📚 Mastering Calculations from the Jan 2021 Unit 5 Mark Scheme | 掌握 2021 年 1 月 Unit 5 评分方案中的计算题型
The January 2021 Edexcel IAL Chemistry Unit 5 paper tested a range of calculation skills essential for top marks. Understanding how marks are allocated in the mark scheme can help you avoid common pitfalls and maximise your score. This article breaks down the key calculation question types from that paper and explains the marking points you need to demonstrate.
2021 年 1 月 Edexcel 国际 A-Level 化学第五单元试卷考察了多种计算技能，这些技能对于取得高分至关重要。了解评分方案中的采分点，可以帮助你避开常见错误并最大化得分。本文将解析该试卷中的关键计算题型，并说明你需要展示的评分要点。
1. Buffer pH Calculations – Tackling Weak Acid/Base Equilibria | 缓冲溶液 pH 计算——攻克弱酸/弱碱平衡
In the Jan 2021 Unit 5 exam, a typical buffer question asked you to calculate the pH after partial neutralisation. The mark scheme rewards accurate mole calculations, correct use of the Ka expression, and the final pH value to an appropriate number of significant figures.
在 2021 年 1 月的 Unit 5 考试中，一道典型的缓冲溶液题要求计算部分中和后的 pH。评分方案对准确的摩尔计算、正确使用 Ka 表达式以及最终 pH 值的有效数字有明确采分点。
To solve, you first determine the moles of acid and base mixed. Excess weak acid remains, and the salt (conjugate base) is formed. Using [H⁺] = Kₐ × [HA] / [A⁻], or pH = pKₐ + log₁₀([A⁻]/[HA]), you can find the hydrogen ion concentration. The mark scheme often gives one mark for moles of HA and A⁻, another for substituting into the expression, and a final mark for the correct pH. For instance, mixing 50.0 cm³ of 0.100 mol dm⁻³ ethanoic acid with 25.0 cm³ of 0.100 mol dm⁻³ NaOH yields equal concentrations of HA and A⁻ after reaction, making pH = pKₐ = 4.76 (given Kₐ = 1.74 × 10⁻⁵ mol dm⁻³).
解题时，首先要计算混合的酸和碱的物质的量。过量的弱酸剩余，并生成盐（共轭碱）。利用 [H⁺] = Kₐ × [HA] / [A⁻] 或 pH = pKₐ + log₁₀([A⁻]/[HA])，可以求出氢离子浓度。评分方案通常会为计算 HA 和 A⁻ 的物质的量设置一个评分点，将数值代入表达式再给一个评分点，最后为正确的 pH 值给分。例如，将 50.0 cm³ 0.100 mol dm⁻³ 乙酸与 25.0 cm³ 0.100 mol dm⁻³ NaOH 混合，反应后 HA 和 A⁻ 浓度相等，因此 pH = pKₐ = 4.76（已知 Kₐ = 1.74 × 10⁻⁵ mol dm⁻³）。
Always show working with units and state assumptions (e.g., no volume change) if required. The mark scheme may penalise missing units or incorrect significant figures.
务必展示带有单位的计算过程，并在需要时说明假设（如忽略体积变化）。评分方案可能因缺少单位或有效数字错误而扣分。
2. Redox Titration Calculations – Manganate(VII)/Fe(II) System | 氧化还原滴定计算——高锰酸钾/铁(II)体系
Redox titrations are a staple of Unit 5, and the Jan 2021 paper included a calculation based on the reaction between acidified manganate(VII) ions and iron(II) ions. The mark scheme focuses on the stoichiometric ratio, mole calculations, and concentration determination.
氧化还原滴定是 Unit 5 的常考内容，2021 年 1 月的试卷包含基于酸化高锰酸根离子与铁(II)离子反应的计算题。评分方案侧重考查化学计量比、摩尔计算以及浓度确定。
The balanced equation is:
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
A typical task: 25.0 cm³ of Fe²⁺ solution required 23.50 cm³ of 0.0200 mol dm⁻³ KMnO₄ to reach the endpoint. To find the iron(II) concentration, you calculate moles of MnO₄⁻ (0.0200 × 23.50/1000 = 4.70 × 10⁻⁴ mol), then apply the 1:5 ratio to get moles of Fe²⁺ (2.35 × 10⁻³ mol). The concentration is mol / volume (dm³) = 0.0940 mol dm⁻³. The mark scheme awards marks for: correct calculation of MnO₄⁻ moles, use of 1:5 ratio, dividing by the correct volume, and final answer with units.
