📚 Mastering Calculation Questions in IAL Chemistry Unit 5 (9620-CH05): Specimen Paper 2016 Walkthrough | 攻克IAL化学Unit 5 (9620-CH05) 计算题型：2016年样卷精讲
The Unit 5 paper for International A‑Level Chemistry (9620-CH05) is notorious among students for its heavy emphasis on quantitative problem‑solving. The 2016 specimen paper exemplifies this: roughly half of the available marks demand structured calculations spanning thermodynamics, kinetics, equilibria, electrochemistry, and organic analysis. This article dissects the calculation styles that appear in that specimen, providing step‑by‑step reasoning, common pitfalls, and strategic advice so you can approach every numerical problem with confidence.
IAL化学第五单元（9620-CH05）因其对定量解题的着重考查而令许多学生头疼。2016年的样卷就是典型代表：近一半的分数要求完成涵盖热力学、动力学、平衡、电化学和有机分析的结构化计算。本文将剖析该样卷中出现的计算题型，提供逐步推理、常见错误与策略建议，帮助你自信应对每一道数值题。
1. Standard Enthalpy Changes from Experimental Data | 由实验数据求标准焓变
One staple calculation in CH05 involves determining a standard enthalpy change, often ΔHneut or ΔHsoln, from a simple calorimetry experiment. You are given masses, temperatures, and the specific heat capacity of water (4.18 J g⁻¹ K⁻¹). The key formula is q = mcΔT, where m is the total mass of the solution (usually approximated as the volume in cm³ of water plus any solid dissolved), c is 4.18 J g⁻¹ K⁻¹, and ΔT is the temperature rise. After calculating the heat energy q in joules, you must scale it to the number of moles of the limiting reactant to obtain ΔH in kJ mol⁻¹.
CH05中一个固定计算题型是根据简单量热实验数据求标准焓变，通常是中和焓或溶解焓。题目会给出质量、温度和水的比热容（4.18 J g⁻¹ K⁻¹）。关键公式为 q = mcΔT，其中 m 是溶液总质量（通常近似为水的体积 cm³ 加上溶解固体的质量），c 取 4.18 J g⁻¹ K⁻¹，ΔT 是温度升高值。算出热量 q（单位 J）后，必须将其换算到极限反应物的物质的量，从而得到以 kJ mol⁻¹ 为单位的 ΔH。
In the 2016 specimen, a typical problem asks for ΔHsoln of an anhydrous salt and its hydrate. The mass of solid is recorded, the volume of water is given, and the temperature change is noted. After working out q, you divide by the moles of solid used. Important: remember that the process is exothermic if the temperature rises, so ΔH must be negative. Also, if the salt is a hydrate, the number of moles is based on the anhydrous equivalent—you may need to subtract the mass of water from the hydrated salt’s mass to find the moles of anhydrous salt correctly.
2016年样卷中，一个典型问题要求计算某种无水盐及其水合物的溶解焓。题目记录固体质量、水的体积，并注明温度变化。算出 q 后，除以所用固体的物质的量。须注意：若温度升高，过程为放热，ΔH 必须为负值。另外，如果盐是水合物，物质的量应基于无水盐的当量——你可能需要从水合盐的总质量中减去水的质量，才能正确求得无水盐的物质的量。
A common error is forgetting to convert q from joules to kilojoules before comparing with ΔH values in kJ mol⁻¹. Always divide q/1000 then divide by moles. Also, watch out for significant figures: the temperature difference usually limits your answer to 2 or 3 significant figures. In the specimen mark scheme, answers are accepted within a small range to allow for rounding.
