pH Calculations for AQA A-Level Chemistry | A-Level AQA 化学：pH计算考点精讲

📚 pH Calculations for AQA A-Level Chemistry | A-Level AQA 化学：pH计算考点精讲

A strong grasp of pH calculations is essential for AQA A-Level Chemistry. This topic brings together the ionic product of water Kw, dissociation constants Ka and Kb, buffer theory, and the interpretation of titration curves. In this article, we break down every type of pH problem you may encounter, from strong acids and bases to buffer solutions, dilution, and the effect of temperature on Kw. Each section presents the key equations and approximations you must know, and the worked-through logic will help you tackle even the trickiest exam questions with confidence.

熟练掌握 pH 计算是 AQA A-Level 化学的核心要求之一。这一专题将水的离子积 Kw、酸解离常数 Ka 和碱解离常数 Kb、缓冲理论以及滴定曲线的解读融合在一起。本文详细拆解考试中可能出现的每一种 pH 计算类型，从强酸强碱到缓冲溶液、稀释问题以及温度对 Kw 的影响。每个部分都呈现了必须掌握的关键方程和近似条件，并通过清晰的推理过程帮助你充满信心地应对最棘手的试题。

1. The Ionic Product of Water, Kw | 水的离子积 Kw

Water undergoes self-ionisation to a very small extent according to the equilibrium: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). The equilibrium constant for this process is called the ionic product of water, Kw = [H⁺][OH⁻]. At 25 °C (298 K), Kw has the value 1.0 × 10⁻¹⁴ mol² dm⁻⁶. In pure water, [H⁺] = [OH⁻] = √Kw, giving a neutral pH of 7.00 at this temperature. Because [H₂O] is essentially constant, it is incorporated into Kw, so we never include the concentration of water in the expression.

水会发生极微弱的自耦电离：H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)。这一过程的平衡常数称为水的离子积，Kw = [H⁺][OH⁻]。在 25 °C（298 K）时，Kw 的值为 1.0 × 10⁻¹⁴ mol² dm⁻⁶。在纯水中，[H⁺] = [OH⁻] = √Kw，因此该温度下中性 pH 为 7.00。由于 [H₂O] 基本保持不变，它被并入 Kw 中，所以表达式中从不出现水的浓度。

2. Calculating pH of Strong Acids | 强酸 pH 计算

Strong monoprotic acids such as HCl, HNO₃ and HClO₄ dissociate completely in aqueous solution. Therefore, the hydrogen ion concentration is equal to the acid concentration: [H⁺] = c(acid). The pH is then calculated using pH = –log₁₀[H⁺]. For example, a 0.050 mol dm⁻³ solution of HCl has [H⁺] = 0.050 mol dm⁻³, giving pH = –log₁₀(0.050) = 1.30. For the diprotic strong acid H₂SO₄, the first proton is fully dissociated; however, the second dissociation is not always complete at A-Level. You should treat the first proton as giving [H⁺] = c(H₂SO₄) and only consider the second ionisation if Ka₂ data are provided. Dilute strong acids approaching neutral pH require simultaneous consideration of Kw, but this is rarely tested except when extremely dilute (ca. 10⁻⁷ mol dm⁻³).

HCl、HNO₃、HClO₄ 等一元强酸在水溶液中完全解离，因此氢离子浓度等于酸的浓度：[H⁺] = c(酸)。然后用 pH = –log₁₀[H⁺] 计算 pH。例如，0.050 mol dm⁻³ 的 HCl 溶液中 [H⁺] = 0.050 mol dm⁻³，pH = –log₁₀(0.050) = 1.30。对于二元强酸 H₂SO₄，第一个质子完全解离；但在 A-Level 中，第二步解离并不总是完全的。通常只需将第一个质子的贡献计为 [H⁺] = c(H₂SO₄)，除非题目提供了 Ka₂ 数据。极稀的强酸接近中性 pH 时需同时考虑 Kw，但在 ≈10⁻⁷ mol dm⁻³ 时才可能考到。

3. Calculating pH of Strong Bases | 强碱 pH 计算

Strong bases such as NaOH and KOH dissociate fully to give OH⁻ ions. First, find the hydroxide ion concentration: for a monobasic strong base, [OH⁻] = c(base). Then use Kw to convert [OH⁻] to [H⁺]: [H⁺] = Kw / [OH⁻]. Finally, pH = –log₁₀[H⁺]. At 25 °C, this simplifies to pH = 14.00 + log₁₀[OH⁻] (since pOH = –log₁₀[OH⁻] and pH + pOH = 14.00). For dibasic strong bases like Ba(OH)₂, [OH⁻] = 2 × c(Ba(OH)₂). For instance, a 0.020 mol dm⁻³ Ba(OH)₂ solution has [OH⁻] = 0.040 mol dm⁻³, giving [H⁺] = 1.0 × 10⁻¹⁴ / 0.040 = 2.5 × 10⁻¹³ mol dm⁻³, so pH = 12.60.

