📚 pH Calculations for A-Level WJEC Chemistry: Key Points | A-Level WJEC 化学：pH计算 考点精讲
Understanding pH calculations is fundamental to mastering acid-base chemistry in the WJEC A-Level specification. This article covers the key concepts, formulas, and problem-solving techniques required for exam success, from strong acids to buffers and titration curves.
掌握pH计算是精通WJEC A-Level化学酸碱部分的基础。本文涵盖考试必备的核心概念、公式和解题技巧，从强酸到缓冲溶液及滴定曲线，助你高效备考。
1. Introduction to pH and the Ionic Product of Water | pH和水的离子积简介
The pH scale is a logarithmic measure of hydrogen ion concentration, defined by pH = -log₁₀[H⁺]. A change of one pH unit represents a tenfold change in [H⁺]. Pure water at 298 K has [H⁺] = 1.0 × 10⁻⁷ mol dm⁻³, giving a neutral pH of 7.
pH是对数标度，用于表示氢离子浓度，公式为 pH = -log₁₀[H⁺]。pH值每变化1个单位，[H⁺]浓度就变化10倍。298 K时纯水的[H⁺] = 1.0 × 10⁻⁷ mol dm⁻³，中性pH为7。
Water undergoes autoionisation: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq), often simplified as H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). The ionic product of water, Kw, is central to all aqueous acid–base calculations. At 298 K, Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶. This relationship allows us to interconvert [H⁺] and [OH⁻] for any aqueous solution.
水发生自耦电离：2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)，常简写为 H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)。水的离子积Kw是所有水溶液酸碱计算的基础。298 K时，Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶。利用这一关系，我们可以相互换算任何水溶液中的[H⁺]与[OH⁻]。
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ (at 298 K)
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ (298 K时)
From Kw, we can also define pOH: pOH = -log₁₀[OH⁻], and hence pH + pOH = 14 at 298 K. This is extremely useful when dealing with alkaline solutions.
通过Kw，还可以定义pOH：pOH = -log₁₀[OH⁻]，因此在298 K时pH + pOH = 14。在处理碱性溶液时这一点非常实用。
2. Strong Acids and Bases | 强酸和强碱
Strong acids such as HCl, HNO₃ and H₂SO₄ are assumed to dissociate completely in aqueous solution. For a monoprotic strong acid, [H⁺] equals the initial acid concentration: [H⁺] = c(acid). Therefore, pH = -log₁₀[c(acid)].
强酸如HCl、HNO₃和H₂SO₄在水溶液中完全电离。对于一元强酸，[H⁺]等于酸的初始浓度：[H⁺] = c(acid)。因此，pH = -log₁₀[c(acid)]。
For diprotic sulfuric acid, WJEC usually treats the first proton as completely dissociated and the second proton as effectively fully dissociated for the purpose of pH calculations at typical concentrations. Thus [H⁺] ≈ 2 × c(H₂SO₄). However, always check the question’s instructions; sometimes they expect you to consider the second dissociation step only partially, but in almost all A-level problems, H₂SO₄ is assumed to supply two H⁺ ions.
对于二元强酸硫酸，在WJEC的典型浓度计算中，通常认为第一步完全电离，第二步也视为完全电离，因此[H⁺] ≈ 2 × c(H₂SO₄)。但务必留意题目要求；在绝大多数A-Level题目中，硫酸被当作提供两个H⁺处理。
Strong bases like NaOH and KOH dissociate completely to give OH⁻ ions: [OH⁻] = c(base). To find pH, first calculate pOH = -log₁₀[OH⁻], then use pH = 14 – pOH. Alternatively, [H⁺] = Kw / [OH⁻].
强碱如NaOH和KOH完全电离产生OH⁻离子：[OH⁻] = c(base)。求pH时，可先计算pOH = -log₁₀[OH⁻]，再利用pH = 14 – pOH；或直接用[H⁺] = Kw / [OH⁻]。
pHstrong base = 14 + log₁₀[OH⁻]
强碱的pH = 14 + log₁₀[OH⁻]
3. Weak Acids and Acid Dissociation Constant (Ka) | 弱酸和酸解离常数(Ka)
Weak acids only partially dissociate, establishing an equilibrium: HA(aq) ⇌ H⁺(aq) + A⁻(aq). The acid dissociation constant Kₐ is given by:
弱酸仅部分电离，建立平衡：HA(aq) ⇌ H⁺(aq) + A⁻(aq)。酸解离常数Kₐ表达式如下：
Kₐ = [H⁺][A⁻] / [HA]
For a weak acid starting with concentration [HA]₀, we assume [H⁺] ≈ [A⁻] and that [HA] at equilibrium ≈ [HA]₀, provided the degree of dissociation is less than 5% (or Kₐ/[HA]₀ ≤ 10⁻²). Then the simplified formula becomes:
对于初始浓度为[HA]₀的弱酸，假设[H⁺] ≈ [A⁻]，且平衡时[HA] ≈ [HA]₀，前提是电离度小于5%（或Kₐ/[HA]₀ ≤ 10⁻²）。此时简化公式为：
[H⁺] = √(Kₐ × [HA]₀)
Always verify the approximation after calculation. If the assumption is invalid (degree of ionisation > 5%), a quadratic equation must be solved using the exact Kₐ expression. Most WJEC problems permit the approximation, but you should state it explicitly in your working.
