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

📚 Buffer Solutions: Key Points for IB Chemistry | IB 化学：缓冲溶液 考点精讲

Buffer solutions are a cornerstone of acid-base chemistry and a vital topic in the IB Chemistry syllabus, especially at Higher Level. They are systems that resist changes in pH when small amounts of acid or alkali are added, or when the solution is diluted. Understanding buffers is not just about memorising the Henderson-Hasselbalch equation; it requires a deep grasp of equilibrium principles, acid-base theories, and the common ion effect. This article dissects the key concepts, calculations, and applications you need to master for your IB exams.

缓冲溶液是酸碱化学的基石，也是IB化学大纲中的核心主题，尤其是在高水平课程中。缓冲溶液能够抵抗因加入少量酸或碱或稀释而引起的pH变化。理解缓冲溶液不仅仅是记忆亨德森-哈塞尔巴尔赫方程，还需要深入掌握化学平衡原理、酸碱理论和同离子效应。本文将剖析IB考试中需要掌握的关键概念、计算方法和实际应用。

1. Definition and Composition of a Buffer | 缓冲溶液的定义与组成

A buffer solution is defined as a system that minimises pH changes when small amounts of an acid or a base are added, or upon dilution. An acidic buffer consists of a weak acid and its conjugate base (usually added as a soluble salt). For example, ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COONa). A basic buffer comprises a weak base and its conjugate acid, such as ammonia (NH₃) and ammonium chloride (NH₄Cl). The key is that both the weak acid/base and its conjugate partner must be present in significant concentrations to effectively ‘mop up’ added H⁺ or OH⁻ ions.

缓冲溶液是指能够将因加入少量酸或碱或稀释引起的pH变化降至最低的体系。酸性缓冲液由弱酸及其共轭碱（常以可溶盐的形式加入）组成。例如，乙酸（CH₃COOH）和乙酸钠（CH₃COONa）。碱性缓冲液由弱碱及其共轭酸组成，如氨水（NH₃）和氯化铵（NH₄Cl）。关键在于弱酸/弱碱与其共轭伙伴必须以相当浓度共存，才能有效清除加入的H⁺或OH⁻离子。

It is crucial to note that buffers are not made from strong acids and their salts (e.g., HCl and NaCl) because strong acids completely dissociate, leaving no reservoir of undissociated acid to counteract added base. A buffer relies on the equilibrium between the weak species and its conjugate.

需要特别注意的是，缓冲溶液不能用强酸及其盐（例如HCl和NaCl）来制备，因为强酸完全电离，没有未电离的酸分子储备来对抗加入的碱。缓冲作用依赖弱电解质与其共轭物之间的平衡。

2. How Buffers Work: The Common Ion Effect | 缓冲溶液的工作原理：同离子效应

The buffering action is primarily explained by the common ion effect and Le Châtelier’s principle. In an acidic buffer like CH₃COOH/CH₃COONa, the weak acid exists in equilibrium: CH₃COOH ⇌ CH₃COO⁻ + H⁺. The sodium ethanoate provides a high concentration of the common ion CH₃COO⁻. This shifts the equilibrium to the left, suppressing the ionisation of the acid and keeping [H⁺] low. When a small amount of acid (H⁺) is added, these H⁺ ions combine with the abundant CH₃COO⁻ to form more CH₃COOH, so the pH remains nearly unchanged. When a base (OH⁻) is added, OH⁻ reacts with H⁺ to form water, but the equilibrium shifts right to replenish H⁺ from the undissociated CH₃COOH, again resisting pH change.

缓冲作用主要用同离子效应和勒夏特列原理来解释。在如CH₃COOH/CH₃COONa的酸性缓冲液中，弱酸存在平衡：CH₃COOH ⇌ CH₃COO⁻ + H⁺。乙酸钠提供了高浓度的同离子CH₃COO⁻，使平衡向左移动，抑制了酸的电离，使[H⁺]保持较低。当加入少量酸（H⁺）时，这些H⁺与大量的CH₃COO⁻结合生成更多的CH₃COOH，因此pH几乎不变。当加入碱（OH⁻）时，OH⁻与H⁺结合生成水，此时平衡向右移动，由未电离的CH₃COOH补充H⁺，同样抵抗pH变化。

3. The Henderson-Hasselbalch Equation | 亨德森-哈塞尔巴尔赫方程

The quantitative description of buffer pH is given by the Henderson-Hasselbalch equation, derived from the acid dissociation constant expression. For an acidic buffer: pH = pKₐ + log₁₀([A⁻]/[HA]), where [A⁻] is the concentration of the conjugate base and [HA] is the concentration of the weak acid. For a basic buffer, the equation is often expressed as pOH = pK_b + log₁₀([HB⁺]/[B]) and then converted to pH = 14 – pOH. The equation is valid when the approximations of the initial concentrations are close to equilibrium concentrations, which holds if the buffer is not extremely dilute and the ratio [A⁻]/[HA] is between 0.1 and 10.

