Case Study Practical Exercises for Year 13 CAIE Chemistry | Year 13 CAIE 化学：案例分析实战演练

📚 Case Study Practical Exercises for Year 13 CAIE Chemistry | Year 13 CAIE 化学：案例分析实战演练

Case studies bridge the gap between theoretical knowledge and practical application, a skill essential for CAIE A2 Chemistry. They mirror the demands of Paper 5 (Planning, Analysis and Evaluation) and sharpen analytical thinking. This article presents ten carefully crafted case studies covering redox titrations, kinetics, organic synthesis, electrochemistry, transition metal colourimetry, buffer design, Born–Haber cycles, polymer degradation, spectroscopic identification, and data evaluation. Each scenario is paired with detailed worked solutions, error analysis, and links to key syllabus concepts.

案例分析弥合了理论知识与实际应用之间的鸿沟，是掌握 CAIE A2 化学的关键技能。它们模拟了试卷五（实验设计与分析评估）的要求，能够显著提升分析思维能力。本文精心设计了十个案例，涵盖氧化还原滴定、动力学、有机合成、电化学、过渡金属比色法、缓冲液设计、玻恩-哈伯循环、聚合物降解、波谱解析以及数据评估。每个场景均配有详细计算过程与误差分析，紧贴核心考纲内容。

1. Case Study 1: Redox Titration – Determining Vitamin C in Tablets | 案例分析1：氧化还原滴定——测定药片中的维生素C

Vitamin C (ascorbic acid, C₆H₈O₆) can be quantified by generating iodine in situ and titrating against standardised thiosulfate. A crushed vitamin C tablet is dissolved in dilute sulfuric acid. An excess of potassium iodate(V) and potassium iodide is added: IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O. The iodine reacts with ascorbic acid (C₆H₈O₆ + I₂ → C₆H₆O₆ + 2I⁻ + 2H⁺). Unreacted iodine is then titrated with 0.100 mol dm⁻³ Na₂S₂O₃ solution, using starch as an indicator near the end point.

维生素C（抗坏血酸，C₆H₈O₆）可通过原位生成碘并用标定过的硫代硫酸钠滴定来定量。将研碎的维生素C片溶解于稀硫酸中，加入过量的碘酸钾和碘化钾：IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O。生成的碘将抗坏血酸氧化（C₆H₈O₆ + I₂ → C₆H₆O₆ + 2I⁻ + 2H⁺），剩余的碘再用 0.100 mol dm⁻³ Na₂S₂O₃ 溶液滴定，临近终点时加入淀粉指示剂。

In a typical run, 25.0 cm³ of the tablet solution (total volume 250 cm³) was treated with 50.0 cm³ of 0.0200 mol dm⁻³ KIO₃ and excess KI. The final titre of thiosulfate was 18.40 cm³. The number of moles of I₂ generated initially is (50.0/1000) × 0.0200 × 3 = 0.00300 mol. Moles of S₂O₃²⁻ used = 0.100 × 0.01840 = 0.00184 mol, which corresponds to I₂ reacted = 0.00184 / 2 = 0.000920 mol. Hence I₂ consumed by ascorbic acid = 0.00300 – 0.000920 = 0.00208 mol, so mass of vitamin C = 0.00208 × 176.1 g mol⁻¹ = 0.366 g per tablet. This method yields a relative precision of around ±0.5% if repeated.

某次实验中，取 25.0 cm³ 药片溶液（总配制体积 250 cm³），加入 50.0 cm³ 0.0200 mol dm⁻³ KIO₃ 和过量 KI。终点时消耗硫代硫酸钠 18.40 cm³。初始生成的 I₂ 的物质的量 = (50.0/1000) × 0.0200 × 3 = 0.00300 mol。消耗的 S₂O₃²⁻ 物质的量 = 0.100 × 0.01840 = 0.00184 mol，对应的剩余 I₂ = 0.00184 / 2 = 0.000920 mol。因此与维生素C反应的 I₂ = 0.00300 – 0.000920 = 0.00208 mol，每片维生素C的质量 = 0.00208 × 176.1 g mol⁻¹ = 0.366 g。多次重复的相对精密度可达 ±0.5%。

Key sources of error include loss of iodine by evaporation, premature starch addition leading to a diffuse end point, and incomplete dissolution of the tablet. Improvement strategies involve working in a cool environment, adding starch only when the solution is pale yellow, and filtering the tablet solution to remove binders. A blank titration can correct for iodine consumed by side reactions.

主要误差来源包括碘挥发、过早加入淀粉导致终点拖尾、药片未完全溶解。改进措施包括在较冷环境下操作、仅在溶液变为浅黄色时加入淀粉、过滤除去药片中的赋形剂。可通过空白滴定校正副反应消耗的碘。

2. Case Study 2: Kinetics – Activation Energy of the Peroxodisulfate–Iodide Reaction | 案例分析2：动力学——过二硫酸盐与碘离子反应的活化能

The oxidation of iodide by peroxodisulfate, S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂, can be followed by quenching samples at timed intervals and titrating the iodine produced. Alternatively, a continuous method using a colorimeter to monitor the appearance of I₂ (absorbance at 470 nm) is more accurate. The initial rate is proportional to the reciprocal of the time taken for the absorbance to reach a fixed value, which allows relative rate determination at different temperatures.

过二硫酸根氧化碘离子（S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂）可通过定时取样并滴定生成的碘来跟踪。更精确的方法是用比色计连续监测 470 nm 处碘的吸光度。初始速率与吸光度达到某一固定值所用时间的倒数成正比，由此可测得不同温度下的相对速率常数。

Using the Arrhenius equation in logarithmic form, ln k = ln A – Eₐ/(RT), a graph of ln(1/time) against 1/T yields a straight line with slope = –Eₐ/R. Convert temperatures to Kelvin (293 K, 303 K, 313 K, 323 K) and calculate 1/T. The slope is approximately –7500 K, giving Eₐ = 7500 × 8.31 J mol⁻¹ ≈ 62.3 kJ mol⁻¹. This is plausible for a reaction involving bond breaking in the peroxo linkage.

使用对数形式的阿伦尼乌斯方程 ln k = ln A – Eₐ/(RT)，以 ln(1/时间) 对 1/T 作图，斜率 = –Eₐ/R。将温度转换为开尔文，计算 1/T，所得斜率约为 –7500 K，求得 Eₐ ≈ 7500 × 8.31 J mol⁻¹ = 62.3 kJ mol⁻¹。该数值与过氧键断裂所需的能量相符。

Sources of error include temperature fluctuations during the reaction, the delay between mixing and starting the clock, and the arbitrary choice of fixed absorbance. Using a thermostatic water bath, rapid mixing devices, and recording the entire absorbance curve to extract true initial rates are highlighted improvements. This mirrors the requirements for error analysis in Paper 5.

误差来源包括反应过程中温度波动、混合与计时的延迟以及固定吸光度值的随意选取。改进措施有：使用恒温水浴、快速混合装置，以及记录完整吸光度曲线以提取真实的初始速率。这完全符合试卷五对误差分析的要求。

3. Case Study 3: Organic Synthesis – Preparation and Analysis of Aspirin | 案例分析3：有机合成——阿司匹林的制备与分析

Aspirin (2-ethanoyloxybenzenecarboxylic acid) is synthesised from salicylic acid and ethanoic anhydride. In a typical school preparation, 3.0 g of salicylic acid, 6 cm³ of ethanoic anhydride, and a few drops of concentrated phosphoric acid are warmed at 60 °C for 20 minutes. The mixture is poured into cold water, and the crude aspirin is filtered, washed, and rec

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