📚 Calculation Questions in A2 Inorganic Chemistry (Oxford AQA International A-Level) | A2 无机化学计算题型（牛津 AQA 国际 A-Level）
Mastering the numerical aspects of A2 Inorganic Chemistry is essential for achieving top grades in the Oxford AQA International A-Level exam. This article covers the most common calculation-based topics, including thermodynamics, redox equilibria, pH, and transition metal chemistry, with clear step-by-step methods and examples. You will learn how to apply Born–Haber cycles, calculate cell potentials under non-standard conditions, determine buffer pH, and work with solubility products and colorimetry data – all of which are regularly tested in topic tests and final papers.
掌握 A2 无机化学中的计算题型，对于在牛津 AQA 国际 A-Level 考试中取得高分至关重要。本文涵盖最常见的计算类主题，包括热力学、氧化还原平衡、pH 和过渡金属化学，提供清晰的逐步解题方法和实例。你将学会如何运用玻恩–哈伯循环、计算非标准条件下的电池电动势、确定缓冲溶液的 pH，以及处理溶度积和比色法数据——这些内容经常出现在单元测试和正式考卷中。
1. Born–Haber Cycles and Lattice Enthalpy | 玻恩–哈伯循环与晶格焓
A Born–Haber cycle is an energy cycle used to calculate the lattice enthalpy of an ionic compound. It links the enthalpy change of formation to other enthalpy changes such as atomisation, ionisation, electron affinity, and the lattice enthalpy itself. The general equation is:
玻恩–哈伯循环是用于计算离子化合物晶格焓的能量循环。它将生成焓变与原子化、电离、电子亲和能及晶格焓等其他焓变联系起来。通用方程为：
ΔH_f = ΔH_at(M) + IE₁ + ΔH_at(X) + EA + ΔH_lattice
where ΔH_f is the standard enthalpy of formation, ΔH_at(M) is the atomisation enthalpy of the metal, IE₁ is the first ionisation energy, ΔH_at(X) is the atomisation enthalpy of the non-metal, EA is the electron affinity, and ΔH_lattice is the lattice enthalpy (exothermic, so it is usually assigned a negative value in the equation). When solving for the lattice enthalpy, rearrange:
其中 ΔH_f 是标准生成焓，ΔH_at(M) 是金属的原子化焓，IE₁ 是第一电离能，ΔH_at(X) 是非金属的原子化焓，EA 是电子亲和能，ΔH_lattice 是晶格焓（放热，在方程中通常取负值）。当求解晶格焓时，移项得：
ΔH_lattice = ΔH_f – [ΔH_at(M) + IE₁ + ΔH_at(X) + EA]
For example, to calculate the lattice enthalpy of NaCl using the data: ΔH_f(NaCl) = –411 kJ mol⁻¹, ΔH_at(Na) = +108 kJ mol⁻¹, IE₁(Na) = +496 kJ mol⁻¹, ΔH_at(Cl) = +122 kJ mol⁻¹, EA(Cl) = –349 kJ mol⁻¹. Then ΔH_lattice = –411 – (108 + 496 + 122 – 349) = –788 kJ mol⁻¹. Always check the sign; lattice enthalpy is exothermic, so it should be negative.
