📚 Acid-Base Theories for WJEC A-Level Chemistry | WJEC A-Level 化学：酸碱理论考点精讲
Understanding acid-base chemistry is fundamental to the WJEC A-Level syllabus. This topic brings together several important models, from the historical Arrhenius definition to the more generalised Brønsted-Lowry and Lewis theories. You will also need to apply these concepts to pH calculations, buffers, and titration curves, making it essential to grasp both the underlying principles and the quantitative side. This article walks you through every key point, with paired English-Chinese explanations designed to reinforce your learning and exam readiness.
理解酸碱化学是 WJEC A-Level 课程的基础。这个主题融合了多个重要模型，从历史上的阿伦尼乌斯定义到更普适的布朗斯特-劳里和路易斯理论。你还需要将这些概念应用于 pH 计算、缓冲溶液和滴定曲线，因此掌握基本原理和定量分析方法都至关重要。本文将逐点精讲，通过英中对照的解释帮助巩固理解，为考试做好充分准备。
1. Introduction to Acids and Bases | 酸碱概述
Acids and bases are classified according to their behaviour in chemical reactions. Historically, their properties were described by observable characteristics, but modern theories focus on the species involved in proton or electron transfers.
酸和碱是根据它们在化学反应中的行为进行分类的。历史上，它们的性质是通过可观察的特征来描述的，但现代理论则关注参与质子或电子转移的物种。
Common properties of acids include a sour taste, the ability to turn blue litmus red and the capacity to neutralise bases. Bases feel slippery, turn red litmus blue and also neutralise acids. However, these empirical properties are not sufficient for quantitative chemistry, so we rely on theoretical frameworks.
酸的常见性质包括酸味、使蓝色石蕊变红以及中和碱的能力。碱摸起来有滑腻感，使红色石蕊变蓝，同样也能中和酸。然而，这些经验性质对于定量化学来说是不够的，因此我们需要依靠理论框架。
2. Arrhenius Theory | 阿伦尼乌斯酸碱理论
The Arrhenius theory, proposed by Svante Arrhenius in the late 19th century, defines an acid as a substance that dissociates in water to produce hydrogen ions, H⁺. A base is defined as a substance that dissociates in water to produce hydroxide ions, OH⁻.
阿伦尼乌斯理论由斯万特·阿伦尼乌斯在 19 世纪末提出，将酸定义为在水中解离产生氢离子 H⁺ 的物质，将碱定义为在水中解离产生氢氧根离子 OH⁻ 的物质。
For example, hydrochloric acid ionises completely in water: HCl → H⁺ + Cl⁻. Sodium hydroxide dissociates to give OH⁻: NaOH → Na⁺ + OH⁻. Neutralisation is simply the reaction of H⁺ with OH⁻ to form water: H⁺ + OH⁻ → H₂O.
例如，盐酸在水中完全电离：HCl → H⁺ + Cl⁻。氢氧化钠解离产生 OH⁻：NaOH → Na⁺ + OH⁻。中和反应就是 H⁺ 与 OH⁻ 反应生成水：H⁺ + OH⁻ → H₂O。
This theory works well for aqueous solutions but has limitations. It cannot explain the basic nature of ammonia, NH₃, which does not contain OH⁻, nor does it account for acid-base reactions in non-aqueous solvents. For WJEC, you must know these limitations and understand why the Brønsted-Lowry theory is preferred.
这个理论在水溶液中解释得很好，但有局限性。它无法解释氨（NH₃）的碱性，因为氨不含 OH⁻，也不能解释非水溶剂中的酸碱反应。在 WJEC 考试中，你需要知道这些局限，并理解为什么布朗斯特-劳里理论更受青睐。
3. Brønsted-Lowry Theory | 布朗斯特-劳里酸碱理论
The Brønsted-Lowry theory defines an acid as a proton (H⁺) donor and a base as a proton acceptor. This model does not require water or the presence of OH⁻, so it applies to a much wider range of reactions.
布朗斯特-劳里理论将酸定义为质子（H⁺）给予体，将碱定义为质子接受体。这个模型不要求水的存在或 OH⁻ 的存在，因此适用于更广泛的反应。
When hydrogen chloride gas dissolves in water, HCl donates a proton to H₂O, forming H₃O⁺ and Cl⁻. Here, HCl is the acid and H₂O acts as a base. In the reaction between ammonia and water, NH₃ accepts a proton from H₂O to form NH₄⁺ and OH⁻, so NH₃ is a base and H₂O acts as an acid.
