Typical Example Questions Explained for A-Level CIE Chemistry | A-Level CIE 化学：典型例题详解

📚 Typical Example Questions Explained for A-Level CIE Chemistry | A-Level CIE 化学：典型例题详解

This article breaks down ten classic worked-example types that frequently appear in CIE A-Level Chemistry papers. Each section mirrors the paired English–Chinese format found on TutorHao, with detailed reasoning to help you master technique.

本文拆解了 CIE A-Level 化学试卷中频繁出现的十种经典例题类型。每个章节都采用配对的中英对照讲解，并附有详细推理，助你掌握解题技巧。

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1. Moles, Molar Mass and Gas Volume (RTP) | 摩尔、摩尔质量与气体体积（RTP）计算

Question: 0.60 g of magnesium ribbon is added to excess dilute sulfuric acid. Calculate the volume of hydrogen gas evolved at RTP, where the molar gas volume is 24.0 dm³ mol⁻¹.

题目：将 0.60 g 镁条加入过量的稀硫酸中。计算在 RTP（气体摩尔体积为 24.0 dm³ mol⁻¹）下产生的氢气体积。

Write the balanced equation: Mg + H₂SO₄ → MgSO₄ + H₂.

写出配平的化学方程式：Mg + H₂SO₄ → MgSO₄ + H₂。

Moles of Mg = mass / Ar = 0.60 g / 24.3 g mol⁻¹ ≈ 0.0247 mol.

镁的物质的量 = 质量 / 相对原子质量 = 0.60 g / 24.3 g mol⁻¹ ≈ 0.0247 mol。

From the equation, 1 mol Mg produces 1 mol H₂, so n(H₂) = 0.0247 mol.

由方程式可知，1 mol Mg 生成 1 mol H₂，因此 n(H₂) = 0.0247 mol。

Volume of H₂ = n × 24.0 dm³ mol⁻¹ = 0.0247 × 24.0 ≈ 0.593 dm³ (593 cm³).

氢气体积 = n × 24.0 dm³ mol⁻¹ = 0.0247 × 24.0 ≈ 0.593 dm³（即 593 cm³）。

Key trap: Always check that the reagent that determines the amount of product is the limiting reactant; here the acid is in excess, so Mg is the limiting reactant.

常见陷阱：务必确认决定产物量的试剂是限制试剂；本题中酸过量，因此镁是限制试剂。

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2. Titration and Back Titration Techniques | 滴定与返滴定技巧

Typical direct titration: 25.0 cm³ of 0.100 mol dm⁻³ HCl is neutralised by 20.0 cm³ of NaOH solution. Find the concentration of NaOH.

典型直接滴定：25.0 cm³ 0.100 mol dm⁻³ 的盐酸被 20.0 cm³ 氢氧化钠溶液中和。求氢氧化钠的浓度。

Equation: HCl + NaOH → NaCl + H₂O. Moles of HCl = 0.0250 dm³ × 0.100 mol dm⁻³ = 0.00250 mol.

方程式：HCl + NaOH → NaCl + H₂O。HCl 物质的量 = 0.0250 dm³ × 0.100 mol dm⁻³ = 0.00250 mol。

Moles of NaOH = 0.00250 mol, so c(NaOH) = 0.00250 mol / 0.0200 dm³ = 0.125 mol dm⁻³.

NaOH 物质的量 = 0.00250 mol，因此 c(NaOH) = 0.00250 mol / 0.0200 dm³ = 0.125 mol dm⁻³。

Back titration question: 0.500 g of impure CaCO₃ is added to 50.0 cm³ of 0.200 mol dm⁻³ HCl (an excess). The unreacted HCl requires 18.0 cm³ of 0.100 mol dm⁻³ NaOH for neutralisation. Calculate the percentage purity of CaCO₃.

返滴定问题：将 0.500 g 不纯的 CaCO₃ 加入 50.0 cm³ 0.200 mol dm⁻³ HCl（过量）中。未反应的 HCl 需要用 18.0 cm³ 0.100 mol dm⁻³ NaOH 中和。计算 CaCO₃ 的纯度。

Equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Total moles of HCl added = 0.0500 dm³ × 0.200 = 0.0100 mol.

方程式：CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂。加入的 HCl 总物质的量 = 0.0500 dm³ × 0.200 = 0.0100 mol。

Moles of excess HCl = moles of NaOH used = 0.0180 dm³ × 0.100 = 0.00180 mol.

过量 HCl 的物质的量 = 所用 NaOH 的物质的量 = 0.0180 dm³ × 0.100 = 0.00180 mol。

Moles of HCl that reacted with CaCO₃ = 0.0100 – 0.00180 = 0.00820 mol.