例如，利用以下数据计算 NaCl 的晶格焓：ΔH_f(NaCl) = –411 kJ mol⁻¹，ΔH_at(Na) = +108 kJ mol⁻¹，IE₁(Na) = +496 kJ mol⁻¹，ΔH_at(Cl) = +122 kJ mol⁻¹，EA(Cl) = –349 kJ mol⁻¹。则 ΔH_lattice = –411 – (108 + 496 + 122 – 349) = –788 kJ mol⁻¹。务必检查符号；晶格焓是放热的，应为负值。
2. Enthalpy of Solution and Hydration | 溶解焓与水合焓
When an ionic compound dissolves in water, the overall enthalpy change of solution is the sum of the lattice dissociation enthalpy (endothermic, breaking the lattice) and the hydration enthalpies of the ions (exothermic). The cycle is:
当离子化合物溶于水时，总溶解焓变等于晶格解离焓（吸热，破坏晶格）和离子水合焓（放热）之和。其循环为：
ΔH_sol = –ΔH_lattice + ΣΔH_hyd
where ΣΔH_hyd is the sum of the hydration enthalpies of the cation and anion. Alternatively, using lattice dissociation enthalpy (ΔH_LED = –ΔH_lattice): ΔH_sol = ΔH_LED + ΔH_hyd(cation) + ΔH_hyd(anion). In calculations, you may be asked to find an unknown hydration enthalpy or lattice enthalpy from solution calorimetry data. For instance, given ΔH_sol(NaCl) = +4 kJ mol⁻¹, lattice enthalpy = –788 kJ mol⁻¹ (so lattice dissociation enthalpy = +788 kJ mol⁻¹), and ΔH_hyd(Cl⁻) = –381 kJ mol⁻¹, calculate ΔH_hyd(Na⁺): +4 = +788 + ΔH_hyd(Na⁺) + (–381) → ΔH_hyd(Na⁺) = –403 kJ mol⁻¹.
其中 ΣΔH_hyd 是阳离子和阴离子水合焓的总和。或者，使用晶格解离焓（ΔH_LED = –ΔH_lattice）：ΔH_sol = ΔH_LED + ΔH_hyd(cation) + ΔH_hyd(anion)。在计算中，你可能需要利用量热数据求解未知的水合焓或晶格焓。例如，已知 ΔH_sol(NaCl) = +4 kJ mol⁻¹，晶格焓 = –788 kJ mol⁻¹（因此晶格解离焓 = +788 kJ mol⁻¹），ΔH_hyd(Cl⁻) = –381 kJ mol⁻¹，计算 ΔH_hyd(Na⁺)：+4 = +788 + ΔH_hyd(Na⁺) + (–381) → ΔH_hyd(Na⁺) = –403 kJ mol⁻¹。
3. Entropy and Gibbs Free Energy in Inorganic Reactions | 无机反应中的熵与吉布斯自由能
The feasibility of an inorganic reaction is determined by the Gibbs free energy change: ΔG = ΔH – TΔS. For a reaction to be thermodynamically feasible, ΔG must be negative. In calculations, you will often use standard entropy values (S°) and standard enthalpy of formation data to find ΔG°. The equation is:
无机反应的自发性由吉布斯自由能变决定：ΔG = ΔH – TΔS。当 ΔG 为负时，反应在热力学上可行。计算中，常使用标准熵值（S°）和标准生成焓数据求 ΔG°。公式为：
ΔG° = ΔH° – TΔS°
where ΔH° = ΣΔH_f°(products) – ΣΔH_f°(reactants) and ΔS° = ΣS°(products) – ΣS°(reactants). Temperature T is in kelvin, and ΔS° is usually in J K⁻¹ mol⁻¹, so you must convert to kJ by dividing by 1000. For example, the decomposition of calcium carbonate: CaCO₃(s) → CaO(s) + CO₂(g). Given ΔH° = +178 kJ mol⁻¹, ΔS° = +160.6 J K⁻¹ mol⁻¹. At 298 K, ΔG° = 178 – (298 × 0.1606) = +130 kJ mol⁻¹, so the reaction is not feasible at 298 K. Find the temperature at which the reaction becomes feasible (ΔG° = 0): T = ΔH°/ΔS° = 178 / 0.1606 = 1108 K. Above this temperature, ΔG° < 0.
