📚 Acids and Bases for A-Level OCR Science | A-Level OCR 科学：酸与碱 考点精讲
Acids and bases are a cornerstone of the OCR A-Level Science (Chemistry) specification. A clear grasp of Bronsted-Lowry theory, equilibrium constants, buffer action and titration analysis is essential for top marks. This revision guide distils the key concepts, equations and common pitfalls, paired with Chinese explanations to support bilingual learners.
酸与碱是 OCR A-Level 科学（化学）考试的核心板块。透彻理解布朗斯特-劳里理论、平衡常数、缓冲作用以及滴定分析是取得高分的关键。本文聚焦考点，采用中英对照方式梳理重要概念、方程式和常见失分点，帮助双语学习者高效备考。
1. Bronsted-Lowry Theory | 布朗斯特-劳里酸碱理论
In the Bronsted-Lowry model, an acid is a proton (H⁺) donor and a base is a proton acceptor. This definition applies to all A-Level acid-base reactions, whether aqueous or gaseous.
根据布朗斯特-劳里理论，酸是质子（H⁺）给予体，碱是质子接受体。该定义适用于 A-Level 阶段所有酸碱反应，无论在溶液中还是气相中。
For example, when hydrogen chloride gas dissolves in water: HCl + H₂O → H₃O⁺ + Cl⁻. Here HCl donates a proton to H₂O, so HCl is the acid and H₂O acts as a base.
例如，氯化氢气体溶于水时：HCl + H₂O → H₃O⁺ + Cl⁻。HCl 把质子转移给 H₂O，因此 HCl 是酸，H₂O 是碱。
Ammonia accepting a proton from water also fits the model: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Water donates a proton, making it the acid, while NH₃ is the base.
氨从水中接受质子的反应同样适用：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。水给出质子，充当酸的角色，NH₃ 则是碱。
2. Conjugate Acid-Base Pairs | 共轭酸碱对
A conjugate acid-base pair consists of two species that differ by exactly one proton. The acid has one more proton than its conjugate base.
共轭酸碱对由相差一个质子的两个物种组成。酸比其共轭碱多一个质子。
In the equilibrium HA + H₂O ⇌ H₃O⁺ + A⁻, the pairs are HA / A⁻ and H₂O / H₃O⁺. HA donates a proton to become A⁻; A⁻ can accept a proton to revert to HA.
在平衡体系 HA + H₂O ⇌ H₃O⁺ + A⁻ 中，HA / A⁻ 和 H₂O / H₃O⁺ 是两对共轭酸碱对。HA 失去质子变为 A⁻，A⁻ 也可以结合质子变回 HA。
A strong acid has a very weak conjugate base, and vice versa. For instance, the conjugate base of HCl is Cl⁻, which has negligible tendency to accept a proton.
强酸的共轭碱非常弱，强碱的共轭酸也非常弱。例如，HCl 的共轭碱 Cl⁻ 几乎没有结合质子的倾向。
3. Strong and Weak Acids and Bases | 强酸强碱与弱酸弱碱
Strong acids and strong bases dissociate completely in aqueous solution. Their ionisation goes to completion, so the concentration of H⁺ or OH⁻ equals the original concentration (for monoprotic acids and monoacidic bases).
强酸和强碱在水溶液中完全电离。电离反应进行到底，因此溶液中 H⁺ 或 OH⁻ 的浓度就等于酸或碱的初始浓度（对于一元酸和一元碱）。
Common strong acids: HCl, HNO₃, H₂SO₄ (first proton). Common strong bases: NaOH, KOH. Weak acids, such as ethanoic acid (CH₃COOH), partially ionise, establishing an equilibrium. Weak bases like NH₃ also set up an equilibrium with water.
