Buffer Solutions: IB & CIE Chemistry Key Points | 缓冲溶液：IB 与 CIE 化学考点精讲

📚 Buffer Solutions: IB & CIE Chemistry Key Points | 缓冲溶液：IB 与 CIE 化学考点精讲

Buffer solutions are one of the most fascinating and practical topics in acid–base chemistry. They appear everywhere from biological blood regulation to industrial processes, and mastering them is essential for top marks in both IB and CIE examinations. This article breaks down every key concept, equation, and exam trick you need to know.

缓冲溶液是酸碱化学中最有趣也最实用的主题之一。从生物体内血液的调控到工业生产，处处可见它们的身影，而要在 IB 和 CIE 考试中取得高分，彻底掌握缓冲溶液至关重要。本文将拆解每一个关键概念、方程式和应试技巧，助你轻松迎考。

1. What is a Buffer Solution? | 什么是缓冲溶液？

A buffer solution is a special aqueous system that resists changes in pH when small amounts of acid or alkali are added, or when the solution is diluted. Unlike ordinary solutions, buffers maintain a nearly constant pH because they contain species capable of neutralising both added H⁺ and OH⁻ ions.

缓冲溶液是一种特殊的水溶液体系，当加入少量酸或碱，或者进行稀释时，它能抵抗 pH 的变化。与普通溶液不同，缓冲溶液能够保持几乎不变的 pH 值，因为它含有能同时中和外加 H⁺ 和 OH⁻ 的物质。

In the IB and CIE syllabuses, you need to be able to define a buffer, identify its components, explain its action, and calculate the pH of buffer mixtures. Both acidic and alkaline buffers are covered.

在 IB 和 CIE 考纲中，你需要能够定义缓冲溶液、识别其组成、解释其作用原理，并计算缓冲混合物的 pH。酸性和碱性缓冲液均属考查范围。

2. Components of a Buffer | 缓冲溶液的组成

Every buffer contains a weak acid and its conjugate base, or a weak base and its conjugate acid. These two species must be present in significant amounts and in equilibrium with each other. The most common arrangement is a weak acid mixed with its salt (which provides the conjugate base), for example ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COONa).

每一种缓冲溶液都含有一对弱酸与其共轭碱，或一对弱碱与其共轭酸。这两个组分必须大量存在并处于相互平衡的状态。最常见的组合是弱酸与其盐混合（由盐提供共轭碱），例如乙酸 (CH₃COOH) 和乙酸钠 (CH₃COONa)。

Similarly, an alkaline buffer can be made from a weak base and its salt, such as ammonia (NH₃) and ammonium chloride (NH₄Cl). The presence of both the weak acid and its conjugate base (or weak base and its conjugate acid) allows the system to ‘soak up’ excess H⁺ or OH⁻ without a large shift in pH.

类似地，碱性缓冲液可由弱碱及其盐配制，例如氨 (NH₃) 和氯化铵 (NH₄Cl)。弱酸和其共轭碱（或弱碱与其共轭酸）同时存在，使得体系能“吸收”多余的 H⁺ 或 OH⁻，而不会引起 pH 大幅波动。

3. How Buffers Work: The Equilibrium Perspective | 缓冲溶液的作用原理：平衡视角

Consider an acidic buffer made of a weak acid HA and its salt MA, which fully dissociates to give A⁻. The equilibrium established is: HA ⇌ H⁺ + A⁻. The salt provides a high concentration of A⁻, suppressing the dissociation of the weak acid via the common ion effect.

考虑由弱酸 HA 及其盐 MA（完全电离出 A⁻）构成的酸性缓冲液。建立的平衡为：HA ⇌ H⁺ + A⁻。盐提供了高浓度的 A⁻，通过同离子效应抑制了弱酸的电离。

When a small amount of strong acid is added, the added H⁺ combines with the conjugate base A⁻ to form more HA: H⁺ + A⁻ → HA. The equilibrium shifts left, and the pH barely changes. When a small amount of strong base is added, OH⁻ reacts with HA: HA + OH⁻ → A⁻ + H₂O. Again, the pH remains relatively stable.

当加入少量强酸时，外加的 H⁺ 与共轭碱 A⁻ 结合生成更多的 HA：H⁺ + A⁻ → HA。平衡向左移动，pH 几乎不变。当加入少量强碱时，OH⁻ 与 HA 反应：HA + OH⁻ → A⁻ + H₂O。pH 再次保持相对稳定。

For an alkaline buffer, such as NH₃ / NH₄⁺, the equilibrium is NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Added acid is neutralised by NH₃, and added base is removed by NH₄⁺.