一个常见错误是在与以 kJ mol⁻¹ 为单位的 ΔH 比较之前，忘记将 q 从焦耳转换为千焦。务必先将 q/1000 再除以物质的量。此外，注意有效数字：温差通常将答案限制在 2 或 3 位有效数字。在样卷的评分方案中，答案允许在一个小范围内波动以考虑四舍五入。
2. Born‑Haber Cycles and Lattice Energy | 玻恩–哈伯循环与晶格能
Born‑Haber cycle questions in CH05 require you to calculate a missing energy term—usually lattice energy or electron affinity—by applying Hess’s law. Given the standard enthalpy of formation and a series of atomisation enthalpies, ionisation energies, and bond dissociation energy, you construct the cycle algebraically. The key relationship is: ΔHf = Σ(atomisation energies) + Σ(ionisation energies) + electron affinity(ies) + lattice energy. For a simple ionic solid MX, the cycle arrows from elements in their standard states to gaseous ions to the solid lattice.
CH05中的玻恩–哈伯循环题要求通过应用盖斯定律计算缺失的能量项——通常是晶格能或电子亲和势。题目给出标准生成焓以及一系列原子化焓、电离能和键解离能，你需要用代数方法构建循环。关键关系式为：ΔHf = Σ(原子化能) + Σ(电离能) + 电子亲和势 + 晶格能。对于简单的离子型固体 MX，循环箭头从处于标准状态的单质元素指向气态离子，再指向固体晶格。
In the specimen paper, you might be given the formation enthalpy of an ionic compound and the first two ionisation energies of a metal, plus the atomisation enthalpies of the metal and non‑metal, and the bond dissociation energy of the non‑metal molecule. You must then determine the lattice energy. Set up the equation carefully: start with elements in standard states, then add steps that absorb energy (positive) and release energy (negative). Lattice energy is usually exothermic and therefore negative. If the calculated value turns out positive, you have likely reversed a sign or misplaced an arrow.
样卷中可能给出某个离子化合物的生成焓、金属的前两级电离能、金属和非金属的原子化焓，以及非金属分子的键解离能，要求确定晶格能。仔细建立方程：从标准状态的单质开始，然后加上吸收能量的步骤（正值）和释放能量的步骤（负值）。晶格能通常为放热过程，因此为负值。如果算出的值为正，很可能你弄错了一个符号或放错了箭头方向。
Many candidates confuse atomisation enthalpy of a non‑metal (e.g., ½Cl₂(g) → Cl(g)) with bond dissociation energy (Cl–Cl bond breaking). For chlorine, bond dissociation energy per mole of bonds is for Cl₂(g) → 2Cl(g), so you need to halve it to get the atomisation enthalpy of chlorine. Always check the definitions provided in the data booklet. Using the wrong value is a frequent source of error.
许多考生将非金属的原子化焓（例如 ½Cl₂(g) → Cl(g)）与键解离能（断裂 Cl–Cl 键）混淆。对于氯气，每摩尔键的解离能对应 Cl₂(g) → 2Cl(g)，因此需要将其减半才能得到氯的原子化焓。务必核验数据手册中给出的定义。使用错误的数值是常见的错误来源。
3. Entropy and Gibbs Free Energy | 熵与吉布斯自由能
Calculations of ΔSsystem, ΔSsurroundings, ΔStotal, and ΔG are core to Unit 5. Using the data booklet, you find standard entropies (S°) for reactants and products. ΔS°system = ΣS°(products) – ΣS°(reactants). ΔS°surroundings = –ΔH/T, where T must be in kelvin, and ΔH is in J mol⁻¹ (not kJ!). ΔS°total = ΔS°system + ΔS°surroundings. Gibbs free energy change is ΔG = ΔH – TΔS, again with units consistent: ΔH in J mol⁻¹, T in K, ΔS in J K⁻¹ mol⁻¹.
计算 ΔSsystem、ΔSsurroundings、ΔStotal 和 ΔG 是 Unit 5 的核心内容。利用数据手册，查出反应物和产物的标准熵（S°）。ΔS°system = ΣS°(产物) – ΣS°(反应物)。ΔS°surroundings = –ΔH/T，其中 T 必须以开尔文为单位，ΔH 以 J mol⁻¹ 为单位（不是 kJ！）。ΔS°total = ΔS°system + ΔS°surroundings。吉布斯自由能变 ΔG = ΔH – TΔS，同样要保持单位一致：ΔH 为 J mol⁻¹，T 为 K，ΔS 为 J K⁻¹ mol⁻¹。
The 2016 specimen includes a question where you calculate ΔStotal at a given temperature and decide if the reaction is feasible. You might be given ΔH and told to use S° values from the data sheet. Pay attention to the stoichiometric coefficients: multiply each S° by the coefficient in the balanced equation. A common slip is using ΔH in kJ mol⁻¹ in the entropy calculation—this gives an answer 1000 times too small and leads to an incorrect conclusion about feasibility. Always convert ΔH to J mol⁻¹ before computing ΔSsurroundings.