计算后务必验证近似是否成立。若电离度大于5%则近似无效，需利用精确的Kₐ表达式求解二次方程。多数WJEC题目允许使用近似，但应在解题过程中明确写明该假设。
Example: Calculate the pH of 0.10 mol dm⁻³ CH₃COOH (Kₐ = 1.8 × 10⁻⁵). [H⁺] = √(1.8×10⁻⁵ × 0.10) = 1.34×10⁻³ mol dm⁻³, pH = 2.87. Checks: ionisation = 1.34×10⁻³/0.10 = 1.34%, valid.
示例：计算0.10 mol dm⁻³ CH₃COOH的pH（Kₐ = 1.8 × 10⁻⁵）。[H⁺] = √(1.8×10⁻⁵ × 0.10) = 1.34×10⁻³ mol dm⁻³，pH = 2.87。验证：电离度 = 1.34×10⁻³/0.10 = 1.34%，近似有效。
4. pKa and its Relationship to pH | pKa及其与pH的关系
The pKₐ value is a more convenient way of expressing acid strength: pKₐ = -log₁₀Kₐ. A smaller Kₐ (weaker acid) gives a larger pKₐ. The relationship between pH, pKₐ and the ratio of conjugate base to acid is the Henderson–Hasselbalch equation:
pKₐ是一种更便捷的表示酸强度的方法：pKₐ = -log₁₀Kₐ。Kₐ越小（酸越弱），pKₐ越大。pH、pKₐ与共轭碱/酸浓度比之间的关系由亨德森-哈塞尔巴尔赫方程给出：
pH = pKₐ + log₁₀([A⁻] / [HA])
This equation is derived directly from the Kₐ expression after taking negative logarithms. It shows that when [A⁻] = [HA], pH = pKₐ. This is critically important for buffer solutions and for interpreting titration curves.
该方程由Kₐ表达式取负对数直接推导得出。它表明当[A⁻] = [HA]时，pH = pKₐ。这一点对缓冲溶液和解读滴定曲线至关重要。
In a titration of a weak acid with strong base, the halfway point to equivalence has pH = pKₐ. This provides an experimental method to determine Kₐ from a pH curve.
在用强碱滴定弱酸时，半等价点的pH = pKₐ。这为通过pH曲线实验测定Kₐ提供了方法。
5. Weak Bases and Kb | 弱碱和Kb
Weak bases partially react with water: B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq). The base dissociation constant is:
弱碱与水部分反应：B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)。碱解离常数Kb为：
Kb = [BH⁺][OH⁻] / [B]
Analogous to weak acids, for an initial concentration [B]₀, we can approximate [OH⁻] ≈ √(Kb × [B]₀) if dissociation is < 5%. The pOH is then calculated, and pH = 14 – pOH.
与弱酸类似，对于初始浓度[B]₀，若电离度小于5%，可用[OH⁻] ≈ √(Kb × [B]₀)近似。然后求出pOH，再计算pH = 14 – pOH。
Moreover, for any conjugate acid-base pair, the relationship Kₐ × Kb = Kw holds at a given temperature. This means if you are given Kₐ for a conjugate acid, you can find Kb for the base (or vice versa). For example, ammonia NH₃ has a conjugate acid NH₄⁺ with Kₐ = 5.6 × 10⁻¹⁰, so Kb(NH₃) = Kw / Kₐ(NH₄⁺) = 1.0×10⁻¹⁴ / 5.6×10⁻¹⁰ = 1.8×10⁻⁵.
另外，对于任何共轭酸碱对，在特定温度下恒有关系 Kₐ × Kb = Kw。这意味着若已知共轭酸的Kₐ，便可求出相应碱的Kb（反之亦然）。例如氨NH₃的共轭酸NH₄⁺的Kₐ = 5.6 × 10⁻¹⁰，则Kb(NH₃) = Kw / Kₐ(NH₄⁺) = 1.0×10⁻¹⁴ / 5.6×10⁻¹⁰ = 1.8×10⁻⁵。
6. Buffer Solutions: Principles and Calculations | 缓冲溶液：原理与计算
A buffer solution resists changes in pH upon addition of small amounts of acid or base. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in significant concentrations. The classic example is a mixture of ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COONa).
缓冲溶液能抵抗因加入少量酸或碱而引起的pH变化。它由弱酸与其共轭碱（或弱碱与其共轭酸）以较高浓度混合而成。典型例子是乙酸（CH₃COOH）与乙酸钠（CH₃COONa）的混合溶液。
To calculate the pH of a buffer, rearrange the Kₐ expression directly or use the Henderson–Hasselbalch equation. The most practical form for an acidic buffer is:
计算缓冲液的pH时，可直接变换Kₐ表达式，或使用亨德森-哈塞尔巴尔赫方程。对于酸性缓冲液，最实用的形式为：
pH = pKₐ + log₁₀([salt] / [acid])
pH = pKₐ + log₁₀([盐] / [酸])
It is crucial to use the concentrations of the base component (the salt or conjugate base) and the acid component in the mixture after any dilution. Since both species are in the same total volume, the volume cancels, and the ratio of moles can be used directly.
