pH Calculations Master Guide | pH 计算完全指南

📚 pH Calculations Master Guide | pH 计算完全指南

pH is one of the most examined topics in Cambridge A-Level Chemistry. Mastering pH calculations requires a clear understanding of equilibrium, logarithms, and the distinction between strong and weak species. This guide covers every calculation type you are likely to meet in Paper 4 and Paper 5 questions.

pH 是剑桥 A-Level 化学中考查频率最高的主题之一。掌握 pH 计算需要清晰理解平衡、对数以及强物种与弱物种之间的区别。本指南涵盖你在 Paper 4 和 Paper 5 中可能遇到的每一种计算题型。

1. The pH Scale and Water Ionization | pH 标度与水的电离

pH is defined as the negative base-10 logarithm of the hydrogen ion concentration. A change of one pH unit corresponds to a tenfold change in [H⁺]. This logarithmic scale allows very small concentrations to be expressed as convenient numbers.

pH 定义为氢离子浓度的负常用对数。pH 每变化 1 个单位，对应 [H⁺] 变化 10 倍。这种对数标度让我们能用方便的数字来表示很小的浓度。

pH = −log₁₀[H⁺] ⇒ [H⁺] = 10⁻ᵖᴴ

Pure water undergoes autoionization: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). The equilibrium constant is called the ionic product of water, Kw. At 25°C, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

纯水发生自偶电离：H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)。其平衡常数称为水的离子积 Kw。在 25°C 时，Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶。

Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 25°C

In pure water, [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³, so pH = 7.00. Because the solution contains equal concentrations of H⁺ and OH⁻, it is neutral. You should also know the related definitions: pOH = −log₁₀[OH⁻] and pKw = −log₁₀ Kw = 14.00 at 25°C, giving pH + pOH = pKw = 14.00.

在纯水中，[H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³，因此 pH = 7.00。由于 H⁺ 和 OH⁻ 浓度相等，溶液为中性。你还应知道相关定义：pOH = −log₁₀[OH⁻]，pKw = −log₁₀ Kw = 14.00（25°C），即 pH + pOH = pKw = 14.00。

2. Strong Acids and Strong Bases | 强酸与强碱

A strong acid is fully dissociated in aqueous solution. For a monoprotic acid such as HCl, the equilibrium lies completely to the right: HCl(aq) → H⁺(aq) + Cl⁻(aq). Therefore the hydrogen ion concentration is simply equal to the original acid concentration.

强酸在水中完全电离。对于 HCl 这样的单质子酸，平衡完全向右移动：HCl(aq) → H⁺(aq) + Cl⁻(aq)。因此，氢离子浓度就等于原始酸的浓度。

Worked example | 示例： Calculate the pH of 0.0100 mol dm⁻³ HCl. Since [H⁺] = 0.0100 mol dm⁻³, pH = −log₁₀(0.0100) = 2.00.

计算示例：计算 0.0100 mol dm⁻³ HCl 的 pH。由于 [H⁺] = 0.0100 mol dm⁻³，pH = −log₁₀(0.0100) = 2.00。

For strong bases, first find [OH⁻], then convert to [H⁺] using Kw. A base with n hydroxide ions per formula unit produces n times the molar concentration of OH⁻.

对于强碱，先求 [OH⁻]，再通过 Kw 换算为 [H⁺]。每摩尔碱式单元含有 n 个氢氧根离子时，会产生 n 倍于碱浓度的 OH⁻。

Worked example | 示例： Calculate the pH of 0.0500 mol dm⁻³ Ba(OH)₂. Ba(OH)₂ → Ba²⁺ + 2OH⁻, so [OH⁻] = 2 × 0.0500 = 0.100 mol dm⁻³. Then [H⁺] = Kw/[OH⁻] = 1.00 × 10⁻¹⁴/0.100 = 1.00 × 10⁻¹³ mol dm⁻³. pH = 13.00.

