A-Level Chemistry: Core Concepts of Acid–Base Equilibria | A-Level 化学：酸碱平衡核心考点

📚 A-Level Chemistry: Core Concepts of Acid–Base Equilibria | A-Level 化学：酸碱平衡核心考点

Acid–base equilibria form one of the most examinable and conceptually rich areas of A-Level Chemistry. From Brønsted–Lowry definitions to pH curves, buffer systems, and indicator selection, this topic tests both your quantitative precision and your qualitative reasoning. This guide condenses the core ideas into a structured revision pathway tailored for CIE A-Level students.

酸碱平衡是 A-Level 化学中考试出题频率最高、概念综合性最强的模块之一。从 Brønsted–Lowry 酸碱定义到 pH 曲线、缓冲溶液体系和指示剂选择，这一专题既考查你的定量计算能力，也考查你的定性分析思维。本指南将核心知识浓缩为系统的复习路径，专为 CIE A-Level 考生设计。

1. Brønsted–Lowry Theory and Conjugate Pairs | Brønsted–Lowry 理论与共轭对

A Brønsted–Lowry acid is a proton donor, and a Brønsted–Lowry base is a proton acceptor. This definition applies not only to aqueous solutions but also to reactions in other solvents and even in the gas phase, making it far more general than the old Arrhenius model. Every acid–base reaction involves two conjugate pairs: when an acid donates a proton, it forms its conjugate base; when a base accepts a proton, it forms its conjugate acid.

Brønsted–Lowry 酸是质子的给予体，Brønsted–Lowry 碱是质子的接受体。这一定义不仅适用于水溶液，也适用于其他溶剂中的反应，甚至气相反应，因此比传统的 Arrhenius 模型更具普适性。每一个酸碱反应都涉及两对共轭酸碱对：酸给出质子后形成其共轭碱；碱接受质子后形成其共轭酸。

For example, in the reaction between hydrochloric acid and water:

例如，在盐酸与水的反应中：

HCl + H₂O → H₃O⁺ + Cl⁻

HCl donates a proton to H₂O. The conjugate acid–base pairs are HCl/Cl⁻ and H₃O⁺/H₂O. The stronger the acid, the weaker its conjugate base, and vice versa. This inverse relationship is crucial for predicting the direction of acid–base reactions and for understanding relative strengths of acids and bases.

HCl 将质子传递给 H₂O。两对共轭酸碱对分别是 HCl/Cl⁻ 和 H₃O⁺/H₂O。酸越强，其共轭碱越弱；反之亦然。这种反向关系对判断酸碱反应方向、理解酸碱相对强弱至关重要。

In addition to the concept of conjugate pairs, some species can act as both an acid and a base depending on the reaction conditions. These are called amphoteric species. Water itself is amphoteric: it accepts a proton when reacting with HCl but donates a proton when reacting with NH₃. Hydrogencarbonate ions (HCO₃⁻) also exhibit this dual behaviour, which appears frequently in exam questions involving acid–base character.

此外，有些物质既能作为酸又能作为碱，取决于反应条件，这类物质称为两性物种。水本身是两性的：与 HCl 反应时接受质子，与 NH₃ 反应时给出质子。碳酸氢根离子（HCO₃⁻）也表现出这种双重性质，这在涉及酸碱性质的考题中经常出现。

2. Strong and Weak Acids and Bases | 强酸强碱与弱酸弱碱

A strong acid is one that fully dissociates in aqueous solution. Common examples include hydrochloric acid (HCl), nitric acid (HNO₃), and sulfuric acid (H₂SO₄). For a strong monoprotic acid, the concentration of H⁺ ions is equal to the initial concentration of the acid: [H⁺] = [HA]₀. In contrast, a weak acid only partially dissociates, establishing an equilibrium between the intact acid molecule and its ions. Ethanoic acid, CH₃COOH, is the classic example.

