Mastering pH Calculations for IB AQA Chemistry | IB AQA 化学：pH计算考点精讲

📚 Mastering pH Calculations for IB AQA Chemistry | IB AQA 化学：pH计算考点精讲

pH calculations form the backbone of acid–base chemistry in the IB AQA specification. From strong acids to buffers, mastering these numerical problems is essential for achieving top grades. This guide breaks down every key topic, provides step-by-step methods, and highlights common pitfalls.

pH 计算是 IB AQA 化学课程中酸碱化学的核心。从强酸到缓冲溶液，掌握这些数值计算对于取得高分至关重要。本文分解每一个关键考点，提供逐步解题方法，并指出常见易错点。

1. Understanding pH and the Ionic Product of Water | 理解 pH 与水的离子积常数

pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration: pH = −log₁₀[H⁺]. Similarly, pOH = −log₁₀[OH⁻]. In aqueous solutions at 25 °C, water self-ionises: H₂O ⇌ H⁺ + OH⁻, and the ionic product K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶. This relationship underpins all pH calculations.

pH 定义为氢离子浓度的负对数（底数为 10）：pH = −log₁₀[H⁺]。类似地，pOH = −log₁₀[OH⁻]。在 25 °C 的水溶液中，水发生自耦电离：H₂O ⇌ H⁺ + OH⁻，离子积常数 K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶。这一关系是所有 pH 计算的基础。

For pure water at 25 °C, [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³, giving pH = 7. As temperature changes, K_w increases, altering the neutral pH, but pH + pOH = 14 still holds at 25 °C in dilute solutions.

在 25 °C 的纯水中，[H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³，因此 pH = 7。当温度改变时，K_w 增大，中性 pH 值发生变化，但在 25 °C 的稀溶液中，pH + pOH = 14 这一关系仍然成立。

2. pH of Strong Acids and Strong Bases | 强酸和强碱的 pH 计算

Strong acids such as HCl, HNO₃, and H₂SO₄ fully dissociate in water. For a monoprotic strong acid of concentration c, [H⁺] = c, so pH = −log₁₀(c). For example, 0.050 mol dm⁻³ HCl gives pH = −log₁₀(0.050) ≈ 1.30.

强酸如 HCl、HNO₃ 和 H₂SO₄ 在水中完全电离。对于浓度为 c 的一元强酸，[H⁺] = c，因此 pH = −log₁₀(c)。例如，0.050 mol dm⁻³ HCl 的 pH = −log₁₀(0.050) ≈ 1.30。

For diprotic strong acids like H₂SO₄, the first dissociation is complete, but the second (HSO₄⁻ ⇌ H⁺ + SO₄²⁻) is partial; however, at A-level, we often assume it is also strong in dilute solutions, giving [H⁺] ≈ 2c. For example, 0.010 mol dm⁻³ H₂SO₄ gives [H⁺] = 0.020 mol dm⁻³, pH ≈ 1.70. Always check the question’s assumption.

对于二元强酸如 H₂SO₄，第一步电离是完全的，而第二步（HSO₄⁻ ⇌ H⁺ + SO₄²⁻）是部分的；但在 A-level 阶段，我们通常假定在稀溶液中它也是强电离的，因此 [H⁺] ≈ 2c。例如，0.010 mol dm⁻³ H₂SO₄ 的 [H⁺] = 0.020 mol dm⁻³，pH ≈ 1.70。解题时务必注意题目给出的假设。

Strong bases like NaOH and KOH fully dissociate, providing [OH⁻] = c. Calculate pOH = −log₁₀[OH⁻], then pH = 14 − pOH (at 25 °C). For 0.200 mol dm⁻³ NaOH, pOH = −log₁₀(0.200) ≈ 0.699, so pH = 13.30.

强碱如 NaOH 和 KOH 完全电离，提供 [OH⁻] = c。先求 pOH = −log₁₀[OH⁻]，再用 pH = 14 − pOH（25 °C 时）计算。0.200 mol dm⁻³ NaOH 溶液的 pOH = −log₁₀(0.200) ≈ 0.699，因此 pH = 13.30。

3. pH of Weak Acids: Using Kₐ | 弱酸的 pH 计算：使用 Kₐ

Weak acids partially dissociate: HA ⇌ H⁺ + A⁻. The acid dissociation constant Kₐ = [H⁺][A⁻]/[HA]. For a monoprotic weak acid, if the initial concentration is c and the degree of dissociation is small, [H⁺] ≈ √(Kₐ × c). This approximation holds when c / Kₐ > 100.

