Buffer Solutions in IB Edexcel Chemistry: Key Exam Points | IB Edexcel 化学：缓冲溶液考点精讲

📚 Buffer Solutions in IB Edexcel Chemistry: Key Exam Points | IB Edexcel 化学：缓冲溶液考点精讲

A buffer solution is a system that minimises pH changes when small amounts of acid or base are added. In IB and Edexcel Chemistry, buffers are a high‑frequency topic, linking equilibrium theory, acid‑base calculations and real‑world applications. Mastering buffers means understanding how conjugate pairs work, how to use the Henderson‑Hasselbalch equation, and how to interpret buffer capacity and range. This article covers all the essential exam points, step by step.

缓冲溶液是一种能够抵抗外加少量酸或碱引起的 pH 变化的体系。在 IB 和 Edexcel 化学中，缓冲溶液是高频考点，它将平衡理论、酸碱计算和实际应用串联起来。真正掌握缓冲溶液，意味着要理解共轭酸碱对如何工作，熟练运用亨德森‑哈塞尔巴尔赫方程，并能分析缓冲容量和缓冲范围。本文循序渐进，覆盖所有关键考点。

1. Definition and Basic Composition | 定义与基本组成

A buffer solution maintains an almost constant pH when small amounts of strong acid or strong base are added. It is typically composed of a weak acid and its conjugate base, or a weak base and its conjugate acid. Both components must be present in appreciable concentrations.

缓冲溶液能在加入少量强酸或强碱时维持 pH 几乎不变。它通常由一种弱酸及其共轭碱，或一种弱碱及其共轭酸组成。两种组分都必须以可观的浓度存在。

The weak acid neutralises any added OH⁻ while the conjugate base neutralises any added H⁺. In an acidic buffer, the equilibrium HA ⇌ H⁺ + A⁻ is central; the large reservoir of HA and A⁻ allows the equilibrium to shift without significant pH alteration.

弱酸中和外加的 OH⁻，而共轭碱中和外加的 H⁺。在酸性缓冲液中，平衡 HA ⇌ H⁺ + A⁻ 是核心；大量的 HA 和 A⁻ 储备使得平衡能够移动而 pH 不会显著变化。

2. Types of Buffer Solutions | 缓冲溶液的类型

Acidic buffers are made from a weak acid and its salt with a strong base, for example ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COO⁻Na⁺). The solution contains both the neutral acid molecules and the conjugate base ions.

酸性缓冲液由弱酸及其强碱盐制成，例如乙酸 (CH₃COOH) 和乙酸钠 (CH₃COO⁻Na⁺)。溶液中同时存在中性酸分子和共轭碱离子。

Basic buffers are made from a weak base and its salt with a strong acid, for example ammonia (NH₃) and ammonium chloride (NH₄⁺Cl⁻). The conjugate acid NH₄⁺ provides the H⁺‑accepting ability.

碱性缓冲液由弱碱及其强酸盐制成，例如氨 (NH₃) 与氯化铵 (NH₄⁺Cl⁻)。共轭酸 NH₄⁺ 提供了接受 H⁺ 的能力。

3. How Buffers Work – The Equilibrium View | 缓冲作用原理 – 从平衡视角看

Consider an acidic buffer: HA ⇌ H⁺ + A⁻. Adding a small amount of strong acid increases [H⁺]; the equilibrium shifts to the left as A⁻ reacts with the added H⁺ to form HA, so [H⁺] is kept low. Adding a strong base removes H⁺ as it forms H₂O; the equilibrium shifts to the right, HA dissociates to replenish H⁺.

考虑一个酸性缓冲液：HA ⇌ H⁺ + A⁻。加入少量强酸时 [H⁺] 增加；平衡向左移动，A⁻ 与外加 H⁺ 结合生成 HA，所以 [H⁺] 得以维持在低水平。加入强碱时，碱消耗 H⁺ 生成水；平衡向右移动，HA 解离来补充 H⁺。

The key is that the concentrations of the weak acid and its conjugate base are large relative to the amount of added strong acid or base. Therefore, the ratio [A⁻]/[HA] changes only slightly, and the pH remains nearly unchanged.