强酸是指在水溶液中完全电离的酸。常见例子包括盐酸（HCl）、硝酸（HNO₃）和硫酸（H₂SO₄）。对于强一元酸，H⁺ 离子浓度等于酸的初始浓度：[H⁺] = [HA]₀。相反，弱酸仅部分电离，在未电离的酸分子与其离子之间建立平衡。乙酸 CH₃COOH 是最经典的例子。

CH₃COOH ⇌ CH₃COO⁻ + H⁺

The equilibrium position lies far to the left, meaning that only about 1–5% of ethanoic acid molecules are ionised at typical concentrations. Similarly, strong bases such as NaOH and KOH fully dissociate to produce OH⁻ ions, while weak bases such as ammonia, NH₃, only partially accept protons from water:

该平衡的位置远偏向左方，意味着在典型浓度下，仅有约 1%–5% 的乙酸分子发生电离。类似地，NaOH 和 KOH 等强碱完全电离产生 OH⁻ 离子，而 NH₃ 这类弱碱仅部分地从水中接受质子：

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

It is essential to distinguish between the terms “concentration” and “strength.” Strength is an intrinsic property determined by the extent of ionisation, whereas concentration refers to the amount of solute dissolved per unit volume. A dilute strong acid still has a lower [H⁺] than a concentrated weak acid, but the strong acid is nevertheless fully ionised. Exam questions often test this distinction.

必须区分”浓度”和”强度”这两个概念。强度是物质本身的性质，由电离程度决定；而浓度是指单位体积内溶质的量。稀的强酸中[H⁺]可能低于浓的弱酸，但强酸仍然完全电离。考试题经常考查这一区别。

3. The Ionic Product of Water and pH Scale | 水的离子积与 pH 标度

Water undergoes a very slight self-ionisation: H₂O ⇌ H⁺ + OH⁻. The equilibrium constant for this process is called the ionic product of water, K_w. At 25°C, K_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶. In pure water, [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³, giving a neutral pH of 7.00. The ionic product is temperature-dependent because self-ionisation is endothermic; at 50°C, K_w rises to about 5.48 × 10⁻¹⁴ and neutral pH becomes approximately 6.63.

水会发生极微弱的自电离：H₂O ⇌ H⁺ + OH⁻。该过程的平衡常数称为水的离子积 K_w。在 25°C 时，K_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶。在纯水中，[H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³，对应的中性 pH 为 7.00。离子积随温度变化而变化，因为水的自电离是吸热过程；在 50°C 时，K_w 升高至约 5.48 × 10⁻¹⁴，此时中性 pH 约为 6.63。

The pH scale is defined as pH = −log₁₀[H⁺]. Because of the logarithmic nature of the scale, a decrease of one pH unit corresponds to a ten-fold increase in hydrogen ion concentration. When the pH of a solution is higher than 7, it is alkaline; when lower than 7, it is acidic. At temperatures other than 25°C, the neutral point shifts because K_w changes. A solution at 60°C with pH = 6.5 is neutral, not acidic, even though its pH is below 7.

pH 标度定义为 pH = −log₁₀[H⁺]。由于标度取了对数，pH 每降低 1 个单位，对应 H⁺ 浓度增大为原来的 10 倍。当溶液 pH 大于 7 时为碱性；小于 7 时为酸性。在非 25°C 条件下，中性点会因 K_w 的变化而移动。60°C 时 pH = 6.5 的溶液实际上是中性的，而不是酸性，尽管其 pH 低于 7。

Students sometimes confuse pOH with pH. Recall that in any aqueous solution at 25°C, pH + pOH = 14. This relationship is directly derived from K_w. For a strong base such as 0.01 mol dm⁻³ NaOH, pOH = 2 and therefore pH = 12. For a weak base, you cannot simply take the negative logarithm of the base concentration; you must use K_b.

学生有时会混淆 pOH 与 pH。请记住，在 25°C 的任何水溶液中，pH + pOH = 14。这一关系直接由 K_w 推导而来。对于 0.01 mol dm⁻³ NaOH 这样的强碱，pOH = 2，因此 pH = 12。对于弱碱，不能简单地取碱浓度的负对数，而必须使用 K_b 进行计算。

4. Acid Dissociation Constant Kₐ and pKₐ | 酸解离常数 Kₐ 与 pKₐ

For a weak acid HA in water, the equilibrium is represented as HA ⇌ H⁺ + A⁻. The acid dissociation constant is:

对于水中的弱酸 HA，其平衡可以表示为 HA ⇌ H⁺ + A⁻。酸解离常数为：

Kₐ = [H⁺][A⁻] / [HA]

