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

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

Mastering pH calculations is a cornerstone of success in both IB and Edexcel Chemistry. This guide consolidates the essential concepts, from the definition of pH and Kw to the more challenging buffer calculations and titration curves. By understanding the underlying principles and practising the key problem types, you will approach any pH question with confidence.

掌握pH计算是IB和Edexcel化学取得高分的关键。本指南系统梳理了从pH定义、水的离子积到较难的缓冲溶液计算和滴定曲线等核心考点。无论是强酸强碱的直接计算，还是弱酸弱碱的近似求解，理解原理并熟练运用公式，你就能轻松应对任何pH相关的考题。

1. Defining pH, pOH and Kw | pH、pOH 与 Kw 的定义

pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration.

pH 定义为氢离子浓度的负对数（以10为底）。

pH = –log₁₀[H⁺]

Similarly, pOH is defined as the negative logarithm of the hydroxide ion concentration.

同样地，pOH 定义为氢氧根离子浓度的负对数。

pOH = –log₁₀[OH⁻]

In any aqueous solution at 25 °C, the ion product of water, Kw, is always 1.0 × 10⁻¹⁴ mol² dm⁻⁶.

在25 °C的任何水溶液中，水的离子积 Kw 恒为 1.0 × 10⁻¹⁴ mol² dm⁻⁶。

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴

Taking the negative logarithm of this expression yields a fundamental relationship: pH + pOH = 14.00 at 25 °C.

对该式两边取负对数可得重要关系式：25 °C 时，pH + pOH = 14.00。

2. Strong Acids and Strong Bases | 强酸与强碱的计算

Strong acids, such as HCl, HNO₃ and H₂SO₄ (first proton), dissociate completely in water. Therefore, the concentration of H⁺ ions equals the initial concentration of the acid, after accounting for basicity.

强酸（如 HCl、HNO₃ 和 H₂SO₄ 的第一级电离）在水中完全解离。因此，H⁺ 离子浓度等于酸的初始浓度乘以酸分子能提供的质子数。

For a monoprotic strong acid of concentration c, [H⁺] = c, and pH = –log₁₀(c).

对于浓度为 c 的一元强酸，[H⁺] = c，pH 直接通过 pH = –log₁₀(c) 求得。

Strong bases, such as NaOH and KOH, dissociate fully, releasing one mole of OH⁻ per mole of base. The [OH⁻] equals the base concentration, pOH is calculated, and pH is found via pH = 14 – pOH.

强碱（如 NaOH 和 KOH）完全解离，每摩尔碱释放一摩尔 OH⁻。[OH⁻] 等于碱的浓度，可先计算 pOH，再利用 pH = 14 – pOH 换算。

For Group 2 metal hydroxides like Ba(OH)₂, two OH⁻ ions are produced per formula unit, so [OH⁻] = 2 × c (where c is the concentration of Ba(OH)₂).

对于 Ba(OH)₂ 等第II族氢氧化物，每个分子提供两个 OH⁻，因此 [OH⁻] = 2 × c，再计算 pOH 和 pH。

3. Weak Acids and the Acid Dissociation Constant Ka | 弱酸与酸解离常数 Ka

Weak acids only partially dissociate in water, establishing an equilibrium described by the acid dissociation constant, Ka.

弱酸在水中仅部分解离，建立了由酸解离常数 Ka 描述的平衡。

For a generic weak acid HA: HA ⇌ H⁺ + A⁻, the expression is:

对于一般弱酸 HA：HA ⇌ H⁺ + A⁻，其表达式为：

Ka = [H⁺][A⁻] / [HA]

At equilibrium, [H⁺] = [A⁻] and the concentration of undissociated HA is approximately equal to the initial concentration c, provided the acid is weak and the dissociation is negligible (c/Ka > 100). This leads to the approximation:

在平衡时，[H⁺] = [A⁻]，且未解离的 HA 浓度约等于初始浓度 c，前提是酸很弱且解离度极小（通常判断标准 c/Ka > 100）。由此得到近似公式：

[H⁺] = √(Ka × c)

Then pH = –log₁₀[H⁺]. If c/Ka is not sufficiently large, the quadratic formula must be used to solve for [H⁺].

然后计算 pH。若 c/Ka 不够大，则不能使用近似，需解二次方程求精确的 [H⁺]。

4. Weak Bases and the Base Dissociation Constant Kb | 弱碱与碱解离常数 Kb

Weak bases, such as ammonia (NH₃) and amines, react partially with water to produce OH⁻ ions, characterised by Kb.

弱碱（如氨 NH₃ 和胺类）与水部分反应生成 OH⁻，其特征常数是 Kb。

For a weak base B: B + H₂O ⇌ BH⁺ + OH⁻, the equilibrium constant is:

对于弱碱 B：B + H₂O ⇌ BH⁺ + OH⁻，其平衡常数为：

Kb = [BH⁺][OH⁻] / [B]

Analogous to weak acids, the hydroxide concentration can be approximated as:

与弱酸类似，氢氧根浓度可近似为：

[OH⁻] = √(Kb × c)

Then pOH = –log₁₀[OH⁻] and pH = 14 – pOH. The relationship Ka × Kb = Kw for a conjugate acid-base pair is frequently tested.

