IB and OCR Chemistry: Detailed Solutions to Typical Example Problems | IB OCR 化学：典型例题详解

📚 IB and OCR Chemistry: Detailed Solutions to Typical Example Problems | IB OCR 化学：典型例题详解

Both IB and OCR A Level Chemistry examinations demand strong problem-solving skills. This article walks through a series of carefully chosen worked examples, covering stoichiometry, energetics, kinetics, equilibrium, acids and bases, redox, organic chemistry, and spectroscopy. Each solution is accompanied by bilingual step-by-step reasoning to help you master the core techniques.

IB 和 OCR A Level 化学考试都非常注重解题能力。本文通过一系列精心挑选的典型例题，涵盖化学计量、热力学、动力学、平衡、酸碱、氧化还原、有机化学和光谱解析，每个解答都配有中英双语的逐步推理，帮助你掌握核心解题技巧。

1. Stoichiometry and Empirical Formula | 化学计量与经验式

Problem: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its molar mass is about 180 g mol⁻¹. Determine its empirical and molecular formula.

问题：某化合物含碳 40.0%、氢 6.7%、氧 53.3%（质量分数），其摩尔质量约为 180 g mol⁻¹。推算它的经验式和分子式。

Step 1: Assume 100 g of compound, so masses are C: 40.0 g, H: 6.7 g, O: 53.3 g.

步骤1：假设样品为 100 g，则各元素质量分别为 C: 40.0 g，H: 6.7 g，O: 53.3 g。

Step 2: Convert each mass to moles.

步骤2：将质量转换为摩尔数。

mol C = 40.0 g ÷ 12.01 g mol⁻¹ = 3.33 mol

mol H = 6.7 g ÷ 1.008 g mol⁻¹ = 6.65 mol

mol O = 53.3 g ÷ 16.00 g mol⁻¹ = 3.33 mol

Step 3: Divide by the smallest number of moles (3.33) to get the ratio.

步骤3：除以最小摩尔数（3.33）得到最简比。

C : H : O = 3.33/3.33 : 6.65/3.33 : 3.33/3.33 ≈ 1 : 2 : 1

Step 4: Empirical formula is CH₂O. The empirical formula mass = 12.01 + 2×1.008 + 16.00 = 30.03 g mol⁻¹.

步骤4：经验式为 CH₂O，经验式质量 = 30.03 g mol⁻¹。

Step 5: Molecular formula: n = molar mass / empirical mass = 180 / 30.03 ≈ 6, so molecular formula = C₆H₁₂O₆.

步骤5：分子式 n = 摩尔质量 / 经验式质量 = 180 / 30.03 ≈ 6，因此分子式为 C₆H₁₂O₆。

2. Gas Laws and Molar Volume | 气体定律与摩尔体积

Problem: At 25 °C and 100 kPa, 0.500 g of a volatile liquid is vaporised in a 250 cm³ flask. What is the molar mass of the liquid? (R = 8.31 J K⁻¹ mol⁻¹)

问题：在 25 °C、100 kPa 下，将 0.500 g 挥发性液体在一个 250 cm³ 烧瓶中完全气化，求该液体的摩尔质量。（R = 8.31 J K⁻¹ mol⁻¹）

Step 1: Use the ideal gas equation pV = nRT, rearranged to n = pV / RT.

步骤1：运用理想气体方程 pV = nRT，变形为 n = pV / RT。

Step 2: Convert units: p = 100 kPa = 100 × 10³ Pa, V = 250 cm³ = 250 × 10⁻⁶ m³, T = 25 + 273 = 298 K.

步骤2：单位换算：p = 100 kPa = 1.00×10⁵ Pa，V = 250 cm³ = 2.50×10⁻⁴ m³，T = 298 K。

n = (1.00×10⁵ Pa × 2.50×10⁻⁴ m³) / (8.31 J K⁻¹ mol⁻¹ × 298 K) ≈ 0.0101 mol

Step 3: Molar mass M = mass / n = 0.500 g / 0.0101 mol ≈ 49.5 g mol⁻¹.

