pH Calculations and Key Concepts in A-Level Chemistry | A-Level化学：pH值的计算方法与要点

📚 pH Calculations and Key Concepts in A-Level Chemistry | A-Level化学：pH值的计算方法与要点

The concept of pH is fundamental to acid-base chemistry and appears in almost every A-Level Chemistry examination paper. A thorough understanding of how to calculate pH for strong acids, weak acids, strong bases, weak bases and buffer solutions is essential for achieving top marks. This article systematically presents the calculation methods, underlying assumptions and common pitfalls that students should master.

pH值是酸碱化学中的核心概念，几乎出现在每一份A-Level化学试卷中。深入掌握强酸、弱酸、强碱、弱碱及缓冲溶液pH值的计算方法，是取得高分的关键。本文将系统地讲解各类体系的计算思路、隐含假设和常见易错点，帮助考生构建完整的知识框架。

---

1. The Definition of pH | pH的定义

The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration, measured in mol dm⁻³. The formula is expressed as:

pH = −log₁₀[H⁺]

For pure water at 25°C, the hydrogen ion concentration is 1.0 × 10⁻⁷ mol dm⁻³, giving a pH of exactly 7. It is important to remember that pH is a dimensionless quantity, although it is derived from a concentration expressed in mol dm⁻³.

溶液的pH值定义为氢离子浓度（以 mol dm⁻³ 为单位）的常用对数的负值，表达式为：

pH = −log₁₀[H⁺]

在25°C时，纯水的氢离子浓度为 1.0 × 10⁻⁷ mol dm⁻³，因此pH恰好为7。需要特别注意的是，pH本身是一个无量纲的量，虽然它来源于以 mol dm⁻³ 表示的浓度。

---

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

The self-ionisation of water is an equilibrium process described by the equation:

H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)

The equilibrium constant for this process is called the ionic product of water, Kw. At 25°C, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶. The expression is:

Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ (at 25°C)

This relationship allows us to interconvert between [H⁺] and [OH⁻]. For any aqueous solution at 25°C, if we know the concentration of one ion, we can calculate the concentration of the other using Kw.

水的自偶电离是一个平衡过程，其方程式为：

H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)

该平衡常数称为水的离子积，记作 Kw。在25°C时，Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶，表达式为：

Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴（25°C时）

这一关系式使我们能够在 [H⁺] 与 [OH⁻] 之间进行换算。在25°C下的任何水溶液中，只要知道其中一种离子的浓度，即可利用 Kw 计算出另一种离子的浓度。

---

3. Strong Acid pH Calculations | 强酸溶液的pH计算

Strong acids, such as HCl, HNO₃ and H₂SO₄, are assumed to ionise completely in aqueous solution. For a monoprotic strong acid of concentration c mol dm⁻³, the hydrogen ion concentration is simply equal to c:

[H⁺] = c

Worked Example: Calculate the pH of a 0.0250 mol dm⁻³ solution of HCl.

pH = −log₁₀(0.0250) = −log₁₀(2.50 × 10⁻²) = 1.60

The pH should be reported to the same number of decimal places as the number of significant figures in the concentration. Since 0.0250 has three significant figures, the pH is reported as 1.60 (two decimal places).

强酸（如 HCl、HNO₃ 和 H₂SO₄）在水中被视为完全电离。对于浓度为 c mol dm⁻³ 的一元强酸，氢离子浓度直接等于 c：

[H⁺] = c

例题：计算 0.0250 mol dm⁻³ HCl 溶液的pH值。

pH = −log₁₀(0.0250) = −log₁₀(2.50 × 10⁻²) = 1.60

pH值的小数位数应与浓度的有效数字位数一致。0.0250 有三位有效数字，因此pH报告为 1.60（两位小数）。

---

4. Strong Base pH Calculations | 强碱溶液的pH计算

Strong bases, such as NaOH and KOH, dissociate completely in water. For a strong base of concentration c mol dm⁻³, the hydroxide ion concentration is equal to c. To calculate the pH, we first determine the pOH and then use the relationship:

pH + pOH = 14 (at 25°C)

Worked Example: Calculate the pH of a 0.0100 mol dm⁻³ solution of NaOH at 25°C.