其中 ΔH° = ΣΔH_f°(产物) – ΣΔH_f°(反应物)，ΔS° = ΣS°(产物) – ΣS°(反应物)。温度 T 以开尔文为单位，ΔS° 通常以 J K⁻¹ mol⁻¹ 给出，因此需除以 1000 转换为 kJ。例如，碳酸钙分解：CaCO₃(s) → CaO(s) + CO₂(g)。已知 ΔH° = +178 kJ mol⁻¹，ΔS° = +160.6 J K⁻¹ mol⁻¹。在 298 K 时，ΔG° = 178 – (298 × 0.1606) = +130 kJ mol⁻¹，故 298 K 下反应不可行。求反应变得可行的温度（ΔG° = 0）：T = ΔH°/ΔS° = 178 / 0.1606 = 1108 K。高于此温度，ΔG° < 0。
4. Redox Titrations | 氧化还原滴定
Redox titrations are frequently used to determine the concentration of metal ions or oxidising agents. The most common example involves manganate(VII) ions oxidising iron(II) under acidic conditions. The half-equations and overall equation must be combined to find the stoichiometric ratio:
氧化还原滴定常用于测定金属离子或氧化剂的浓度。最常见的例子是在酸性条件下用高锰酸根离子氧化铁(II)。必须结合半反应和总反应得出生量关系：
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Fe²⁺ → Fe³⁺ + e⁻
The overall ratio is 1 MnO₄⁻ reacts with 5 Fe²⁺. In a typical calculation, 25.0 cm³ of Fe²⁺ solution required 22.50 cm³ of 0.0200 mol dm⁻³ KMnO₄ to reach the endpoint. Moles of MnO₄⁻ = 0.02250 × 0.0200 = 4.50 × 10⁻⁴ mol. Moles of Fe²⁺ = 5 × 4.50 × 10⁻⁴ = 2.25 × 10⁻³ mol. Concentration of Fe²⁺ = (2.25 × 10⁻³) / 0.0250 = 0.0900 mol dm⁻³. Similar calculations apply to dichromate titrations (Cr₂O₇²⁻ : Fe²⁺ = 1 : 6) or thiosulfate-iodine titrations.
总反应比为 1 MnO₄⁻ 与 5 Fe²⁺ 反应。典型计算中，25.0 cm³ Fe²⁺ 溶液需要 22.50 cm³ 0.0200 mol dm⁻³ KMnO₄ 滴定至终点。MnO₄⁻ 的物质的量 = 0.02250 × 0.0200 = 4.50 × 10⁻⁴ mol。Fe²⁺ 的物质的量 = 5 × 4.50 × 10⁻⁴ = 2.25 × 10⁻³ mol。Fe²⁺ 浓度 = (2.25 × 10⁻³) / 0.0250 = 0.0900 mol dm⁻³。类似计算适用于重铬酸盐滴定（Cr₂O₇²⁻ : Fe²⁺ = 1 : 6）或硫代硫酸盐-碘滴定。
5. Electrode Potentials and Cell EMF | 电极电势与电池电动势
The EMF of an electrochemical cell is calculated under standard conditions as E°_cell = E°_right – E°_left, where both half-cell potentials are written as reduction potentials. When conditions are non-standard, the Nernst equation is used. For a half-cell with reduced species Red and oxidised species Ox: aOx + ne⁻ ⇌ bRed, the Nernst equation at 298 K simplifies to:
在标准条件下，电池的电动势计算为 E°_cell = E°_right – E°_left，其中两个半电池电势均以还原电势形式给出。当条件非标准时，使用能斯特方程。对于还原型为 Red、氧化型为 Ox 的半反应 aOx + ne⁻ ⇌ bRed，298 K 下的能斯特方程简化为：
E = E° – (0.0592 / n) log₁₀([Red]ᵇ / [Ox]ᵃ)
For example, for the Fe³⁺/Fe²⁺ half-cell: Fe³⁺ + e⁻ ⇌ Fe²⁺, E° = +0.77 V. If [Fe³⁺] = 0.10 mol dm⁻³ and [Fe²⁺] = 1.0 mol dm⁻³, then E = 0.77 – (0.0592/1) log(1.0/0.10) = 0.77 – 0.0592 = 0.71 V. When two non-standard half-cells are connected, calculate each half-cell potential using the Nernst equation, then find cell EMF = E(cathode) – E(anode).