典型题目：25.0 cm³ Fe²⁺ 溶液耗用 23.50 cm³ 0.0200 mol dm⁻³ KMnO₄ 到达终点。计算铁(II)浓度，先求 MnO₄⁻ 物质的量（0.0200 × 23.50/1000 = 4.70 × 10⁻⁴ mol），然后利用 1:5 比例得到 Fe²⁺ 物质的量（2.35 × 10⁻³ mol）。浓度 = mol / 体积（dm³）= 0.0940 mol dm⁻³。评分方案采分点包括：正确计算 MnO₄⁻ 物质的量、使用 1:5 比例、除以正确体积、得出带有单位的最终答案。
Remember to convert volumes to dm³ and express the concentration to an appropriate number of significant figures (often 3 s.f. here). Another variant involves iodine-thiosulfate titrations to determine copper(II) or bleach content; the same structured approach applies.
记得将体积换算为 dm³，并以恰当的有效数字（此处常为 3 位有效数字）表示浓度。另一种变体是碘-硫代硫酸盐滴定，用来测定铜(II)或漂白剂含量；同样采用结构化的计算方法。
3. Cell EMF and Thermodynamics – ΔG = –nFE | 电池电动势与热力学——ΔG = –nFE
Unit 5 frequently links electrochemistry with thermodynamics. The Jan 2021 mark scheme showed that you must be able to calculate the standard cell potential E⁰_cell from half-cell potentials, then use ΔG⁰ = –nFE⁰_cell to find the Gibbs free energy change.
Unit 5 经常将电化学与热力学联系起来。2021 年 1 月的评分方案表明，你必须能够从半电池电势计算标准电池电动势 E⁰_cell，然后利用 ΔG⁰ = –nFE⁰_cell 求得吉布斯自由能变。
For a cell Cu|Cu²⁺ || Ag⁺|Ag, with E⁰ Ag⁺/Ag = +0.80 V and E⁰ Cu²⁺/Cu = +0.34 V, E⁰_cell = 0.80 – 0.34 = +0.46 V. With n = 2 mol electrons, ΔG⁰ = –2 × 96500 × 0.46 = –88780 J mol⁻¹ (≈ –89 kJ mol⁻¹). Marks are allocated for the correct E⁰_cell, recognition of n, correct Faraday constant, and conversion to kJ if requested. The negative ΔG⁰ confirms thermodynamic feasibility.
对于电池 Cu|Cu²⁺ || Ag⁺|Ag，已知 E⁰ Ag⁺/Ag = +0.80 V，E⁰ Cu²⁺/Cu = +0.34 V，则 E⁰_cell = 0.80 – 0.34 = +0.46 V。n = 2 mol 电子，ΔG⁰ = –2 × 96500 × 0.46 = –88780 J mol⁻¹（≈ –89 kJ mol⁻¹）。评分点包括：正确计算 E⁰_cell、识别 n 值、正确使用法拉第常数，以及按要求换算为 kJ。负的 ΔG⁰ 证实热力学可行性。
Some mark scheme details also assess the relationship between ΔG⁰ and the equilibrium constant via ΔG⁰ = –RT ln K. Being able to convert between E⁰_cell and ln K is a valued skill.
评分方案的某些细则还会考察 ΔG⁰ 与平衡常数的关系 ΔG⁰ = –RT ln K。能够在 E⁰_cell 与 ln K 之间转换是一项重要技能。
4. Thermodynamic Feasibility – Entropy and Gibbs Free Energy | 热力学可行性——熵与吉布斯自由能
Another calculation area in the Jan 2021 paper required you to use ΔS⁰_system and ΔH⁰ to determine the temperature at which a reaction becomes feasible. The mark scheme consistently awards points for converting entropy units, correct substitution into ΔG = ΔH – TΔS, and solving for T when ΔG = 0.
2021 年 1 月试卷的另一个计算领域需要利用 ΔS⁰_system 和 ΔH⁰ 确定反应变为可行的温度。评分方案一贯给分点包括：熵单位换算、正确代入 ΔG = ΔH – TΔS，以及当 ΔG = 0 时求解温度 T。
For example, a reaction has ΔH = +120 kJ mol⁻¹ and ΔS = +250 J K⁻¹ mol⁻¹. At the crossover temperature, ΔG = 0 → T = ΔH / ΔS = 120,000 J mol⁻¹ / 250 J K⁻¹ mol⁻¹ = 480 K. Marks are given for converting kJ to J, setting ΔG = 0, and calculating the temperature in Kelvin. Many candidates lose marks by forgetting to convert entropy to kJ or mismatching units.