2016年样卷中有一道题要求计算给定温度下的 ΔStotal 并判断反应是否自发。题目可能给出 ΔH，并要求使用数据表中的 S° 值。注意化学计量系数：将每个 S° 值乘以配平方程中的系数。一个常见失误是在熵计算中使用以 kJ mol⁻¹ 为单位的 ΔH——这将导致答案缩小 1000 倍，并得出关于可行性的错误结论。在计算 ΔSsurroundings 之前，务必先将 ΔH 转换为 J mol⁻¹。
Finally, when ΔG = 0, the system is at equilibrium. You may be asked to find the temperature at which this occurs: T = ΔH / ΔS, provided ΔH and ΔS are known and assumed independent of temperature. Set up the equation, but again ensure ΔH is in J mol⁻¹ to match ΔS in J K⁻¹ mol⁻¹, giving temperature in K.
最后，当 ΔG = 0 时，体系处于平衡状态。你可能需要求出达到这一状态的温度：T = ΔH / ΔS，前提是已知 ΔH 和 ΔS，并假设它们不随温度变化。列出方程，但务必确保 ΔH 以 J mol⁻¹ 为单位，以与 ΔS 的 J K⁻¹ mol⁻¹ 匹配，从而得到以 K 为单位的温度。
4. Equilibrium Constant Kc from Homogeneous Equilibria | 由均相平衡求平衡常数 Kc
Calculation of Kc often involves determining equilibrium concentrations from initial amounts and an equilibrium amount of one species. You construct a table showing initial moles, change in moles, and equilibrium moles, then convert to concentrations by dividing by the total volume. The expression for Kc is then assembled and evaluated. The CH05 specimen paper includes a classic esterification equilibrium: acid + alcohol ⇌ ester + water, where the initial moles of acid and alcohol are known, and the equilibrium moles of ester (or water) are given.
Kc 的计算通常涉及由初始量和某一物种的平衡量来确定平衡浓度。你需要构建一个表格，展示初始物质的量、变化量和平衡物质的量，然后除以总体积得到浓度。接着写出 Kc 的表达式并计算数值。CH05样卷中包含了经典的酯化平衡：酸 + 醇 ⇌ 酯 + 水，已知酸和醇的初始物质的量，并给出酯（或水）的平衡物质的量。
A typical pitfall is forgetting that the volume in the denominator of the concentration terms appears in the Kc expression. For a reaction with an equal number of moles on both sides (like the esterification example), the volume cancels, so you can use moles directly. But for reactions where the number of moles changes, you must divide by volume. The mark scheme penalises the omission of volume even when it cancels; you should show the division step to demonstrate understanding.
一个典型陷阱是忘记了浓度项分母中的体积会出现在 Kc 表达式中。对于反应前后物质的量相等的反应（如酯化例子），体积可以约去，因此可以直接使用物质的量。但对于物质的量发生变化的反应，必须除以体积。评分方案会对遗漏体积步骤的做法扣分，即使它可以约去；你应当展示除以体积的步骤以证明理解。
In the specimen, you might also be asked to find Kc for a dissociation like N₂O₄ ⇌ 2NO₂, where the total pressure and degree of dissociation α are given, and you need to work out partial pressures and Kp. The relationship between Kp and Kc (Kp = Kc(RT)
在样卷中，你可能还会遇到类似 N₂O₄ ⇌ 2NO₂ 这样的解离反应，给出总压和离解度 α，要求计算分压和 Kp。然后可以探讨 Kp 与 Kc 的关系（Kp = Kc(RT)
5. pH and Buffer Calculations | pH与缓冲溶液计算
CH05 features strong acid‑strong base pH calculations, weak acid pH using Ka, and buffer pH using the Henderson‑Hasselbalch approach. For a weak acid HA, Ka = [H⁺][A⁻]/[HA], and if the acid is pure, [H⁺] = [A⁻] = x, so Ka = x²/(c – x). Often the approximation c >> x allows simplification to Ka ≈ x²/c, giving [H⁺] = √(Ka × c). The specimen paper expects you to use this approximation and then calculate pH = –log₁₀[H⁺].