[OH⁻] = 0.0100 mol dm⁻³
pOH = −log₁₀(0.0100) = 2.00
pH = 14.00 − 2.00 = 12.00

For dibasic strong bases such as Ba(OH)₂, each mole of the base produces two moles of hydroxide ions, so [OH⁻] = 2c.

强碱（如 NaOH 和 KOH）在水中完全解离。对于浓度为 c mol dm⁻³ 的强碱，氢氧根离子浓度等于 c。计算pH时，先求出pOH，再利用以下关系式：

pH + pOH = 14（25°C时）

例题：计算 0.0100 mol dm⁻³ NaOH 溶液在25°C时的pH值。

[OH⁻] = 0.0100 mol dm⁻³
pOH = −log₁₀(0.0100) = 2.00
pH = 14.00 − 2.00 = 12.00

对于二元强碱如 Ba(OH)₂，每摩尔碱产生两摩尔氢氧根离子，因此 [OH⁻] = 2c。

---

5. Weak Acid pH Calculations | 弱酸溶液的pH计算

Weak acids, such as ethanoic acid (CH₃COOH), only partially ionise in water. The equilibrium is represented as:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

The acid dissociation constant Ka is given by:

Ka = [H⁺][A⁻] / [HA]

In the calculation of pH for a weak acid, three simplifying assumptions are made:

• Since ionisation is very slight, [HA] at equilibrium is approximately equal to the initial concentration c.
• [H⁺] ≈ [A⁻], because each molecule of HA that ionises produces one H⁺ and one A⁻.
• The contribution of H⁺ from the self-ionisation of water is negligible.

With these assumptions, the expression simplifies to:

Ka ≈ [H⁺]² / c

Rearranging this gives:

[H⁺] = √(Ka × c)

Therefore, the pH of a weak acid is calculated using:

pH = ½(pKa − log₁₀ c)

Worked Example: Calculate the pH of a 0.100 mol dm⁻³ ethanoic acid solution, given Ka = 1.74 × 10⁻⁵ mol dm⁻³.

[H⁺] = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³
pH = −log₁₀(1.32 × 10⁻³) = 2.88

弱酸（如乙酸 CH₃COOH）在水中仅部分电离，其平衡可表示为：

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

酸解离常数 Ka 的表达式为：

Ka = [H⁺][A⁻] / [HA]

在计算弱酸pH时，需要作三项简化假设：

• 由于电离程度极小，平衡时的 [HA] 近似等于初始浓度 c。
• [H⁺] ≈ [A⁻]，因为每电离一个 HA 分子就产生一个 H⁺ 和一个 A⁻。
• 水的自偶电离所贡献的 H⁺ 可以忽略不计。

据此，Ka 的表达式可简化为：

Ka ≈ [H⁺]² / c

整理后得到：

[H⁺] = √(Ka × c)

因此，弱酸的pH可通过以下公式计算：

pH = ½(pKa − log₁₀ c)

例题：已知 Ka = 1.74 × 10⁻⁵ mol dm⁻³，计算 0.100 mol dm⁻³ 乙酸溶液的pH值。

[H⁺] = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³
pH = −log₁₀(1.32 × 10⁻³) = 2.88

---

6. Weak Base pH Calculations | 弱碱溶液的pH计算

Weak bases, such as ammonia (NH₃), only partially ionise in water. The equilibrium is:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

The base dissociation constant Kb is defined as:

Kb = [NH₄⁺][OH⁻] / [NH₃]

Using the same assumptions as for weak acids, we obtain:

[OH⁻] = √(Kb × c)

Once [OH⁻] is calculated, the pOH and then the pH can be determined. Alternatively, the relationship between Ka and Kb for a conjugate acid-base pair is:

Ka × Kb = Kw = 1.00 × 10⁻¹⁴ (at 25°C)

Worked Example: Calculate the pH of a 0.150 mol dm⁻³ ammonia solution, given 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 = −log₁₀(1.64 × 10⁻³) = 2.79
pH = 14.00 − 2.79 = 11.21

弱碱（如氨 NH₃）在水中仅部分电离，其平衡为：

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

碱解离常数 Kb 定义为：

Kb = [NH₄⁺][OH⁻] / [NH₃]

采用与弱酸相同的假设，可得：

[OH⁻] = √(Kb × c)

