Buffer Solutions | 缓冲溶液

📚 Buffer Solutions | 缓冲溶液

Buffer solutions are an essential topic in Cambridge International AS & A Level Chemistry. They explain how many chemical and biological systems keep a nearly constant pH even when small quantities of acid or base are introduced. Understanding buffers requires careful use of acid-base equilibria, Le Chatelier’s principle, and equilibrium constants.

缓冲溶液是剑桥国际 AS & A Level 化学中的一个重要主题。它们解释了许多化学和生物系统如何在加入少量酸或碱时仍能保持几乎恒定的 pH。理解缓冲溶液需要灵活运用酸碱平衡、勒夏特列原理和平衡常数。

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

A buffer solution is a mixture that resists changes in pH when small amounts of an acid or a base are added, or when the solution is diluted. It does not make the solution completely immune to pH change, but it greatly reduces the shift in pH compared with unbuffered water.

缓冲溶液是一种当加入少量酸或碱、或溶液被稀释时能抵抗 pH 变化的混合物。它不能使溶液完全不受 pH 变化影响，但与未加缓冲的水相比，它能大大减小 pH 的变化幅度。

A buffer must contain either a weak acid and its conjugate base, or a weak base and its conjugate acid. Strong acids and strong bases cannot produce buffers because they dissociate completely and therefore cannot provide the two reservoirs needed to remove both added H⁺ and added OH⁻.

缓冲溶液必须含有弱酸及其共轭碱，或弱碱及其共轭酸。强酸和强碱不能形成缓冲溶液，因为它们完全解离，无法同时提供用于消耗加入的 H⁺ 和 OH⁻ 的两种储备组分。

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2. Types of Buffer Systems | 缓冲体系的类型

An acidic buffer is formed from a weak acid and its salt with a strong base. A typical example is ethanoic acid, CH₃COOH, mixed with sodium ethanoate, CH₃COONa. The salt provides the conjugate base, CH₃COO⁻.

酸性缓冲溶液由弱酸和弱酸与强碱形成的盐组成。典型例子是乙酸 CH₃COOH 与乙酸钠 CH₃COONa 的混合物。盐提供共轭碱 CH₃COO⁻。

A basic buffer is formed from a weak base and its salt with a strong acid. A common example is ammonia, NH₃, mixed with ammonium chloride, NH₄Cl. The salt provides the conjugate acid, NH₄⁺.

碱性缓冲溶液由弱碱和弱碱与强酸形成的盐组成。常见例子是氨 NH₃ 与氯化铵 NH₄Cl 的混合物。盐提供共轭酸 NH₄⁺。

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3. The Acidic Buffer Equilibrium | 酸性缓冲平衡

In an ethanoic acid buffer, the weak acid establishes the equilibrium: CH₃COOH ⇌ H⁺ + CH₃COO⁻. Sodium ethanoate dissociates fully: CH₃COONa → Na⁺ + CH₃COO⁻, producing a high concentration of ethanoate ions.

在乙酸盐缓冲溶液中，弱酸建立平衡：CH₃COOH ⇌ H⁺ + CH₃COO⁻。乙酸钠完全解离：CH₃COONa → Na⁺ + CH₃COO⁻，产生高浓度的乙酸根离子。

The high concentration of CH₃COO⁻ from the salt shifts the weak-acid equilibrium far to the left. This is an example of the common ion effect: the ionisation of CH₃COOH is suppressed, so the solution contains a large reserve of unreacted CH₃COOH molecules and a large reserve of CH₃COO⁻ ions.

来自盐的高浓度 CH₃COO⁻ 使弱酸平衡强烈向左移动。这是同离子效应的一个例子：CH₃COOH 的电离被抑制，因此溶液中含有大量未反应的 CH₃COOH 分子储备和大量 CH₃COO⁻ 离子储备。

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4. How an Acidic Buffer Resists pH Change | 酸性缓冲如何抵抗 pH 变化

When a small amount of strong acid, H⁺, is added, the added H⁺ reacts with the ethanoate ion reservoir: H⁺ + CH₃COO⁻ → CH₃COOH. The equilibrium shifts left, consuming most of the added H⁺, so the pH falls only very slightly.

当加入少量强酸 H⁺ 时，加入的 H⁺ 与乙酸根离子储备反应：H⁺ + CH₃COO⁻ → CH₃COOH。平衡向左移动，大部分加入的 H⁺ 被消耗，因此 pH 只非常轻微地下降。

When a small amount of strong base, OH⁻, is added, the OH⁻ reacts with H⁺ from the weak-acid equilibrium: OH⁻ + H⁺ → H₂O. This removes H⁺, so the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ shifts right. The weak acid molecules dissociate to replace the removed H⁺, and the pH rises only very slightly.

当加入少量强碱 OH⁻ 时，OH⁻ 与弱酸平衡中的 H⁺ 反应：OH⁻ + H⁺ → H₂O。这会移除 H⁺，因此平衡 CH₃COOH ⇌ H⁺ + CH₃COO⁻ 向右移动。弱酸分子解离以补充被移除的 H⁺，pH 只非常轻微地上升。

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5. The Basic Buffer Equilibrium and Action | 碱性缓冲平衡及其作用

An ammonia buffer contains the weak base equilibrium: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Ammonium chloride dissociates fully: NH₄Cl → NH₄⁺ + Cl⁻, giving a high concentration of NH₄⁺ ions.

氨缓冲溶液含有弱碱平衡：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。氯化铵完全解离：NH₄Cl → NH₄⁺ + Cl⁻，提供高浓度的 NH₄⁺ 离子。

When H⁺ is added, it reacts with OH⁻ to form water: H⁺ + OH⁻ → H₂O. The removal of OH⁻ shifts the equilibrium to the right, so more NH₃ reacts with water and the added H⁺ is neutralised. When OH⁻ is added, the increased OH⁻ concentration shifts the equilibrium to the left, so NH₄⁺ reacts with OH⁻ to form NH₃ and H₂O.

当加入 H⁺ 时，它与 OH⁻ 反应生成水：H⁺ + OH⁻ → H₂O。OH⁻ 的移除使平衡向右移动，因此更多的 NH₃ 与水反应，加入的 H⁺ 被中和。当加入 OH⁻ 时，OH⁻ 浓度增大使平衡向左移动，因此 NH₄⁺ 与 OH⁻ 反应生成 NH₃ 和 H₂O。

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6. The Henderson-Hasselbalch Equation | 亨德森-哈塞尔巴赫方程

For an acidic buffer containing a weak acid HA and its conjugate base A⁻, the equilibrium constant is written as:

对于含有弱酸 HA 及其共轭碱 A⁻ 的酸性缓冲溶液，平衡常数可写为：

Kₐ = [H⁺] × [A⁻] ÷ [HA]

Rearranging this expression gives [H⁺] = Kₐ × [HA] ÷ [A⁻]. Taking the negative logarithm to base 10 gives the Henderson-Hasselbalch equation:

重新整理该表达式得到 [H⁺] = Kₐ × [HA] ÷ [A⁻]。取以 10 为底的负对数，即可得到亨德森-哈塞尔巴赫方程：

pH = pKₐ + log₁₀([A⁻] ÷ [HA])

This equation assumes that the concentration of HA at equilibrium is approximately equal to the initial concentration of the weak acid, and that the concentration of A⁻ is approximately equal to the concentration of the added salt. These assumptions are valid because the common ion effect suppresses the ionisation of

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