计算示例：计算 0.0500 mol dm⁻³ Ba(OH)₂ 的 pH。Ba(OH)₂ → Ba²⁺ + 2OH⁻，所以 [OH⁻] = 2 × 0.0500 = 0.100 mol dm⁻³。则 [H⁺] = Kw/[OH⁻] = 1.00 × 10⁻¹⁴/0.100 = 1.00 × 10⁻¹³ mol dm⁻³。pH = 13.00。

3. Weak Acids: Ka and the ICE Method | 弱酸：Ka 与 ICE 法

A weak acid only partially dissociates in water. The equilibrium for a generic weak acid HA is: HA(aq) ⇌ H⁺(aq) + A⁻(aq). The acid dissociation constant Ka is defined as:

弱酸在水中仅部分电离。一般弱酸 HA 的平衡为：HA(aq) ⇌ H⁺(aq) + A⁻(aq)。酸解离常数 Ka 定义为：

Ka = [H⁺][A⁻] / [HA]

For calculations, two assumptions are usually made. First, because the acid is weak, the amount of dissociation is negligible, so [HA] at equilibrium equals the initial concentration c. Second, each HA that dissociates produces one H⁺ and one A⁻, so [H⁺] = [A⁻]. This gives the simplified expression:

计算时通常做两个近似。第一，由于酸很弱，解离量可忽略不计，因此平衡时 [HA] 等于初始浓度 c。第二，每个 HA 解离产生一个 H⁺ 和一个 A⁻，所以 [H⁺] = [A⁻]。由此得到简化表达式：

Ka ≈ [H⁺]² / c ⇒ [H⁺] ≈ √(Ka × c)

Worked example | 示例： A 0.100 mol dm⁻³ solution of ethanoic acid has Ka = 1.74 × 10⁻⁵ mol dm⁻³. Calculate its pH. Using [H⁺] = √(Ka × c) = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³. Therefore pH = −log₁₀(1.32 × 10⁻³) = 2.88.

计算示例：0.100 mol dm⁻³ 乙酸溶液的 Ka = 1.74 × 10⁻⁵ mol dm⁻³，计算其 pH。由 [H⁺] = √(Ka × c) = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³。因此 pH = −log₁₀(1.32 × 10⁻³) = 2.88。

You may also be asked to work backwards, finding Ka from a measured pH. Always use the full ICE table in such cases rather than skipping the equilibrium concentrations.

题目也可能让你由测得的 pH 反推 Ka。此时应使用完整的 ICE 表格，而不是跳过平衡浓度。

	HA	H⁺	A⁻
Initial	c	0	0
Change	−x	+x	+x
Equilibrium	c − x	x	x

4. Weak Bases and Kb | 弱碱与 Kb

A weak base accepts a proton from water. For a generic base B: B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq). The base dissociation constant Kb is:

弱碱从水中接受质子。对于一般碱 B：B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)。碱解离常数 Kb 为：

Kb = [BH⁺][OH⁻] / [B]

Using the same logic as for a weak acid, [BH⁺] = [OH⁻] and [B] ≈ initial concentration, so [OH⁻] ≈ √(Kb × c). Once [OH⁻] is known, calculate pOH = −log₁₀[OH⁻], then pH = 14.00 − pOH at 25°C.

与弱酸同理，[BH⁺] = [OH⁻]，[B] ≈ 初始浓度，因此 [OH⁻] ≈ √(Kb × c)。求出 [OH⁻] 后，计算 pOH = −log₁₀[OH⁻]，再在 25°C 下用 pH = 14.00 − pOH 得出 pH。

Worked example | 示例： Calculate the pH of 0.150 mol dm⁻³ ammonia solution. For NH₃, Kb = 1.80 × 10⁻⁵ mol dm⁻³. [OH⁻] = √(1.80 × 10⁻⁵ × 0.150) = √(2.70 × 10⁻⁶) = 1.64 × 10⁻³ mol dm⁻³. pOH = 2.78, so pH = 14.00 − 2.78 = 11.22.