Strictly speaking, concentrations in Kₐ expressions should be equilibrium concentrations, and water is omitted because its concentration is effectively constant. The units of Kₐ depend on the number of species in the numerator and denominator; for a simple monoprotic acid, units are mol dm⁻³. In practice, Kₐ values are usually quoted in mol dm⁻³. The value of pKₐ is defined as pKₐ = −log₁₀Kₐ. A smaller pKₐ corresponds to a stronger acid. For example, ethanoic acid has Kₐ ≈ 1.74 × 10⁻⁵ mol dm⁻³ and pKₐ ≈ 4.76, whereas chloroethanoic acid has Kₐ ≈ 1.38 × 10⁻³ mol dm⁻³ and pKₐ ≈ 2.86.

严格而言，Kₐ 表达式中的浓度应当是平衡浓度，水被省略是因为其浓度在稀溶液中实际上保持恒定。Kₐ 的单位取决于分子和分母中物种的数量；对于简单的一元酸，单位为 mol dm⁻³。实际应用中，Kₐ 值通常以 mol dm⁻³ 为单位给出。pKₐ 定义为 pKₐ = −log₁₀Kₐ。pKₐ 越小代表酸越强。例如，乙酸的 Kₐ ≈ 1.74 × 10⁻⁵ mol dm⁻³，pKₐ ≈ 4.76；而氯乙酸的 Kₐ ≈ 1.38 × 10⁻³ mol dm⁻³，pKₐ ≈ 2.86。

To calculate the pH of a weak acid, use the approximation that [H⁺] ≈ [A⁻] because only a tiny fraction of HA dissociates, and that the equilibrium concentration of HA remains very close to its initial concentration. This gives [H⁺]² ≈ Kₐ × [HA]₀. For a 0.100 mol dm⁻³ solution of ethanoic acid:

计算弱酸 pH 时，可以利用两个近似：[H⁺] ≈ [A⁻]，因为只有很少一部分 HA 电离；同时 HA 的平衡浓度近似等于其初始浓度。由此可得 [H⁺]² ≈ Kₐ × [HA]₀。对于 0.100 mol dm⁻³ 的乙酸溶液：

[H⁺] = √(Kₐ × [HA]₀) = √(1.74 × 10⁻⁵ × 0.100) ≈ 1.32 × 10⁻³ mol dm⁻³

pH ≈ −log₁₀(1.32 × 10⁻³) ≈ 2.88

This approximation is valid when Kₐ is very small and the acid is not extremely dilute. If the acid is very dilute or Kₐ is relatively large, the full quadratic equation must be solved instead. An exam-style extension asks for the percentage ionisation, which equals ([H⁺]/[HA]₀) × 100%.

这一近似在 Kₐ 很小且酸浓度不太低时成立。如果酸极稀或 Kₐ 相对较大，则须用完整的一元二次方程求解。考试中常见的延伸问题要求计算电离度百分比，即 ([H⁺]/[HA]₀) × 100%。

5. Base Dissociation Constant K_b and K_a × K_b = K_w | 碱解离常数 K_b 与 K_a × K_b = K_w

Weak bases such as ammonia establish an equilibrium in water. The base dissociation constant K_b is defined for the reaction B + H₂O ⇌ BH⁺ + OH⁻ as:

氨这类弱碱在水中建立平衡。碱解离常数 K_b 定义为反应 B + H₂O ⇌ BH⁺ + OH⁻ 的平衡常数：

K_b = [BH⁺][OH⁻] / [B]

There is an extremely useful relationship between K_a and K_b for a conjugate acid–base pair: K_a × K_b = K_w. At 25°C, this means pK_a + pK_b = 14. This relationship allows you to interconvert between the acidity of a weak acid and the basicity of its conjugate base. For instance, the K_b of the ethanoate ion, CH₃COO⁻, is K_w / K_a(CH₃COOH) = 1.00 × 10⁻¹⁴ / 1.74 × 10⁻⁵ ≈ 5.75 × 10⁻¹⁰ mol dm⁻³.

