AS Chemistry: pH Calculations Key Points | AS 化学：pH 计算 考点精讲

📚 AS Chemistry: pH Calculations Key Points | AS 化学：pH 计算 考点精讲

pH calculations are a central topic in AS Chemistry, linking quantitative concentration work with the principles of acid-base equilibria. Students must be confident in handling strong and weak acids, strong bases, the ionic product of water (Kw), acid dissociation constants (Ka), pKa and buffer solutions. Every calculation ultimately depends on the concentration of hydrogen ions, [H⁺], which governs the pH through a simple logarithmic relationship.

pH 计算是 AS 化学的核心专题，将定量浓度运算与酸碱平衡原理紧密结合。学生必须熟练掌握强酸、弱酸、强碱、水的离子积 (Kw)、酸解离常数 (Ka)、pKa 以及缓冲溶液的计算。所有计算最终都取决于氢离子浓度 [H⁺]，并通过简洁的对数关系决定 pH 值。

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1. Introduction to pH | pH 简介

The pH scale is a logarithmic scale used to express the acidity or basicity of an aqueous solution. It is defined as pH = -log[H⁺] (logarithm to base 10), where [H⁺] is the hydrogen ion concentration measured in mol dm⁻³.

pH 标度是对数标度，用于表示水溶液的酸碱度。定义为 pH = -log[H⁺]（以 10 为底的对数），其中 [H⁺] 是氢离子浓度，单位为 mol dm⁻³。

A low pH (<7) indicates an acidic solution; a high pH (>7) indicates a basic solution. At 25°C, pure water has a pH of exactly 7 and is neutral because [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³.

低 pH (<7) 表示酸性溶液，高 pH (>7) 表示碱性溶液。在 25°C 下，纯水的 pH 正好为 7，呈中性，因为 [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³。

Owing to the logarithmic scale, a change of one pH unit corresponds to a tenfold change in [H⁺]. For instance, a solution with pH = 3 has [H⁺] = 10⁻³ mol dm⁻³, ten times more concentrated than a pH = 4 solution.

由于对数的特性，pH 每改变 1 个单位，[H⁺] 就变化 10 倍。例如，pH = 3 的溶液 [H⁺] = 10⁻³ mol dm⁻³，其浓度是 pH = 4 溶液的 10 倍。

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2. Calculating pH of Strong Acids | 强酸 pH 计算

A strong acid is one that completely dissociates in water, releasing all of its acidic protons. Common strong monoprotic acids include hydrochloric acid (HCl) and nitric acid (HNO₃). For these, [H⁺] = concentration of the acid.

强酸在水中完全离解，释放出所有酸性质子。常见的一元强酸有盐酸 (HCl) 和硝酸 (HNO₃)。对于这类酸，[H⁺] = 酸的浓度。

Example: Calculate the pH of 0.050 mol dm⁻³ HCl.
[H⁺] = 0.050 mol dm⁻³, so pH = -log(0.050) = 1.30.

示例：计算 0.050 mol dm⁻³ HCl 的 pH。
[H⁺] = 0.050 mol dm⁻³，因此 pH = -log(0.050) = 1.30。

Sulfuric acid (H₂SO₄) is a strong diprotic acid. In AS calculations, it is usually assumed that both protons fully dissociate, giving [H⁺] = 2 × [H₂SO₄]. For example, 0.050 mol dm⁻³ H₂SO₄ gives [H⁺] = 0.10 mol dm⁻³, pH = -log(0.10) = 1.00.

硫酸 (H₂SO₄) 是二元强酸。在 AS 计算中通常假定两个质子完全电离，因此 [H⁺] = 2 × [H₂SO₄]。例如 0.050 mol dm⁻³ H₂SO₄ 给出 [H⁺] = 0.10 mol dm⁻³，pH = -log(0.10) = 1.00。

Always check whether the acid is mono- or diprotic and use the correct stoichiometry.

务必确认酸是一元还是二元，并使用正确的化学计量关系。

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3. Calculating pH of Strong Bases | 强碱 pH 计算

Strong bases, such as sodium hydroxide (NaOH) and potassium hydroxide (KOH), dissociate completely to yield hydroxide ions (OH⁻). The concentration of OH⁻ is used to find pOH: pOH = -log[OH⁻].