弱酸部分电离：HA ⇌ H⁺ + A⁻。酸解离常数 Kₐ = [H⁺][A⁻]/[HA]。对于一元弱酸，若初始浓度为 c 且解离度很小，则 [H⁺] ≈ √(Kₐ × c)。当 c / Kₐ > 100 时，此近似公式成立。

For example, ethanoic acid (CH₃COOH) has Kₐ = 1.8 × 10⁻⁵ mol dm⁻³. To find the pH of 0.100 mol dm⁻³ solution: [H⁺] = √(1.8×10⁻⁵ × 0.100) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ mol dm⁻³, pH = −log₁₀(1.34×10⁻³) ≈ 2.87. Always check the approximation: (1.34×10⁻³ / 0.100) × 100% = 1.34%, which is less than 5%, so the approximation is valid.

例如，乙酸（CH₃COOH）的 Kₐ = 1.8 × 10⁻⁵ mol dm⁻³。计算 0.100 mol dm⁻³ 溶液的 pH：[H⁺] = √(1.8×10⁻⁵ × 0.100) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ mol dm⁻³，pH = −log₁₀(1.34×10⁻³) ≈ 2.87。务必检验近似条件：(1.34×10⁻³ / 0.100) × 100% = 1.34%，小于 5%，因此近似有效。

If the approximation is invalid, solve the quadratic equation: Kₐ = x²/(c − x). Rearrange to x² + Kₐx − Kₐc = 0, then find [H⁺] = x using the quadratic formula.

若近似不成立，需解二次方程：Kₐ = x²/(c − x)。整理得 x² + Kₐx − Kₐc = 0，然后用求根公式计算出 [H⁺] = x。

4. pH of Weak Bases and K_b | 弱碱的 pH 与 K_b

Weak bases like ammonia (NH₃) partially accept protons: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. The base dissociation constant K_b = [NH₄⁺][OH⁻]/[NH₃]. Analogous to weak acids, [OH⁻] ≈ √(K_b × c) when the dissociation is small.

像氨（NH₃）这样的弱碱部分接受质子：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。碱解离常数 K_b = [NH₄⁺][OH⁻]/[NH₃]。与弱酸类似，当解离度很小时，[OH⁻] ≈ √(K_b × c)。

For 0.150 mol dm⁻³ NH₃ with K_b = 1.8 × 10⁻⁵: [OH⁻] = √(1.8×10⁻⁵ × 0.150) ≈ 1.64×10⁻³ mol dm⁻³, pOH = 2.79, pH = 14 − 2.79 = 11.21. Remember to convert through pOH.

对于 0.150 mol dm⁻³ NH₃，K_b = 1.8 × 10⁻⁵：[OH⁻] = √(1.8×10⁻⁵ × 0.150) ≈ 1.64×10⁻³ mol dm⁻³，pOH = 2.79，pH = 14 − 2.79 = 11.21。记住要通过 pOH 换算。

The relationship between Kₐ and K_b for a conjugate acid–base pair is Kₐ × K_b = K_w. This is particularly useful when given Kₐ of a weak acid and asked to find the pH of its conjugate base.

共轭酸碱对之间的 Kₐ 和 K_b 满足关系式 Kₐ × K_b = K_w。当题目给出弱酸的 Kₐ 并要求计算其共轭碱的 pH 时，这一关系尤为有用。

5. Converting between pKₐ and Kₐ | pKₐ 与 Kₐ 的相互转换

pKₐ is defined as −log₁₀(Kₐ). A smaller pKₐ means a stronger acid. The conversion is straightforward: Kₐ = 10⁻ᵖᴷᵅ. For example, if pKₐ = 4.75, then Kₐ = 10⁻⁴·⁷⁵ = 1.8 × 10⁻⁵ mol dm⁻³.

pKₐ 定义为 −log₁₀(Kₐ)。pKₐ 越小，酸性越强。转换公式非常简单：Kₐ = 10⁻ᵖᴷᵅ。例如，若 pKₐ = 4.75，则 Kₐ = 10⁻⁴·⁷⁵ = 1.8 × 10⁻⁵ mol dm⁻³。

pK_b and K_b follow the same rules, and pKₐ + pK_b = 14 for conjugate pairs at 25 °C. This is often used to deduce the strength of a conjugate base from a given pKₐ.

pK_b 和 K_b 遵循相同的规则，且对于共轭酸碱对，在 25 °C 时 pKₐ + pK_b = 14。这常被用来从已知的 pKₐ 推断其共轭碱的强度。

6. pH of Buffer Solutions | 缓冲溶液的 pH 计算

A buffer solution resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid). The pH is calculated using the Henderson–Hasselbalch equation: pH = pKₐ + log₁₀([A⁻]/[HA]).