关键是要使用混合后碱组分（盐或共轭碱）与酸组分的浓度。由于两者处于同一总体积中，体积可抵消，因此可直接使用摩尔比。
Example: A buffer prepared from 50 cm³ of 0.10 mol dm⁻³ CH₃COOH and 25 cm³ of 0.10 mol dm⁻³ CH₃COONa. Moles of acid = 0.0050, moles of salt = 0.0025; ratio salt/acid = 0.5. pH = pKₐ + log₁₀(0.5) = 4.76 – 0.30 = 4.46. (pKₐ of ethanoic acid ≈ 4.76).
示例：用50 cm³ 0.10 mol dm⁻³ CH₃COOH和25 cm³ 0.10 mol dm⁻³ CH₃COONa配制缓冲液。酸的物质的量 = 0.0050，盐的物质的量 = 0.0025；盐/酸比 = 0.5。pH = 4.76 + log₁₀(0.5) = 4.76 – 0.30 = 4.46。
7. Henderson–Hasselbalch Equation | 亨德森-哈塞尔巴尔赫方程
The Henderson–Hasselbalch equation is central to buffer problem-solving. It is derived from the logarithmic form of the Kₐ expression:
亨德森-哈塞尔巴尔赫方程是解决缓冲问题的核心。它由Kₐ表达式的对数形式推导而来：
pH = pKₐ + log₁₀([conjugate base] / [weak acid])
pH = pKₐ + log₁₀([共轭碱] / [弱酸])
This equation is valid when the concentrations of the acid and its conjugate base are significantly larger than [H⁺] and [OH⁻] from water, which is true for all typical buffer problems.
当酸及其共轭碱的浓度远大于水本身给出的[H⁺]和[OH⁻]时，该方程成立，而所有典型缓冲问题皆满足此条件。
When small amounts of strong acid are added to a buffer, the added H⁺ reacts with the conjugate base (A⁻), decreasing [A⁻] and increasing [HA]. The new ratio is used in the Henderson–Hasselbalch equation. Similarly, added OH⁻ reacts with HA, converting it to A⁻. Always calculate the new moles of HA and A⁻ after reaction, then substitute.
当向缓冲液中加入少量强酸时，加入的H⁺与共轭碱A⁻反应，使[A⁻]减少、[HA]增加。将新比值代入亨德森方程即可。类似地，加入的OH⁻与HA反应，将HA转化为A⁻。始终在反应后计算新的HA与A⁻的物质的量再代入。
For basic buffers (e.g., NH₃ / NH₄Cl), it is often easier to work with Kb and pOH, or convert to the Kₐ of the conjugate acid and use pH = pKₐ + log₁₀([base]/[acid]). Both approaches give the same result.
对于碱性缓冲液（如NH₃ / NH₄Cl），通常用Kb和pOH计算更为简便，或转化为共轭酸的Kₐ，再用pH = pKₐ + log₁₀([碱]/[酸])计算。两种方法结果一致。
8. pH Curves and Titrations | pH曲线和滴定
pH titration curves show how pH changes as a titrant is added. The shape depends on the strengths of the acid and base involved. Four main types are examined:
pH滴定曲线展示加入滴定剂时pH的变化。曲线形状取决于所涉及的酸和碱的强度。主要考察以下四种类型：
- Strong acid – strong base: steep vertical jump from pH ~3 to ~11 at equivalence; equivalence point pH = 7.
- 强酸 – 强碱：等价点附近出现pH ~3至~11的陡直突跃；等价点pH = 7。
- Weak acid – strong base: buffer region before equivalence; equivalence point pH > 7 (due to hydrolysis of the conjugate base). Half-equivalence pH = pKₐ.
- 弱酸 – 强碱：等价点前存在缓冲区；等价点pH > 7（因共轭碱水解）。半等价点pH = pKₐ。
- Strong acid – weak base: equivalence point pH < 7.
- 强酸 – 弱碱：等价点pH < 7。
- Weak acid – weak base: very gradual pH change, no sharp vertical rise; rarely used for titrations with indicators.
- 弱酸 – 弱碱：pH变化非常平缓，无陡直突跃；很少用指示剂进行此类滴定。
The equivalence point is where moles of H⁺ supplied exactly equal moles of OH⁻ supplied (or stoichiometric equivalents). In weak/strong combinations, the salt formed undergoes hydrolysis, making the solution acidic or alkaline.
等价点是指提供的H⁺的物质的量与OH⁻的物质的量恰好按化学计量比相等。在弱/强组合中，生成的盐发生水解，使溶液呈酸性或碱性。
For polyprotic acids (e.g., H₃PO₄) or bases, multiple equivalence points and buffer regions appear. Each segment can be analysed with appropriate Kₐ values.
对于多元酸（如H₃PO₄）或多元碱，会出现多个等价点和缓冲区。每一段可用相应的Kₐ值进行分析。
9. Indicators and Endpoint Selection |
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