Buffer Solutions: Composition and Action | 缓冲溶液的组成与作用

📚 Buffer Solutions: Composition and Action | 缓冲溶液的组成与作用

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added, or when it is diluted. This property, known as buffer action, is of fundamental importance in many chemical and biological systems, from maintaining the pH of blood to enabling precise industrial and laboratory reactions.

缓冲溶液是一种能够抵抗因加入少量酸或碱，或发生稀释而引起pH值显著变化的溶液。这种性质称为缓冲作用，在众多化学和生物体系中具有根本性的重要意义——从维持血液的pH到实现精确的工业和实验室反应，都离不开缓冲溶液。

1. Definition and Core Concept | 定义与核心概念

A buffer solution is defined as a solution that maintains a nearly constant pH despite the addition of small amounts of strong acid (H⁺) or strong base (OH⁻). The pH of a buffer does change slightly upon addition of acid or base, but the change is very small compared to that of an equal volume of pure water.

缓冲溶液的定义是：在加入少量强酸（H⁺）或强碱（OH⁻）时，能保持pH几乎恒定的溶液。加入酸或碱后，缓冲溶液的pH确实会发生微小的变化，但与等体积纯水的pH变化相比，这种变化非常小。

The key to this resistance lies in the presence of two components: a weak acid and its conjugate base, or a weak base and its conjugate acid. These two species work together to neutralise added H⁺ or OH⁻ ions.

这种抵抗能力的关键在于溶液中存在两种组分：一种弱酸及其共轭碱，或者一种弱碱及其共轭酸。这两种物质协同作用，中和外加的H⁺或OH⁻离子。

For a buffer to function effectively, both components must be present in appreciable, comparable concentrations. If one component is depleted by excessive addition of acid or base, the buffer capacity is exceeded and the pH will change dramatically.

缓冲溶液要有效发挥作用，两种组分必须以可观的、相近的浓度同时存在。如果加入的酸或碱过多，导致某一组分被耗尽，缓冲容量就会被突破，pH将发生剧烈变化。

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

There are two main types of buffer solutions: acidic buffers and alkaline buffers. Each type is distinguished by its pH range and the nature of its components.

缓冲溶液主要分为两类：酸性缓冲溶液和碱性缓冲溶液。每种类型根据其pH范围及组分性质加以区分。

Acidic buffers have a pH below 7. They are formed by mixing a weak acid with a salt of that weak acid. For example, a mixture of ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COONa) produces an acidic buffer with a pH of approximately 4.76.

酸性缓冲溶液的pH低于7。它们由弱酸与该弱酸的盐混合而成。例如，乙酸（CH₃COOH）和乙酸钠（CH₃COONa）的混合物可形成pH约为4.76的酸性缓冲溶液。

Alkaline buffers have a pH above 7. They are formed by mixing a weak base with a salt of that weak base. A classic example is a mixture of ammonia (NH₃) and ammonium chloride (NH₄Cl), which produces an alkaline buffer with a pH of approximately 9.25.

碱性缓冲溶液的pH高于7。它们由弱碱与该弱碱的盐混合而成。经典例子是氨水（NH₃）和氯化铵（NH₄Cl）的混合物，可形成pH约为9.25的碱性缓冲溶液。

It is crucial to recognise that a buffer is not simply a solution of a weak acid or base alone; it requires both the weak acid/base and its conjugate partner in appreciable amounts.

必须认识到，缓冲溶液不仅仅是弱酸或弱碱的溶液；它必须以可观的量同时含有弱酸/弱碱及其共轭配偶体。

3. Composition of an Acidic Buffer | 酸性缓冲溶液的组成

An acidic buffer consists of a weak acid (HA) and its conjugate base (A⁻) in the form of a soluble salt. The weak acid provides molecules of HA, while the salt provides a reservoir of A⁻ ions.