当氯化氢气体溶于水时，HCl 向 H₂O 提供一个质子，形成 H₃O⁺ 和 Cl⁻。在此反应中，HCl 是酸，H₂O 充当碱。在氨和水的反应中，NH₃ 从 H₂O 接受一个质子，形成 NH₄⁺ 和 OH⁻，因此 NH₃ 是碱，H₂O 充当酸。
This ability of water to act as either an acid or a base makes it an amphoteric substance. You will often see the term ‘amphiprotic’ used specifically for species that can donate or accept a proton, as in the WJEC specification.
水既能作为酸又能作为碱的能力使其成为一种两性物质。在 WJEC 考纲中，你常会看到 ‘amphiprotic’ 这个术语，专指那些能给出或接受质子的物种。
4. Conjugate Acid-Base Pairs | 共轭酸碱对
A conjugate acid-base pair consists of two species that differ by one proton. The acid member donates a proton to become its conjugate base, while the base member accepts a proton to become its conjugate acid.
共轭酸碱对由相差一个质子的两个物种组成。酸成员失去一个质子变成其共轭碱，碱成员得到一个质子变成其共轭酸。
For the reaction CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺, the acid CH₃COOH and its conjugate base CH₃COO⁻ form one pair, while the base H₂O and its conjugate acid H₃O⁺ form the other pair. Recognising these pairs is essential for writing correct equilibrium expressions and for buffer calculations.
对于反应 CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺，酸 CH₃COOH 和其共轭碱 CH₃COO⁻ 组成一对，碱 H₂O 和其共轭酸 H₃O⁺ 组成另一对。识别这些共轭对对于正确书写平衡表达式和进行缓冲溶液计算至关重要。
In the Brønsted-Lowry framework, every acid-base reaction involves two conjugate pairs. The stronger the acid, the weaker its conjugate base, and vice versa. This concept helps to predict the direction of equilibrium in proton-transfer reactions.
在布朗斯特-劳里框架中，每个酸碱反应都包含两个共轭对。酸越强，其共轭碱就越弱，反之亦然。这一概念有助于预测质子转移反应中平衡的方向。
5. Lewis Theory | 路易斯酸碱理论
The Lewis theory takes a broader view by defining an acid as an electron-pair acceptor and a base as an electron-pair donor. This model covers reactions that do not involve protons at all and is particularly useful in organic and coordination chemistry.
路易斯理论采用更广阔的视角，将酸定义为电子对接受体，碱定义为电子对给予体。这个模型涵盖了完全不涉及质子的反应，在有机化学和配位化学中尤其有用。
For example, boron trifluoride, BF₃, is a Lewis acid because it can accept an electron pair from ammonia, NH₃, a Lewis base, forming a dative covalent bond. Another classic example is the reaction between a metal cation (Lewis acid) and ligands (Lewis bases) to form complex ions.
例如，三氟化硼 BF₃ 是路易斯酸，因为它能从氨 NH₃（路易斯碱）那里接受一对电子，形成配位共价键。另一个经典例子是金属阳离子（路易斯酸）与配体（路易斯碱）反应生成配离子。
While the WJEC specification focuses primarily on Brønsted-Lowry theory, you should be aware that all Brønsted-Lowry bases are also Lewis bases because they donate an electron pair to a proton. However, not all Lewis acids are Brønsted-Lowry acids, as they may not donate a proton.
尽管 WJEC 考纲主要关注布朗斯特-劳里理论，但你应知道所有布朗斯特-劳里碱同时也是路易斯碱，因为它们向质子提供电子对。然而，并非所有路易斯酸都是布朗斯特-劳里酸，因为它们可能不提供质子。
6. Strong vs Weak Acids and Bases | 强酸强碱与弱酸弱碱
A strong acid, such as HCl or HNO₃, dissociates completely in aqueous solution, meaning almost every molecule donates its proton to water. A weak acid, like ethanoic acid CH₃COOH, only partially dissociates, setting up an equilibrium where most molecules remain undissociated.
像 HCl 或 HNO₃ 这样的强酸在水溶液中完全解离，意味着几乎每个分子都将其质子给予水。像乙酸 CH₃COOH 这样的弱酸只部分解离，建立起一个平衡，其中大多数分子保持未解离状态。
Similarly, strong bases such as NaOH and KOH fully dissociate to release OH⁻ ions, while weak bases such as NH₃ only react partially with water, forming an equilibrium mixture containing NH₃, NH₄⁺ and OH⁻.
同样，像 NaOH 和 KOH 这样的强碱完全解离释放出 OH⁻ 离子，而像 NH₃ 这样的弱碱只部分与水反应，形成含有 NH₃、NH₄⁺ 和 OH⁻ 的平衡混合物。
The strength of an acid or base is measured by its dissociation constant, not by its concentration. For the WJEC exam, never confuse the terms ‘strong’ and ‘concentrated’ – a strong acid is one that ionises fully, while a concentrated solution simply contains a large amount of solute per unit volume.