常见的强酸有 HCl、HNO₃ 和 H₂SO₄（第一个质子完全电离）。常见强碱有 NaOH、KOH。弱酸如乙酸（CH₃COOH）只能部分电离，存在平衡。弱碱如 NH₃ 也与水形成平衡。
4. The pH Scale and the Ionic Product of Water, Kw | pH 标度与水的离子积
pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration:
pH 定义为氢离子物质的量浓度的负对数（以 10 为底）：
pH = -log[H⁺]
Similarly, pOH = -log[OH⁻]. For any aqueous solution at 298 K, pH + pOH = 14.
类似地，pOH = -log[OH⁻]。在 298 K 的任何水溶液中都有 pH + pOH = 14。
The ionic product of water, Kw, is derived from the self-ionisation of water: 2H₂O ⇌ H₃O⁺ + OH⁻. Its expression and value at 298 K are:
水的离子积 Kw 源自水的自解离：2H₂O ⇌ H₃O⁺ + OH⁻。298 K 时的表达式和数值为：
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶
Kw increases with temperature because the forward reaction is endothermic. This means the pH of pure water falls as temperature rises, but the solution remains neutral because [H⁺] = [OH⁻].
Kw 随温度升高而增大，因为水的自解离是吸热反应。这意味着纯水的 pH 随温度上升而降低，但溶液仍为中性，因为 [H⁺] = [OH⁻]。
5. Calculating pH of Strong Acids and Bases | 强酸强碱的 pH 计算
For a strong monoprotic acid, [H⁺] equals the acid concentration. The pH is obtained directly from pH = -log[H⁺].
对于一元强酸，[H⁺] 等于酸的分析浓度。直接用公式 pH = -log[H⁺] 即可求得。
Example: A 0.050 mol dm⁻³ solution of HCl has [H⁺] = 0.050 mol dm⁻³, giving pH = -log(0.050) ≈ 1.30.
示例：0.050 mol dm⁻³ 的 HCl 溶液，[H⁺] = 0.050 mol dm⁻³，pH = -log(0.050) ≈ 1.30。
For a strong monoacidic base such as NaOH, [OH⁻] equals the base concentration. First calculate [H⁺] using Kw: [H⁺] = Kw / [OH⁻], then find pH.
对于一元强碱如 NaOH，[OH⁻] 等于碱分析浓度。先用 Kw 求出 [H⁺]：[H⁺] = Kw / [OH⁻]，再计算 pH。
Example: 0.020 mol dm⁻³ NaOH at 298 K gives [OH⁻] = 0.020 mol dm⁻³, [H⁺] = 1.0 × 10⁻¹⁴ / 0.020 = 5.0 × 10⁻¹³ mol dm⁻³, pH ≈ 12.30.
示例：298 K 时 0.020 mol dm⁻³ NaOH，[OH⁻] = 0.020 mol dm⁻³，[H⁺] = 1.0 × 10⁻¹⁴ / 0.020 = 5.0 × 10⁻¹³ mol dm⁻³，pH ≈ 12.30。
6. Acid Dissociation Constant, Ka and pKa | 酸离解常数 Ka 与 pKa
A weak acid HA partially dissociates: HA(aq) ⇌ H⁺(aq) + A⁻(aq). The acid dissociation constant is:
弱酸 HA 发生部分解离：HA(aq) ⇌ H⁺(aq) + A⁻(aq)。酸离解常数 Ka 的表达为：
Ka = [H⁺][A⁻] / [HA]
Units of Ka are mol dm⁻³. A larger Ka indicates a stronger weak acid. pKa = -log Ka; a smaller pKa means a stronger acid.
Ka 的单位是 mol dm⁻³。Ka 越大，弱酸越强。pKa = -log Ka；pKa 越小，酸性越强。
For ethanoic acid at 298 K, Ka = 1.7 × 10⁻⁵ mol dm⁻³, so pKa = 4.76. This value is often used in buffer calculations.
