📚 A-Level OCR Chemistry: Mastering pH Calculations | A-Level OCR 化学：pH计算考点精讲
pH calculations form a cornerstone of the OCR A-Level Chemistry physical chemistry topics, appearing in both AS and A2 units. This article provides a thorough revision guide on every type of pH problem you could encounter — from strong acids and bases to weak acid/base equilibria, buffer solutions, and titration curves. Master these and you will secure valuable marks on your exams.
pH 计算是 OCR A-Level 化学物理化学部分的核心内容，贯穿 AS 和 A2 单元。本文为你全面梳理所有可能出现的 pH 计算类型——从强酸强碱到弱酸弱碱平衡、缓冲溶液、滴定曲线。牢牢掌握这些考点，你将在考试中稳稳拿分。
1. Introduction to pH and the pH Scale | pH 与 pH 标度简介
It is defined as the negative logarithm (base 10) of the hydrogen ion concentration: pH = – H⁺] and a more acidic solution; a higher pH means lower [H⁺] and a more alkaline solution.
pH 标度用来衡量水溶液的酸碱度。其定义为氢离子浓度的负对数（以10为底）：pH = –log₁₀[H⁺]。在 25 °C 的纯水中，[H⁺] = 1.0 × 10⁻⁷ mol dm⁻³，因此 pH = 7。pH 越小，[H⁺] 越大，溶液酸性越强；pH 越大，[H⁺] 越小，溶液碱性越强。
You must also be able to convert from pH back to [H⁺] using [H⁺] = 10⁻ᵖᴴ r that pH is logarithmic: a change of one pH unit corresponds to a tenfold change in [H⁺].
你还需掌握从 pH 倒推 [H⁺] 的公式：[H⁺] = 10⁻ᵖᴴ。例如，pH = 3.0 的溶液 [H⁺] = 1.0 × 10⁻³ mol dm⁻³。注意 pH 是对数标度：pH 每改变一个单位，[H⁺] 就变化十倍。
2. Calculating pH of Strong Acids | 强酸 pH 计算
ng acid, [H⁺] equals the acid concentration (provided the acid is not too dilute for water's autoionisation to interfere). So for 0.100 mol dm⁻³ HCl, [H⁺] = 0.100 mol dm⁻³, and pH = –log(0.100) = 1.00.
像 HCl、HNO₃ 和 H₂SO₄ 这样的强酸在水中完全电离。对于一元强酸，[H⁺] 等于酸的浓度（前提是酸不是太稀，以免水的自耦电离干扰）。因此 0.100 mol dm⁻³ HCl 的 [H⁺] = 0.100 mol dm⁻³，pH = –log(0.100) = 1.00。
at moderate concentrations. Thus 0.100 mol dm⁻³ H₂SO₄ gives [H⁺] = 0.200 mol dm⁻³, so pH = – lute solutions (below 10⁻⁶ mol dm⁻³) require consideration of the contribution from water.
对二元强酸如 H₂SO₄，第一个质子完全电离，第二个质子在中等浓度下也完全电离。因此 0.100 mol dm⁻³ H₂SO₄ 产生的 [H⁺] = 0.200 mol dm⁻³，pH = –log(0.200) = 0.70。但对于极稀溶液（低于 10⁻⁶ mol dm⁻³），必须考虑水的贡献。
3. Calculating pH of Strong Bases | 强碱 pH 计算
the ionic product of water, Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C, to find [H⁺] = Kw / [OH⁻]. Finally apply pH = –log[H⁺].
像 NaOH 和 KOH 这样的强碱完全电离产生 OH⁻ 离子。要计算 pH，首先由碱的浓度求出 [OH⁻]，然后利用水的离子积常数 Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶（25 °C），求出 [H⁺] = Kw / [OH⁻]，最后应用 pH = –log[H⁺]。
NaOH gives [OH⁻] = 0.050 mol dm⁻³. [H⁺] = 1.0×10⁻¹⁴ / 0.050 = 2.0×10⁻¹³ mol dm⁻³, so pH = –log(2.0×10⁻¹³) ≈ 12.70. Alternatively, you can calculate pOH = –log[OH⁻] = –log(0.050) = 1.30, then use pH = 14.00 – pOH = 12.70.
例如：0.050 mol dm⁻³ NaOH 中 [OH⁻] = 0.050 mol dm⁻³。[H⁺] = 1.0×10⁻¹⁴ / 0.050 = 2.0×10⁻¹³ mol dm⁻³，因此 pH = –log(2.0×10⁻¹³) ≈ 12.70。或者先计算 pOH = –log[OH⁻] = –log(0.050) = 1.30，再用 pH = 14.00 – pOH = 12.70。
4. The Ionic Product of Water, Kw | 水的离子积常数 Kw
Water undergoes autoprotolysis: 2H₂O ⇌ eases, Kw increases, meaning the pH of pure water decreases (becomes less than 7) even though the water remains neutral because [H⁺] = [OH⁻].