酸性缓冲溶液由弱酸（HA）及其共轭碱（A⁻）组成，其中共轭碱以可溶性盐的形式存在。弱酸提供HA分子，而盐提供A⁻离子的储备。

Taking the ethanoic acid-sodium ethanoate system as an example, the components are:

以乙酸-乙酸钠体系为例，其组分为：

Ethanoic acid (CH₃COOH) — a weak acid that partially dissociates: CH₃COOH ⇌ CH₃COO⁻ + H⁺
乙酸（CH₃COOH）——一种部分电离的弱酸：CH₃COOH ⇌ CH₃COO⁻ + H⁺
Sodium ethanoate (CH₃COONa) — a soluble salt that fully dissociates: CH₃COONa → CH₃COO⁻ + Na⁺
乙酸钠（CH₃COONa）——一种完全电离的可溶性盐：CH₃COONa → CH₃COO⁻ + Na⁺

In this buffer, the concentration of CH₃COOH is high (from the weak acid) and the concentration of CH₃COO⁻ is also high (from both the dissociation of the acid and the complete dissociation of the salt). The equilibrium position is governed by the acid dissociation constant Kₐ.

在此缓冲溶液中，CH₃COOH的浓度很高（来自弱酸），CH₃COO⁻的浓度也很高（既来自酸的电离，也来自盐的完全电离）。平衡位置由酸电离常数Kₐ控制。

4. Composition of an Alkaline Buffer | 碱性缓冲溶液的组成

An alkaline buffer consists of a weak base (B) and its conjugate acid (BH⁺) in the form of a soluble salt. The weak base provides molecules of B, while the salt provides a reservoir of BH⁺ ions.

碱性缓冲溶液由弱碱（B）及其共轭酸（BH⁺）组成，其中共轭酸以可溶性盐的形式存在。弱碱提供B分子，而盐提供BH⁺离子的储备。

Taking the ammonia-ammonium chloride system as an example, the components are:

以氨-氯化铵体系为例，其组分为：

Ammonia (NH₃) — a weak base that partially dissociates: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
氨（NH₃）——一种部分电离的弱碱：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Ammonium chloride (NH₄Cl) — a soluble salt that fully dissociates: NH₄Cl → NH₄⁺ + Cl⁻
氯化铵（NH₄Cl）——一种完全电离的可溶性盐：NH₄Cl → NH₄⁺ + Cl⁻

The equilibrium is governed by the base dissociation constant K_b. The presence of a high concentration of NH₄⁺ from the salt suppresses the ionisation of the weak base through the common ion effect, ensuring that an adequate reservoir of NH₃ molecules remains available to neutralise added H⁺.

该平衡由碱电离常数K_b控制。盐提供的大量NH₄⁺通过同离子效应抑制了弱碱的电离，确保溶液中始终保有一定量的NH₃分子储备，用以中和外加的H⁺。

5. How a Buffer Resists Added Acid | 缓冲溶液如何抵抗外加酸

When a small amount of a strong acid (e.g., HCl) is added to an acidic buffer, the added H⁺ ions react with the conjugate base (A⁻) present in the buffer:

当向酸性缓冲溶液中加入少量强酸（如HCl）时，外加的H⁺离子会与缓冲溶液中的共轭碱（A⁻）发生反应：

H⁺(aq) + A⁻(aq) → HA(aq)

This reaction consumes the added H⁺ ions and converts them into the weak acid HA. Since HA is a weak acid, it only partially dissociates, so the H⁺ concentration — and therefore the pH — remains almost unchanged.

该反应消耗了外加的H⁺离子，将其转化为弱酸HA。由于HA是弱酸，只能部分电离，因此H⁺浓度——进而pH值——几乎保持不变。

For the ethanoate buffer specifically: CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq). The CH₃COO⁻ ions are replenished by the reservoir provided by the fully-dissociated sodium ethanoate salt.

具体到乙酸盐缓冲体系：CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq)。CH₃COO⁻离子由完全电离的乙酸钠盐这一储备持续补充。

In an alkaline buffer, the added H⁺ reacts directly with the weak base NH₃: NH₃(aq) + H⁺(aq) → NH₄⁺(aq). This removes the added acid from solution without significantly changing the pH.

