📚 Worked Examples in A-Level AQA Chemistry | A-Level AQA 化学典型例题详解
Mastering A-level Chemistry requires more than memorising facts — it demands the ability to apply concepts to unfamiliar problems. In this article, we walk through four carefully selected worked examples that mirror the style and difficulty of AQA exam questions. Each example is broken down step by step, with common pitfalls highlighted, so you can build confidence and precision. The topics covered include buffer pH calculations, rate equations from initial rates, transition metal colour and substitution reactions, and organic structure elucidation using mass and IR spectra.
精通 A-level 化学不仅需要记忆事实，更需要将概念应用于陌生问题的能力。本文精选了四道例题，完全模拟 AQA 考试的风格与难度，并逐步拆解，指出常见误区，帮助你建立信心和准确性。涵盖的主题包括缓冲溶液 pH 计算、从初始速率确定速率方程、过渡金属颜色与取代反应，以及利用质谱和红外光谱解析有机结构。
1. Example 1: Buffer pH Calculation | 例题1：缓冲溶液pH计算
A buffer solution is prepared by mixing 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid (CH₃COOH) with 50.0 cm³ of 0.200 mol dm⁻³ sodium ethanoate (CH₃COONa). The Ka of ethanoic acid is 1.74 × 10⁻⁵ mol dm⁻³ at 298 K. Calculate the pH of this buffer solution.
将 50.0 cm³ 0.200 mol dm⁻³ 的乙酸 (CH₃COOH) 与 50.0 cm³ 0.200 mol dm⁻³ 的乙酸钠 (CH₃COONa) 混合，制备缓冲溶液。已知乙酸在 298 K 时的 Ka = 1.74 × 10⁻⁵ mol dm⁻³。计算该缓冲溶液的 pH。
2. Step-by-Step Working for Buffer | 缓冲液计算分步详解
Because the solutions are mixed in equal volumes, the total volume doubles and both concentrations are halved. After mixing, [CH₃COOH] = 0.100 mol dm⁻³ and [CH₃COO⁻] = 0.100 mol dm⁻³.
由于两种溶液等体积混合，总体积加倍，两者的浓度均减半。混合后，[CH₃COOH] = 0.100 mol dm⁻³，[CH₃COO⁻] = 0.100 mol dm⁻³。
Write the expression for the acid dissociation constant: Ka = [H⁺][CH₃COO⁻] / [CH₃COOH].
写出酸解离常数表达式：Ka = [H⁺][CH₃COO⁻] / [CH₃COOH]。
Rearrange to find [H⁺]: [H⁺] = Ka × [CH₃COOH] / [CH₃COO⁻].
整理求得 [H⁺]：[H⁺] = Ka × [CH₃COOH] / [CH₃COO⁻]。
Substitute the values: [H⁺] = (1.74 × 10⁻⁵) × (0.100) / (0.100) = 1.74 × 10⁻⁵ mol dm⁻³.
代入数值：[H⁺] = (1.74 × 10⁻⁵) × (0.100) / (0.100) = 1.74 × 10⁻⁵ mol dm⁻³。
Then pH = –log₁₀[H⁺] = –log₁₀(1.74 × 10⁻⁵) ≈ 4.76.
因此 pH = –log₁₀[H⁺] = –log₁₀(1.74 × 10⁻⁵) ≈ 4.76。
When the acid and salt concentrations are equal, pH = pKa. Indeed, pKa = –log(1.74 × 10⁻⁵) ≈ 4.76.
当酸和盐的浓度相等时，pH = pKa。实际上，pKa = –log(1.74 × 10⁻⁵) ≈ 4.76。
If a small amount of strong acid is added, the added H⁺ reacts with CH₃COO⁻ to form CH₃COOH, slightly changing the ratio and allowing the new pH to be calculated using the same formula.
如果加入少量强酸，加入的 H⁺ 会与 CH₃COO⁻ 反应生成 CH₃COOH，略微改变比例，并可使用同一公式计算新的 pH。
3. Common Pitfalls in Buffer Calculations | 缓冲计算常见误区
Pitfall 1: Forgetting to account for dilution when mixing equal volumes. Always calculate the new concentrations after mixing.
误区1：等体积混合时忘记考虑稀释。务必重新计算混合后的浓度。
Pitfall 2: Using the original undiluted concentrations directly in the Ka expression. This leads to an overestimated buffering capacity.
误区2：将混合前未稀释的浓度直接代入 Ka 表达式。这会导致高估缓冲能力。
Pitfall 3: Assuming that adding a strong acid does not shift the equilibrium. Remember that the added H⁺ is consumed by the conjugate base, so [HA] increases and [A⁻] decreases.