水存在自耦电离平衡：2H₂O ⇌ H₃O⁺ + OH⁻。该平衡的平衡常数即为 Kw = [H⁺][OH⁻]。25 °C 时 Kw = 1.0 × 10⁻¹⁴。Kw 的数值受温度影响：温度升高，Kw 增大，这意味着纯水的 pH 会降低（小于7），但由于 [H⁺] = [OH⁻]，水仍然呈中性。
H + pOH = 14.00 at 25 °C. It is essential for connecting strong base concentrations to pH.
计算时请留意题目给出的温度。若无特别说明，则假定为 25 °C。Kw 把 pH 和 pOH 联系起来：25 °C 时 pH + pOH = 14.00。这是连接强碱浓度与 pH 的关键。
5. Calculating pH of Weak Acids | 弱酸 pH 计算
ak acids such as ethanoic acid (CH₃COOH) only partially dissociate. The equilibrium is: HA ⇌ one, [H⁺] = [A⁻], and if dissociation is small, [HA]≈ initial concentration C. Then [H⁺] ≈ √Ka×C.
弱酸如乙酸 (CH₃COOH) 仅部分电离。平衡为：HA ⇌ H⁺ + A⁻。酸的解离常数 Ka = [H⁺][A⁻]/[HA]。对于单纯的弱酸溶液，[H⁺] = [A⁻]，若电离度很小，[HA] ≈ 初始浓度 C，则 [H⁺] ≈ √Ka×C。
Example: For 0.10 mol dm⁻³ CH₃COOH with Ka = 1.7 × 10⁻⁵ mol dm⁻³, [H⁺] = √1.7×10⁻⁵ × 0.10 = √1.7×10⁻⁶ = 1.3 × 10⁻³ mol dm⁻³, giving pH = –log(1.3×10⁻³) ≈ 2.89. Always check the percent dissociation; if more than 5%, you may need to solve a quadratic, though OCR tends to avoid this at A-level.
例如：对于 0.10 mol dm⁻³ CH₃COOH，Ka = 1.7 × 10⁻⁵ mol dm⁻³，[H⁺] = √1.7×10⁻⁵ × 0.10 = √1.7×10⁻⁶ = 1.3 × 10⁻³ mol dm⁻³，pH = –log(1.3×10⁻³) ≈ 2.89。需检查电离度百分比；若大于5%，原则上应求解二次方程，但 OCR A-level 通常回避这种复杂情况。
6. Calculating pH of Weak Bases | 弱碱 pH 计算
Weak bases like ammonia (NH₃) accept protons. The equilibrium is B + H₂O ⇌ constant Kb = [BH⁺][OH⁻]/[B]. For a weak base alone, [BH⁺] = [OH⁻] and [B] ≈ C, so [OH⁻] ≈ √Kb×C. Then use Kw to find [H⁺] and pH.
弱碱如氨 (NH₃) 可与质子结合。平衡为：B + H₂O ⇌ BH⁺ + OH⁻。碱解离常数 Kb = [BH⁺][OH⁻]/[B]。对于单纯的弱碱溶液，[BH⁺] = [OH⁻]，[B] ≈ C，则 [OH⁻] ≈ √Kb×C，再通过 Kw 求得 [H⁺] 和 pH。
For 0.20 mol dm⁻³ NH₃ (Kb = 1.8 × 10⁻⁵), [OH⁻] = √1.8×10⁻⁵ × 0.20 = √3.6×10⁻⁶ mol dm⁻³. Then [H⁺] = 1.0×10⁻¹⁴ / 1.90×10⁻³ = 5.26×10⁻¹², pH = 11.28. Alternatively, pOH = –log(1.90×10⁻³) = 2.72, pH = 14.00 – 2.72 = 11.28.
例如 0.20 mol dm⁻³ NH₃ (Kb = 1.8 × 10⁻⁵)，[OH⁻] = √1.8×10⁻⁵ × 0.20 = √3.6×10⁻⁶ = 1.90×10⁻³ mol dm⁻³，则 [H⁺] = 1.0×10⁻¹⁴ / 1.90×10⁻³ = 5.26×10⁻¹²，pH = 11.28。或计算 pOH = –log(1.90×10⁻³) = 2.72，pH = 14.00 – 2.72 = 11.28。
7. Buffer Solutions: Acidic Buffers | 缓冲溶液：酸性缓冲液
An acidic buffer is made from a weak acid and its conjugate base (salt). Common examples: CH₃COOH/CH₃COONa. It resists changes in pH upon addition of small amounts of acid or base. The pH can be calculated using the equilibrium expression or the Henderson–Hasselbalch equation.