计算示例：计算 0.150 mol dm⁻³ 氨水的 pH。NH₃ 的 Kb = 1.80 × 10⁻⁵ mol dm⁻³。[OH⁻] = √(1.80 × 10⁻⁵ × 0.150) = √(2.70 × 10⁻⁶) = 1.64 × 10⁻³ mol dm⁻³。pOH = 2.78，所以 pH = 14.00 − 2.78 = 11.22。

Remember the relationship between Ka and Kb for a conjugate acid-base pair: Ka × Kb = Kw. This is essential when dealing with salt hydrolysis in Section 8.

记住共轭酸碱对的 Ka 与 Kb 关系：Ka × Kb = Kw。这在第 8 节处理盐类水解时至关重要。

5. Water Ionization and Temperature Effects | 水的离子积与温度效应

The value of Kw is not a universal constant — it depends on temperature. The self-ionization of water is endothermic, so increasing the temperature shifts the equilibrium right and increases Kw.

Kw 不是一个普遍常数，它依赖于温度。水的自偶电离是吸热过程，因此升高温度会使平衡右移，Kw 增大。

For example, at 50°C, Kw = 5.48 × 10⁻¹⁴ mol² dm⁻⁶. In pure water at this temperature, [H⁺] = [OH⁻] = √(5.48 × 10⁻¹⁴) = 2.34 × 10⁻⁷ mol dm⁻³, giving pH = 6.63. The water is still neutral because [H⁺] = [OH⁻], even though the pH is below 7.

例如，在 50°C 时，Kw = 5.48 × 10⁻¹⁴ mol² dm⁻⁶。在该温度下纯水中，[H⁺] = [OH⁻] = √(5.48 × 10⁻¹⁴) = 2.34 × 10⁻⁷ mol dm⁻³，因此 pH = 6.63。此时水仍是中性的，因为 [H⁺] = [OH⁻]，尽管 pH 低于 7。

Exam questions often ask you to explain why the pH of pure water is 7.00 only at 25°C. Always link pH neutrality to the equality [H⁺] = [OH⁻], not to a fixed pH value of 7.

考题常要求解释为什么纯水的 pH 只有在 25°C 时才等于 7.00。务必把中性条件联系到 [H⁺] = [OH⁻] 这一等式，而不是固定 pH 等于 7。

6. Buffer Solutions | 缓冲溶液

A buffer solution resists changes in pH when small amounts of acid or base are added. There are two common types: an acid buffer consists of a weak acid and its conjugate base (usually as a sodium or potassium salt); a basic buffer consists of a weak base and its conjugate acid (usually as a chloride or nitrate salt).

缓冲溶液能在加入少量酸或碱时抵抗 pH 的变化。常见类型有两种：酸性缓冲液由弱酸及其共轭碱（通常以钠盐或钾盐形式）组成；碱性缓冲液由弱碱及其共轭酸（通常以氯化物或硝酸盐形式）组成。

An acid buffer maintains a fairly constant pH because the weak acid neutralizes added OH⁻, while the conjugate base neutralizes added H⁺. The pH depends on the ratio of salt to acid concentrations.

酸性缓冲液之所以能维持 pH 基本恒定，是因为弱酸能中和加入的 OH⁻，而共轭碱能中和加入的 H⁺。其 pH 取决于盐与酸的浓度之比。

For the buffer calculation, substitute equilibrium concentrations into the Ka expression. For a weak acid HA with added salt NaA, [H⁺] = Ka × [HA]/[A⁻]. The larger the concentration of the buffer components, the greater its buffer capacity.