298 K 时乙酸的 Kₐ = 1.7 × 10⁻⁵ mol dm⁻³，因此 pKₐ = 4.76。这一数值常用于缓冲溶液计算。
7. pH of Weak Acids | 弱酸的 pH 计算
To find the pH of a weak acid solution, we use the Ka expression and an ICE (Initial, Change, Equilibrium) table. Usually the dissociation α is small, so [HA] at equilibrium ≈ initial concentration, c.
计算弱酸 pH 时需使用 Kₐ 表达式并建立 ICE（初始、变化、平衡）表格。通常电离度 α 很小，因此平衡时 [HA] ≈ 初始浓度 c。
Under the approximation that [H⁺] = [A⁻] and [HA] ≈ c, the expression simplifies to Ka ≈ [H⁺]² / c, giving:
在 [H⁺] = [A⁻] 且 [HA] ≈ c 的近似下，表达式简化为 Kₐ ≈ [H⁺]² / c，从而得到：
[H⁺] = √(Ka c)
This approximation is acceptable when c / Ka > 100, or when the percentage ionisation is below 5 %. The pH then follows from pH = -log[H⁺].
当 c / Kₐ > 100 或电离百分数低于 5% 时，此近似成立。然后由 pH = -log[H⁺] 得到 pH。
Example: Find the pH of 0.10 mol dm⁻³ CH₃COOH (Kₐ = 1.7 × 10⁻⁵). [H⁺] = √(1.7 × 10⁻⁵ × 0.10) = 1.3 × 10⁻³ mol dm⁻³, pH = 2.89.
示例：计算 0.10 mol dm⁻³ CH₃COOH 的 pH（Kₐ = 1.7 × 10⁻⁵）。[H⁺] = √(1.7 × 10⁻⁵ × 0.10) = 1.3 × 10⁻³ mol dm⁻³，pH = 2.89。
8. Weak Bases and Base Dissociation | 弱碱与碱离解
Weak bases such as NH₃ accept a proton from water: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. The base dissociation constant Kb is defined as:
弱碱如 NH₃ 从水中接受质子：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。碱离解常数 Kb 定义为：
Kb = [NH₄⁺][OH⁻] / [NH₃]
Using a similar approximation, [OH⁻] = √(Kb c), where c is the initial base concentration. [H⁺] is then found from Kw to calculate pH.
使用类似近似，[OH⁻] = √(Kb c)，其中 c 为碱的初始浓度。然后通过 Kw 求算 [H⁺] 并计算 pH。
For ammonia, Kb = 1.8 × 10⁻⁵ mol dm⁻³ at 298 K. A 0.10 mol dm⁻³ NH₃ solution yields [OH⁻] ≈ 1.34 × 10⁻³ mol dm⁻³, pOH = 2.87, pH = 11.13.
298 K 时氨的 Kb = 1.8 × 10⁻⁵ mol dm⁻³。0.10 mol dm⁻³ 的 NH₃ 溶液中 [OH⁻] ≈ 1.34 × 10⁻³ mol dm⁻³，pOH = 2.87，pH = 11.13。
Note that Kw = Ka × Kb for a conjugate acid-base pair. This relationship allows conversion between Ka of NH₄⁺ and Kb of NH₃.
注意，对于共轭酸碱对，Kw = Kₐ × Kb。利用这一关系可以在 NH₄⁺ 的 Kₐ 和 NH₃ 的 Kb 之间进行换算。
9. Buffer Solutions | 缓冲溶液
A buffer solution resists changes in pH when small amounts of acid or base are added. It is typically made from a weak acid and its conjugate base (in roughly equal concentrations), or a weak base and its conjugate acid.
缓冲溶液能抵御少量外加酸碱引起的 pH 变化。它通常由近似等浓度的弱酸及其共轭碱（或弱碱及其共轭酸）组成。
The pH of an acidic buffer can be calculated directly from the equilibrium expression Ka = [H⁺][A⁻] / [HA]. Rearranging gives:
酸性缓冲液的 pH 可直接从平衡表达式 Kₐ = [H⁺][A⁻] / [HA] 计算。整理后得到：
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