在碱性缓冲溶液中，外加的H⁺直接与弱碱NH₃反应：NH₃(aq) + H⁺(aq) → NH₄⁺(aq)。这一过程将外加酸从溶液中移除，而pH值不会发生显著变化。

6. How a Buffer Resists Added Base | 缓冲溶液如何抵抗外加碱

When a small amount of a strong base (e.g., NaOH) is added to an acidic buffer, the added OH⁻ ions react with the weak acid HA:

当向酸性缓冲溶液中加入少量强碱（如NaOH）时，外加的OH⁻离子与弱酸HA发生反应：

OH⁻(aq) + HA(aq) → A⁻(aq) + H₂O(l)

This reaction consumes the added OH⁻ ions and converts the weak acid HA into its conjugate base A⁻. The reservoir of HA molecules in the buffer ensures that sufficient weak acid remains to neutralise the added base.

该反应消耗了外加的OH⁻离子，将弱酸HA转化为其共轭碱A⁻。缓冲溶液中HA分子的储备确保有足够的弱酸来中和外加的碱。

In an alkaline buffer, the added OH⁻ reacts with the conjugate acid NH₄⁺:

在碱性缓冲溶液中，外加的OH⁻与共轭酸NH₄⁺发生反应：

NH₄⁺(aq) + OH⁻(aq) → NH₃(aq) + H₂O(l)

The NH₄⁺ ions are supplied by the ammonium chloride salt, and the NH₃ produced simply adds to the reservoir of weak base already present. The pH therefore remains effectively constant.

NH₄⁺离子由氯化铵盐提供，生成的NH₃只是加入溶液中已有的弱碱储备。因此pH值保持基本恒定。

7. The Role of the Salt Component | 盐组分的作用

The salt in a buffer serves two essential functions. First, it provides a large reservoir of the conjugate acid or conjugate base species through complete dissociation. Second, it establishes the common ion effect, which suppresses the ionisation of the weak acid or weak base.

缓冲溶液中的盐具有两个基本功能。第一，它通过完全电离提供大量共轭酸或共轭碱物种的储备。第二，它建立同离子效应，抑制弱酸或弱碱的电离。

Without the salt, a solution of a weak acid alone would behave very differently. If H⁺ is added to a pure weak acid solution, the equilibrium HA ⇌ H⁺ + A⁻ shifts left, but the concentration of A⁻ is very low, so only a limited amount of H⁺ can be removed. Furthermore, the weak acid alone provides an extremely low concentration of A⁻, and the buffer capacity is minimal.

如果没有盐，单独的弱酸溶液行为将非常不同。如果向纯弱酸溶液中加入H⁺，平衡HA ⇌ H⁺ + A⁻左移，但A⁻浓度非常低，因此只能移除有限量的H⁺。此外，单独的弱酸提供的A⁻浓度极低，缓冲容量微乎其微。

The salt is therefore not merely a spectator; it is an integral component that defines the buffer’s capacity and its pH. The pH of an acidic buffer is calculated using the Henderson-Hasselbalch equation:

因此，盐不仅仅是旁观者；它是定义缓冲容量和pH的不可或缺的组成部分。酸性缓冲溶液的pH使用亨德森-哈塞尔巴尔赫方程计算：

pH = pKₐ + log₁₀([A⁻]/[HA])

where [A⁻] is dominated by the concentration of the salt, and [HA] is the concentration of the weak acid.

其中[A⁻]主要由盐的浓度决定，[HA]是弱酸的浓度。

8. Buffer Capacity and Effective Range | 缓冲容量与有效范围

Buffer capacity is defined as the amount of acid or base that a buffer can neutralise before its pH begins to change significantly. The capacity depends on the absolute concentrations of the buffer components: the higher the concentrations of HA and A⁻ (or NH₃ and NH₄⁺), the greater the buffer capacity.

缓冲容量定义为缓冲溶液在pH开始显著变化之前所能中和的酸或碱的量。容量取决于缓冲组分绝对浓度：HA和A⁻（或NH₃和NH₄⁺）的浓度越高，缓冲容量越大。

A buffer is most effective when the concentrations of the weak acid and its conjugate base are roughly equal. Under these conditions, the buffer has maximum capacity to respond to both added acid and added base, because neither component is present in limiting amounts.

当弱酸与其共轭碱的浓度大致相等时，缓冲溶液最为有效。在这种条件下，缓冲溶液对加入酸和碱都具有最大的响应能力，因为两种组分均不构成限制因素。

The effective pH range of a buffer is generally considered to be pKₐ ± 1. Outside this range, the ratio [A⁻]/[HA] becomes either too large or too small, and the buffer’s ability to resist pH changes diminishes rapidly.