CH05包含强酸强碱的pH计算、利用Ka的弱酸pH计算，以及使用亨德森–哈塞尔巴尔赫方法的缓冲溶液pH计算。对于弱酸HA，Ka = [H⁺][A⁻]/[HA]，如果为纯酸溶液，[H⁺] = [A⁻] = x，因此Ka = x²/(c – x)。通常采用近似 c >> x 可简化为 Ka ≈ x²/c，从而得到 [H⁺] = √(Ka × c)。样卷希望学生使用这一近似，然后计算 pH = –log₁₀[H⁺]。
For a buffer made from a weak acid and its salt, the equation is [H⁺] = Ka × ([acid]/[salt]) or pH = pKa + log₁₀([salt]/[acid]). In the 2016 specimen, a typical buffer problem might ask: “Calculate the pH of a buffer formed by mixing 50 cm³ of 0.10 mol dm⁻³ ethanoic acid with 25 cm³ of 0.20 mol dm⁻³ sodium ethanoate.” You calculate moles of acid and salt after mixing, divide by total volume (which cancels in the ratio), and plug into the equation. Watch units: Ka for ethanoic acid is 1.74 × 10⁻⁵ mol dm⁻³; pKa = 4.76. The final pH will be around 4.5–5.0.
对于由弱酸及其盐组成的缓冲溶液，方程为 [H⁺] = Ka × ([酸]/[盐]) 或 pH = pKa + log₁₀([盐]/[酸])。在2016年样卷中，一个典型的缓冲问题可能是：“计算将50 cm³ 0.10 mol dm⁻³ 乙酸与25 cm³ 0.20 mol dm⁻³ 乙酸钠混合后所得缓冲溶液的pH。”先计算混合后酸和盐的物质的量，除以总体积（在比值中会约去），然后代入方程。注意单位：乙酸的Ka = 1.74 × 10⁻⁵ mol dm⁻³；pKa = 4.76。最终pH值大约在4.5到5.0之间。
When acid or alkali is added to a buffer, do not just apply the Henderson equation directly. You must determine the new moles of acid and conjugate base after the added H⁺ or OH⁻ reacts stoichiometrically. The specimen includes a two‑step calculation: first find the new mole ratio, then compute pH. Careless omission of this reaction step leads to a wrong buffer pH.
当向缓冲溶液中加入酸或碱时，不要直接套用亨德森方程。你必须先确定加入的H⁺或OH⁻按化学计量反应后，酸和共轭碱的新物质的量。样卷中常出现两步计算：先求出新的摩尔比，再计算pH。疏忽这一反应步骤会导致缓冲pH计算错误。
6. Electrode Potentials and Cell EMF | 电极电势与电池电动势
Calculating the standard cell EMF is straightforward: E°cell = E°(right‑hand electrode) – E°(left‑hand electrode), where the cell diagram convention must be followed. The specimen paper may also ask you to calculate the cell EMF under non‑standard conditions using the Nernst equation: E = E° – (RT/nF) ln Q, or at 298 K: E = E° – (0.0257/n) ln Q. Typical problems involve a concentration cell or a metal/metal ion half‑cell with a different concentration.