例如，对于 Fe³⁺/Fe²⁺ 半电池：Fe³⁺ + e⁻ ⇌ Fe²⁺，E° = +0.77 V。若 [Fe³⁺] = 0.10 mol dm⁻³，[Fe²⁺] = 1.0 mol dm⁻³，则 E = 0.77 – (0.0592/1) log(1.0/0.10) = 0.77 – 0.0592 = 0.71 V。当连接两个非标准半电池时，先分别用能斯特方程计算各半电池电势，然后求电池电动势 EMF = E(阴极) – E(阳极)。
6. Equilibrium Constants Kc and Kp | 平衡常数 Kc 与 Kp
Many inorganic equilibria require calculation of the equilibrium constant in terms of concentration (Kc) or partial pressure (Kp). The general expression for the reaction aA + bB ⇌ cC + dD is:
许多无机平衡需要计算用浓度（Kc）或分压（Kp）表示的平衡常数。对于反应 aA + bB ⇌ cC + dD，其通式为：
Kc = [C]ᶜ [D]ᵈ / ([A]ᵃ [B]ᵇ)
For heterogeneous equilibria, solids are omitted from the expression. In a common question, you might be given the initial moles, equilibrium moles, and total volume to calculate Kc. For Kp problems, you need the mole fraction of each gas multiplied by the total pressure. Remember that Kp uses partial pressures: p_A = (mole fraction of A) × total pressure.
对于多相平衡，表达式省略固体。常见题型给出初始物质的量、平衡物质的量和总体积，要求计算 Kc。对于 Kp 问题，需要将各气体的摩尔分数乘以总压。记住 Kp 使用分压：p_A = (A 的摩尔分数) × 总压。
Example: N₂O₄(g) ⇌ 2 NO₂(g). 1.00 mol N₂O₄ is placed in a 10.0 dm³ vessel; at equilibrium 0.20 mol N₂O₄ remains. Then moles NO₂ = 2 × (1.00 – 0.20) = 1.60 mol. [N₂O₄] = 0.20/10 = 0.020 mol dm⁻³, [NO₂] = 1.60/10 = 0.160 mol dm⁻³. Kc = (0.160)² / 0.020 = 1.28 mol dm⁻³. If the total pressure at equilibrium is 200 kPa, mole fraction N₂O₄ = 0.20/(0.20+1.60)=0.111, p(N₂O₄)=22.2 kPa; mole fraction NO₂=0.889, p(NO₂)=177.8 kPa. Kp = (177.8)² / 22.2 = 1424 kPa.
示例：N₂O₄(g) ⇌ 2 NO₂(g)。将 1.00 mol N₂O₄ 放入 10.0 dm³ 容器；平衡时剩余 0.20 mol N₂O₄。则 NO₂ 物质的量 = 2 × (1.00 – 0.20) = 1.60 mol。[N₂O₄] = 0.20/10 = 0.020 mol dm⁻³，[NO₂] = 1.60/10 = 0.160 mol dm⁻³。Kc = (0.160)² / 0.020 = 1.28 mol dm⁻³。若平衡总压为 200 kPa，N₂O₄ 摩尔分数 = 0.20/(0.20+1.60) = 0.111，p(N₂O₄) = 22.2 kPa；NO₂ 摩尔分数 = 0.889，p(NO₂) = 177.8 kPa。Kp = (177.8)² / 22.2 = 1424 kPa。
7. pH and Buffer Calculations | pH 与缓冲溶液计算
pH calculations for strong and weak acids are essential. For a strong monoprotic acid, pH = –log₁₀[H⁺], and [H⁺] equals the acid concentration. For a weak acid HA, the dissociation constant Ka is used:
强酸和弱酸的 pH 计算是基础内容。对于一元强酸，pH = –log₁₀[H⁺]，且 [H⁺] 等于酸的浓度。对于弱酸 HA，使用解离常数 Ka：
Ka = [H⁺][A⁻] / [HA]
Assuming [H⁺] = [A⁻] and the dissociation is small, [HA] ≈ initial concentration. Then [H⁺] = √(Ka × [HA]). For buffers, the Henderson–Hasselbalch equation is convenient: pH = pKa + log₁₀([salt]/[acid]), where pKa = –log₁₀Ka. For example, a buffer made from 0.50 mol dm⁻³ CH₃COOH (Ka = 1.8 × 10⁻⁵) and 0.50 mol dm⁻³ CH₃COONa: pKa = 4.74, pH = 4.74 + log(0.50/0.50) = 4.74. If the concentrations differ, such as 0.20 mol dm⁻³ acid and 0.80 mol dm⁻³ salt, pH = 4.74 + log(0.80/0.20) = 5.34.