缓冲溶液的有效pH范围通常认为是pKₐ ± 1。超出此范围时，[A⁻]/[HA]的比值变得过大或过小，缓冲溶液抵抗pH变化的能力迅速减弱。

When the ratio [A⁻]/[HA] = 1, the pH equals pKₐ. This is the centre of the buffer’s effective range and the point of maximum buffer capacity.

当[A⁻]/[HA] = 1时，pH等于pKₐ。这是缓冲有效范围的中心，也是缓冲容量最大的点。

9. Calculating the pH of a Buffer | 计算缓冲溶液的pH

The Henderson-Hasselbalch equation provides a direct method for calculating the pH of a buffer solution:

亨德森-哈塞尔巴尔赫方程提供了计算缓冲溶液pH的直接方法：

pH = pKₐ + log₁₀([A⁻]/[HA])

Worked example: Calculate the pH of a buffer containing 0.20 mol dm⁻³ ethanoic acid (Kₐ = 1.74 × 10⁻⁵ mol dm⁻³) and 0.50 mol dm⁻³ sodium ethanoate.

例题：计算含有0.20 mol dm⁻³乙酸（Kₐ = 1.74 × 10⁻⁵ mol dm⁻³）和0.50 mol dm⁻³乙酸钠的缓冲溶液的pH。

Step 1: Calculate pKₐ = −log₁₀(1.74 × 10⁻⁵) = 4.76
Step 2: Substitute into the equation: pH = 4.76 + log₁₀(0.50/0.20)
Step 3: pH = 4.76 + log₁₀(2.5) = 4.76 + 0.40 = 5.16

步骤1：计算pKₐ = −log₁₀(1.74 × 10⁻⁵) = 4.76
步骤2：代入方程：pH = 4.76 + log₁₀(0.50/0.20)
步骤3：pH = 4.76 + log₁₀(2.5) = 4.76 + 0.40 = 5.16

For an alkaline buffer, the corresponding calculation uses K_b to find pOH, or one may use the relationship pKₐ + pK_b = 14 for the conjugate pair to convert directly to pH.

对于碱性缓冲溶液，相应的计算使用K_b求pOH，或者利用共轭酸碱对的pKₐ + pK_b = 14关系直接换算为pH。

10. Biological Importance of Buffers | 缓冲溶液的生物学意义

Buffers play a vital role in biological systems. Human blood, for example, is maintained at a pH of approximately 7.40 by the carbonic acid-hydrogen carbonate buffer system (H₂CO₃/HCO₃⁻). A deviation of even 0.1 pH unit can have serious physiological consequences.

缓冲溶液在生物体系中扮演着至关重要的角色。例如，人体血液通过碳酸-碳酸氢根缓冲体系（H₂CO₃/HCO₃⁻）维持在约7.40的pH。即使偏离0.1个pH单位也可能导致严重的生理后果。

When CO₂ is produced by cellular respiration, it dissolves in blood plasma and forms carbonic acid:

当细胞呼吸产生CO₂时，CO₂溶于血浆中形成碳酸：

CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)

If the H⁺ concentration rises (blood becomes too acidic), the equilibrium shifts to the left, and the excess H⁺ is removed by combination with HCO₃⁻. If the H⁺ concentration falls (blood becomes too alkaline), the equilibrium shifts to the right, releasing more H⁺. This dynamic equilibrium keeps blood pH within its narrow, life-sustaining range.

如果H⁺浓度升高（血液过酸），平衡左移，过量的H⁺与HCO₃⁻结合而被移除。如果H⁺浓度降低（血液过碱），平衡右移，释放更多H⁺。这一动态平衡使血液pH保持在狭窄而维持生命的范围内。

Enzymes are also extremely sensitive to pH; most enzymes function optimally within a narrow pH range. Buffer solutions in laboratory settings ensure that enzyme activity is studied under controlled, constant pH conditions.