对于碱性缓冲液，如 NH₃ / NH₄⁺，平衡为 NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。外加的酸被 NH₃ 中和，外加的碱被 NH₄⁺ 消耗。

4. Acidic Buffers: Weak Acid + Its Conjugate Base | 酸性缓冲液：弱酸及其共轭碱

The classic acidic buffer is a mixture of a weak acid and its sodium or potassium salt. Examples include CH₃COOH / CH₃COONa, HCOOH / HCOONa, and carbonic acid / hydrogencarbonate. These systems keep the pH in the acidic range, typically below 7.

典型的酸性缓冲液是弱酸与其钠盐或钾盐的混合物。例子包括 CH₃COOH / CH₃COONa、HCOOH / HCOONa 以及碳酸 / 碳酸氢盐。这类体系将 pH 维持在酸性范围，通常低于 7。

When preparing an acidic buffer, you can either mix the weak acid directly with its salt, or partially neutralise the weak acid with a strong base. Both methods ensure that appreciable amounts of HA and A⁻ coexist in solution.

配制酸性缓冲液时，既可将弱酸与其盐直接混合，也可用强碱部分中和弱酸。两种方法都能确保溶液中同时存在足量的 HA 和 A⁻。

5. Alkaline Buffers: Weak Base + Its Conjugate Acid | 碱性缓冲液：弱碱及其共轭酸

Alkaline buffers consist of a weak base and a salt containing its conjugate acid. The most frequently examined example is the ammonia–ammonium chloride buffer: NH₃ and NH₄Cl. Such buffers maintain a pH above 7 and are particularly important in systems like hair dyes and certain biochemical assays.

碱性缓冲液由弱碱和含有其共轭酸的盐组成。考试中最常见的例子是氨–氯化铵缓冲液：NH₃ 和 NH₄Cl。这类缓冲液维持 pH 高于 7，在染发剂和某些生化检测中尤为重要。

The operation principle is analogous: NH₃ neutralises added H⁺ to form NH₄⁺, while NH₄⁺ reacts with added OH⁻ to regenerate NH₃ and water. The common ion NH₄⁺ from the salt suppresses the ionisation of the weak base, making the system robust.

作用原理类似：NH₃ 中和外加的 H⁺ 生成 NH₄⁺，而 NH₄⁺ 与外加的 OH⁻ 反应重新生成 NH₃ 和水。来自盐的 NH₄⁺ 同离子抑制了弱碱的电离，使体系更稳定。

6. The Henderson–Hasselbalch Equation | 亨德森–哈塞尔巴尔赫方程

The quantitative treatment of buffer pH centres on the Henderson–Hasselbalch equation. For an acidic buffer derived from a weak acid HA, it is written as:

缓冲 pH 的定量处理核心是亨德森–哈塞尔巴尔赫方程。对于由弱酸 HA 构成的酸性缓冲液，其形式为：

pH = pKₐ + log₁₀([A⁻] / [HA])

Here, pKₐ = −log₁₀(Kₐ), where Kₐ is the acid dissociation constant. The equation assumes that the concentrations of the conjugate base and the weak acid at equilibrium are approximately equal to the initial concentrations of the salt and the acid, respectively. This approximation holds when the buffer concentrations are reasonably high and the extent of dissociation is small.

这里 pKₐ = −log₁₀(Kₐ)，Kₐ 是酸电离常数。该方程假设共轭碱和弱酸的平衡浓度分别近似等于盐和酸的初始浓度。当缓冲液浓度较高且电离程度很小时，这一近似成立。

For an alkaline buffer, the analogous form uses pKₐ of the conjugate acid:

对于碱性缓冲液，类似形式使用共轭酸的 pKₐ：

pOH = pK_b + log₁₀([conjugate acid] / [weak base])

Or, converting to pH: pH = 14 − pOH. However, most IB and CIE questions can be answered using the primary acidic version of the equation, by identifying the weak acid in the conjugate acid–base pair.

或转换为 pH：pH = 14 − pOH。但大多数 IB 和 CIE 考题可直接使用酸式方程，只需识别共轭酸碱对中的弱酸即可。

7. Calculating the pH of a Buffer | 缓冲溶液 pH 的计算

Let us work through a typical examination-style calculation. Suppose you mix 50 cm³ of 0.10 mol dm⁻³ CH₃COOH with 25 cm³ of 0.10 mol dm⁻³ NaOH. After neutralisation, the solution contains unreacted CH₃COOH and the salt CH₃COONa, forming a buffer. The moles of acid originally are 0.0050, and the moles of OH⁻ added are 0.0025. Reaction leaves 0.0025 mol HA and produces 0.0025 mol A⁻ in a total volume of 75 cm³. Thus the ratio [A⁻]/[HA] = 1. If pKₐ of ethanoic acid is 4.76, then pH = 4.76 + log₁₀(1) = 4.76.