缓冲溶液能在加入少量酸或碱时抵抗 pH 的变化。它由弱酸及其共轭碱（或弱碱及其共轭酸）组成。其 pH 可通过亨德森-哈塞尔巴尔赫方程计算：pH = pKₐ + log₁₀([A⁻]/[HA])。

When the concentrations of the acid and its salt are equal, pH = pKₐ. This is the half-equivalence point in a titration. For example, a buffer containing 0.20 mol dm⁻³ CH₃COOH and 0.10 mol dm⁻³ CH₃COO⁻Na⁺ has pH = 4.75 + log₁₀(0.10/0.20) = 4.75 − 0.30 = 4.45.

当酸和其盐的浓度相等时，pH = pKₐ，这正是滴定中的半等当点。例如，含有 0.20 mol dm⁻³ CH₃COOH 和 0.10 mol dm⁻³ CH₃COO⁻Na⁺ 的缓冲溶液，pH = 4.75 + log₁₀(0.10/0.20) = 4.75 − 0.30 = 4.45。

Buffer problems may involve changes after adding a small amount of strong acid or base. Use stoichiometry to find the new [HA] and [A⁻], then apply the equation. Remember that adding H⁺ converts A⁻ to HA, while adding OH⁻ converts HA to A⁻.

缓冲溶液的题目可能会涉及加入少量强酸或强碱后的变化。先用化学计量法求出新的 [HA] 和 [A⁻]，再代入方程计算。记住加入 H⁺ 会将 A⁻ 转化为 HA，而加入 OH⁻ 会将 HA 转化为 A⁻。

7. Acid–Base Titration Curves and pH Changes | 酸碱滴定曲线与 pH 变化

Plotting pH against volume of titrant added produces a titration curve. For a strong acid–strong base titration, the curve has a steep vertical section around pH 7. The equivalence point is at pH 7, and any indicator with a range within the steep part (e.g., phenolphthalein, methyl orange) is suitable.

以 pH 对加入滴定剂体积作图可得到滴定曲线。强酸-强碱滴定的曲线在 pH 7 附近有一个陡峭的垂直段。等当点位于 pH 7，任何变色范围落在此陡峭部分的指示剂（如酚酞、甲基橙）都适用。

For a weak acid–strong base titration (e.g., CH₃COOH with NaOH), the equivalence point is above pH 7 (typically 8–9) because the conjugate base hydrolyses water to produce OH⁻. The curve has an initial steep rise then a gradual buffer region, followed by a sharp rise. Phenolphthalein (range pH 8.3–10.0) is ideal.

对于弱酸-强碱滴定（如 CH₃COOH 用 NaOH 滴定），等当点位于 pH > 7（通常 8–9），因为共轭碱水解产生 OH⁻。曲线初始上升较快，随后进入逐渐变化的缓冲区域，最后陡升。酚酞（变色范围 pH 8.3–10.0）是理想指示剂。

Strong acid–weak base titrations have an equivalence point below pH 7. Methyl orange (range 3.1–4.4) works well. The half-equivalence point on a weak acid curve gives pH = pKₐ of the acid, a key method for determining Kₐ experimentally.

强酸-弱碱滴定的等当点位于 pH < 7。甲基橙（变色范围 3.1–4.4）非常适用。在弱酸滴定曲线上，半等当点处 pH = pKₐ，这是实验测定 Kₐ 的关键方法。

8. Indicator Selection and pH Range | 指示剂的选择与 pH 范围

Indicators are weak acids themselves, with a characteristic pKₐ. The colour change occurs roughly over pH = pKₐ ± 1. You must choose an indicator whose colour change interval lies entirely within the steepest part of the titration curve.

指示剂本身就是弱酸，具有特征的 pKₐ。颜色变化大约发生在 pH = pKₐ ± 1 的范围内。必须选择变色区间完全落在滴定曲线最陡峭部分的指示剂。

Common indicators: phenolphthalein (pKₐ ≈ 9.4, colourless in acid, pink in base), methyl orange (pKₐ ≈ 3.5, red in acid, yellow in base), and bromothymol blue (pKₐ ≈ 7.0, yellow in acid, blue in base). For titrations involving weak species, avoid indicators that change colour far from the equivalence point.