酶对pH也极为敏感；大多数酶在狭窄的pH范围内具有最佳活性。实验室中的缓冲溶液确保酶的活性在受控的恒定pH条件下进行研究。

11. Preparation of Buffer Solutions | 缓冲溶液的配制

Buffers can be prepared in several ways, and the method chosen depends on the desired pH and the available materials.

缓冲溶液可以通过多种方法配制，所选择的方法取决于所需pH和可用的试剂。

Method 1: Mixing a weak acid with its salt. For example, mixing ethanoic acid with sodium ethanoate. The ratio of the two concentrations determines the pH according to the Henderson-Hasselbalch equation.

方法一：混合弱酸及其盐。例如，混合乙酸和乙酸钠。两种浓度的比例根据亨德森-哈塞尔巴尔赫方程决定pH。

Method 2: Partial neutralisation of a weak acid with a strong base. For example, adding NaOH to excess ethanoic acid. Some of the CH₃COOH is converted to CH₃COO⁻, leaving a mixture of HA and A⁻ in solution.

方法二：用强碱部分中和弱酸。例如，向过量的乙酸中加入NaOH。部分CH₃COOH被转化为CH₃COO⁻，溶液中留下HA和A⁻的混合物。

Method 3: Mixing a weak base with its salt. For example, mixing ammonia with ammonium chloride to give an alkaline buffer.

方法三：混合弱碱及其盐。例如，混合氨水和氯化铵以制得碱性缓冲溶液。

The table below summarises the key differences between the two types of buffers:

下表总结了两种类型缓冲溶液的主要区别：

Property / 性质	Acidic Buffer / 酸性缓冲液	Alkaline Buffer / 碱性缓冲液
Components / 组分	Weak acid + salt of weak acid	Weak base + salt of weak base
Example / 实例	CH₃COOH / CH₃COONa	NH₃ / NH₄Cl
pH range / pH范围	Below 7 (typically 3–6)	Above 7 (typically 8–11)
Key equilibrium / 关键平衡	HA ⇌ H⁺ + A⁻	B + H₂O ⇌ BH⁺ + OH⁻

12. Common Errors and Exam Tips | 常见错误与考试要点

Students often confuse a buffer with a simple weak acid solution. A weak acid alone does not constitute a buffer because it lacks a sufficient reservoir of the conjugate base. Remember: a buffer always requires two species — the weak acid/base and its conjugate partner.

学生常将缓冲溶液与简单的弱酸溶液混淆。单独的弱酸不构成缓冲溶液，因为它缺乏足量的共轭碱储备。记住：缓冲溶液始终需要两种物种——弱酸/弱碱及其共轭配偶体。

Another common error is neglecting the contribution of the salt to [A⁻]. In calculations, students sometimes use only the initial concentration of the weak acid and ignore the fact that the salt fully dissociates to provide A⁻. Always include both sources of the conjugate species in your calculations.

另一个常见错误是忽略盐对[A⁻]的贡献。在计算中，学生有时只使用弱酸的初始浓度，忽略盐完全电离提供A⁻的事实。计算时务必同时考虑共轭物种的两个来源。

For CIE A-Level examinations, be sure to:

针对CIE A-Level考试，务必做到：

State clearly that buffers resist pH change upon addition of small amounts of acid or base
清晰说明缓冲溶液在加入少量酸或碱时能抵抗pH变化
Write the correct ionic equations for the reaction of added H⁺ and OH⁻ with the buffer components
正确写出外加H⁺和OH⁻与缓冲组分反应的离子方程式
Use the Henderson-Hasselbalch equation correctly, including the ratio [A⁻]/[HA]
正确使用亨德森-哈塞尔巴尔赫方程，包括比值[A⁻]/[HA]
Explain, not merely describe, why the pH remains constant by referencing equilibrium shifts
通过平衡移动来解释（而非仅仅描述）为什么pH保持恒定

Finally, when answering questions about buffer capacity, emphasise that capacity depends on the absolute concentrations of both components, while the pH depends on the ratio of their concentrations. These two concepts are distinct and must not be confused.

最后，在回答关于缓冲容量的问题时，强调容量取决于两种组分的绝对浓度，而pH取决于它们的浓度比。这两个概念是不同的，不可混淆。

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Buffer Solutions: Composition and Action | 缓冲溶液的组成与作用