与 CaCO₃ 反应的 HCl 物质的量 = 0.0100 – 0.00180 = 0.00820 mol。

From the stoichiometry, 2 mol HCl react with 1 mol CaCO₃, so n(CaCO₃) = 0.00820 / 2 = 0.00410 mol.

根据化学计量比，2 mol HCl 与 1 mol CaCO₃ 反应，因此 n(CaCO₃) = 0.00820 / 2 = 0.00410 mol。

Mass of pure CaCO₃ = 0.00410 mol × 100.1 g mol⁻¹ ≈ 0.410 g. Purity = (0.410 / 0.500) × 100% = 82.0%.

纯 CaCO₃ 的质量 = 0.00410 mol × 100.1 g mol⁻¹ ≈ 0.410 g。纯度 = (0.410 / 0.500) × 100% = 82.0%。

Back titrations are essential when the sample is insoluble or the reaction is slow; always identify the sequence of acid consumption carefully.

当样品不溶或反应缓慢时，返滴定至关重要；务必仔细理清酸消耗的顺序。

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3. Equilibrium Constant Kc Calculations | 平衡常数 Kc 计算

Question: 0.40 mol of PCl₅ is placed in a 2.0 dm³ container and heated. At equilibrium, 0.20 mol of PCl₅ remains. Calculate Kc for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).

题目：将 0.40 mol PCl₅ 放入 2.0 dm³ 容器中加热。平衡时，剩余 0.20 mol PCl₅。计算反应 PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) 的 Kc。

Initial moles: PCl₅ = 0.40, PCl₃ = 0, Cl₂ = 0. Change: –x, +x, +x. At equilibrium, PCl₅ = 0.40 – x = 0.20 → x = 0.20 mol.

初始物质的量：PCl₅ = 0.40，PCl₃ = 0，Cl₂ = 0。变化：–x，+x，+x。平衡时，PCl₅ = 0.40 – x = 0.20 → x = 0.20 mol。

Equilibrium moles: PCl₅ = 0.20, PCl₃ = 0.20, Cl₂ = 0.20. Concentrations: [PCl₅] = 0.20/2.0 = 0.10 mol dm⁻³, [PCl₃] = 0.10 mol dm⁻³, [Cl₂] = 0.10 mol dm⁻³.

平衡物质的量：PCl₅ = 0.20，PCl₃ = 0.20，Cl₂ = 0.20。浓度：[PCl₅] = 0.20/2.0 = 0.10 mol dm⁻³，[PCl₃] = 0.10 mol dm⁻³，[Cl₂] = 0.10 mol dm⁻³。

Kc = ([PCl₃][Cl₂]) / [PCl₅] = (0.10 × 0.10) / 0.10 = 0.10 mol dm⁻³ (units must be stated).

Kc = ([PCl₃][Cl₂]) / [PCl₅] = (0.10 × 0.10) / 0.10 = 0.10 mol dm⁻³（需注明单位）。

Tip: Always draw a RICE table (Reaction, Initial, Change, Equilibrium) to avoid arithmetic mistakes. Also remember that solids and pure liquids are omitted from the Kc expression.

提示：务必列出 RICE 表格（反应式、初始、变化、平衡），避免计算错误。同时记住，固体和纯液体不出现在 Kc 表达式中。

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4. pH and Ka of Weak Acids | 弱酸的 pH 与 Ka

Problem: Ethanoic acid has Ka = 1.8 × 10⁻⁵ mol dm⁻³ at 298 K. Calculate the pH of a 0.100 mol dm⁻³ solution of CH₃COOH.

例题：298 K 下乙酸的电离常数 Ka = 1.8 × 10⁻⁵ mol dm⁻³。计算 0.100 mol dm⁻³ CH₃COOH 溶液的 pH。

For a weak acid, [H⁺] ≈ √(Ka × c) as long as c/Ka > 500. Check: 0.100 / 1.8×10⁻⁵ ≈ 5556, so the approximation holds.

对于弱酸，当 c/Ka > 500 时，可用近似式 [H⁺] ≈ √(Ka × c)。验证：0.100 / 1.8×10⁻⁵ ≈ 5556，因此近似成立。

[H⁺] = √(1.8×10⁻⁵ × 0.100) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ mol dm⁻³.

[H⁺] = √(1.8×10⁻⁵ × 0.100) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ mol dm⁻³。

pH = –log₁₀(1.34×10⁻³) ≈ 2.87. Remember to give the answer to two decimal places as pH is a logarithmic scale.

pH = –log₁₀(1.34×10⁻³) ≈ 2.87。记得将答案保留两位小数，因为 pH 是对数标度。

If a salt of the weak acid is added (a buffer), use the Henderson–Hasselbalch equation: pH = pKa + log₁₀([salt]/[acid]). For instance, if [CH₃COONa] = 0.150 mol dm⁻³ and [CH₃COOH] = 0.100 mol dm⁻³, then pH = –log(1.8×10⁻⁵) + log(0.150/0.100) ≈ 4.74 + 0.176 = 4.92.