NaOH、KOH 等强碱完全解离产生 OH⁻ 离子。首先计算氢氧根浓度：对一元强碱，[OH⁻] = c(碱)。然后利用 Kw 转换为 [H⁺]：[H⁺] = Kw / [OH⁻]。最后 pH = –log₁₀[H⁺]。在 25 °C 下，这可以简化为 pH = 14.00 + log₁₀[OH⁻]（因为 pOH = –log₁₀[OH⁻]，且 pH + pOH = 14.00）。对于二元强碱如 Ba(OH)₂，[OH⁻] = 2 × c(Ba(OH)₂)。例如，0.020 mol dm⁻³ Ba(OH)₂ 溶液中 [OH⁻] = 0.040 mol dm⁻³，则 [H⁺] = 1.0 × 10⁻¹⁴ / 0.040 = 2.5 × 10⁻¹³ mol dm⁻³，pH = 12.60。

4. Weak Acids and Ka | 弱酸与酸解离常数 Ka

A weak acid (HA) partially dissociates in water: HA(aq) ⇌ H⁺(aq) + A⁻(aq). The acid dissociation constant is Ka = [H⁺][A⁻] / [HA]. To find [H⁺], assume that the amount dissociated is small compared with the initial concentration c, so [HA] ≈ c at equilibrium. Since [H⁺] = [A⁻] = x, we get Ka = x² / c, hence x = √(Ka c). The approximation is valid when c / Ka > 100. The pH is then pH = –log₁₀(√(Ka c)). You may also be required to calculate Ka from measured pH: if pH is known, then [H⁺] = 10⁻^pH, and Ka = [H⁺]² / (c – [H⁺]). The percentage dissociation can be calculated as ([H⁺] / c) × 100%.

弱酸 HA 在水中部分解离：HA(aq) ⇌ H⁺(aq) + A⁻(aq)。酸解离常数 Ka = [H⁺][A⁻] / [HA]。计算 [H⁺] 时常假设解离量远小于初始浓度 c，因此平衡时 [HA] ≈ c。由于 [H⁺] = [A⁻] = x，可得 Ka = x² / c，从而 x = √(Ka c)。当 c / Ka > 100 时，该近似合理。然后 pH = –log₁₀(√(Ka c))。也可能需要由测得的 pH 反算 Ka：若已知 pH，则 [H⁺] = 10⁻^pH，Ka = [H⁺]² / (c – [H⁺])。解离度可由 ([H⁺] / c) × 100% 求出。

5. Weak Bases and Kb | 弱碱与碱解离常数 Kb

Weak bases such as ammonia react with water: B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq). The base dissociation constant is Kb = [BH⁺][OH⁻] / [B]. By the same small-dissociation approximation, [OH⁻] = √(Kb c). Then [H⁺] is found via Kw, and pH = –log₁₀(Kw / √(Kb c)). At 25 °C this is equivalent to pOH = –log₁₀(√(Kb c)) and pH = 14.00 – pOH. Many AQA problems involve NH₃ (Kb = 1.8 × 10⁻⁵) or amines. Remember that for a conjugate acid-base pair, Ka × Kb = Kw, so you can convert between Ka of the conjugate acid and Kb of the base.

氨等弱碱与水反应：B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)。碱解离常数 Kb = [BH⁺][OH⁻] / [B]。采用同样的低解离近似，[OH⁻] = √(Kb c)。接着通过 Kw 求出 [H⁺]，pH = –log₁₀(Kw / √(Kb c))。在 25 °C 下，等同于 pOH = –log₁₀(√(Kb c)) 且 pH = 14.00 – pOH。许多 AQA 题目涉及 NH₃（Kb = 1.8 × 10⁻⁵）或胺类。需牢记共轭酸碱对满足 Ka × Kb = Kw，因此可在酸的 Ka 与共轭碱的 Kb 之间灵活转换。

6. pKa, pKb and pKw | pKa, pKb 和 pKw 概念

The ‘p’ notation means the negative logarithm to base 10: pKa = –log₁₀Ka, pKb = –log₁₀Kb, and pKw = –log₁₀Kw. At 25 °C, pKw = 14.00. Because Ka × Kb = Kw, taking negative logs gives pKa + pKb = pKw. This relationship is extremely useful for finding the strength of a conjugate base if you know the Ka of the acid, and vice versa. For example, if ethanoic acid has pKa = 4.76, then the pKb of its conjugate base, the ethanoate ion, is 14.00 – 4.76 = 9.24. pKa and pKb values directly indicate acid/base strength: the smaller the pKa, the stronger the acid; the smaller the pKb, the stronger the base.