求出 [OH⁻] 后，即可确定 pOH 和 pH。此外，对于共轭酸碱对，Ka 与 Kb 的关系为：

Ka × Kb = Kw = 1.00 × 10⁻¹⁴（25°C时）

例题：已知 Kb = 1.80 × 10⁻⁵ mol dm⁻³，计算 0.150 mol dm⁻³ 氨溶液的pH值。

[OH⁻] = √(1.80 × 10⁻⁵ × 0.150) = √(2.70 × 10⁻⁶) = 1.64 × 10⁻³ mol dm⁻³
pOH = −log₁₀(1.64 × 10⁻³) = 2.79
pH = 14.00 − 2.79 = 11.21

---

7. Buffer Solutions | 缓冲溶液

A buffer solution resists changes in pH when small quantities of acid or base are added. There are two types of buffers: acidic buffers, which consist of a weak acid and its conjugate base (e.g. CH₃COOH / CH₃COO⁻), and basic buffers, which consist of a weak base and its conjugate acid (e.g. NH₃ / NH₄⁺).

缓冲溶液能在加入少量酸或碱时抵抗pH的显著变化。缓冲溶液分为两类：酸性缓冲液由弱酸及其共轭碱组成（如 CH₃COOH / CH₃COO⁻）；碱性缓冲液由弱碱及其共轭酸组成（如 NH₃ / NH₄⁺）。

The pH of an acidic buffer can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log₁₀([A⁻] / [HA])

For a basic buffer, the corresponding equation in terms of pOH is:

pOH = pKb + log₁₀([BH⁺] / [B])

酸性缓冲液的pH可用亨德森-哈塞尔巴尔赫方程计算：

pH = pKa + log₁₀([A⁻] / [HA])

对于碱性缓冲液，对应的pOH方程为：

pOH = pKb + log₁₀([BH⁺] / [B])

Worked Example: A buffer solution is prepared by mixing 25.0 cm³ of 0.200 mol dm⁻³ CH₃COOH with 20.0 cm³ of 0.150 mol dm⁻³ CH₃COONa. Calculate the pH of this buffer. (Ka for CH₃COOH = 1.74 × 10⁻⁵ mol dm⁻³)

Total volume = 45.0 cm³ = 0.0450 dm³
[HA] = (0.200 × 0.0250) / 0.0450 = 0.111 mol dm⁻³
[A⁻] = (0.150 × 0.0200) / 0.0450 = 0.0667 mol dm⁻³
pKa = −log₁₀(1.74 × 10⁻⁵) = 4.76
pH = 4.76 + log₁₀(0.0667 / 0.111) = 4.76 + (−0.221) = 4.54

例题：将 25.0 cm³ 的 0.200 mol dm⁻³ CH₃COOH 与 20.0 cm³ 的 0.150 mol dm⁻³ CH₃COONa 混合制备缓冲溶液，计算该缓冲液的pH值。（CH₃COOH 的 Ka = 1.74 × 10⁻⁵ mol dm⁻³）

总体积 = 45.0 cm³ = 0.0450 dm³
[HA] = (0.200 × 0.0250) / 0.0450 = 0.111 mol dm⁻³
[A⁻] = (0.150 × 0.0200) / 0.0450 = 0.0667 mol dm⁻³
pKa = −log₁₀(1.74 × 10⁻⁵) = 4.76
pH = 4.76 + log₁₀(0.0667 / 0.111) = 4.76 + (−0.221) = 4.54

---

8. pH of Water and Temperature Effects | 水的pH与温度效应

The ionisation of water is an endothermic process. According to Le Chatelier’s principle, an increase in temperature shifts the equilibrium to the right, increasing both [H⁺] and [OH⁻]. Consequently, the value of Kw increases with temperature.

水的电离是一个吸热过程。根据勒夏特列原理，温度升高会使平衡向右移动，[H⁺] 与 [OH⁻] 同时增大。因此，Kw 的值随温度升高而增大。

At 25°C, Kw = 1.00 × 10⁻¹⁴ and pure water has a pH of 7.00. However, at 50°C, Kw = 5.48 × 10⁻¹⁴, and the pH of pure water becomes:

[H⁺] = √Kw = √(5.48 × 10⁻¹⁴) = 2.34 × 10⁻⁷ mol dm⁻³
pH = −log₁₀(2.34 × 10⁻⁷) = 6.63

It is crucial to understand that although the pH of pure water at 50°C is 6.63, the solution is still neutral because [H⁺] = [OH⁻]. Neutrality is defined by equal concentrations of H⁺ and OH⁻, not by pH = 7.