对于一对共轭酸碱，存在一个非常重要的关系：K_a × K_b = K_w。在 25°C 时，即 pK_a + pK_b = 14。该关系使你能够在弱酸的酸性与其共轭碱的碱性之间相互转换。例如，乙酸根离子 CH₃COO⁻ 的 K_b = K_w / K_a(CH₃COOH) = 1.00 × 10⁻¹⁴ / 1.74 × 10⁻⁵ ≈ 5.75 × 10⁻¹⁰ mol dm⁻³。

To calculate the pH of a weak base, write the equilibrium in terms of OH⁻ production, apply the approximation [OH⁻]² ≈ K_b × [B]₀, then use pOH = −log₁₀[OH⁻] and pH = 14 − pOH. For 0.0500 mol dm⁻³ ammonia, K_b = 1.80 × 10⁻⁵ mol dm⁻³.

计算弱碱 pH 时，写出产生 OH⁻ 的平衡式，利用近似 [OH⁻]² ≈ K_b × [B]₀，然后 pOH = −log₁₀[OH⁻]，pH = 14 − pOH。对于 0.0500 mol dm⁻³ 的氨水，K_b = 1.80 × 10⁻⁵ mol dm⁻³。

[OH⁻] = √(1.80 × 10⁻⁵ × 0.0500) ≈ 9.49 × 10⁻⁴ mol dm⁻³

pOH ≈ 3.02, pH ≈ 10.98

Notice that the pH of this weak base is significantly lower than the pH of a strong base of similar concentration. A 0.0500 mol dm⁻³ NaOH solution would have pH = 12.70. This difference illustrates how incomplete dissociation limits the concentration of hydroxide ions.

注意，这个弱碱溶液的 pH 明显低于同浓度强碱的 pH。0.0500 mol dm⁻³ NaOH 溶液的 pH 为 12.70。这一差异说明不完全电离限制了氢氧根离子的浓度。

6. Buffer Solutions: Composition and Action | 缓冲溶液：组成与作用机制

A buffer solution is a system that resists changes in pH when small amounts of acid or base are added or when it is diluted. A buffer is typically composed of a weak acid and its conjugate base (for example, ethanoic acid and sodium ethanoate), or a weak base and its conjugate acid (for example, ammonia and ammonium chloride). The key to buffering action lies in the presence of both a reservoir of the weak acid to neutralise added OH⁻ and a reservoir of the conjugate base to neutralise added H⁺.

缓冲溶液是一种当加入少量酸或碱或进行稀释时能够抵抗 pH 显著变化的体系。缓冲溶液通常由弱酸及其共轭碱组成（如乙酸与乙酸钠），或由弱碱及其共轭酸组成（如氨与氯化铵）。缓冲作用的关键在于同时存在足量的弱酸用于中和外加的 OH⁻，以及足量的共轭碱用于中和外加的 H⁺。

When a small amount of strong acid is added to an ethanoate buffer, the H⁺ ions react with CH₃COO⁻ to form CH₃COOH. The equilibrium shifts to the left, but the change in [H⁺] is small. When a small amount of strong base is added, OH⁻ ions react with CH₃COOH to form CH₃COO⁻ and water, again minimising pH change. The buffer only works effectively within a limited pH range and only for moderate additions of acid or base; exceeding the buffering capacity exhausts one component and the pH then changes rapidly.

当少量强酸加入乙酸缓冲体系时，H⁺ 离子与 CH₃COO⁻ 反应生成 CH₃COOH。平衡向左移动，但[H⁺]的变化很小。当加入少量强碱时，OH⁻ 离子与 CH₃COOH 反应生成 CH₃COO⁻ 和水，同样使 pH 变化最小化。缓冲溶液的缓冲能力仅在一定 pH 范围内有效，且只能应对中等量的酸碱添加；一旦超过缓冲容量，某一组分被耗尽，pH 便会迅速变化。

The Henderson–Hasselbalch equation provides a convenient way to calculate the pH of a buffer:

Henderson–Hasselbalch 方程为计算缓冲溶液的 pH 提供了便捷方法：

pH = pK_a + log₁₀([A⁻] / [HA])

For a buffer made from a weak acid HA and its salt NaA, the ratio [A⁻]/[HA] is approximately equal to the ratio of salt concentration to acid concentration. Note that this equation is derived from the K_a expression and assumes that the equilibrium concentrations are close to the initial analytical concentrations.