‘p’ 标记表示以 10 为底的负对数：pKa = –log₁₀Ka，pKb = –log₁₀Kb，pKw = –log₁₀Kw。25 °C 时，pKw = 14.00。由 Ka × Kb = Kw 两端取负对数可得 pKa + pKb = pKw。这一关系在已知酸 Ka 求其共轭碱强度时非常有用，反之亦然。例如，乙酸的 pKa = 4.76，则其共轭碱乙酸根的 pKb = 14.00 – 4.76 = 9.24。pKa 和 pKb 数值直接反映酸碱强弱：pKa 越小酸性越强，pKb 越小碱性越强。

7. Buffer Solutions and Henderson-Hasselbalch Equation | 缓冲溶液与 Henderson-Hasselbalch 方程

A buffer solution resists changes in pH upon the addition of small amounts of acid or base. Acidic buffers consist of a weak acid and its conjugate base (usually added as a salt), while basic buffers contain a weak base and its conjugate acid. The pH of an acidic buffer is given by the Henderson-Hasselbalch equation: pH = pKa + log₁₀([A⁻] / [HA]), where [A⁻] is the concentration of the conjugate base and [HA] is the concentration of the weak acid. This equation assumes that the volume of the solution is the same for both components (so the ratio of moles equals the ratio of concentrations). When small amounts of H⁺ are added, they react with A⁻ to form HA; when OH⁻ is added, it reacts with HA to form A⁻, both with minimal pH change. You must be able to calculate the pH of a buffer prepared from given masses or volumes, and to predict the pH after adding a known quantity of strong acid or base.

缓冲溶液能够抵抗少量外加酸或碱引起的 pH 变化。酸性缓冲液由弱酸及其共轭碱（常以盐的形式加入）组成，碱性缓冲液由弱碱及其共轭酸组成。酸性缓冲液的 pH 可用 Henderson-Hasselbalch 方程计算：pH = pKa + log₁₀([A⁻] / [HA])，其中 [A⁻] 是共轭碱浓度，[HA] 是弱酸浓度。该方程假设两种组分处于同一总体积中，因此摩尔比等于浓度比。加入少量 H⁺ 时，H⁺ 与 A⁻ 反应生成 HA；加入 OH⁻ 时，OH⁻ 与 HA 反应生成 A⁻，这两种情况下 pH 变化均很小。考生必须能够根据给定的质量或体积计算缓冲液的 pH，并能预测加入已知量强酸或强碱后的 pH。

8. Dilution and Mixing pH Problems | 稀释与混合 pH 问题

Diluting an acidic or alkaline solution changes the concentration of H⁺ or OH⁻, and pH adjusts accordingly. For strong acids, a ten-fold dilution increases pH by 1 unit, provided the solution is not extremely dilute. When the concentration approaches 10⁻⁷ mol dm⁻³, the contribution from water auto-ionisation becomes significant, and pH tends towards 7.00. In mixing problems, first write a balanced neutralisation equation and determine which reactant is in excess. Calculate the excess moles of H⁺ or OH⁻, divide by the total volume to get the concentration, then find pH or pOH. If neither is in excess, the solution is neutral at the equivalence point (for a strong acid-strong base mixing) and pH = 7.00 at 25 °C.

稀释酸性或碱性溶液会改变 H⁺ 或 OH⁻ 浓度，pH 随之变化。对于强酸，每稀释十倍 pH 增大 1 个单位，前提是溶液并非极稀。当浓度接近 10⁻⁷ mol dm⁻³ 时，水的自耦电离贡献不可忽略，pH 逐渐趋近 7.00。在混合问题中，首先写出平衡的中和方程式，判断何种反应物过量。计算过量的 H⁺ 或 OH⁻ 的物质的量，除以总体积得到浓度，再求 pH 或 pOH。若两者均无过量，则在等当点时溶液呈中性（强酸强碱混合），25 °C 下 pH = 7.00。