常用指示剂：酚酞（pKₐ ≈ 9.4，酸中无色，碱中粉红）、甲基橙（pKₐ ≈ 3.5，酸中红色，碱中黄色）、溴百里酚蓝（pKₐ ≈ 7.0，酸中黄色，碱中蓝色）。对于涉及弱酸或弱碱的滴定，避免选用变色点远离等当点的指示剂。

9. Effect of Dilution on pH | 稀释对 pH 的影响

Diluting a strong acid by a factor of 10 increases the pH by 1 unit, because [H⁺] drops by the same factor. However, for weak acids, dilution increases the degree of dissociation, so the change in pH is less than 1 unit. Calculations should use the Kₐ expression with the new concentration.

将强酸稀释 10 倍，pH 增加 1 个单位，因为 [H⁺] 同比例下降。然而对于弱酸，稀释会增大解离度，因此 pH 变化小于 1 个单位。计算时需用新浓度代入 Kₐ 表达式求解。

For extremely dilute strong acids (c < 10⁻⁶ mol dm⁻³), the autoionisation of water cannot be ignored. [H⁺] = c + 10⁻⁷ must be considered, and pH will approach but not exceed 7 when diluting an acid.

对于极稀的强酸（c < 10⁻⁶ mol dm⁻³），水的自耦电离不可忽略。需考虑 [H⁺] = c + 10⁻⁷，此时稀释酸液时 pH 会接近但不会超过 7。

10. Mixtures and Multi-Step pH Calculations | 混合溶液与多步计算

When mixing an acid and a base, first determine the limiting reactant and the resulting species. If an excess of strong acid remains, calculate [H⁺] from the leftover concentration. If strong base is in excess, calculate [OH⁻] and then pOH and pH.

将酸与碱混合时，首先确定限制反应物及生成物。若强酸过量，则根据剩余浓度计算 [H⁺]。若强碱过量，则计算 [OH⁻]，再求 pOH 和 pH。

For a mixture of weak acid and strong base at the equivalence point, the solution contains only the conjugate base; use its K_b to find [OH⁻] and then pH. Multi-step problems may involve back-titrations or indirect measurements – always write balanced equations and track moles.

对于弱酸与强碱在等当点处的混合液，溶液中仅含共轭碱；需用其 K_b 求出 [OH⁻] 再算 pH。多步问题可能涉及返滴定或间接测量——务必先写出配平的化学方程式并追踪物质的量。

11. Common Mistakes and Checkpoints | 常见错误与检查点

Forgetting to convert between pOH and pH: always check whether you have [H⁺] or [OH⁻] before taking the log.

忘记 pOH 与 pH 之间的转换：在取对数之前，务必检查你得到的是 [H⁺] 还是 [OH⁻]。
Using the Kₐ approximation blindly: always verify that c / Kₐ > 100, or check the percent dissociation.

盲目使用 Kₐ 近似公式：务必验证 c / Kₐ > 100，或检查解离百分数。
Ignoring the autoionisation of water in very dilute solutions: if calculated pH exceeds 7 for an acid, you’ve likely made this mistake.

忽略极稀溶液中水的自耦电离：若算得的酸溶液 pH 超过 7，很可能就是这个错误。
Miscalculating buffer pH after adding H⁺/OH⁻: use stoichiometry to adjust [HA] and [A⁻], not just the added volume.

加入 H⁺/OH⁻ 后缓冲溶液 pH 计算错误：应用化学计量法调整 [HA] 和 [A⁻]，而非仅考虑体积变化。
Using the wrong indicator: an indicator must change colour entirely on the steep portion of the curve.

选错指示剂：指示剂必须在曲线的陡峭段完全变色。

12. Exam Tips Summary | 考试技巧总结

Always write the expression for Kₐ or K_b before substituting numbers. Show clearly when you are using the approximation, and state the assumption. Include units for all constants, and round final pH values to two decimal places. Finally, check if your answer makes chemical sense – an acid should have pH < 7, a base pH > 7 at 25 °C.

代入数字前，务必先写出 Kₐ 或 K_b 的表达式。明确说明何时使用近似处理，并陈述相关假设。所有常数均需注明单位，最终 pH 值保留两位小数。最后，检查答案在化学上是否合理——25 °C 时酸溶液的 pH 应小于 7，碱溶液则应大于 7。

With consistent practice, pH calculations become a reliable source of marks. Master these core methods, and you will approach any IB AQA chemistry paper with confidence.

通过持续练习，pH 计算将成为稳固的得分点。掌握这些核心方法，你将能自信应对任何 IB AQA 化学试卷。

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Mastering pH Calculations for IB AQA Chemistry | IB AQA 化学：pH计算 考点精讲