步骤3：摩尔质量 M = 质量 / 物质的量 = 0.500 g / 0.0101 mol ≈ 49.5 g mol⁻¹。

3. Titration and Back Titration | 滴定与返滴定

Problem: 25.0 cm³ of 0.200 mol dm⁻³ HCl is neutralised by 20.0 cm³ of NaOH solution. Find the concentration of NaOH.

问题：25.0 cm³ 0.200 mol dm⁻³ 的 HCl 被 20.0 cm³ NaOH 溶液中和，求 NaOH 溶液的浓度。

Step 1: Write the neutralisation reaction: HCl + NaOH → NaCl + H₂O. Mole ratio = 1:1.

步骤1：写出中和反应：HCl + NaOH → NaCl + H₂O，摩尔比为 1:1。

Step 2: Calculate moles of HCl used: n(HCl) = c × V = 0.200 mol dm⁻³ × (25.0/1000) dm³ = 0.00500 mol.

步骤2：计算 HCl 物质的量：n(HCl) = 0.200 × 0.0250 = 0.00500 mol。

Step 3: By the 1:1 ratio, n(NaOH) = 0.00500 mol.

步骤3：根据 1:1 摩尔比，n(NaOH) = 0.00500 mol。

Step 4: Concentration of NaOH = n / V = 0.00500 mol / (20.0/1000) dm³ = 0.250 mol dm⁻³.

步骤4：NaOH 浓度 = 0.00500 mol / 0.0200 dm³ = 0.250 mol dm⁻³。

4. Hess’s Law and Enthalpy Cycles | 赫斯定律与焓循环

Problem: Given: (1) C(s) + O₂(g) → CO₂(g) ΔH = −394 kJ mol⁻¹; (2) CO(g) + ½O₂(g) → CO₂(g) ΔH = −283 kJ mol⁻¹. Calculate ΔH for: C(s) + ½O₂(g) → CO(g).

问题：已知 (1) C(s) + O₂(g) → CO₂(g) ΔH = −394 kJ mol⁻¹；(2) CO(g) + ½O₂(g) → CO₂(g) ΔH = −283 kJ mol⁻¹。计算反应 C(s) + ½O₂(g) → CO(g) 的 ΔH。

Step 1: Construct an enthalpy cycle with CO₂ as a common intermediate.

步骤1：构建以 CO₂ 为共同中间体的焓循环。

Step 2: Path A: C(s) + O₂(g) → CO₂(g) ΔH₁ = −394 kJ.

步骤2：途径 A：C(s) + O₂(g) → CO₂(g) ΔH₁ = −394 kJ。

Path B: C(s) + ½O₂(g) → CO(g) [unknown ΔH] followed by CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = −283 kJ.

途径 B：C(s) + ½O₂(g) → CO(g) [未知 ΔH]，然后 CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = −283 kJ。

Step 3: According to Hess’s Law, ΔH₁ = ΔH + ΔH₂, so ΔH = ΔH₁ − ΔH₂ = −394 − (−283) = −111 kJ mol⁻¹.

步骤3：由赫斯定律，ΔH₁ = ΔH + ΔH₂，因此 ΔH = −394 − (−283) = −111 kJ mol⁻¹。

5. Rate Equations and Initial Rates | 速率方程与初始速率法

Problem: For the reaction 2A + B → C, the following initial rate data were obtained. Find the rate equation and rate constant k.

问题：对于反应 2A + B → C，测得如下初始速率数据。确定速率方程和速率常数 k。

[A] (mol dm⁻³)	[B] (mol dm⁻³)	Initial rate (mol dm⁻³ s⁻¹)
0.10	0.10	2.0 × 10⁻⁴
0.20	0.10	8.0 × 10⁻⁴
0.10	0.20	4.0 × 10⁻⁴

Step 1: Assume rate = k[A]^m[B]ⁿ. Compare experiments 1 and 2: [A] doubles, [B] constant, rate quadruples (×4). Thus 2^m = 4 → m = 2.

步骤1：设速率方程 rate = k[A]^m[B]ⁿ。对比实验1和2：[A]加倍，[B]不变，速率变为4倍，故 2^m = 4，m = 2。

Step 2: Compare experiments 1 and 3: [B] doubles, [A] constant, rate doubles (×2). Thus 2ⁿ = 2 → n = 1.