酸或碱的强弱由其解离常数来衡量，而不是由其浓度决定。在 WJEC 考试中，切勿混淆 ‘strong’ 和 ‘concentrated’ 这两个术语——强酸指完全电离的酸，而浓溶液仅指单位体积中含有大量溶质。
7. The Ionic Product of Water, Kw | 水的离子积 Kw
Water undergoes self-ionisation to a very small extent: 2H₂O ⇌ H₃O⁺ + OH⁻. The equilibrium constant for this reaction is known as the ionic product of water, Kw = [H⁺][OH⁻], where [H⁺] is often used as an abbreviation for [H₃O⁺].
水在极小程度上发生自偶电离：2H₂O ⇌ H₃O⁺ + OH⁻。该反应的平衡常数称为水的离子积 Kw = [H⁺][OH⁻]，其中 [H⁺] 常作为 [H₃O⁺] 的缩写。
At 298 K, Kw has a value of 1.0 × 10⁻¹⁴ mol² dm⁻⁶. This means that in pure water at this temperature, [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³, giving a neutral pH of 7. Remember that Kw, like all equilibrium constants, is temperature dependent; thus the pH of neutrality changes with temperature.
在 298 K 时，Kw 的值为 1.0 × 10⁻¹⁴ mol² dm⁻⁶。这意味着在该温度下的纯水中，[H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³，中性 pH 为 7。请记住，Kw 与所有平衡常数一样，依赖于温度；因此中性的 pH 会随温度改变而变化。
Using Kw, you can calculate the [H⁺] of a strong base solution if you know its hydroxide ion concentration, and vice versa. For instance, for 0.10 mol dm⁻³ NaOH, [OH⁻] = 0.10 mol dm⁻³, so [H⁺] = Kw / [OH⁻] = 1.0 × 10⁻¹³ mol dm⁻³, giving a pH of 13.
利用 Kw，如果你知道强碱溶液的氢氧根浓度，就可以计算出 [H⁺]，反之亦然。例如，对于 0.10 mol dm⁻³ NaOH，[OH⁻] = 0.10 mol dm⁻³，因此 [H⁺] = Kw / [OH⁻] = 1.0 × 10⁻¹³ mol dm⁻³，pH 为 13。
8. pH and pOH Calculations | pH 与 pOH 的计算
The pH of a solution is defined as the negative logarithm to base 10 of the hydrogen ion concentration:
pH = -log₁₀[H⁺]
溶液的 pH 定义为氢离子浓度的以 10 为底的负对数：
pH = -log₁₀[H⁺]
For a strong monoprotic acid, such as HCl at 0.05 mol dm⁻³, [H⁺] = 0.05 mol dm⁻³ and therefore pH = -log₁₀(0.05) ≈ 1.30. For diprotic strong acids like H₂SO₄, the first proton is fully dissociated, and the second proton must be considered carefully, though at A-Level the approximation is often made that both protons contribute equally to [H⁺] for strong diprotic acids, giving [H⁺] = 2 × acid concentration.
对于强的一元酸，如 0.05 mol dm⁻³ 的 HCl，[H⁺] = 0.05 mol dm⁻³，因此 pH = -log₁₀(0.05) ≈ 1.30。对于二元强酸如 H₂SO₄，第一个质子完全解离，第二个质子须仔细考虑；不过在 A-Level 中，常作近似认为两个质子对 [H⁺] 的贡献相等，因此 [H⁺] = 2 × 酸的浓度。
The pOH scale is similarly defined as pOH = -log₁₀[OH⁻]. Since pH + pOH = 14 at 298 K, you can easily convert between the two. While not always required directly, understanding pOH is helpful when dealing with alkaline solutions and buffer preparations.
pOH 类似地定义为 pOH = -log₁₀[OH⁻]。由于在 298 K 时 pH + pOH = 14，两者之间容易转换。尽管并不总是直接要求，但理解 pOH 在处理碱性溶液和配制缓冲液时很有帮助。
9. Acid Dissociation Constant, Ka and pKa | 酸解离常数 Ka 与 pKa
For a weak acid HA, the equilibrium HA ⇌ H⁺ + A⁻ is described by the acid dissociation constant:
Ka = [H⁺][A⁻] / [HA]
对于弱酸 HA，平衡 HA ⇌ H⁺ + A⁻ 由酸解离常数描述：
Ka = [H⁺][A⁻] / [HA]
Ka has units of mol dm⁻³ and its magnitude indicates the strength of the acid. The smaller the Ka, the weaker the acid. To simplify comparison, pKa is used, defined as pKa = -log₁₀Ka. A small pKa corresponds to a stronger acid.