随后计算 pOH，再换算为 pH。共轭酸碱对满足 Ka × Kb = Kw，这条关系常常成为考点。

5. pH of Salt Solutions | 盐溶液的 pH

Salts formed from a strong acid and a strong base (e.g., NaCl) dissolve to give neutral solutions (pH = 7 at 25 °C).

由强酸和强碱生成的盐（如 NaCl）溶于水呈中性（25 °C 时 pH = 7）。

A salt from a strong acid and a weak base (e.g., NH₄Cl) produces an acidic solution because the ammonium ion hydrolyses water: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺.

强酸与弱碱生成的盐（如 NH₄Cl）因铵根离子水解呈酸性：NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺。

The pH can be calculated using the Ka of the conjugate acid, where Ka(NH₄⁺) = Kw / Kb(NH₃). Then treat the solution as a weak acid.

计算时利用共轭酸的 Ka（Ka = Kw / Kb），然后按弱酸近似公式求 [H⁺] 和 pH。

A salt from a weak acid and a strong base (e.g., CH₃COONa) is basic due to the hydrolysis of the anion: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, using Kb = Kw / Ka.

弱酸与强碱生成的盐（如 CH₃COONa）因阴离子水解呈碱性。计算时先求 Kb = Kw / Ka，再按弱碱方式求 [OH⁻] 和 pH。

6. Buffer Solutions and the Henderson–Hasselbalch Equation | 缓冲溶液与亨德森-哈塞尔巴尔赫方程

A buffer solution resists changes in pH upon the 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.

缓冲溶液能够抵抗少量外加酸碱引起的 pH 变化。它通常由弱酸与其共轭碱，或弱碱与其共轭酸组成。

The pH of an acidic buffer is given by the Henderson–Hasselbalch equation:

酸性缓冲溶液的 pH 可通过亨德森-哈塞尔巴尔赫方程计算：

pH = pKa + log₁₀([A⁻] / [HA])

This equation assumes that the concentrations of the acid and salt at equilibrium are approximately equal to their initial concentrations. Buffer effectiveness is greatest when pH is close to pKa (±1 range).

该式假设平衡时酸和盐的浓度与初始浓度近似相等。缓冲能力在 pH 接近 pKa（±1 范围内）时最强。

For a basic buffer, the corresponding expression in terms of pOH and pKb is often used, or it can be converted via pKa of the conjugate acid.

对于碱性缓冲溶液，常用 pOH 和 pKb 的类似形式，或转化为共轭酸的 pKa 后再计算 pH。

7. Polyprotic Acids and Stepwise Dissociation | 多元酸与分步解离

Polyprotic acids, such as H₂SO₄ and H₃PO₄, release protons in steps. The first proton of H₂SO₄ is completely dissociated, making the first step strong.

多元酸（如 H₂SO₄ 和 H₃PO₄）分步释放质子。H₂SO₄ 的第一步解离是完全的，可当作强酸处理。

The second dissociation of H₂SO₄ is weak (Ka₂ ≈ 1.2 × 10⁻²), while for phosphoric acid all steps are weak, with Ka₁ ≫ Ka₂ ≫ Ka₃.

H₂SO₄ 的第二步解离是弱酸电离（Ka₂ ≈ 1.2 × 10⁻²）；磷酸三步皆是弱电离，且 Ka₁ ≫ Ka₂ ≫ Ka₃。

For calculating pH in such systems, the first dissociation usually dominates, but for quantitative work on H₂SO₄, ICE tables or quadratic equations are needed to account for second dissociation.

计算这类体系的 pH 时，第一步解离通常起主导作用；但对 H₂SO₄ 的精确计算，需借助 ICE 表格或二次方程考虑第二步解离的贡献。

8. Titration Curves and Choice of Indicators | 滴定曲线与指示剂的选择

A pH titration curve shows how pH changes as a titrant is added. Key regions include the initial pH, the buffer region, the equivalence point, and the post-equivalence plateau.

pH 滴定曲线反映了滴定过程中 pH 随滴定剂加入量的变化。关键区域包括初始 pH、缓冲区域、化学计量点和过量的平台区。

For a strong acid–strong base titration, the equivalence point occurs at pH 7, and the steep vertical portion spans a large pH range. Suitable indicators include phenolphthalein and methyl orange.

强酸强碱滴定的化学计量点出现在 pH 7，曲线突跃范围宽，可选酚酞或甲基橙作为指示剂。

For a weak acid–strong base titration, the equivalence point is basic (pH > 7). Only indicators that change colour in the basic range, such as phenolphthalein, are appropriate.

弱酸-强碱滴定的化学计量点呈碱性（pH > 7），应选在碱性范围内变色的指示剂，如酚酞。

For a weak base–strong acid titration, the equivalence point is acidic, making methyl orange a better choice than phenolphthalein.