酸性缓冲液由弱酸及其共轭碱（盐）组成，常见的有 CH₃COOH/CH₃COONa。当加入少量酸或碱时，它能抵抗 pH 的改变。pH 可利用平衡表达式或亨德森-哈塞尔巴赫方程进行计算。
ves [H⁺] = Ka × [HA]/[A⁻]. Taking logs: pH = pKa + log₁₀([A⁻]/[HA]). This is the Henderson– d its salt. For a buffer containing 0.50 mol dm⁻³ CH₃COOH and 0.50 mol dm⁻³ CH₃COONa, pH = –log(1.7×10⁻⁵) + log(0.50/0.50) = 4.77.
由 Ka = [H⁺][A⁻]/[HA] 变形得 [H⁺] = Ka × [HA]/[A⁻]。两边取对数：pH = pKa + log₁₀([A⁻]/[HA])，即亨德森-哈塞尔巴赫方程。需知道弱酸及其盐的浓度。对于含 0.50 mol dm⁻³ CH₃COOH 和 0.50 mol dm⁻³ CH₃COONa 的缓冲液，pH = –log(1.7×10⁻⁵) + log(0.50/0.50) = 4.77。
8. Buffer Solutions: Basic Buffers | 缓冲溶液：碱性缓冲液
e and its conjugate acid (often as a salt). Example: NH₃/NH₄Cl. The equilibrium is B + H₂O ⇌ ]. The pH can be found by first finding pOH from pKb and the ratio, then converting to pH.
碱性缓冲液由弱碱及其共轭酸（常以盐的形式）组成，如 NH₃/NH₄Cl。平衡为 B + H₂O ⇌ BH⁺ + OH⁻，Kb = [BH⁺][OH⁻]/[B]。可先利用 pKb 与浓度比求得 pOH，再转换为 pH。
From Kb: [OH⁻] = Kb × [B]/[BH⁺]; pOH = pKb + log₁₀([BH⁺]/[B]); then pH = 14.00 – pOH. Example: buffer with 0.10 mol dm⁻³ NH₃ and 0.20 mol dm⁻³ NH₄Cl, pKb = –log(1.8×10⁻⁵) = 4.74. pOH = 4.74 + log(0.20/0.10) = 4.74 + 0.30 = 5.04, pH = 14.00 – 5.04 = 8.96.
由 Kb 可得：[OH⁻] = Kb × [B]/[BH⁺]；pOH = pKb + log₁₀([BH⁺]/[B])；则 pH = 14.00 – pOH。例如：含 0.10 mol dm⁻³ NH₃ 和 0.20 mol dm⁻³ NH₄Cl 的缓冲液，pKb = –log(1.8×10⁻⁵) = 4.74，pOH = 4.74 + log(0.20/0.10) = 4.74 + 0.30 = 5.04，pH = 14.00 – 5.04 = 8.96。
9. Buffer Action and Henderson–Hasselbalch | 缓冲作用与亨德森-哈塞尔巴赫方程
relatively constant. Added OH⁻ reacts with HA to form A⁻ and water. Buffers work best when [HA] ≈ [A⁻], i.e., pH ≈ pKa, and when concentrations are high compared to the added acid/base.
当向酸性缓冲液中加入少量 H⁺ 时，A⁻ 离子与 H⁺ 反应生成 HA，[HA] 略有增加，[A⁻] 略有减少，比值和 pH 保持相对稳定。加入 OH⁻ 则会与 HA 反应生成 A⁻ 和水。缓冲液在 [HA] ≈ [A⁻]、即 pH ≈ pKa 时，且浓度远大于外加酸碱时，缓冲效果最佳。
The Henderson– ining 0.50 mol dm⁻³ HA and 0.50 mol dm⁻³ A⁻ changes [HA] to 0.51, [A⁻] to 0.49. New pH = pKa + log(0.49/0.51) ≈ pKa – 0.02, a tiny change.
亨德森-哈塞尔巴赫方程也可用于计算少量加入后的 pH 变化。例如，向含 0.50 mol dm⁻³ HA 和 0.50 mol dm⁻³ A⁻ 的 1 dm³ 缓冲液中加入 0.010 mol HCl，[HA] 变为 0.51，[A⁻] 变为 0.49。新 pH = pKa + log(0.49/0.51) ≈ pKa – 0.02，变化极小。
10. pH Curves and Indicators | pH 曲线与指示剂
pH titration curves show how pH changes during an acid–base titration. Shapes differ for strong acid–strong base, strong acid–weak base, weak acid–strong base, and weak acid–weak base. The equivalence point occurs when moles of acid equal moles of base. You must be able to sketch and interpret these curves.
pH 滴定曲线表示酸碱滴定过程中 pH 的变化。强酸-强碱、强酸-弱碱、弱酸-强碱和弱酸-弱碱的曲线形状各不相同。等当点在酸和碱的摩尔数相等时到达。你必须能够绘制和解读这些曲线。
Key features: strong acid–strong base has an equivalence point at pH 7 with a very steep vertical section. Weak acid–strong base equivalence point is >7; strong acid–weak base equivalence point is <7. The vertical region where pH changes rapidly is used to select a suitable indicator, whose pKin should lie within this range.