对于缓冲液计算，将平衡浓度代入 Ka 表达式。对于加入盐 NaA 的弱酸 HA，[H⁺] = Ka × [HA]/[A⁻]。缓冲组分的浓度越大，缓冲容量就越大。

7. The Henderson-Hasselbalch Equation | 亨德森-哈塞尔巴尔赫方程

Taking the negative logarithm of the Ka expression gives the Henderson-Hasselbalch equation, which is extremely useful for buffer pH calculations:

对 Ka 表达式取负对数，得到亨德森-哈塞尔巴尔赫方程，该方程在缓冲液 pH 计算中非常有用：

pH = pKa + log₁₀([A⁻] / [HA])

Worked example | 示例： A buffer contains 0.20 mol dm⁻³ ethanoic acid and 0.15 mol dm⁻³ sodium ethanoate. Given pKa = 4.76, calculate the pH. Using the equation: pH = 4.76 + log₁₀(0.15/0.20) = 4.76 + log₁₀(0.75) = 4.76 − 0.125 = 4.64.

计算示例：某缓冲液含 0.20 mol dm⁻³ 乙酸和 0.15 mol dm⁻³ 乙酸钠。已知 pKa = 4.76，计算 pH。代入方程：pH = 4.76 + log₁₀(0.15/0.20) = 4.76 + log₁₀(0.75) = 4.76 − 0.125 = 4.64。

For a basic buffer, the analogous form uses pKb: pOH = pKb + log₁₀([BH⁺]/[B]), then pH = 14.00 − pOH at 25°C. Note that in the buffer region of a titration curve, pH changes very slowly, and at the half-equivalence point pH = pKa.

对于碱性缓冲液，对应形式使用 pKb：pOH = pKb + log₁₀([BH⁺]/[B])，再在 25°C 下用 pH = 14.00 − pOH。注意，在滴定曲线的缓冲区域内，pH 变化非常缓慢；在半中和点时，pH = pKa。

8. Salt Hydrolysis | 盐类水解

Many salts dissolve to form solutions that are not neutral. A salt of a strong base and a weak acid, such as sodium ethanoate, produces a basic solution because the conjugate base hydrolyses: CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq).

许多盐溶于水后形成的溶液并不呈中性。强碱与弱酸形成的盐，如乙酸钠，会产生碱性溶液，因为其共轭碱发生水解：CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq)。

To calculate the pH of such a salt solution, first find Kb from the weak acid’s Ka using Kb = Kw/Ka. Then treat the salt solution as a weak base with initial concentration equal to the salt concentration.

要计算此类盐溶液的 pH，先用 Kb = Kw/Ka 从弱酸的 Ka 求出 Kb，然后把盐溶液当作初始浓度等于盐浓度的弱碱来处理。

Worked example | 示例： Calculate the pH of 0.100 mol dm⁻³ sodium ethanoate. For ethanoic acid, Ka = 1.74 × 10⁻⁵, so Kb = 1.00 × 10⁻¹⁴/1.74 × 10⁻⁵ = 5.75 × 10⁻¹⁰. Then [OH⁻] = √(Kb × c) = √(5.75 × 10⁻¹⁰ × 0.100) = 7.58 × 10⁻⁶ mol dm⁻³. pOH = 5.12, pH = 14.00 − 5.12 = 8.88.

计算示例：计算 0.100 mol dm⁻³ 乙酸钠溶液的 pH。乙酸的 Ka = 1.74 × 10⁻⁵，因此 Kb = 1.00 × 10⁻¹⁴/1.74 × 10⁻⁵ = 5.75 × 10⁻¹⁰。则 [OH⁻] = √(Kb × c) = √(5.75 × 10⁻¹⁰ × 0.100) = 7.58 × 10⁻⁶ mol dm⁻³。pOH = 5.12，pH = 14.00 − 5.12 = 8.88。

Similarly, salts of a strong acid and a weak base, such as NH₄Cl, produce an acidic solution because the conjugate acid hydrolyses: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq). Use Ka = Kw/Kb for the conjugate acid and treat it as a weak acid.

类似地，强酸与弱碱形成的盐，如 NH₄Cl，会产生酸性溶液，因为其共轭酸发生水解：NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)。用 Ka = Kw/Kb 求共轭酸的 Ka，再按弱酸处理。

9. Titration Curves and Indicator Choice | 滴定曲线与指示剂选择

Understanding pH calculations helps you interpret titration curves. A strong acid-strong base curve starts at very low pH and ends at very high pH, with a sharp vertical rise around pH 7. A weak acid-strong base curve starts higher, has a buffer region, and the equivalence point is above pH 7. A strong acid-weak base curve has an equivalence point below pH 7.