误区3：认为加强酸不会移动平衡。记住加入的 H⁺ 会被共轭碱消耗，因此 [HA] 增加，[A⁻] 减少。
Pitfall 4: Misplacing the Ka expression — it is always [H⁺][A⁻]/[HA], not [HA]/[H⁺][A⁻].
误区4：Ka 表达式出错 — 总是 [H⁺][A⁻]/[HA]，而非 [HA]/[H⁺][A⁻]。
4. Example 2: Determining Rate Equation from Initial Rates | 例题2：由初始速率确定速率方程
The reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) is studied by the initial rates method at a fixed temperature. The table below shows the results:
在固定温度下，用初始速率法研究反应 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g)。下表为结果：
| Experiment | [NO] / mol dm⁻³ | [H₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
| 1 | 0.10 | 0.10 | 1.2 × 10⁻³ |
| 2 | 0.10 | 0.20 | 2.4 × 10⁻³ |
| 3 | 0.20 | 0.10 | 4.8 × 10⁻³ |
Determine the rate equation and calculate the value of the rate constant, k, giving its units.
确定速率方程，并计算速率常数 k 的值与单位。
5. Working Out the Orders and Rate Constant | 推导反应级数与速率常数
The rate equation can be expressed as: Rate = k[NO]ᵐ[H₂]ⁿ. Compare experiments 1 and 2: [NO] is constant, [H₂] doubles from 0.10 to 0.20 mol dm⁻³, and the rate doubles from 1.2 × 10⁻³ to 2.4 × 10⁻³. Therefore, the reaction is first order with respect to H₂ (n = 1).
速率方程可表示为： Rate = k[NO]ᵐ[H₂]ⁿ。对比实验 1 和 2：[NO] 恒定，[H₂] 从 0.10 加倍到 0.20 mol dm⁻³，速率从 1.2 × 10⁻³ 加倍到 2.4 × 10⁻³。因此，对 H₂ 为一级反应 (n = 1)。
Compare experiments 1 and 3: [H₂] is constant at 0.10 mol dm⁻³, [NO] doubles from 0.10 to 0.20 mol dm⁻³, and the rate increases by a factor of 4 (from 1.2 × 10⁻³ to 4.8 × 10⁻³). So, 2ᵐ = 4, thus m = 2. The reaction is second order with respect to NO.
对比实验 1 和 3：[H₂] 恒定为 0.10 mol dm⁻³，[NO] 加倍，速率增大为原来的 4 倍。所以 2ᵐ = 4，因此 m = 2。反应对 NO 为二级。
The overall rate equation is: Rate = k[NO]²[H₂].
总速率方程为：Rate = k[NO]²[H₂]。
To find k, use any experiment, e.g., experiment 1: 1.2 × 10⁻³ = k × (0.10)² × (0.10). Thus k = (1.2 × 10⁻³) / (1.0 × 10⁻³) = 1.2 dm⁶ mol⁻² s⁻¹.
求 k，取任一实验数据，如实验 1：1.2 × 10⁻³ = k × (0.10)² × (0.10)。因此 k = (1.2 × 10⁻³) / (1.0 × 10⁻³) = 1.2 dm⁶ mol⁻² s⁻¹。
Units: dm⁶ mol⁻² s⁻¹ come from (mol dm⁻³)⁻² (mol dm⁻³)⁻¹ s⁻¹ = mol⁻³ dm⁹ s⁻¹? Let’s verify: [k] = rate / ([NO]²[H₂]) = (mol dm⁻³ s⁻¹) / ((mol dm⁻³)² (mol dm⁻³)) = (mol dm⁻³ s⁻¹) / (mol³ dm⁻⁹) = mol⁻² dm⁶ s⁻¹. The unit is dm⁶ mol⁻² s⁻¹.
单位：dm⁶ mol⁻² s⁻¹，由 rate / ([NO]²[H₂]) 推导，得到 mol⁻² dm⁶ s⁻¹。
6. Deducing a Mechanism Consistent with Rate Equation | 推断与速率方程相符的机理
A proposed two-step mechanism: (1) 2NO ⇌ N₂O₂ (fast, equilibrium); (2) N₂O₂ + H₂ → N₂O + H₂O (slow); followed by a fast step N₂O + H₂ → N₂ + H₂O. The rate-determining step is the slow step, so rate = k₂[N₂O₂][H₂]. From the fast equilibrium, K = [N₂O₂]/[NO]², so [N₂O₂] = K[NO]². Substituting gives rate = k₂K[NO]²[H₂], which matches the experimental rate equation.