关键在于弱酸及其共轭碱的浓度相对外加的强酸或强碱的量是很大的。因此，[A⁻]/[HA] 的比值只会发生微小的变化，pH 几乎保持不变。

4. The Henderson‑Hasselbalch Equation | 亨德森‑哈塞尔巴尔赫方程

For an acidic buffer, the pH can be quickly estimated using:

对于酸性缓冲液，可用下式快速估算 pH：

pH = pKₐ + log₁₀([A⁻]/[HA])

This equation is derived from the acid dissociation constant expression Kₐ = [H⁺][A⁻]/[HA]. Taking the negative logarithm and rearranging gives the Henderson‑Hasselbalch form. It is valid when the approximations [HA] ≈ initial [HA] and [A⁻] ≈ initial [A⁻] hold, i.e. when the buffer is not overwhelmed.

该方程由酸解离常数表达式 Kₐ = [H⁺][A⁻]/[HA] 推导而来。取负对数并整理即得到亨德森‑哈塞尔巴尔赫形式。当 [HA] ≈ 初始 [HA] 和 [A⁻] ≈ 初始 [A⁻] 近似成立时（即缓冲液未被击穿），该方程有效。

For a basic buffer, the analogous form uses pKₐ of the conjugate acid: pOH = pK_b + log₁₀([BH⁺]/[B]), or pH = 14 – pOH. In IB and Edexcel exams, the acid‑buffer form is more frequently tested.

对于碱性缓冲液，类似的形式使用共轭酸的 pKₐ：pOH = pK_b + log₁₀([BH⁺]/[B])，或 pH = 14 – pOH。在 IB 和 Edexcel 考试中，酸性缓冲液形式考得更多。

5. Calculating the pH of a Buffer | 缓冲液 pH 的计算

Example: A buffer is prepared by mixing 50.0 cm³ of 0.10 mol dm⁻³ CH₃COOH with 50.0 cm³ of 0.10 mol dm⁻³ CH₃COONa. Calculate the pH. (pKₐ of ethanoic acid = 4.76)

例题：将 50.0 cm³ 0.10 mol dm⁻³ CH₃COOH 与 50.0 cm³ 0.10 mol dm⁻³ CH₃COONa 混合制成缓冲液。计算 pH。（乙酸的 pKₐ = 4.76）

After mixing, total volume = 100 cm³, so [HA] = 0.050 mol dm⁻³, [A⁻] = 0.050 mol dm⁻³. Ratio [A⁻]/[HA] = 1. pH = 4.76 + log₁₀(1) = 4.76. If 1.0 cm³ of 1.0 mol dm⁻³ HCl is added, recalculate the new pH using the stoichiometric change in moles of HA and A⁻.

混合后，总体积 = 100 cm³，因此 [HA] = 0.050 mol dm⁻³，[A⁻] = 0.050 mol dm⁻³。比值 [A⁻]/[HA] = 1。pH = 4.76 + log₁₀(1) = 4.76。若加入 1.0 cm³ 1.0 mol dm⁻³ HCl，先计算 HA 和 A⁻ 摩尔量的变化，再求新的 pH。

6. Buffer Capacity | 缓冲容量

Buffer capacity is a measure of how much strong acid or base a buffer can neutralise before its pH changes significantly. It depends on two factors: the total concentration of the buffer components and the ratio [A⁻]/[HA]. A buffer has maximum capacity when [A⁻] = [HA], i.e. pH = pKₐ.

缓冲容量衡量缓冲液在 pH 发生显著变化前能中和多少强酸或强碱。它取决于两个因素：缓冲组分总浓度，以及比值 [A⁻]/[HA]。当 [A⁻] = [HA]，即 pH = pKₐ 时，缓冲容量最大。

Higher total concentration (e.g. 0.5 M HA/A⁻ vs 0.1 M) gives a larger capacity because more species are available to react with added H⁺ or OH⁻.

总浓度越高（例如 0.5 M HA/A⁻ 与 0.1 M 相比），容量越大，因为有更多物种可与外加的 H⁺ 或 OH⁻ 反应。

7. Effective Buffer Range | 有效缓冲范围

A buffer works effectively within pH ≈ pKₐ ± 1. Outside this range, the ratio [A⁻]/[HA] becomes too extreme (less than 0.1 or greater than 10), and the buffer capacity drops sharply.