我们做一个典型考题计算。假设将 50 cm³ 0.10 mol dm⁻³ CH₃COOH 与 25 cm³ 0.10 mol dm⁻³ NaOH 混合。中和后，溶液中残留未反应的 CH₃COOH 和生成的盐 CH₃COONa，形成缓冲液。酸的初始摩尔数为 0.0050，加入的 OH⁻ 摩尔数为 0.0025。反应后剩下 0.0025 mol HA，生成 0.0025 mol A⁻，总体积 75 cm³。因此 [A⁻]/[HA] 比值 = 1。若乙酸的 pKₐ = 4.76，则 pH = 4.76 + log₁₀(1) = 4.76。

When the ratio [A⁻]/[HA] is not 1:1, you must compute the concentrations carefully. Remember, using moles directly in the log term is acceptable only if the total volume is the same for both species, since concentration ∝ moles in the same solution. The most common pitfall is failing to convert volumes and concentrations correctly. Always check units.

当 [A⁻]/[HA] 比值不是 1:1 时，必须仔细计算浓度。切记，只有当两种物质处于同一溶液（总体积相同）时，才可直接用摩尔数代替浓度取对数，因为浓度与摩尔数成正比。最常见的失分点就是未能正确转换体积和浓度。请务必检查单位。

8. Buffer Capacity and Range | 缓冲容量与缓冲范围

Buffer capacity (β) is a measure of a buffer’s ability to resist pH change. It is defined as the number of moles of strong acid or strong base required to change the pH of 1 dm³ of the buffer solution by one unit. A high buffer capacity means the pH changes very little upon addition of acid or base.

缓冲容量 (β) 衡量缓冲液抵抗 pH 变化的能力。其定义为使 1 dm³ 缓冲溶液的 pH 改变一个单位所需加强酸或强碱的摩尔数。缓冲容量高，意味着加入酸或碱后 pH 变化极小。

Maximum buffer capacity is achieved when [A⁻] = [HA], i.e. when the pH equals pKₐ. At this point, the buffer is equally effective against added acid and added base. As a rule of thumb, a buffer works effectively within the pH range of pKₐ ± 1. Outside this range, the buffer capacity drops significantly because one component is nearly exhausted.

当 [A⁻] = [HA]，即 pH 等于 pKₐ 时，缓冲容量最大。此时缓冲液对加入的酸和碱同样有效。经验规则是，缓冲液在 pH = pKₐ ± 1 的范围内有效工作。超出此范围，由于某一组分近乎耗尽，缓冲容量明显下降。

For alkaline buffers, the effective range is pK_b ± 1 around the pK_b of the weak base (or pH around 14 − pK_b ± 1). In exam questions, you may be asked to choose the best acid–base pair for a desired pH; always select the one whose pKₐ is as close as possible to the target pH.

对于碱性缓冲液，有效范围是弱碱 pK_b 附近的 pK_b ± 1（或 pH 约为 14 − pK_b ± 1）。考题中可能会让你为特定 pH 选择最佳的酸碱对，务必选择 pKₐ 尽可能接近目标 pH 的组合。

9. Preparing a Buffer Solution | 缓冲溶液的配制

In the laboratory, buffers are prepared by two main methods: (1) mixing a weak acid with its salt directly, or (2) partially neutralising a weak acid with a strong base. The second method allows precise control over the [A⁻]/[HA] ratio by adjusting the volume of base added.

在实验室中，配制缓冲溶液主要有两种方法：(1) 将弱酸与其盐直接混合，或 (2) 用强碱部分中和弱酸。第二种方法可通过调整加入碱的体积，精准控制 [A⁻]/[HA] 比值。

For an alkaline buffer, one can mix a weak base with its salt, or partially neutralise a weak base with a strong acid. When doing calculations for preparation, the Henderson–Hasselbalch equation is used to determine the required ratio. For example, to prepare a buffer at pH 5.00 using ethanoic acid (pKₐ = 4.76), you need log₁₀([A⁻]/[HA]) = 5.00 − 4.76 = 0.24, so [A⁻]/[HA] ≈ 1.74. This means the conjugate base concentration must be 1.74 times that of the weak acid.

配制碱性缓冲液时，可混合弱碱与其盐，或用强酸部分中和弱碱。进行计算时，利用亨德森–哈塞尔巴尔赫方程求所需比例。例如，用乙酸（pKₐ = 4.76）配制 pH 5.00 的缓冲液，需要 log₁₀([A⁻]/[HA]) = 5.00 − 4.76 = 0.24，因此 [A⁻]/[HA] ≈ 1.74。即共轭碱浓度须为弱酸浓度的 1.74 倍。

10. Buffer Action in Biological Systems: Blood pH | 生物系统中的缓冲作用：血液 pH

One of the most important biological buffers is the carbonic acid–hydrogencarbonate system that maintains human blood pH at approximately 7.40. The equilibrium involved is: CO₂(aq) + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻. This system is open, with CO₂ constantly being exchanged in the lungs.