假设 [H⁺] = [A⁻] 且解离度很小，[HA] ≈ 初始浓度。则 [H⁺] = √(Ka × [HA])。对于缓冲溶液，亨德森-哈塞尔巴尔赫方程很方便：pH = pKa + log₁₀([盐]/[酸])，其中 pKa = –log₁₀Ka。例如，由 0.50 mol dm⁻³ CH₃COOH (Ka = 1.8 × 10⁻⁵) 和 0.50 mol dm⁻³ CH₃COONa 组成的缓冲液：pKa = 4.74，pH = 4.74 + log(0.50/0.50) = 4.74。若浓度不同，如酸 0.20 mol dm⁻³、盐 0.80 mol dm⁻³，则 pH = 4.74 + log(0.80/0.20) = 5.34。
8. Solubility Product Ksp | 溶度积 Ksp
The solubility product is used for sparingly soluble ionic compounds. For a salt with formula MₓAᵧ, the equilibrium is MₓAᵧ(s) ⇌ x Mʸ⁺(aq) + y Aˣ⁻(aq), and Ksp = [Mʸ⁺]ˣ [Aˣ⁻]ʸ. Solubility s (in mol dm⁻³) is related to Ksp through the stoichiometric coefficients. For a 1:1 salt like AgCl, Ksp = s², so s = √Ksp. For a 1:2 salt like PbI₂, Ksp = [Pb²⁺][I⁻]² = (s)(2s)² = 4s³, so s = ³√(Ksp/4). In calculations, you may need to predict precipitation by comparing the ionic product Q with Ksp. If Q > Ksp, precipitation occurs.
溶度积用于难溶离子化合物。对于化学式为 MₓAᵧ 的盐，平衡 MₓAᵧ(s) ⇌ x Mʸ⁺(aq) + y Aˣ⁻(aq)，Ksp = [Mʸ⁺]ˣ [Aˣ⁻]ʸ。溶解度 s（mol dm⁻³）通过化学计量系数与 Ksp 关联。对于 1:1 型盐如 AgCl，Ksp = s²，故 s = √Ksp。对于 1:2 型盐如 PbI₂，Ksp = [Pb²⁺][I⁻]² = (s)(2s)² = 4s³，故 s = ³√(Ksp/4)。在计算中，你可能需要比较离子积 Q 与 Ksp 来判断沉淀。若 Q > Ksp，则发生沉淀。
Example: Ksp of CaF₂ is 3.9 × 10⁻¹¹. Find its solubility in pure water: s = ³√(3.9 × 10⁻¹¹ / 4) = 2.1 × 10⁻⁴ mol dm⁻³. If the water already contains 0.010 mol dm⁻³ NaF, then [F⁻] ≈ 0.010 mol dm⁻³, and a new solubility s’ is calculated: Ksp = (s’)(0.010)² = 3.9 × 10⁻¹¹ → s’ = 3.9 × 10⁻⁷ mol dm⁻³. This illustrates the common ion effect.