对于由弱酸 HA 及其盐 NaA 配制的缓冲溶液，[A⁻]/[HA] 约等于盐浓度与酸浓度之比。注意该方程由 K_a 表达式推导而来，其条件是平衡浓度与初始分析浓度接近。

For example, calculate the pH of a buffer containing 0.200 mol dm⁻³ ethanoic acid (pK_a = 4.76) and 0.150 mol dm⁻³ sodium ethanoate.

例如，计算含 0.200 mol dm⁻³ 乙酸（pK_a = 4.76）和 0.150 mol dm⁻³ 乙酸钠的缓冲溶液的 pH。

pH = 4.76 + log₁₀(0.150 / 0.200) = 4.76 − 0.125 = 4.64

The pH of this buffer is slightly below pK_a because the acid concentration exceeds the salt concentration. In exam questions, you may also be asked to calculate the change in pH after adding a known amount of strong acid or base. In such cases, first calculate how the addition changes the number of moles of HA and A⁻, then substitute the new concentrations into the Henderson–Hasselbalch equation.

该缓冲溶液的 pH 略低于 pK_a，因为酸的浓度高于盐的浓度。在考试中，还可能要求计算加入一定量强酸或强碱后 pH 的变化。此时应先计算出添加过程对 HA 和 A⁻ 物质的量的影响，再将新的浓度代入 Henderson–Hasselbalch 方程。

7. pH Curves and Indicator Selection | pH 曲线与指示剂选择

A pH curve shows how the pH of a solution changes as a titrant is added. The shape of the curve depends on whether the acid and base are strong or weak. Four classic combinations exist: strong acid–strong base, strong acid–weak base, weak acid–strong base, and weak acid–weak base.

pH 曲线表示随着滴定剂加入，溶液 pH 如何变化。曲线的形状取决于酸和碱是强还是弱。存在四种经典组合：强酸–强碱、强酸–弱碱、弱酸–强碱、弱酸–弱碱。

For a strong acid titrated with a strong base, the initial pH is very low. The pH rises slowly at first, then steeply near the equivalence point (a vertical section typically from pH 3 to pH 11), and finally levels off at a high pH. The equivalence point occurs at pH 7. For a weak acid titrated with a strong base, the initial pH is higher, and the curve shows a buffer region where pH changes only slowly before the steep rise. The equivalence point for this combination is above 7 because the conjugate base of the weak acid hydrolyses to produce OH⁻.

对于强酸滴定强碱，起始 pH 很低。pH 先缓慢升高，在化学计量点附近急剧上升（垂直段通常从 pH 3 到 pH 11），最后在高 pH 区趋于平缓。化学计量点的 pH 为 7。对于弱酸滴定强碱，起始 pH 较高，曲线存在一个缓冲平台区，pH 变化缓慢，随后才出现急剧上升。该组合的化学计量点 pH 大于 7，因为弱酸的共轭碱水解产生 OH⁻。

The equivalence point is different in each combination:

不同组合的化学计量点 pH 各不相同：

Acid–Base Combination	Equivalence Point pH	Indicator Options
Strong acid – Strong base	7	Methyl orange, phenolphthalein, bromothymol blue
Strong acid – Weak base	Below 7	Methyl orange
Weak acid – Strong base	Above 7	Phenolphthalein
Weak acid – Weak base	Approximately 7	No sharp vertical region; no suitable single indicator

Indicator selection depends on matching the indicator’s colour-change interval to the vertical section of the pH curve. The indicator must change colour within the steep portion so that the endpoint is close to the equivalence point. Methyl orange has a pH range of approximately 3.1–4.4, while phenolphthalein changes between 8.3 and 10.0. Neither works for weak acid–weak base titrations because the pH change around the equivalence point is too gradual and small to trigger a sharp colour change.