标准电池电动势的计算很简单：E°cell = E°(右电极) – E°(左电极)，必须遵循电池图式惯例。样卷还可能要求使用能斯特方程计算非标准条件下的电池电动势：E = E° – (RT/nF) ln Q，或在298 K时：E = E° – (0.0257/n) ln Q。典型题目涉及浓差电池或不同浓度的金属/金属离子半电池。
In the CH05 specimen, you might see a question such as: “A cell is set up with Fe²⁺/Fe³⁺ half‑cell and a standard hydrogen electrode. The concentration of Fe²⁺ is 0.10 mol dm⁻³ and Fe³⁺ is 0.010 mol dm⁻³. Calculate the cell EMF at 25 °C.” You need the standard potential for Fe³⁺ + e⁻ → Fe²⁺ (+0.77 V), then apply the Nernst equation with n = 1. Q = [Fe²⁺]/[Fe³⁺] = 0.10/0.010 = 10. So E = 0.77 – (0.059/1) × log₁₀(10) = 0.77 – 0.059 = 0.71 V (using the common 0.059 V constant). Remember to check if the reaction quotient is written correctly based on the spontaneous direction; sometimes the question asks for the E of the half‑cell vs SHE, and you simply plug in the concentrations.
在CH05样卷中，你可能会看到类似这样的题目：“一个电池由Fe²⁺/Fe³⁺半电池和标准氢电极组成。Fe²⁺浓度为0.10 mol dm⁻³，Fe³⁺浓度为0.010 mol dm⁻³。计算25°C时的电池电动势。”你需要知道 Fe³⁺ + e⁻ → Fe²⁺ 的标准电势 (+0.77 V)，然后应用能斯特方程，其中 n = 1。Q = [Fe²⁺]/[Fe³⁺] = 0.10/0.010 = 10。所以 E = 0.77 – (0.059/1) × log₁₀(10) = 0.77 – 0.059 = 0.71 V（使用常见的0.059 V常数）。注意根据自发方向正确书写反应商；有时题目会要求计算半电池相对于SHE的电势，你只需代入浓度即可。
Another calculation involves the relationship between ΔG and E: ΔG = –nFE. The specimen may ask you to determine the equilibrium constant K using the link E°cell = (RT/nF) ln K. At 298 K, log₁₀ K = (nE°cell)/0.059. These equations are found at the bottom of the data booklet; make sure you use them correctly, with n being the number of electrons transferred in the balanced redox equation.
另一个计算涉及 ΔG 与 E 的关系：ΔG = –nFE。样卷可能要求学生利用关系式 E°cell = (RT/nF) ln K 确定平衡常数 K。在298 K时，log₁₀ K = (nE°cell)/0.059。这些公式可在数据手册底部找到；务必正确使用，其中 n 为配平的氧化还原方程中转移的电子数。
7. Rate Equations and the Arrhenius Equation | 速率方程与阿伦尼乌斯方程
Determining the order of reaction from initial rate data is a classic CH05 calculation. You compare experiments where one reactant concentration changes while others are constant. The rate doubles when [A] doubles → first order in A. The rate quadruples when [B] doubles → second order in B. From the orders, the rate constant k is calculated using rate = k [A]ᵐ[B]ⁿ. Pay attention to units: k’s units depend on the overall order. The 2016 specimen includes a table of three experiments and asks for orders and k.
根据初始速率数据确定反应级数是CH05中的经典计算。你需要比较只有一个反应物浓度变化而其他保持不变的实验。当[A]加倍时速率加倍 → 对A为一级。当[B]加倍时速率变为四倍 → 对B为二级。根据级数，使用速率方程 rate = k [A]ᵐ[B]ⁿ 计算速率常数 k。注意单位：k 的单位取决于总级数。2016年样卷提供一个含三个实验的表格，要求求出级数和 k。
Be systematic: write the ratio of rates, factor out constants. For example, if exp 1: [A]=0.1, [B]=0.1, rate=1.0×10⁻³; exp 2: [A]=0.2, [B]=0.1, rate=2.0×10⁻³. Rate ratio = 2, concentration ratio for A=2, so order with respect to A is 1. Then use another pair to find the order in B. Once orders are known, pick any experiment, plug in numbers and solve for k, including its units (e.g., mol⁻¹ dm³ s⁻¹ for second order overall).