提议的两步机理：(1) 2NO ⇌ N₂O₂（快，平衡）；(2) N₂O₂ + H₂ → N₂O + H₂O（慢）；随后是快步骤 N₂O + H₂ → N₂ + H₂O。决速步为慢步骤，因此 rate = k₂[N₂O₂][H₂]。由快速平衡可得 K = [N₂O₂]/[NO]²，即 [N₂O₂] = K[NO]²。代入得 rate = k₂K[NO]²[H₂]，与实验速率方程一致。
This mechanism is plausible because the molecularity of the slow step matches the orders (termolecular with respect to N₂O₂ and H₂), and the equilibrium provides the [NO]² dependence.
该机理合理，因为慢步骤的分子数对应级数（涉及 N₂O₂ 和 H₂ 的三分子反应），且平衡给出了对 [NO]² 的依赖关系。
7. Example 3: Transition Metal Complex Colour and Reactions | 例题3：过渡金属配合物的颜色与反应
A student has a pale blue aqueous solution containing the hexaaquacopper(II) ion, [Cu(H₂O)₆]²⁺. When concentrated hydrochloric acid is added dropwise, the solution turns green and then yellow-brown. When excess aqueous ammonia is added to the original blue solution, a deep blue solution forms. Explain these observations in terms of electronic structure and ligand substitution.
某学生有一种淡蓝色的水溶液，含有六水合铜(II)离子 [Cu(H₂O)₆]²⁺。滴加浓盐酸时，溶液变为绿色，最后呈黄棕色。向原蓝色溶液中加入过量氨水，则形成深蓝色溶液。依据电子结构和配体取代解释这些观察现象。
8. Explaining Colour Changes using d-Orbital Splitting | 用d轨道分裂解释颜色变化
The [Cu(H₂O)₆]²⁺ ion has a d⁹ configuration, with one unpaired electron in the dₓ²₋ᵧ² orbital. In an octahedral field of water ligands, the d-orbitals split into two sets: lower energy t₂g and higher energy eg. The energy gap Δoct corresponds to absorption in the red/orange region, transmitting a pale blue colour.
[Cu(H₂O)₆]²⁺ 离子具有 d⁹ 构型，在 dₓ²₋ᵧ² 轨道上有一个未成对电子。在水分子的八面体场中，d 轨道分裂为两组：能量较低的 t₂g 和较高的 eg。能隙 Δoct 对应于红/橙光的吸收，透过淡蓝色。
Chloride ions are weaker-field ligands than water, but when concentrated HCl is added, Cl⁻ ligands replace water molecules stepwise, eventually forming [CuCl₄]²⁻, which has a tetrahedral geometry. Tetrahedral splitting Δtet is smaller than Δoct, so the absorption shifts to lower energy (longer wavelength), giving a yellow-brown colour. The intermediate green colour is due to a mixture of complexes with different ligands.
氯离子是比水弱的场配体，但加入浓 HCl 时，Cl⁻ 配体逐步取代水分子，最终形成四氯合铜(II)离子 [CuCl₄]²⁻，呈四面体形。四面体分裂 Δtet 小于八面体分裂，因此吸收红移至更低能量（更长波长），呈现黄棕色。中间的绿色是由于不同配体配合物的混合。
9. Ligand Substitution and Identification | 配体取代与鉴定
Adding excess ammonia to the blue [Cu(H₂O)₆]²⁺ causes a ligand substitution reaction: four ammonia molecules replace four water molecules, forming [Cu(NH₃)₄(H₂O)₂]²⁺. Ammonia is a stronger-field ligand than water, which increases the crystal field splitting Δ. The absorption moves to higher energy, so the complex appears violet/blue, but the observed deep blue is actually due to the specific Δ of this complex. The equation: [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O.
向蓝色 [Cu(H₂O)₆]²⁺ 中加入过量氨水，发生配体取代反应：四个氨分子取代四个水分子，形成 [Cu(NH₃)₄(H₂O)₂]²⁺。氨是比水更强的场配体，增大了晶体场分裂能 Δ。吸收蓝移至更高能量，因此配合物本应呈紫/蓝色，但实际观察到的深蓝色是该配合物特定 Δ 的结果。方程式：[Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O。
This substitution is incomplete – water ligands remain in the axial positions due to Jahn–Teller distortion. The reaction is a classic test for Cu²⁺ ions, producing a distinctive deep blue colour.
这种取代不完全 — 由于姜-泰勒畸变，轴向位置仍留有水配体。该反应是检验 Cu²⁺ 离子的经典方法，生成特征的深蓝色。
10. Example 4: Organic Structure Elucidation using Spectra | 例题4：利用波谱解析有机结构
An unknown organic compound X has a molecular ion peak at m/z = 72 in its mass spectrum. The infrared spectrum shows a strong, sharp absorption at 1715 cm⁻¹ and no broad absorption around 3400 cm⁻¹. Using the mass spectral data below, deduce the structure of X and give its systematic name. Mass spectrum (major peaks): m/z 72 (molecular ion, small), 57 (base peak), 43, 29.