弱碱-强酸滴定的化学计量点呈酸性，此时甲基橙比酚酞更合适。

9. Dilution and Mixing Effects on pH | 稀释与混合对 pH 的影响

When a solution is diluted, the concentration of H⁺ or OH⁻ decreases. For strong acids, a tenfold dilution raises the pH by one unit (e.g., from 1 to 2). However, for very dilute solutions (near 10⁻⁷ mol dm⁻³), the autoionisation of water must be considered.

稀释时 H⁺ 或 OH⁻ 浓度降低。对于强酸，每稀释10倍 pH 升高1个单位；但在极稀溶液（接近 10⁻⁷ mol dm⁻³）时，必须考虑水的自耦电离的影响。

When mixing two acidic solutions, the resulting [H⁺] is the sum of the moles of H⁺ divided by total volume, assuming complete dissociation.

混合两种酸性溶液时，假设它们完全解离，最终 [H⁺] 等于氢离子总物质的量除以总体积。

If the two solutions react (acid + base), calculate the excess moles of either H⁺ or OH⁻ after neutralisation, then determine the concentration in the new total volume and compute pH.

若酸与碱混合，要先计算中和后的过量 H⁺ 或 OH⁻ 物质的量，再除以总体积求浓度，最后计算 pH。

10. The Effect of Temperature on Kw and pH | 温度对 Kw 和 pH 的影响

The dissociation of water is endothermic: 2H₂O ⇌ H₃O⁺ + OH⁻, ΔH > 0. As temperature increases, Kw increases, meaning the product [H⁺][OH⁻] becomes larger.

水的解离是吸热过程：2H₂O ⇌ H₃O⁺ + OH⁻，ΔH > 0。温度升高时 Kw 增大，[H⁺][OH⁻] 的积随之增大。

Consequently, the pH of pure water is only 7.00 at 25 °C. At higher temperatures, neutral pH is lower than 7 (e.g., pH 6.63 at 50 °C), but the solution remains neutral because [H⁺] = [OH⁻].

因此，纯水的 pH 仅在 25 °C 时为 7.00。高温下中性 pH 低于 7（如 50 °C 时约为 6.63），但由于 [H⁺] = [OH⁻]，溶液仍为中性。

Exam questions often ask to calculate the new Kw or the pH of neutral water at a different temperature using given [H⁺] data.

考题常要求根据给定的 [H⁺] 数据计算不同温度下的 Kw 或此时中性水的 pH。

11. Common Pitfalls and Exam Tips | 常见误区与应试技巧

Always check whether the acid or base is strong or weak before applying a formula. Using [H⁺] = √(Ka·c) for a strong acid will give a completely wrong result.

在套用公式前务必判断酸碱的强弱。对强酸使用 [H⁺] = √(Ka·c) 会得出完全错误的结果。

Make sure to use consistent units, especially when converting between pH and concentration. Never forget that pH < 7 is acidic and pH > 7 is basic at 25 °C, but this scale shifts with temperature.

计算时注意保持一致的单位，尤其是在 pH 和浓度换算时。牢记 25 °C 时 pH < 7 为酸性，pH > 7 为碱性，但该界限随温度变化。

For buffer calculations, do not forget that the ratio is [conjugate base]/[acid]; applying it upside down is a frequent mistake. Also, always check the assumptions: if c/Ka < 100, the approximation is invalid and the full quadratic solution is required.

缓冲溶液计算中勿将 [共轭碱]/[酸] 的比例弄反，这是常见错误。同时要检验近似条件：若 c/Ka < 100，则不能使用近似，必须求解二次方程。

In titration curve questions, link the half-equivalence point (where pH = pKa) to buffer recognition and indicator selection. At half-equivalence, [HA] = [A⁻].

在滴定曲线题中，半等当点（此时 pH = pKa）常用来识别缓冲区域和选择指示剂。半等当点处 [HA] = [A⁻]。

12. Summary of Key Formulas and Final Review | 核心公式总结与回顾

The essential toolkit for pH calculations includes: pH = –log₁₀[H⁺]; pOH = –log₁₀[OH⁻]; pH + pOH = pKw = 14.00 at 25 °C. For weak acids: [H⁺] = √(Ka·c); for weak bases: [OH⁻] = √(Kb·c). Buffer: pH = pKa + log₁₀([conjugate base]/[acid]).

pH 计算的核心公式工具箱：pH = –log₁₀[H⁺]；pOH = –log₁₀[OH⁻]；25 °C 时 pH + pOH = pKw = 14.00。弱酸：[H⁺] = √(Ka·c)；弱碱：[OH⁻] = √(Kb·c)。缓冲：pH = pKa + log₁₀([共轭碱]/[酸])。

Internalise the concept of Kw and the interplay between Ka, Kb, and Kw. By methodically identifying the type of system, making valid approximations, and cross-checking the limitations, you will master pH calculations for both IB and Edexcel exams.

深入理解 Kw 以及 Ka、Kb、Kw 之间的相互转换。通过系统性地判断体系类型、合理近似并验证前提条件，你就能在 IB 和 Edexcel 化学考试中彻底攻克 pH 计算。

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