亨德森-哈塞尔巴尔赫方程定量描述了缓冲溶液的pH，它由酸解离常数表达式推导而来。对于酸性缓冲溶液：pH = pKₐ + log₁₀([A⁻]/[HA])，其中[A⁻]为共轭碱的浓度，[HA]为弱酸的浓度。对于碱性缓冲液，通常表示为pOH = pK_b + log₁₀([HB⁺]/[B])，然后换算为pH = 14 – pOH。当初始浓度近似等于平衡浓度时，该方程成立；只要缓冲溶液不是极度稀释，且比值[A⁻]/[HA]在0.1到10之间，这一近似是有效的。

4. Calculating the pH of Buffer Solutions | 计算缓冲溶液的pH

IB exams frequently require students to calculate the pH of a buffer either directly from the Henderson-Hasselbalch equation or through an ICE table approach. Example: Calculate the pH of a buffer made by mixing 50.0 cm³ of 0.100 mol dm⁻³ CH₃COOH (Kₐ = 1.8 × 10⁻⁵) with 50.0 cm³ of 0.100 mol dm⁻³ CH₃COONa. After mixing, both concentrations are halved: [HA] = 0.0500 mol dm⁻³, [A⁻] = 0.0500 mol dm⁻³. pKₐ = –log₁₀(1.8×10⁻⁵) = 4.74. Then pH = 4.74 + log₁₀(0.0500/0.0500) = 4.74 + log₁₀(1) = 4.74. When the ratio is 1, pH = pKₐ.

IB考试时常要求学生利用亨德森-哈塞尔巴尔赫方程或ICE表格法计算缓冲溶液的pH。例题：将50.0 cm³ 0.100 mol dm⁻³的CH₃COOH（Kₐ = 1.8 × 10⁻⁵）与50.0 cm³ 0.100 mol dm⁻³的CH₃COONa混合，求缓冲液的pH。混合后，两种物质的浓度均减半：[HA] = 0.0500 mol dm⁻³，[A⁻] = 0.0500 mol dm⁻³。pKₐ = –log₁₀(1.8×10⁻⁵) = 4.74。则pH = 4.74 + log₁₀(0.0500/0.0500) = 4.74 + log₁₀(1) = 4.74。当比值为1时，pH = pKₐ。

If 1.0 cm³ of 1.0 mol dm⁻³ HCl is added to 100 cm³ of the above buffer, the added H⁺ reacts with CH₃COO⁻, changing the mole ratio. Recalculate the new moles of HA and A⁻, determine the new ratio, and apply the equation. The pH should change by only a small amount, demonstrating buffer action. For the same addition to pure water, the pH would plummet from 7 to around 2, showing the protective power of the buffer.

若向100 cm³上述缓冲液中加入1.0 cm³ 1.0 mol dm⁻³ HCl，加入的H⁺与CH₃COO⁻反应，改变其物质的量比。重新计算HA和A⁻的物质的量，得出新的比值并代入方程。pH应只发生微小变化，从而验证缓冲作用。若将等量酸加入纯水中，pH将从7骤降至大约2，突显缓冲液的保护能力。

5. Buffer Capacity | 缓冲容量

Buffer capacity (β) is a measure of the efficiency of a buffer to resist pH change upon addition of acid or base. It is defined as the amount of strong acid or base (in moles) required to change the pH of 1 dm³ of the buffer by 1 unit. A buffer is most effective when the concentrations of the weak acid and its conjugate base are high and their ratio is close to 1. Mathematically, buffer capacity depends on the total concentration (c_total = [HA] + [A⁻]) and the ratio. The maximum buffer capacity for a given total concentration occurs at pH = pKₐ, i.e., when [HA] = [A⁻].

缓冲容量（β）衡量缓冲溶液抵抗pH变化的能力。定义为使1 dm³缓冲溶液的pH改变1个单位所需的强酸或强碱的物质的量。当弱酸及其共轭碱的浓度较高且比值接近1时，缓冲效率最高。从数学上讲，缓冲容量取决于总浓度（c_total = [HA] + [A⁻]）和比值。对于给定总浓度，在pH = pKₐ时，即[HA] = [A⁻]时，缓冲容量最大。

6. Buffer Range | 缓冲范围

The effective pH range of a buffer is generally considered to be pKₐ ± 1. Within this range, the ratio [A⁻]/[HA] lies between 0.1 and 10, and the buffer can still effectively neutralise added acid or base. Beyond this range, one of the buffer components becomes too depleted to maintain equilibrium, and the pH will change significantly. When selecting a buffer for a desired pH, choose an acid with a pKₐ value as close as possible to the target pH.

缓冲溶液的有效pH范围通常认为是pKₐ ± 1。在此范围内，[A⁻]/[HA]比值介于0.1到10，缓冲液仍能有效中和加入的酸或碱。超出此范围，某一缓冲组分浓度过低，无法维持平衡，pH将发生显著变化。为特定pH选择缓冲液时，应选择pKₐ尽可能接近目标pH的酸。

7. Preparing a Buffer Solution | 缓冲溶液的制备

更多咨询请联系16621398022（同微信）