示例：CaF₂ 的 Ksp = 3.9 × 10⁻¹¹。求其在纯水中的溶解度：s = ³√(3.9 × 10⁻¹¹ / 4) = 2.1 × 10⁻⁴ mol dm⁻³。若水中已含 0.010 mol dm⁻³ NaF，则 [F⁻] ≈ 0.010 mol dm⁻³，新溶解度 s’ 计算如下：Ksp = (s’)(0.010)² = 3.9 × 10⁻¹¹ → s’ = 3.9 × 10⁻⁷ mol dm⁻³。这体现了同离子效应。
9. Colorimetry and Transition Metal Ion Concentration | 比色法与过渡金属离子浓度
Colorimetry measures the absorbance of coloured transition metal complexes to determine their concentration using the Beer–Lambert law: A = ε c l, where A is absorbance, ε is the molar absorptivity, c is concentration, and l is the path length. If ε and l are constant, absorbance is directly proportional to concentration. A calibration graph of absorbance vs. concentration for standard solutions is plotted, and the unknown concentration is read from the graph. For example, a series of standard Cu²⁺(aq) solutions give absorbances; the unknown sample has A = 0.45, and the calibration curve yields c = 0.112 mol dm⁻³. In dilution calculations, C₁V₁ = C₂V₂ is used.
比色法通过测量有色过渡金属配合物的吸光度，利用比尔-朗伯定律确定其浓度：A = ε c l，其中 A 为吸光度，ε 为摩尔吸光系数，c 为浓度，l 为光程长度。若 ε 和 l 恒定，吸光度与浓度成正比。绘制标准溶液的吸光度-浓度校正曲线，从图中读出未知浓度。例如，一系列 Cu²⁺(aq) 标准溶液产生相应吸光度；未知样品 A = 0.45，校正曲线得出 c = 0.112 mol dm⁻³。稀释计算中用到 C₁V₁ = C₂V₂。
In some exam questions, you must calculate the mass of a metal in a sample: from the concentration found by colorimetry, multiply by the volume and molar mass. For instance, a 250 cm³ solution of nickel(II) sulfate has a concentration of 0.085 mol dm⁻³ Ni²⁺. Mass of nickel = 0.085 × 0.250 × 58.7 = 1.25 g.
在部分考题中，必须计算样品中金属的质量：由比色法得到的浓度乘以体积和摩尔质量。例如，250 cm³ 硫酸镍(II) 溶液中 Ni²⁺ 浓度为 0.085 mol dm⁻³，则镍的质量 = 0.085 × 0.250 × 58.7 = 1.25 g。
10. Stoichiometry in Inorganic Synthesis and Gravimetric Analysis | 无机合成与重量分析中的化学计量
Stoichiometric calculations are used to determine the yield of a product or the percentage composition of a mixture. For a multi-step synthesis, atom economy and percentage yield are calculated. In gravimetric analysis, the mass of a precipitate is used to find the amount of a particular ion. For example, the sulfate content of a fertiliser is determined by precipitating BaSO₄. Given that 1.20 g of BaSO₄ (molar mass 233.4 g mol⁻¹) was obtained from a 2.50 g sample, moles of BaSO₄ = 1.20 / 233.4 = 0.00514 mol. Each mole BaSO₄ contains one mole of SO₄²⁻, so mass of sulfur = 0.00514 × 32.1 = 0.165 g. The percentage sulfur by mass = (0.165/2.50) × 100% = 6.6%.
化学计量计算用于确定产物产率或混合物的百分组成。多步合成中计算原子经济性和百分产率。重量分析法中，利用沉淀的质量来求特定离子的量。例如，通过沉淀 BaSO₄ 测定肥料中的硫酸盐含量。假设从 2.50 g 样品中得到 1.20 g BaSO₄（摩尔质量 233.4 g mol⁻¹），BaSO₄ 的物质的量 = 1.20 / 233.4 = 0.00514 mol。每摩尔 BaSO₄ 含 1 摩尔 SO₄²⁻，因此硫的质量 = 0.00514 × 32.1 = 0.165 g。硫的质量百分数 = (0.165/2.50) × 100% = 6.6%。
Another common question involves back-titration: an excess of a known reagent is added, and the unreacted excess is titrated. This is useful for determining the amount of an insoluble metal carbonate or a slow-reacting metal. The difference in moles gives the amount that reacted.
另一种常见题型是返滴定：加入过量已知试剂，然后滴定剩余的部分。这对于测定难溶金属碳酸盐或反应缓慢的金属特别有效。通过摩尔差值即可得出已反应的量。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply