AS Chemistry: Buffer Solutions Key Points | AS 化学：缓冲溶液 考点精讲

📚 AS Chemistry: Buffer Solutions Key Points | AS 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 alkali are added. In AS Chemistry, understanding how buffers work, how to calculate their pH, and how to select suitable weak acid/conjugate base pairs is essential. This article covers all key concepts, equations, and common exam pitfalls.

缓冲溶液是一种当加入少量酸或碱时能最大限度减小 pH 变化的体系。在 AS 化学中，理解缓冲液的工作原理、如何计算其 pH、以及如何选择合适的弱酸/共轭碱对至关重要。本文涵盖所有关键概念、方程式和常见考试陷阱。

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

A buffer solution resists changes in pH upon the addition of small quantities of acid or alkali. It is typically composed of a weak acid and its conjugate base, or a weak base and its conjugate acid, both present in significant concentrations. The ability to maintain a nearly constant pH is crucial for many biological and chemical processes.

缓冲溶液能抵抗因加入少量酸或碱而引起的 pH 变化。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，两者均以显著浓度存在。维持几乎恒定 pH 的能力对于许多生物和化学过程至关重要。

2. Types of Buffer Systems | 缓冲体系的类型

Two main types are encountered at AS level: acidic buffers, made from a weak acid and its salt (e.g., ethanoic acid and sodium ethanoate), and basic buffers, made from a weak base and its salt (e.g., ammonia and ammonium chloride). Acidic buffers maintain pH below 7, while basic buffers work in the alkaline region.

在 AS 阶段主要遇到两种类型：酸性缓冲液，由弱酸及其盐制成（例如乙酸和乙酸钠）；碱性缓冲液，由弱碱及其盐制成（例如氨和氯化铵）。酸性缓冲液维持 pH 低于 7，而碱性缓冲液在碱性范围内工作。

3. How Acidic Buffers Work | 酸性缓冲液如何工作

Consider an ethanoic acid/sodium ethanoate buffer. The weak acid, CH₃COOH, partially dissociates: CH₃COOH ⇌ CH₃COO⁻ + H⁺. The salt provides a high concentration of the conjugate base, CH₃COO⁻. When a small amount of H⁺ is added, it reacts with CH₃COO⁻ to form more CH₃COOH, shifting the equilibrium left and removing most added H⁺. When OH⁻ is added, it reacts with H⁺ from the acid dissociation, forming water; more CH₃COOH then dissociates to replace the H⁺, so the pH hardly changes.

以乙酸/乙酸钠缓冲液为例。弱酸 CH₃COOH 部分解离：CH₃COOH ⇌ CH₃COO⁻ + H⁺。该盐提供了高浓度的共轭碱 CH₃COO⁻。当加入少量 H⁺ 时，它与 CH₃COO⁻ 反应生成更多 CH₃COOH，使平衡向左移动，从而除去大部分加入的 H⁺。当加入 OH⁻ 时，它与酸解离产生的 H⁺ 反应生成水；然后更多的 CH₃COOH 解离以补充 H⁺，因此 pH 几乎不变。

4. The Equilibrium behind Buffer Action | 缓冲作用背后的平衡

The key equilibrium is the dissociation of the weak acid: HA ⇌ H⁺ + A⁻. The acid dissociation constant, Kₐ = [H⁺][A⁻] / [HA]. Rearranging gives [H⁺] = Kₐ × ([HA] / [A⁻]). Taking negative logs leads to the Henderson–Hasselbalch equation: pH = pKₐ + log₁₀([A⁻] / [HA]). This shows that when [A⁻] = [HA], the pH = pKₐ.

关键平衡是弱酸的解离：HA ⇌ H⁺ + A⁻。酸解离常数 Kₐ = [H⁺][A⁻] / [HA]。重新排列可得 [H⁺] = Kₐ × ([HA] / [A⁻])。取负对数后得到 Henderson–Hasselbalch 方程：pH = pKₐ + log₁₀([A⁻] / [HA])。这表明当 [A⁻] = [HA] 时，pH = pKₐ。

5. The Henderson–Hasselbalch Equation | Henderson–Hasselbalch 方程

For an acidic buffer, the pH is given by:

pH = pKₐ + log₁₀([conjugate base] / [weak acid])

This equation is central to buffer calculations. It assumes the concentrations of the weak acid and its conjugate base at equilibrium are approximately equal to their initial concentrations, because dissociation is negligible in the presence of a common ion.

对于酸性缓冲液，pH 由下式给出：

pH = pKₐ + log₁₀([共轭碱] / [弱酸])

这个方程是缓冲液计算的核心。它假设弱酸和其共轭碱在平衡时的浓度大致等于它们的初始浓度，因为在存在同离子时解离可忽略不计。

6. Calculating pH of a Buffer (given concentrations) | 给定浓度计算缓冲液的 pH

Example: A buffer contains 0.50 mol dm⁻³ ethanoic acid (Kₐ = 1.8×10⁻⁵ mol dm⁻³) and 0.30 mol dm⁻³ sodium ethanoate. First, pKₐ = –log₁₀(1.8×10⁻⁵) ≈ 4.74. Then pH = 4.74 + log₁₀(0.30/0.50) = 4.74 + log₁₀(0.60) ≈ 4.74 – 0.22 = 4.52. Always check that the concentrations used refer to the acid and its conjugate base.

例题：某缓冲液含 0.50 mol dm⁻³ 乙酸（Kₐ = 1.8×10⁻⁵ mol dm⁻³）和 0.30 mol dm⁻³ 乙酸钠。首先，pKₐ = –log₁₀(1.8×10⁻⁵) ≈ 4.74。然后 pH = 4.74 + log₁₀(0.30/0.50) = 4.74 + log₁₀(0.60) ≈ 4.74 – 0.22 = 4.52。务必检查所用浓度指的是酸和其共轭碱的浓度。

7. Preparing a Buffer with a Specific pH | 配制具有特定 pH 的缓冲液

To prepare an acidic buffer of a desired pH, select a weak acid whose pKₐ is within ±1 of the target pH. Then adjust the ratio of conjugate base to weak acid according to the Henderson–Hasselbalch equation. If equal concentrations of acid and conjugate base are used, the pH equals pKₐ. For a given pH, the ratio [A⁻]/[HA] = 10^(pH – pKₐ).

要配制所需 pH 的酸性缓冲液，应选择 pKₐ 在目标 pH ±1 范围内的弱酸。然后根据 Henderson–Hasselbalch 方程调整共轭碱与弱酸的比例。如果使用等浓度的酸和共轭碱，pH 等于 pKₐ。对于给定 pH，比例 [A⁻]/[HA] = 10^(pH – pKₐ)。

8. Buffer Capacity | 缓冲容量

Buffer capacity refers to the amount of strong acid or base that can be added before the pH changes significantly. It depends on the absolute concentrations of the buffering species; high concentrations give high capacity. The optimum buffering occurs when [A⁻] = [HA], where the pH = pKₐ and the capacity is maximal for a given total concentration.

缓冲容量指在 pH 发生显著变化之前所能加入的强酸或强碱的量。它取决于缓冲物种的绝对浓度；高浓度提供高容量。当 [A⁻] = [HA] 时缓冲效果最佳，此时 pH = pKₐ，且在给定总浓度下容量最大。

9. Effect of Dilution on Buffer pH | 稀释对缓冲液 pH 的影响

Dilution of an acidic buffer with water does not change its pH significantly because the ratio [A⁻]/[HA] remains almost constant. However, buffer capacity decreases because the total concentration of buffering species drops. In exams, students often confuse pH stability with capacity.

用水稀释酸性缓冲液并不会显著改变其 pH，因为 [A⁻]/[HA] 的比值几乎保持不变。然而，由于缓冲物种的总浓度下降，缓冲容量会减小。考试中，学生经常混淆 pH 稳定性和容量。

10. Basic Buffers and Their Calculations | 碱性缓冲液及其计算

A basic buffer, e.g., NH₃/NH₄Cl, uses the equilibrium: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Calculations can be performed via Kₐ of the conjugate acid (NH₄⁺) or K_b of the weak base. Using the Henderson–Hasselbalch form for basic buffers: pOH = pK_b + log₁₀([conjugate acid]/[weak base]), then pH = 14 – pOH. Alternatively, pH = pKₐ(NH₄⁺) + log₁₀([NH₃]/[NH₄⁺]).

碱性缓冲液，例如 NH₃/NH₄Cl，利用平衡：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。可通过共轭酸（NH₄⁺）的 Kₐ 或弱碱的 K_b 进行计算。对碱性缓冲液使用 Henderson–Hasselbalch 形式：pOH = pK_b + log₁₀([共轭酸]/[弱碱])，然后 pH = 14 – pOH。或者，pH = pKₐ(NH₄⁺) + log₁₀([NH₃]/[NH₄⁺])。

11. Biological and Practical Applications | 生物及实际应用

Blood pH is maintained at about 7.4 by the carbonic acid/hydrogencarbonate buffer: H₂CO₃ ⇌ H⁺ + HCO₃⁻. In the lab, buffers are used to calibrate pH meters, control reaction conditions in enzyme studies, and in shampoos and baby lotions to prevent skin irritation. Industrial processes like electroplating and dyeing also rely on buffer systems.

血液的 pH 通过碳酸/碳酸氢盐缓冲液维持在 7.4 左右：H₂CO₃ ⇌ H⁺ + HCO₃⁻。在实验室中，缓冲液用于校准 pH 计、控制酶研究中的反应条件，以及用于洗发水和婴儿乳液中以防止皮肤刺激。电镀和染色等工业过程也依赖缓冲体系。

12. Common Exam Mistakes and Tips | 常见考试错误与提示

• Forgetting that buffer pH does not equal pKₐ unless [A⁻] = [HA].

  忘记缓冲液 pH 并不等于 pKₐ，除非 [A⁻] = [HA]。

• Using moles instead of concentrations in the Henderson–Hasselbalch ratio — the ratio is dimensionless, so moles can be used if volumes are the same, but check carefully.

  在 Henderson–Hasselbalch 比值中使用摩尔而非浓度 — 该比值无量纲，因此体积相同时可使用摩尔，但须仔细检查。

• Ignoring assumptions: the equation assumes negligible dissociation of the weak acid and negligible hydrolysis of the salt. It fails at very low or very high concentrations.

  忽略假设：该方程假设弱酸解离和盐水解均可忽略。在极低或极高浓度下不适用。

• Confusing buffer action with neutralisation: buffers do not simply neutralise added acid/base; they shift equilibrium.

  将缓冲作用与中和混淆：缓冲液并非简单中和加入的酸/碱；而是移动平衡。

Published by TutorHao | Chemistry Revision Series | aleveler.com

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