例如，某个反应的 ΔH = +120 kJ mol⁻¹，ΔS = +250 J K⁻¹ mol⁻¹。在转折温度处，ΔG = 0 → T = ΔH / ΔS = 120,000 J mol⁻¹ / 250 J K⁻¹ mol⁻¹ = 480 K。将 kJ 换算为 J、设 ΔG = 0 以及计算开尔文温度均有相应分数。不少考生因忘记将熵换算为 kJ 或单位不匹配而失分。
Sometimes you must combine ΔG = –nFE with ΔG = ΔH – TΔS to find entropy changes from electrochemical data, a typical Unit 5 synoptic calculation.
有时你需要结合 ΔG = –nFE 与 ΔG = ΔH – TΔS，从电化学数据求出熵变，这是 Unit 5 典型的综合计算。
5. Rate Equation Determination – Using Initial Rates Data | 速率方程确定——使用初始速率数据
The Jan 2021 mark scheme included a kinetic problem where you deduced orders of reaction from a table of initial rates, then calculated the rate constant k with its units. This is a core Unit 5 skill.
2021 年 1 月评分方案中包含一道动力学问题：根据初始速率表格推导反应级数，然后计算速率常数 k 及其单位。这是 Unit 5 的核心技能。
Given data: when [A] doubles and [B] constant, rate quadruples → order with respect to A = 2; when [B] doubles and [A] constant, rate doubles → order with respect to B = 1. Overall order = 3. To find k, use rate = k[A]²[B], then k = rate / ([A]²[B]) with units mol⁻² dm⁶ s⁻¹. The mark scheme assigns separate marks for each order, the rate equation, calculation of k, and the correct units. Always show the unit derivation.
给定数据：当 [A] 加倍而 [B] 不变，速率变为四倍 → A 的反应级数为 2；当 [B] 加倍而 [A] 不变，速率加倍 → B 的反应级数为 1。总级数 = 3。求 k 时，使用 rate = k[A]²[B]，则 k = rate / ([A]²[B])，单位为 mol⁻² dm⁶ s⁻¹。评分方案对每个级数、速率方程、k 的计算以及正确单位均设有独立采分点。务必展示单位推导过程。
Arrhenius calculations also appeared: using ln k = ln A – Eₐ/(RT), you may be asked to calculate Eₐ from a graph or two-point data. Marks go to correct conversion to temperature in Kelvin, calculating 1/T and ln k, gradient determination, and final Eₐ in kJ mol⁻¹.
阿伦尼乌斯公式的计算也有涉及：使用 ln k = ln A – Eₐ/(RT)，你可能需要从图像或双点数据计算 Eₐ。采分点包括：正确转换为开尔文温度、计算 1/T 与 ln k、确定斜率以及最终的 Eₐ 以 kJ mol⁻¹ 表示。
6. Kp Calculations – Partial Pressures and Mole Fractions | Kp 计算——分压与摩尔分数
Equilibrium constant Kp questions in the Jan 2021 Unit 5 required you to work out mole fractions, then partial pressures, before substituting into the Kp expression. The mark scheme heavily assesses correct use of the total pressure and recognising the effect of units.
2021 年 1 月 Unit 5 中的平衡常数 Kp 题目要求你先计算出摩尔分数，再求分压，最后代入 Kp 表达式。评分方案重点评估正确使用总压以及对单位影响的认知。
For the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at a total pressure of 5000 kPa, given equilibrium moles: N₂ 1.0 mol, H₂ 3.0 mol, NH₃ 2.0 mol, total moles = 6.0. Mole fraction N₂ = 1.0/6.0 = 0.1667; partial pressure = mole fraction × total pressure = 0.1667 × 5000 = 833.5 kPa. Similarly for H₂ and NH₃. Kp = p(NH₃)² / [p(N₂) × p(H₂)³]. The mark scheme gives marks for mole fractions, subtraction from total pressure if needed, setting up the expression, and calculating Kp with its unit (e.g., kPa⁻²).
