strong because it is fully dissociated into ions, whereas ethanoic acid is weak and only partially dissociated, so at the same concentration it has a higher pH, conducts electricity less well and reacts more slowly with magnesium.
同一组数据还可以用来比较强酸和弱酸。盐酸是强酸，完全电离成离子；乙酸是弱酸，只部分电离，因此在相同浓度下 pH 更高、导电性更差、与镁反应更慢。
HCl(aq) → H⁺(aq) + Cl⁻(aq)
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq)
Neutralisation reactions are the practical heart of this topic. Acid plus alkali gives a salt and water; acid plus metal carbonate gives a salt, water and carbon dioxide; acid plus metal gives a salt and hydrogen. Farmers spread lime (calcium hydroxide or calcium carbonate) to raise the pH of acidic soil, and antacid tablets do the same job in the stomach.
中和反应是本主题的实用核心。酸加碱生成盐和水；酸加金属碳酸盐生成盐、水和二氧化碳；酸加金属生成盐和氢气。农民施用石灰（氢氧化钙或碳酸钙）来提高酸性土壤的 pH，抗酸药片则在胃中起同样的作用。
H⁺(aq) + OH⁻(aq) → H₂O(l)
Finally, note the dilution shortcut: diluting an acid tenfold lowers the hydrogen ion concentration by a factor of ten and raises the pH by one unit, so a pH 2 acid diluted tenfold becomes pH 3. This is a very common one-mark question in SQA papers.
最后注意稀释的快捷规律：把酸稀释十倍，氢离子浓度降为原来的十分之一，pH 上升一个单位，所以 pH 2 的酸稀释十倍后变为 pH 3。这是 SQA 试卷中非常常见的一分题。
6. Case Study 5: Equilibrium and Le Chatelier | 案例五：化学平衡与勒沙特列原理
In a classic demonstration, a few drops of iron(III) nitrate solution are mixed with potassium thiocyanate solution. The mixture turns deep red because of the formation of the FeSCN²⁺ ion. Adding more iron(III) ions or more thiocyanate ions intensifies the red colour; adding sodium hydroxide removes iron(III) ions as a precipitate and the colour fades.
在一个经典演示中，把几滴硝酸铁(III)溶液与硫氰酸钾溶液混合，由于生成 FeSCN²⁺ 离子，混合物呈深红色。再加入铁(III)离子或硫氰酸根离子会使红色加深；加入氢氧化钠会把铁(III)离子沉淀移除，颜色随之变浅。
Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq)
Dynamic equilibrium exists in a closed system when the forward and reverse reactions occur at the same rate
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