缓冲液在 pH ≈ pKₐ ± 1 的范围内有效工作。超出此范围，[A⁻]/[HA] 的比值过大或过小（小于 0.1 或大于 10），缓冲容量急剧下降。

When selecting a buffer for a specific pH, choose a weak acid whose pKₐ is within one unit of the target pH. For physiological pH 7.4, the H₂PO₄⁻/HPO₄²⁻ pair (pKₐ₂ ≈ 7.2) is often used.

为特定 pH 选择缓冲液时，应选 pKₐ 在目标 pH 一个单位以内的弱酸。对于生理 pH 7.4，常用 H₂PO₄⁻/HPO₄²⁻ 对（pKₐ₂ ≈ 7.2）。

8. Preparing Buffer Solutions | 缓冲溶液的配制

Method 1: Direct mixing of a weak acid and its conjugate base in the required ratio. For example, weigh a calculated mass of the salt and dissolve it with the weak acid in a volumetric flask.

方法一：直接按所需比例混合弱酸及其共轭碱。例如，称取计算量的盐，与弱酸一起在容量瓶中溶解。

Method 2: Partial neutralisation of a weak acid with a strong base. If 0.1 mol of CH₃COOH is reacted with 0.05 mol of NaOH, the solution contains 0.05 mol of un‑reacted CH₃COOH and 0.05 mol of CH₃COO⁻, forming a buffer with pH = pKₐ.

方法二：用强碱部分中和弱酸。若 0.1 mol CH₃COOH 与 0.05 mol NaOH 反应，溶液中含 0.05 mol 未反应的 CH₃COOH 和 0.05 mol CH₃COO⁻，形成 pH = pKₐ 的缓冲液。

9. Biological and Industrial Applications | 生物与工业应用

In human blood, the carbonic acid‑hydrogencarbonate system (H₂CO₃/HCO₃⁻) keeps pH around 7.4. Enzyme activity is pH‑dependent, so buffers maintain optimal conditions inside cells and in laboratory media.

在人体血液中，碳酸‑碳酸氢根体系 (H₂CO₃/HCO₃⁻) 将 pH 维持在 7.4 左右。酶活性依赖 pH，因此缓冲液在细胞内和实验室培养基中维持最佳条件。

In industry, buffers are used in fermentation, dyeing processes, and in electroplating baths to ensure consistent product quality. In analytical chemistry, buffer solutions calibrate pH meters.

在工业中，缓冲液用于发酵、染色工艺以及电镀槽，以确保产品一致性。在分析化学中，缓冲溶液用于校准 pH 计。

10. Common Exam Questions and Answering Techniques | 常见考题与答题技巧

Explain how a buffer resists pH change: always refer to the equilibrium shift, name the species that react with added H⁺ and OH⁻, and state that the pH change is minimal because the ratio [A⁻]/[HA] stays nearly constant.

解释缓冲液如何抵抗 pH 变化：一定要提到平衡移动，说明与外加 H⁺ 和 OH⁻ 反应的具体物种，并说明因为 [A⁻]/[HA] 比值近乎不变，pH 变化很小。

Calculations: practice using Henderson‑Hasselbalch, but also be prepared to calculate the new concentrations after an addition by stoichiometry, then plug into the equation. Watch units – convert cm³ to dm³ where necessary.

计算题：熟练使用亨德森‑哈塞尔巴尔赫方程，同时也要准备先通过化学计量算出添加后的新浓度，再代入方程。注意单位——必要时将 cm³ 转换为 dm³。

Buffer selection: given a pH, choose an appropriate weak acid by comparing pKₐ values. Justify your choice with the pKₐ ± 1 rule.

缓冲液选择：给定 pH，通过比较 pKₐ 值选择合适的弱酸。用 pKₐ ± 1 规则说明理由。

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Buffer Solutions in IB Edexcel Chemistry: Key Exam Points | IB Edexcel 化学：缓冲溶液 考点精讲