9. Acid-Base Titration pH Curves | 酸碱滴定 pH 曲线

pH titration curves show how pH changes as a base is added to an acid (or vice versa). Key features AQA expects you to recognise and sketch are:

Strong acid – strong base: starts low, relatively flat before the vertical jump. The equivalence point is at pH 7, and the vertical section spans roughly pH 3–11.
Strong acid – weak base: starts low, equivalence point below pH 7 (around 5), vertical jump narrower and in the acidic region.
Weak acid – strong base: starts at a higher pH than strong acid, has a buffer region before the jump, equivalence point above pH 7 (around 9), vertical jump in basic region.
Weak acid – weak base: very gradual change, no sharp vertical section, equivalence point near 7 but not useful for indicator choice.

You must be able to calculate the pH at the half-equivalence point (where pH = pKa for a weak acid titration) and identify the buffer region.

pH 滴定曲线展示了向酸中加碱（或反之）时 pH 的变化。AQA 要求能识别并草绘以下关键特征：

强酸–强碱：起点低，跃迁前较为平坦。等当点在 pH 7，垂直段范围约 pH 3–11。
强酸–弱碱：起点低，等当点低于 7（约 pH 5），垂直跃迁范围较窄且位于酸性区域。
弱酸–强碱：起始 pH 较强酸高，跃迁前存在缓冲区域，等当点高于 7（约 pH 9），垂直跃迁位于碱性区域。
弱酸–弱碱：变化十分平缓，无陡峭垂直段，等当点接近 7 但不适于指示剂选择。

你需要能够计算半等当点的 pH（弱酸滴定时 pH = pKa）并识别缓冲区域。

10. Choosing Indicators for Titrations | 滴定中指示剂的选用

Acid-base indicators are weak acids or bases whose conjugate forms have different colours. The indicator changes colour over a range of approximately pH = pKa(indicator) ± 1. An indicator is suitable for a titration if its colour change range falls entirely or at least partially within the steep vertical part of the titration curve. For a strong acid – strong base titration, both methyl orange (range 3.1–4.4) and phenolphthalein (8.3–10.0) work well. For a weak acid – strong base titration, phenolphthalein is appropriate; for a strong acid – weak base, methyl orange is suitable. A weak acid – weak base titration has no sharp pH change, so a single indicator cannot be used – instead, a pH meter would be required.

酸碱指示剂本身是弱酸或弱碱，其共轭型体具有不同颜色。指示剂的变色范围约为 pH = pKa(指示剂) ± 1。若指示剂的变色区间完全或部分落在滴定曲线陡峭垂直段内，则该指示剂适用于该滴定。强酸–强碱滴定中，甲基橙（范围 3.1–4.4）和酚酞（8.3–10.0）均可使用。弱酸–强碱滴定适用酚酞；强酸–弱碱滴定适用甲基橙。弱酸–弱碱滴定无明显的 pH 突跃，因此不能使用单一指示剂，需借助 pH 计来确定终点。

11. Effect of Temperature on Kw and Neutral pH | 温度对 Kw 与中性 pH 的影响

The self-ionisation of water is endothermic: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) ΔH > 0. According to Le Chatelier’s principle, increasing the temperature shifts the equilibrium to the right, so Kw increases. For example, at 40 °C Kw ≈ 2.92 × 10⁻¹⁴, giving neutral [H⁺] = √Kw ≈ 1.71 × 10⁻⁷ mol dm⁻³ and pH ≈ 6.77. Thus, neutral pH is only 7.00 at 25 °C. At higher temperatures, the neutral point moves to a lower pH. Importantly, the solution is still neutral because [H⁺] = [OH⁻]. You must be aware that pKw varies with temperature, and that pH alone cannot indicate neutrality unless the temperature is known.

水的自耦电离为吸热过程：H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) ΔH > 0。根据勒夏特列原理，升高温度使平衡向右移动，Kw 增大。例如，40 °C 时 Kw ≈ 2.92 × 10⁻¹⁴，中性时的 [H⁺] = √Kw ≈ 1.71 × 10⁻⁷ mol dm⁻³，pH ≈ 6.77。由此可见，中性 pH 仅在 25 °C 时为 7.00。温度升高时，中性点向低 pH 方向移动。重要的是，此时溶液仍为中性，因为 [H⁺] = [OH⁻]。必须清楚 pKw 随温度变化，仅凭 pH 值无法判断中性，除非已知温度。

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pH Calculations for AQA A-Level Chemistry | A-Level AQA 化学：pH计算 考点精讲