指示剂的选择取决于其变色范围是否与 pH 曲线的垂直段相匹配。指示剂必须在突跃范围内变色，从而使滴定终点接近化学计量点。甲基橙的变色范围约为 pH 3.1–4.4，而酚酞的变色范围约为 pH 8.3–10.0。这两种指示剂都不适用于弱酸–弱碱滴定，因为该条件下化学计量点附近的 pH 变化过于平缓且范围较小，无法产生明显的颜色突变。

8. Salt Hydrolysis and Acidic/Basic Salts | 盐类水解与酸性/碱性盐

When a salt dissolves in water, the resulting solution may not be neutral. Salt hydrolysis occurs when one or both of the salt’s ions react with water to produce H⁺ or OH⁻. Salts formed from a strong acid and a weak base produce acidic solutions because the conjugate acid of the weak base hydrolyses, releasing H⁺. For example, ammonium chloride, NH₄Cl, dissociates into NH₄⁺ and Cl⁻. The NH₄⁺ ion acts as a weak acid:

当盐溶于水时，所得溶液不一定呈中性。当盐的一种或两种离子与水反应生成 H⁺ 或 OH⁻ 时，就发生盐类水解。由强酸与弱碱生成的盐溶液呈酸性，因为弱碱的共轭酸发生水解，释放出 H⁺。例如，氯化铵 NH₄Cl 电离出 NH₄⁺ 和 Cl⁻。NH₄⁺ 离子表现弱酸性：

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺

The solution is therefore acidic, with pH typically between 4 and 6 depending on concentration. Conversely, salts formed from a weak acid and a strong base, such as sodium ethanoate, produce alkaline solutions because the conjugate base CH₃COO⁻ hydrolyses:

因此溶液显酸性，pH 通常在 4 到 6 之间，具体取决于浓度。相反，弱酸与强碱生成的盐，如乙酸钠，溶液呈碱性，因为共轭碱 CH₃COO⁻ 发生水解：

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

Salts from a strong acid and a strong base, such as NaCl, do not hydrolyse to any significant degree, and their aqueous solutions are neutral. When both the cation and the anion can hydrolyse, as in ammonium ethanoate, the pH depends on the relative values of Kₐ and K_b for the two ions. In the case of ammonium ethanoate, because Kₐ for NH₄⁺ is about the same as K_b for CH₃COO⁻, the solution is roughly neutral.

由强酸与强碱形成的盐，如 NaCl，几乎不发生水解，其水溶液呈中性。当阳离子和阴离子均能水解时（如乙酸铵），溶液的 pH 取决于两种离子对应的 Kₐ 与 K_b 的相对大小。就乙酸铵而言，NH₄⁺ 的 Kₐ 与 CH₃COO⁻ 的 K_b 大致相等，因此溶液约呈中性。

9. Common Exam Calculations and Pitfalls | 常见考试计算与易错点

The most frequent calculation types in CIE exams involve: finding pH of strong acids and strong bases; calculating pH of weak acids using Kₐ and approximations; determining the degree of ionisation; calculating buffer pH before and after additions of acid or base; identifying the equivalence point and half-equivalence point from titration data; and calculating Kₐ from the pH at the half-equivalence point, where pH = pKₐ.

CIE 考试中最常见的计算类型包括：求强酸强碱的 pH；利用 Kₐ 及近似计算弱酸 pH；求电离度；计算缓冲液在加酸或加碱前后的 pH；从滴定数据中识别化学计量点和半化学计量点；以及利用半化学计量点的 pH 求 Kₐ，因为该点 pH = pKₐ。

One of the most common mistakes students make is treating a weak acid as fully dissociated and writing [H⁺] = [HA]. Another frequent error is using initial concentrations directly into an equilibrium expression without accounting for the change. You must remember that Kₐ expressions require equilibrium concentrations. A third error is forgetting to adjust volumes when mixing solutions; concentration is moles divided by total volume, not moles divided by original volume. Fourthly, when calculating the pH of a strong base, some students forget that [OH⁻] corresponds to pOH, not pH, and that the relationship pH + pOH = 14 must be used at 25°C.