对于平衡 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，总压 5000 kPa，平衡时物质的量：N₂ 1.0 mol，H₂ 3.0 mol，NH₃ 2.0 mol，总物质的量 = 6.0。N₂ 的摩尔分数 = 1.0/6.0 = 0.1667；分压 = 摩尔分数 × 总压 = 0.1667 × 5000 = 833.5 kPa。同理计算 H₂ 和 NH₃。Kp = p(NH₃)² / [p(N₂) × p(H₂)³]。评分方案为摩尔分数、必要时的总压减法、表达式建立以及计算 Kp 及其单位（如 kPa⁻²）分别设分。
Always state the unit, as the mark scheme often penalises missing units. Also, be prepared to calculate the total pressure at equilibrium from initial conditions and the extent of reaction.
务必写出单位，因为评分方案常因缺少单位而扣分。此外，要准备好从初始条件和反应进度计算平衡时的总压。
7. Colorimetry and Transition Metal Stoichiometry | 比色法与过渡金属计量关系
Unit 5 may include a colorimetry-based calculation to determine the formula of a transition metal complex, as seen in the Jan 2021 application questions. The mark scheme looks for accurate reading of absorbance, construction of a calibration curve, and determination of the mole ratio.
Unit 5 可能包含基于比色法的计算，以确定过渡金属配合物的化学式，2021 年 1 月的应用题中就有出现。评分方案关注吸光度的准确读取、校正曲线的绘制以及摩尔比的确定。
A typical scenario: a series of standard solutions containing Ni²⁺ and a ligand L are prepared, their absorbance measured at a specific wavelength. By plotting absorbance versus mole fraction of ligand, the point of maximum absorbance (or intersection of two straight lines) gives the stoichiometric ratio, e.g., NiL₃²⁺. Marks are allocated for plotting points, drawing lines of best fit, identifying the mole fraction at the peak, and deducing the complex formula.
典型场景：配制一系列含有 Ni²⁺ 和配体 L 的标准溶液，在特定波长下测量吸光度。通过绘制吸光度对配体摩尔分数的图形，最高吸光度点（或两直线交点）给出化学计量比，例如 NiL₃²⁺。评分点包括：描点、绘制最佳拟合线、识别峰值处的摩尔分数以及推断配合物化学式。
Sometimes the calculation requires you to convert mass to moles or use dilution factors—the same rigorous, stepwise approach demanded across all Unit 5 calculations.
有时计算需要将质量转换为物质的量或使用稀释因子——与所有 Unit 5 计算一样，要求同样严谨、循序渐进的解题方法。
8. Organic Yield and Limiting Reagent Calculations | 有机产率与限量试剂计算
Organic nitrogen chemistry in Unit 5 often includes multi-step synthesis calculations. The Jan 2021 mark scheme highlighted that you must confidently identify the limiting reagent and calculate percentage yield, showing all working clearly.
Unit 5 中的有机氮化学常常包含多步合成计算。2021 年 1 月的评分方案强调，你必须能够自信地识别限量试剂并计算百分产率，且清晰展示所有过程。
For example, to prepare phenylamine from nitrobenzene, you may be given masses and asked to calculate the theoretical yield and percentage yield. First, calculate moles of each reactant to find the limiting reagent, then use the 1:1 stoichiometry to determine theoretical moles of product. Convert to mass and compare with the actual yield. The mark scheme frequently awards marks for mole calculations, identifying the limiting reagent, correct theoretical mass, and the final percentage yield to an appropriate precision.
例如，由硝基苯制备苯胺时，你可能会得到质量数据，要求计算理论产量和百分产率。首先计算各反应物的物质的量以确定限量试剂，再利用 1:1 化学计量比求出产物的理论物质的量，换算为质量并与实际产量比较。评分方案常为摩尔计算、识别限量试剂、正确理论产量以及最终百分产率的恰当精度分别设分。
Atom economy questions may also appear; the mark scheme gives marks for the correct formula and substitution, so memorise it: % atom economy = (mass of desired product / total mass of reactants) × 100.
原子经济性问题也可能出现；评分方案对正确公式及其代入设有采分点，因此请牢记：% 原子经济性 =（目标产物质量 / 反应物总质量）× 100。
9. Acid–Base Titrations for Amines – pKb Determination | 胺的酸碱滴定——pKb 的测定
The Jan 2021 paper asked you to interpret titration curves for weak bases like amines and calculate pKb
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