Ka 的单位是 mol dm⁻³，其大小指示酸的强弱。Ka 值越小，酸越弱。为方便比较，使用 pKa，定义为 pKa = -log₁₀Ka。pKa 值越小，酸越强。
Calculating the pH of a weak acid involves assuming that the amount dissociated is negligible compared to the initial concentration, c. Then [H⁺] ≈ √(Ka × c) and pH = -log₁₀√(Ka × c). This approximation is valid when c / Ka > 100, a condition often checked in WJEC exam questions.
计算弱酸的 pH 时，假设解离的部分与初始浓度 c 相比可忽略不计。则 [H⁺] ≈ √(Ka × c)，pH = -log₁₀√(Ka × c)。当 c / Ka > 100 时该近似有效，WJEC 考试题中常需检验这一条件。
For example, ethanoic acid has Ka = 1.74 × 10⁻⁵ mol dm⁻³ at 298 K. For a 0.20 mol dm⁻³ solution, [H⁺] = √(1.74 × 10⁻⁵ × 0.20) ≈ 1.86 × 10⁻³ mol dm⁻³, giving pH ≈ 2.73.
例如，乙酸在 298 K 时的 Ka = 1.74 × 10⁻⁵ mol dm⁻³。对于 0.20 mol dm⁻³ 的溶液，[H⁺] = √(1.74 × 10⁻⁵ × 0.20) ≈ 1.86 × 10⁻³ mol dm⁻³，pH ≈ 2.73。
10. Buffer Solutions | 缓冲溶液
A buffer solution resists changes in pH when small amounts of acid or base are added. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid, both present in significant concentrations.
缓冲溶液能够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，且两者均以可观的浓度存在。
The pH of an acidic buffer can be calculated using the equilibrium expression rearranged to give the Henderson-Hasselbalch equation:
pH = pKa + log₁₀([A⁻] / [HA])
酸性缓冲溶液的 pH 可利用平衡表达式重排得到的亨德森-哈塞尔巴尔赫方程计算：
pH = pKa + log₁₀([A⁻] / [HA])
Here [HA] is the equilibrium concentration of the weak acid and [A⁻] that of its conjugate base. Since both come from the acid and its salt, respectively, and dissociation is suppressed, the initial concentrations can be used as good approximations.
这里 [HA] 是弱酸的平衡浓度，[A⁻] 是其共轭碱的浓度。由于两者分别来自酸及其盐，且解离被抑制，初始浓度可用作很好的近似。
For instance, a buffer made from 0.50 mol dm⁻³ CH₃COOH and 0.50 mol dm⁻³ CH₃COONa will have pH = pKa + log₁₀(0.50/0.50) = pKa = -log₁₀(1.74 × 10⁻⁵) ≈ 4.76. Doubling the acid concentration while keeping the salt constant lowers the pH, demonstrating how the ratio controls the pH.
例如，由 0.50 mol dm⁻³ CH₃COOH 和 0.50 mol dm⁻³ CH₃COONa 配制的缓冲液，其 pH = pKa + log₁₀(0.50/0.50) = pKa = -log₁₀(1.74 × 10⁻⁵) ≈ 4.76。将酸的浓度加倍而保持盐浓度不变会降低 pH，这说明了该比值如何控制 pH。
Buffer action relies on the equilibrium shifting to remove added H⁺ or OH⁻. Added acid is mopped up by the conjugate base: A⁻ + H⁺ → HA. Added base is neutralised by the weak acid: HA + OH⁻ → A⁻ + H₂O. As long as the buffer capacity is not exceeded, the pH remains nearly constant.
缓冲作用依赖于平衡移动来消耗加入的 H⁺ 或 OH⁻。加入的酸被共轭碱吸纳：A⁻ + H⁺ → HA。加入的碱被弱酸中和：HA + OH⁻ → A⁻ + H₂O。只要不超过缓冲容量，pH 几乎保持不变。
11. Acid-Base Titrations and Indicators | 酸碱滴定与指示剂
In an acid-base titration, a solution of known concentration is used to determine the concentration of an unknown acid or base. The shape of the titration curve depends on the strengths of the acid and base involved.
在酸碱滴定中，用已知浓度的溶液来测定未知酸或碱的浓度。滴定曲线的形状取决于所用酸和碱的强弱。
For a strong acid-strong base titration, the curve has a vertical section around pH 7, and the equivalence point is at pH 7. With a strong acid and weak base, the equivalence point is below pH 7 because the salt formed hydrolyses to give an acidic solution. With a
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