要有条理：写出速率之比，将常数因子提出。例如，若实验1：[A]=0.1, [B]=0.1, rate=1.0×10⁻³；实验2：[A]=0.2, [B]=0.1, rate=2.0×10⁻³。速率比为2，A的浓度比为2，故对A的级数为1。然后用另一组数据求出对B的级数。一旦知道了级数，选取任一实验，代入数值，解出 k 及其单位（例如对于总二级反应，单位为 mol⁻¹ dm³ s⁻¹）。
The Arrhenius equation often appears in graphical form: ln k = ln A – Eₐ/(RT). A two‑point calculation may be required: ln(k₂/k₁) = –Eₐ/R (1/T₂ – 1/T₁). The specimen paper gives k at two different temperatures and asks for activation energy Eₐ, or vice versa. Use R = 8.31 J K⁻¹ mol⁻¹. Convert temperatures to kelvin. The answer for Eₐ is typically in J mol⁻¹, often expressed in kJ mol⁻¹ in the final answer. Be careful with the sign: if k increases with T, Eₐ is positive.
阿伦尼乌斯方程常以图形形式出现：ln k = ln A – Eₐ/(RT)。也可能要求进行两点计算：ln(k₂/k₁) = –Eₐ/R (1/T₂ – 1/T₁)。样卷给出两个不同温度下的 k，要求计算活化能 Eₐ，或者反过来。使用 R = 8.31 J K⁻¹ mol⁻¹。将温度转换为开尔文。Eₐ 的答案通常以 J mol⁻¹ 为单位，最终答案常以 kJ mol⁻¹ 表示。注意符号：若 k 随 T 升高而增大，则 Eₐ 为正值。
8. Titration Curves and Indicators | 滴定曲线与指示剂
CH05 asks you to calculate the pH at various stages of a titration: initial pH (weak acid), pH after adding some strong base (buffer region), pH at equivalence, and pH beyond equivalence. These step‑wise calculations are marks‑heavy. The specimen paper presents a titration of a weak acid with a strong base, and you must determine the pH at half‑equivalence (which equals pKₐ) and at equivalence (hydrolysis of the salt). For a weak acid titrated with NaOH, at equivalence the solution contains the conjugate base A⁻, which hydrolyses: A⁻ + H₂O ⇌ HA + OH⁻. Kb = Kw/Ka; then [OH⁻] = √(Kb × csalt), and pOH = –log[OH⁻], pH = 14 – pOH.
CH05要求计算滴定各个阶段的pH：初始pH（弱酸）、加入部分强碱后的pH（缓冲区域）、等当点时的pH，以及等当点之后的pH。这些分步计算占分很重。样卷呈现的是用强碱滴定弱酸，你必须确定半等当点时的pH（等于 pKₐ）和等当点时的pH（盐的水解）。对于用NaOH滴定弱酸，等当点时溶液含有共轭碱A⁻，发生水解：A⁻ + H₂O ⇌ HA + OH⁻。Kb = Kw/Ka；然后 [OH⁻] = √(Kb × c盐)，pOH = –log[OH⁻]，pH = 14 – pOH。
Choosing the correct indicator relies on the pH at equivalence: the indicator’s pKₐ should be within ±1 of the equivalence pH. You may need to calculate the vertical jump in pH around equivalence to confirm whether an indicator with a given pH range is suitable. All steps involve careful arithmetic with logs and square roots, so practice using your calculator efficiently.
选择合适的指示剂取决于等当点时的pH：指示剂的pKₐ应在等当点pH的±1范围内。你可能需要计算等当点附近的pH突跃范围，以确认给定pH范围的指示剂是否合适。所有步骤都涉及对数和平方根的细致运算，因此要练习高效使用计算器。
9. Redox Titrations and Percentage Purity | 氧化还原滴定与纯度计算
Redox titration calculations are ubiquitous. In the CH05 specimen, you might find a multi‑step back‑titration where an excess of a known reagent is added, and the unreacted excess is titrated with another solution. A classic example: determination of copper in a coin. The coin is dissolved, an excess of KI is added, and the liberated iodine is titrated with thiosulfate. You must follow the stoichiometric chain: 2Cu²⁺ + 4I⁻ → 2CuI + I₂; then I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. The mole ratio Cu²⁺ : S₂O₃²⁻ is 1 : 1 because 2Cu²⁺ produces 1I₂ which reacts with 2S₂O₃²⁻, so effectively each Cu²⁺ corresponds to one S₂O₃²⁻.