最重要的生物缓冲系统之一是碳酸–碳酸氢盐缓冲对，它使人体血液 pH 维持在约 7.40。相关的平衡是：CO₂(aq) + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻。这是一个开放体系，CO₂ 在肺部不断进行交换。

If the blood becomes too acidic (acidosis), the equilibrium shifts left and breathing rate increases to expel more CO₂. If the blood becomes too alkaline (alkalosis), breathing rate decreases to retain CO₂. The Henderson–Hasselbalch equation applied to this system gives: pH = pKₐ₁ + log₁₀([HCO₃⁻] / [CO₂]). The pKₐ₁ for carbonic acid is about 6.1 at body temperature, but the system works effectively because the body tightly regulates the [HCO₃⁻]/[CO₂] ratio.

若血液过酸（酸中毒），平衡向左移动，呼吸加快以排出更多 CO₂。若血液过碱（碱中毒），呼吸减慢以保留 CO₂。应用亨德森–哈塞尔巴尔赫方程于此体系：pH = pKₐ₁ + log₁₀([HCO₃⁻] / [CO₂])。体温下碳酸的 pKₐ₁ 约为 6.1，但由于人体严格调控 [HCO₃⁻]/[CO₂] 比值，该系统仍能有效工作。

Another intracellular buffer is the dihydrogenphosphate–hydrogenphosphate pair: H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻. IB and CIE papers often link buffer theory to real-world examples, so be prepared to apply the principles in unfamiliar contexts.

另一种细胞内缓冲对是磷酸二氢盐–磷酸氢盐：H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻。IB 和 CIE 试卷常将缓冲理论与现实案例结合，务必准备好将原理应用于陌生情境。

11. Common Exam-Style Questions | 常见考题类型

In both IB and CIE exams, buffer questions typically require you to:

在 IB 和 CIE 考试中，缓冲类的题目通常要求你：

Define a buffer and identify its components from a given mixture.

定义缓冲溶液，并从给定混合物中识别其组分。
Explain how a buffer resists pH changes using equations and Le Chatelier’s principle.

运用化学方程式和勒夏特列原理解释缓冲液如何抵抗 pH 变化。
Calculate the pH of a buffer after mixing known volumes and concentrations.

计算混合已知体积和浓度后的缓冲液 pH。
Determine the mass of salt needed to prepare a buffer of a given pH.

计算配制指定 pH 缓冲液所需的盐的质量。
Evaluate buffer selection and buffer range using pKₐ values.

利用 pKₐ 值评估缓冲液的选择和缓冲范围。

Graphical questions also appear: you may be given a titration curve and asked to identify the buffer region, which is the flat part of the curve before the equivalence point where pH changes slowly. In a weak acid–strong base titration, the buffer region is centred around the half-equivalence point, where [HA] = [A⁻] and pH = pKₐ.

图表题也会出现：你可能会看到一条滴定曲线，并被要求指出缓冲区，即等当点之前 pH 变化缓慢的平台部分。在弱酸–强碱滴定中，缓冲区位于半等当点附近，此时 [HA] = [A⁻]，且 pH = pKₐ。

12. Quick Summary and Key Takeaways | 快速总结与关键要点

A buffer is a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid. It works by neutralising added H⁺ and OH⁻, with the equilibrium shifting to consume the added species. The pH of an acidic buffer is given by pH = pKₐ + log₁₀([A⁻]/[HA]). Maximum buffer capacity is at pH = pKₐ, and the effective range is pKₐ ± 1.

缓冲溶液是弱酸及其共轭碱，或弱碱及其共轭酸的混合物。它通过中和外加的 H⁺ 和 OH⁻ 并移动平衡来发挥作用。酸性缓冲液的 pH 由 pH = pKₐ + log₁₀([A⁻]/[HA]) 给出。最大缓冲容量出现在 pH = pKₐ 处，有效范围为 pKₐ ± 1。

Always show your working clearly in calculations, state any assumptions, and remember that dilution does not change buffer pH appreciably because the ratio of concentrations stays constant. With these fundamentals, you can tackle any buffer problem confidently.

计算时请始终清晰展示运算过程，说明所作假设，并记住稀释不会明显改变缓冲液的 pH，因为浓度比值保持不变。掌握这些基础，你就能自信地解决任何缓冲液问题。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Buffer Solutions: IB & CIE Chemistry Key Points | 缓冲溶液：IB 与 CIE 化学考点精讲