最常见的错误之一是把弱酸当作完全电离，直接写出 [H⁺] = [HA]。另一个常见错误是直接将初始浓度代入平衡表达式而不考虑浓度的变化。必须记住，Kₐ 表达式中使用的是平衡浓度。第三个错误是混合溶液时忘记调整体积；浓度应等于物质的量除以总体积，而不是除以原体积。第四，计算强碱 pH 时，有些学生忘记 [OH⁻] 对应的是 pOH 而不是 pH，必须利用 25°C 时 pH + pOH = 14 的关系。

Consider a typical problem: a 25.0 cm³ sample of 0.100 mol dm⁻³ ethanoic acid is titrated against 0.100 mol dm⁻³ NaOH. At the half-equivalence point, half of the ethanoic acid has been neutralised, and the concentrations of CH₃COOH and CH₃COO⁻ are equal. Consequently, pH = pKₐ = 4.76. At the equivalence point, all the acid has been converted to CH₃COO⁻, and the pH is governed by the hydrolysis of the ethanoate ion. The pH at the equivalence point is always above 7 for a weak acid–strong base titration.

看一个典型问题：25.0 cm³ 0.100 mol dm⁻³ 乙酸用 0.100 mol dm⁻³ NaOH 滴定。在半化学计量点时，一半乙酸被中和，CH₃COOH 与 CH₃COO⁻ 浓度相等，因此 pH = pKₐ = 4.76。在化学计量点时，所有酸都转化为 CH₃COO⁻，pH 由乙酸根的水解决定。对于弱酸–强碱滴定，化学计量点 pH 总是大于 7。

When indicators are discussed, you should also understand that the endpoint colour change occurs over a range, not at a single pH. The indicator itself is a weak acid with its own Kₐ, and it exists in two differently coloured forms. The colour of the indicator depends on the ratio of the two forms, which in turn depends on the pH of the solution relative to pK_ind. In a titration, the indicator is added in very small amounts so that it does not affect the pH of the solution being titrated.

在讨论指示剂时，还应理解指示剂的颜色变化发生在一定的 pH 范围内，而不是在某一个特定 pH 瞬间完成。指示剂本身是弱酸，具有自己的 Kₐ 值，并以两种不同颜色的形式存在。指示剂颜色取决于两种形式的比例，而这又取决于溶液 pH 与 pK_ind 的相对大小。在滴定中，指示剂用量很小，不会影响被滴定溶液的 pH。

10. Strategic Summary and Exam Technique | 策略性总结与答题技巧

Start every acid–base question by identifying the species present in solution and classifying them as strong acid, strong base, weak acid, weak base, salt, or buffer. This classification immediately tells you which formula or approach to use. For strong acids and bases, use direct stoichiometric dissociation. For weak acids and bases, apply Kₐ/K_b with appropriate approximations. For buffers, always use the Henderson–Hasselbalch equation or the full equilibrium method. For salts, determine which ions hydrolyse and whether the solution is acidic, basic, or neutral.

解决每一道酸碱题时，首先需确定溶液中存在的物种，并将其归类为强酸、强碱、弱酸、弱碱、盐或缓冲体系。这一分类会立即告诉你该使用哪种公式或方法。对于强酸强碱，使用直接化学计量电离。对于弱酸弱碱，应用 Kₐ/K_b 并配合合理的近似处理。对于缓冲溶液，使用 Henderson–Hasselbalch 方程或完整的平衡方法。对于盐溶液，判断哪些离子发生水解以及溶液是酸性、碱性还是中性。

Show your working clearly and include units throughout calculations. In CIE exams, method marks are often awarded even when the final numerical answer is wrong, provided the correct sequence of steps is visible. Carry out logarithm operations at the end rather than rounding intermediate values too early. State approximations explicitly, such as “assuming negligible dissociation of the weak acid,” where appropriate, to demonstrate conceptual understanding.

答题时要清晰展示计算过程，并全程包含单位。在 CIE 考试中，即使最终数值答错，只要步骤顺序正确，通常会给予方法分。对数运算应放在最后进行，不要过早舍入中间值。在适当处明确写出所用近似条件，例如”假设弱酸解离可忽略不计”，以展示概念理解程度。

Practise sketching pH curves by hand, marking the initial pH, the buffering region, the vertical section, and the equivalence point. Become comfortable translating between pH, [H⁺], pOH, and [OH⁻] without hesitation. Finally, remember that acid–base equilibria do not exist in isolation: they connect strongly to equilibrium constants, enthalpy changes, and quantitative chemistry, and CIE examiners frequently combine these ideas in multi-part questions.

练习手绘 pH 曲线

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A-Level Chemistry: Core Concepts of Acid–Base Equilibria | A-Level 化学：酸碱平衡核心考点