理解 pH 计算有助于解释滴定曲线。强酸-强碱曲线从很低的 pH 开始，到很高的 pH 结束，在 pH ≈ 7 附近出现陡峭的垂直跃升。弱酸-强碱曲线起始 pH 较高，存在缓冲区域，且等当点高于 pH 7。强酸-弱碱曲线的等当点低于 pH 7。

At the equivalence point, the added titrant exactly neutralizes the analyte. For a weak acid-strong base titration, the solution at equivalence contains the conjugate base, so the pH is determined by salt hydrolysis. For a strong acid-weak base titration, the cation hydrolyses and the pH is below 7.

在等当点，所加滴定剂恰好完全中和被滴定物质。对于弱酸-强碱滴定，等当点溶液含共轭碱，因此 pH 由盐类水解决定。对于强酸-弱碱滴定，阳离子发生水解，pH 低于 7。

An indicator is suitable if its pH range overlaps the steep part of the titration curve. For example, phenolphthalein works for strong acid-strong base and weak acid-strong base titrations, while methyl orange works for strong acid-weak base titrations.

当指示剂的变色范围与滴定曲线的陡峭部分重叠时，该指示剂适用。例如，酚酞适用于强酸-强碱和弱酸-强碱滴定，而甲基橙适用于强酸-弱碱滴定。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Many students lose marks on pH questions not because of difficult mathematics, but because of small procedural errors. The list below covers the most frequent problems.

许多学生在 pH 题目上失分并非因为数学难度，而是因为一些微小的步骤错误。下面列出最常见的问题。

Confusing strong and weak species. A concentrated weak acid is still only slightly dissociated; do not use the strong acid shortcut [H⁺] = c. Conversely, a dilute strong acid is still fully dissociated.

混淆强与弱。浓的弱酸仍然只解离少量，不能使用 [H⁺] = c 这一强酸快捷方式；反之，稀的强酸仍然完全解离。
Forgetting to multiply by n for diprotic bases or acids. H₂SO₄ and Ba(OH)₂ release two particles per formula unit. Check the stoichiometry before calculating [H⁺] or [OH⁻].

忘记对二元酸或二元碱乘以 n。H₂SO₄ 和 Ba(OH)₂ 每单位释放两个粒子。计算 [H⁺] 或 [OH⁻] 之前先检查化学计量关系。
Ignoring dilution when mixing solutions. If you mix 25.0 cm³ of acid with 25.0 cm³ of base, the total volume becomes 50.0 cm³ and all concentrations are halved. Always divide moles by the final total volume.

混合溶液时忽略稀释。若将 25.0 cm³ 酸与 25.0 cm³ 碱混合，总体积变为 50.0 cm³，所有浓度减半。务必用最终总体积去除物质的量。
Using pH = 14 − pOH without checking temperature. This relation holds only when pKw = 14.00, i.e. at 25°C.

不检查温度就使用 pH = 14 − pOH。该关系仅在 pKw = 14.00 时成立，即 25°C。
Rounding too early. When taking logarithms, keep at least three significant figures until the final step. Carrying one extra digit avoids small errors in the final answer.

过早四舍五入。取对数时，至少保留三位有效数字直到最后一步。多保留一位数字可以避免最终答案出现微小误差。
Skipping the balanced equation. A correct stoichiometric equation immediately tells you the mole ratios, the number of H⁺ or OH⁻ per unit, and whether the salt formed will hydrolyse.

跳过配平的化学方程式。正确的化学方程式能直接告诉你摩尔比例、每单位含多少个 H⁺ 或 OH⁻，以及生成的盐是否会发生水解。

Finally, always show your working.

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pH Calculations Master Guide | pH 计算完全指南