关键特征：强酸-强碱滴定等当点在 pH 7 处，且有非常陡峭的垂直段。弱酸-强碱的等当点 >7；强酸-弱碱的等当点 <7。pH 急剧变化的垂直区域可用于选择适当的指示剂，其 pKin 应落在此范围内。
Common indicators: phenolphthalein (pH range 8.2–10.0) suitable for strong base titrations; methyl orange (3.1–4.4) suitable for strong acid titrations. A weak acid–weak base titration does not have a sharp change and is unsuitable for simple indicators.
常用指示剂：酚酞（pH 范围 8.2–10.0）适用于强碱滴定；甲基橙（3.1–4.4）适用于强酸滴定。弱酸-弱碱滴定没有敏锐的 pH 突跃，不适合用普通指示剂。
11. pH of Dilution and Mixing | 稀释与混合的 pH 计算
When a strong acid is diluted, the pH increases because [H⁺] decreases. For a 10-fold dilution of a strong acid, the pH rises by 1 unit. For weak acids, dilution also increases the degree of dissociation, so the pH change is less than predicted by simple dilution. You may be asked to calculate the pH after mixing an acid and an alkali, including excess reactants.
稀释强酸时，pH 上升，因为 [H⁺] 减小。强酸稀释 10 倍，pH 增加 1 个单位。对弱酸而言，稀释还会增大电离度，因此 pH 的变化小于简单稀释所预计的。题目可能要求计算酸碱混合后的 pH，包括有过量反应物的情况。
ompute pH. For example, mixing 50 cm³ of 0.10 mol dm⁻³ HCl with 50 cm³ of 0.080 mol dm⁻³ NaOH: moles H⁺ = 0.0050, OH⁻ = 0.0040, excess H⁺ = 0.0010 mol in 100 cm³, [H⁺] = 0.010 mol dm⁻³, pH = 2.00.
混合时，首先求出 H⁺ 和 OH⁻ 的摩尔数，判断何者过量，计算过量离子在总体积中的浓度，再求 pH。例如，将 50 cm³ 0.10 mol dm⁻³ HCl 与 50 cm³ 0.080 mol dm⁻³ NaOH 混合：H⁺ 摩尔 = 0.0050，OH⁻ 摩尔 = 0.0040，过量 H⁺ = 0.0010 mol 于 100 cm³ 中，[H⁺] = 0.010 mol dm⁻³，pH = 2.00。
For mixing weak acids and strong bases, if the base is in excess, the resulting solution may be a buffer or completely neutralised; careful stoichiometric reasoning is needed.
混合弱酸与强碱时，若碱过量，最终可能是缓冲液或完全中和，需要进行细致的化学计量分析。
12. Summary and Exam Tips | 总结与考试技巧
- Write down the knowns: [acid], [base], Ka, Kb, Kw. Check temperature.
- 写出已知量： [酸]、[碱]、Ka、Kb、Kw。注意温度。
- Identify the system: Strong? Weak? Buffer? Titration mixture?
- 识别体系： 强？弱？缓冲？滴定混合液？
- Use the appropriate formula: Don’t mix up Ka and Kb; remember pH + pOH = 14.00.
- 使用正确的公式： 不要混淆 Ka 和 Kb；记住 pH + pOH = 14.00。
- Show units: [H⁺] in mol dm⁻³; Kw in mol² dm⁻⁶.
- 带单位： [H⁺] 的单位为 mol dm⁻³；Kw 的单位为 mol² dm⁻⁶。
- Check assumptions: For weak acids, dissociation < 5% justifies [HA] ≈ C.
- 检查假设： 对于弱酸，若电离度 < 5%，可认为 [HA] ≈ C。
- Buffers: pH = pKa + log([salt]/[acid]). Learn the derivation.
- 缓冲液： pH = pKa + log([盐]/[酸])。掌握推导过程。
- Titration curves: Know the four shapes and appropriate indicators.
- 滴定曲线： 掌握四种形状及相应指示剂。
- Dilution and mixing: Use mol calculations, not just C1V1.
- 稀释与混合： 用摩尔计算，不要仅用 C1V1。
Practice with past paper questions, paying close attention to the command words and mark schemes. OCR exam questions often require a multi-step approach, so set out your working logically.
勤做历年真题，仔细审题并研究评分标准。OCR 考题常需多步计算，所以作答时务必逻辑清晰、步骤完整。
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