步骤2：对比实验1和3：[B]加倍，[A]不变，速率加倍，故 2ⁿ = 2，n = 1。

Step 3: Rate equation: rate = k[A]²[B]. Use data from experiment 1: 2.0×10⁻⁴ = k (0.10)²(0.10) → k = 2.0×10⁻⁴ / (0.0010) = 0.20 dm⁶ mol⁻² s⁻¹.

步骤3：速率方程为 rate = k[A]²[B]。用实验1数据：2.0×10⁻⁴ = k×(0.10)²×0.10，得 k = 0.20 dm⁶ mol⁻² s⁻¹。

6. Equilibrium Constant K_c | 平衡常数 K_c

Problem: For H₂(g) + I₂(g) ⇌ 2HI(g), at equilibrium the concentrations are [H₂] = 0.10 M, [I₂] = 0.20 M, [HI] = 0.40 M. Calculate K_c and determine the units.

问题：对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，平衡时各物质浓度为 [H₂]=0.10 M，[I₂]=0.20 M，[HI]=0.40 M。计算 K_c 并确定其单位。

Step 1: Write the expression: K_c = [HI]² / ([H₂][I₂]).

步骤1：写出平衡常数表达式：K_c = [HI]² / ([H₂][I₂])。

Step 2: Substitute values: K_c = (0.40)² / (0.10 × 0.20) = 0.16 / 0.020 = 8.0.

步骤2：代入数值：K_c = (0.40)² / (0.10×0.20) = 0.16/0.020 = 8.0。

Step 3: Units: (mol dm⁻³)² / [(mol dm⁻³) × (mol dm⁻³)] = no unit (dimensionless).

步骤3：单位分析：(mol dm⁻³)² / [(mol dm⁻³)(mol dm⁻³)]，分子分母指数相同，故无单位。

7. Acids, Bases and Buffer Calculations | 酸、碱与缓冲溶液计算

Problem: Calculate the pH of a buffer made by mixing 50.0 cm³ of 0.10 mol dm⁻³ CH₃COOH with 25.0 cm³ of 0.20 mol dm⁻³ CH₃COONa. (K_a of acetic acid = 1.8 × 10⁻⁵)

问题：由 50.0 cm³ 0.10 mol dm⁻³ 醋酸和 25.0 cm³ 0.20 mol dm⁻³ 醋酸钠混合制成缓冲溶液，计算其 pH。（醋酸 K_a = 1.8×10⁻⁵）

Step 1: Use the Henderson-Hasselbalch equation: pH = pK_a + log([A⁻]/[HA]).

步骤1：使用 Henderson-Hasselbalch 方程：pH = pK_a + log([A⁻]/[HA])。

Step 2: Calculate pK_a = −log(1.8×10⁻⁵) = 4.74.

步骤2：计算 pK_a = −log(1.8×10⁻⁵) = 4.74。

Step 3: Find moles after mixing: n(HA) = 0.10 × 0.0500 = 0.0050 mol; n(A⁻) = 0.20 × 0.0250 = 0.0050 mol. Total volume = 0.0750 dm³.

步骤3：混合后物质的量：n(HA)=0.10×0.0500=0.0050 mol，n(A⁻)=0.20×0.0250=0.0050 mol。总体积=0.0750 dm³。

Step 4: Ratio [A⁻]/[HA] = moles ratio = 0.0050/0.0050 = 1.0, so log(1) = 0.

步骤4：[A⁻]/[HA] 的比值等于物质的量比 = 1.0，因此 log(1)=0。

Step 5: pH = pK_a + 0 = 4.74.

步骤5：pH = 4.74。

8. Redox Titrations and Electrode Potentials | 氧化还原滴定与电极电势

Problem: A 25.0 cm³ sample of Fe²⁺ solution required 18.5 cm³ of 0.0200 mol dm⁻³ KMnO₄ for complete oxidation in acidic medium. Determine the concentration of Fe²⁺. (MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O)

问题：用 0.0200 mol dm⁻³ KMnO₄ 滴定 25.0 cm³ Fe²⁺ 溶液，在酸性介质中完全氧化需 18.5 cm³。求

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IB and OCR Chemistry: Detailed Solutions to Typical Example Problems | IB OCR 化学：典型例题详解