强碱如氢氧化钠 (NaOH) 和氢氧化钾 (KOH) 完全离解产生 OH⁻ 离子。利用 OH⁻ 浓度可求得 pOH：pOH = -log[OH⁻]。

At 25°C, the relationship pH + pOH = 14 holds. Therefore, pH = 14 – pOH.

在 25°C 时，存在关系 pH + pOH = 14。因此 pH = 14 – pOH。

Example: 0.10 mol dm⁻³ NaOH.
[OH⁻] = 0.10 mol dm⁻³ → pOH = 1.00 → pH = 14 – 1.00 = 13.00.

示例：0.10 mol dm⁻³ NaOH。
[OH⁻] = 0.10 mol dm⁻³ → pOH = 1.00 → pH = 14 – 1.00 = 13.00。

For bases that release two OH⁻ ions per formula unit, such as Ba(OH)₂, [OH⁻] = 2 × [base]. So 0.050 mol dm⁻³ Ba(OH)₂ gives [OH⁻] = 0.10 mol dm⁻³, pOH = 1.00, pH = 13.00.

对于每个化学式可释放两个 OH⁻ 的碱（如 Ba(OH)₂），[OH⁻] = 2 × [碱]。因此 0.050 mol dm⁻³ Ba(OH)₂ 给出 [OH⁻] = 0.10 mol dm⁻³，pOH = 1.00，pH = 13.00。

As an alternative, the pH can be obtained directly from [H⁺] = Kw / [OH⁻], which will be discussed next.

也可直接用 [H⁺] = Kw / [OH⁻] 求得 pH，这一点将在下文讨论。

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

Water undergoes very slight self-ionisation: H₂O ⇌ H⁺ + OH⁻. The equilibrium constant for this process is called the ionic product of water, Kw = [H⁺][OH⁻].

水发生极微弱的自电离：H₂O ⇌ H⁺ + OH⁻。该过程的平衡常数称为水的离子积，Kw = [H⁺][OH⁻]。

At 25°C (298 K), Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶. In pure water, [H⁺] = [OH⁻] = √Kw = 1.0 × 10⁻⁷ mol dm⁻³, hence pH = 7.

在 25°C (298 K)，Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶。纯水中 [H⁺] = [OH⁻] = √Kw = 1.0 × 10⁻⁷ mol dm⁻³，因此 pH = 7。

Kw increases with temperature because the autoionisation of water is endothermic. At higher temperatures, the neutral point shifts to a lower pH. For instance, at 40°C, Kw ≈ 2.9 × 10⁻¹⁴, giving neutral pH ≈ 6.77. The solution remains neutral because [H⁺] = [OH⁻]; only the pH value changes.

Kw 随温度升高而增大，因为水的自电离是吸热过程。温度升高时，中性点的 pH 值下降。例如在 40°C，Kw ≈ 2.9 × 10⁻¹⁴，中性 pH 约为 6.77。溶液仍呈中性，因为 [H⁺] = [OH⁻]；只是 pH 数值发生了变化。

When working with strong bases, the expression [H⁺] = Kw / [OH⁻] is extremely useful for calculating pH directly without first converting to pOH.

在处理强碱时，表达式 [H⁺] = Kw / [OH⁻] 非常实用，可直接计算 pH 而不必先求 pOH。

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

Weak acids, such as ethanoic acid (CH₃COOH), partially dissociate in water, establishing a dynamic equilibrium: CH₃COOH ⇌ H⁺ + CH₃COO⁻.

弱酸如乙酸 (CH₃COOH) 在水中部分电离，建立动态平衡：CH₃COOH ⇌ H⁺ + CH₃COO⁻。

The acid dissociation constant, Ka, is given by Ka = [H⁺][A⁻] / [HA], where [HA] is the equilibrium concentration of the undissociated acid and [A⁻] is the concentration of the conjugate base.

酸解离常数 Ka 定义为 Ka = [H⁺][A⁻] / [HA]，其中 [HA] 是未电离酸的平衡浓度，[A⁻] 是共轭碱的浓度。

To calculate the pH, assume [H⁺] = [A⁻] = x, and if the initial acid concentration is c and Ka is small compared to c, then [HA] ≈ c. This gives Ka = x² / c → x = √(Ka × c).

计算 pH 时，设 [H⁺] = [A⁻] = x，若起始酸浓度为 c 且 Ka 远小于 c，可近似

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