氧化还原滴定计算无处不在。在CH05样卷中，你可能会遇到多步返滴定：先加入过量已知试剂，然后用另一种溶液滴定未反应的部分。一个经典例子：测定硬币中铜的含量。硬币溶解后，加入过量KI，释放出的碘用硫代硫酸钠滴定。必须遵循化学计量关系链：2Cu²⁺ + 4I⁻ → 2CuI + I₂；然后 I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻。Cu²⁺ 与 S₂O₃²⁻ 的摩尔比为 1 : 1，因为2个Cu²⁺产生1个I₂，而1个I₂与2个S₂O₃²⁻反应，所以每个Cu²⁺对应一个S₂O₃²⁻。
From the titre volume and concentration of thiosulfate, you find moles of thiosulfate used, which equals moles of copper(II) ions in the aliquot. Scale to the original sample, find mass of copper, then percentage purity by mass. The specimen mark scheme penalises any missing factor-of‑10 or scaling errors, so always write the calculation steps clearly: moles in titre → moles in aliquot → moles in original sample → mass → % purity.
根据硫代硫酸盐的滴定体积和浓度，求出所用硫代硫酸盐的物质的量，它等于所取试液中铜(II)离子的物质的量。放大到原始样品，求出铜的质量，然后计算质量百分比纯度。样卷的评分方案会对任何遗漏的十倍因子或放大错误扣分，因此要始终清晰地写出计算步骤：滴定剂物质的量 → 试液中物质的量 → 原始样品中物质的量 → 质量 → 纯度百分比。
10. Degree of Hydration and Water of Crystallisation | 水合度与结晶水
Questions asking for the value of x in a hydrated salt, such as Na₂CO₃·xH₂O, often involve a titration or heating to constant mass. In the 2016 specimen, a typical problem might give the mass loss on heating or the titre from an acid‑base titration of the hydrated salt solution. For example, a known mass of hydrated sodium carbonate is dissolved and titrated with standard HCl. Moles of HCl reacted → moles of Na₂CO₃ in the solution. Since you know the mass of the hydrated salt, you can compare the mass of anhydrous Na₂CO₃ (calculated from moles) with the original mass, and the difference is water. Then find the mole ratio of water to Na₂CO₃.
要求确定水合盐中x值的题目，如 Na₂CO₃·xH₂O，常涉及滴定或加热至恒重。2016年样卷中，一个典型问题可能给出加热失重数据，或水合盐溶液的酸碱滴定数据。例如，称取一定质量的水合碳酸钠溶解，用标准HCl滴定。反应的HCl物质的量 → 溶液中Na₂CO₃的物质的量。由于已知水合盐的质量，你可以将计算出的无水Na₂CO₃质量与原始质量进行比较，差值即为水的质量。然后求出水与Na₂CO₃的摩尔比。
Precision matters: if the titration data gives moles of Na₂CO₃ with three significant figures, but the mass of salt was given to two decimal places, your final answer for x must be an integer (or half‑integer) consistent with the formula. Usually, the mark scheme expects x to be a whole number, and any rounding should be justified by experimental error.
精度很重要：如果滴定数据给出三位有效数字的Na₂CO₃物质的量，但盐的质量仅给出两位小数，那么x的最终答案应该是一个与化学式相符的整数（或半整数）。通常评分方案要求x为整数，任何四舍五入都应说明是由实验误差所致。
11. Combining Half‑Equations and Calculating E° for Unknown Couples | 组合半反应与计算未知电对的标准电势
Using the relationship ΔG = –nFE, you can calculate the standard electrode potential for a redox couple that cannot be measured directly. The CH05 specimen often provides two known half‑cell potentials and asks for the third via
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