某未知有机物 X 的质谱中分子离子峰为 m/z = 72。红外光谱在 1715 cm⁻¹ 处有强而尖的吸收，并且在 3400 cm⁻¹ 附近无宽吸收。使用以下质谱数据推断 X 的结构并给出系统命名。质谱（主要峰）：m/z 72（分子离子峰，小峰），57（基峰），43，29。
11. Analysing Mass Spectrum and IR Spectrum | 分析质谱与红外光谱
The strong IR absorption at 1715 cm⁻¹ is characteristic of a C=O stretch, suggesting a carbonyl compound (ketone or aldehyde). The absence of a broad O–H absorption near 3400 cm⁻¹ rules out alcohols and carboxylic acids.
IR 在 1715 cm⁻¹ 的强吸收是 C=O 伸缩振动的特征，表明是羰基化合物（酮或醛）。在 3400 cm⁻¹ 附近无宽 O–H 吸收，排除了醇和羧酸。
Molecular ion peak m/z 72 gives the molecular mass. Possible carbonyl compounds with formula CₙH₂ₙO, where 12n + 2n + 16 = 14n + 16 = 72 → 14n = 56 → n = 4. So the molecular formula is C₄H₈O.
分子离子峰 m/z 72 给出分子质量。可能的羰基化合物通式为 CₙH₂ₙO，12n + 2n + 16 = 14n + 16 = 72 → n = 4。分子式为 C₄H₈O。
Isomers with this formula: butanal (CH₃CH₂CH₂CHO) and butanone (CH₃COCH₂CH₃). The mass spectrum helps distinguish them: base peak at m/z 57 corresponds to loss of a methyl radical (M – 15) from the molecular ion only if the methyl group is attached to the carbonyl (alpha cleavage). In butanone, cleavage of the bond adjacent to the carbonyl yields CH₃CH₂C≡O⁺ (m/z 57) as the acylium ion, which is very stable. Butanal would give a prominent M – 29 (loss of ethyl) peak or M – 43 differently. The presence of m/z 43 (likely CH₃CO⁺) also supports a methyl ketone.
该分子式的异构体有：丁醛 (CH₃CH₂CH₂CHO) 和丁酮 (CH₃COCH₂CH₃)。质谱可区分二者：基峰 m/z 57 对应于从分子离子失去一个甲基自由基 (M – 15)，仅当甲基连接在羰基上（α-裂解）时容易发生。在丁酮中，邻近羰基的键断裂产生 CH₃CH₂C≡O⁺ 酰基正离子 (m/z 57)，非常稳定。丁醛则会给出显著的 M – 29（失乙基）峰或不同的 M – 43 峰。同时，出现 m/z 43（很可能为 CH₃CO⁺）也支持甲基酮。
The peak at m/z 29 is consistent with an ethyl cation (C₂H₅⁺), which can form from butanone. Thus, compound X is butanone (CH₃COCH₂CH₃).
m/z 29 的峰符合乙基正离子 (C₂H₅⁺)，可由丁酮生成。因此，化合物 X 是丁酮 (CH₃COCH₂CH₃)。
12. Putting Together the Structure and Naming | 整合结构并命名
From the combined evidence: molecular formula C₄H₈O, carbonyl group, fragmentation pattern consistent with a methyl ketone and ethyl group. The systematic IUPAC name is butan-2-one. The structure is CH₃–CO–CH₂–CH₃.
综合证据：分子式 C₄H₈O，含羰基，碎片模式与甲基酮及乙基相符。系统 IUPAC 名称为丁-2-酮。结构为 CH₃–CO–CH₂–CH₃。
This example illustrates how integrating IR and mass spectrometry allows unambiguous identification without NMR. The base peak at m/z 57 is key; it is produced via alpha cleavage: CH₃–CO–CH₂CH₃ → CH₃CH₂C≡O⁺ + •CH₃.
该例题说明如何结合 IR 和质谱在无需 NMR 的情况下明确鉴定。基峰 m/z 57 是关键；它由 α-裂解产生：CH₃–CO–CH₂CH₃ → CH₃CH₂C≡O⁺ + •CH₃。
When approaching similar exam problems, always check the molecular formula first, identify functional groups from IR, then use mass fragmentation to distinguish between isomers.
在面对类似考试题时，务必首先确定分子式，从 IR 鉴定官能团，然后利用质谱碎片区分异构体。
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