如果加入该弱酸的盐（形成缓冲溶液），使用 Henderson–Hasselbalch 方程：pH = pKa + log₁₀([盐]/[酸])。例如，[CH₃COONa] = 0.150 mol dm⁻³，[CH₃COOH] = 0.100 mol dm⁻³，则 pH = –log(1.8×10⁻⁵) + log(0.150/0.100) ≈ 4.74 + 0.176 = 4.92。

Always check that your assumptions are valid; examiners expect you to state that the dissociation is negligible compared to the initial concentration.

务必检查假设是否合理；考官期望你明确说明，相对于初始浓度，解离程度可忽略不计。

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5. Enthalpy Changes Using Hess’s Law | 利用赫斯定律求焓变

Question: Calculate the standard enthalpy of formation of ethanol, ΔHf°, given the following standard combustion data: ΔHc°(C) = –394 kJ mol⁻¹, ΔHc°(H₂) = –286 kJ mol⁻¹, ΔHc°(C₂H₅OH) = –1367 kJ mol⁻¹.

问题：已知下列标准燃烧焓数据：ΔHc°(C) = –394 kJ mol⁻¹，ΔHc°(H₂) = –286 kJ mol⁻¹，ΔHc°(C₂H₅OH) = –1367 kJ mol⁻¹，计算乙醇的标准生成焓 ΔHf°。

Construct a Hess cycle: Formation from elements (2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)) can be obtained by combining combustion reactions.

构建赫斯循环：由元素生成乙醇 (2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)) 可通过组合燃烧反应求得。

ΔHf° = [2 × ΔHc°(C) + 3 × ΔHc°(H₂)] – ΔHc°(C₂H₅OH) = [2×(–394) + 3×(–286)] – (–1367) kJ mol⁻¹.

ΔHf° = [2 × ΔHc°(C) + 3 × ΔHc°(H₂)] – ΔHc°(C₂H₅OH) = [2×(–394) + 3×(–286)] – (–1367) kJ mol⁻¹。

Calculate: –788 – 858 + 1367 = –1646 + 1367 = –279 kJ mol⁻¹.

计算：–788 – 858 + 1367 = –1646 + 1367 = –279 kJ mol⁻¹。

The negative sign shows the formation is exothermic. Always draw an energy cycle and check the direction of arrows to avoid sign errors.

负号表明生成过程放热。务必画出能量循环并检查箭头方向，避免符号错误。

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6. Rate Equations and the Arrhenius Equation | 速率方程与阿伦尼乌斯方程

Typical rate-concentration data: For the reaction A + B → products, when [A] doubles (with [B] constant) the rate doubles; when [B] doubles (with [A] constant) the rate quadruples. Deduce the rate equation.

典型的浓度-速率数据：反应 A + B → 产物，当 [A] 加倍（[B] 恒定）时速率加倍；当 [B] 加倍（[A] 恒定）时速率变为原来的四倍。推导速率方程。

Rate ∝ [A]¹ because a doubling of [A] causes a doubling of rate. Rate ∝ [B]² because doubling [B] quadruples the rate. Therefore, rate = k[A][B]².

速率与 [A] 成正比（指数为 1），因为 [A] 加倍导致速率加倍。速率与 [B]² 成正比，因为 [B] 加倍使速率变为四倍。因此速率方程为 rate = k[A][B]²。

To find k, plug in data from one experiment: e.g., at [A] = 0.10 mol dm⁻³, [B] = 0.20 mol dm⁻³, rate = 1.6×10⁻³ mol dm⁻³ s⁻¹. Then k = rate / ([A][B]²) = 1.6×10⁻³ / (0.10 × 0.20²) = 1.6×10⁻³ / 0.004 = 0.40 mol⁻² dm⁶ s⁻¹.

求 k 时，代入某组实验数据：例如 [A] = 0.10 mol dm⁻³，[B] = 0.20 mol dm⁻³，rate = 1.6×10⁻³ mol dm⁻³ s⁻¹。则 k = rate / ([A][B]²) = 1.6×10⁻³ / (0.10 × 0.20²) = 1.6×10⁻³ / 0.004 = 0.40 mol⁻² dm⁶ s⁻¹。

Arrhenius equation is often tested as: ln k = –Ea/RT + ln A. Two-point calculation: ln(k₂/k₁) = –Ea/R (1/T₂ – 1/T₁).A-level students should be able to calculate activation energy from such data.

Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

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