在25°C时，Kw = 1.00 × 10⁻¹⁴，纯水的pH为 7.00。然而在50°C时，Kw = 5.48 × 10⁻¹⁴，纯水的pH变为：

[H⁺] = √Kw = √(5.48 × 10⁻¹⁴) = 2.34 × 10⁻⁷ mol dm⁻³
pH = −log₁₀(2.34 × 10⁻⁷) = 6.63

需要特别强调的是，虽然50°C时纯水的pH为 6.63，但该溶液仍然呈中性，因为 [H⁺] = [OH⁻]。中性的定义是 H⁺ 与 OH⁻ 浓度相等，而非 pH 等于 7。

---

9. pH of Salt Solutions | 盐溶液的pH

Salts can produce acidic, basic or neutral solutions depending on the nature of the parent acid and base. This is due to hydrolysis reactions between the salt ions and water.

盐溶液可能呈酸性、碱性或中性，取决于其母体酸和碱的性质。这是因为盐的离子与水发生水解反应所致。

Type of Salt | 盐的类型	Example | 示例	pH of Solution | 溶液pH
Strong acid + strong base | 强酸 + 强碱	NaCl, KNO₃	Neutral (pH = 7) | 中性（pH = 7）
Weak acid + strong base | 弱酸 + 强碱	CH₃COONa, Na₂CO₃	Basic (pH > 7) | 碱性（pH > 7）
Strong acid + weak base | 强酸 + 弱碱	NH₄Cl, NH₄NO₃	Acidic (pH < 7) | 酸性（pH < 7）
Weak acid + weak base | 弱酸 + 弱碱	CH₃COONH₄	Depends on Ka and Kb | 取决于 Ka 与 Kb 的相对大小

To calculate the pH of a salt solution, we treat the relevant ion as a weak acid or weak base and use the appropriate Ka or Kb value. For example, in a solution of ammonium chloride, the NH₄⁺ ion acts as a weak acid:

NH₄⁺(aq) ⇌ H⁺(aq) + NH₃(aq)

The Ka for NH₄⁺ can be obtained from the relationship Ka × Kb = Kw, where Kb refers to the conjugate base NH₃.

要计算盐溶液的pH，应将相关离子视为弱酸或弱碱，并使用相应的 Ka 或 Kb 值。例如，在氯化铵溶液中，NH₄⁺ 离子充当弱酸：

NH₄⁺(aq) ⇌ H⁺(aq) + NH₃(aq)

NH₄⁺ 的 Ka 可通过关系式 Ka × Kb = Kw 求得，其中 Kb 对应其共轭碱 NH₃。

---

10. Common Pitfalls and Exam Tips | 常见易错点与考试提示

Students frequently lose marks in pH-related questions due to avoidable errors. The following points highlight the most common pitfalls and how to avoid them.

学生在pH相关题目中常因一些可以避免的错误而失分。以下列出最常见的易错点及规避方法。

• Confusing pOH and pH: Always remember that pH + pOH = 14 at 25°C. For bases, calculate pOH first, then subtract from 14.
• Forgetting the factor of 2 for dibasic acids: For acids like H₂SO₄, the first ionisation is complete and the second is partial. In dilute solutions, treat H₂SO₄ as producing 2H⁺ per mole for the purpose of simple calculations, unless the question specifies otherwise.
• Using the initial concentration for a weak acid without checking validity: If the acid is very dilute or Ka is relatively large, the approximation [HA] ≈ c may break down, requiring the quadratic formula.
• Incorrect significant figures: The number of decimal places in pH should equal the number of significant figures in the concentration. A common error is reporting too many decimal places.
• Neglecting the volume change when preparing buffer solutions: When solutions are mixed, concentrations must be recalculated using the total volume.
• Mixing up Ka and Kb: For a conjugate acid-base pair, always use the correct constant. If you are given Kb but the question asks for pH of an acidic buffer containing the conjugate acid, you must convert Kb to Ka first.

常见易错点包括以下方面：

• 混淆pOH与pH：切记25°C时 pH + pOH = 14。计算碱时，先求pOH，再用14减去。
• 忽略二元酸的系数2：对于 H₂SO₄，第一步完全电离，第二步部分电离。在稀溶液中，除非题目另有说明，可按每摩尔产生 2 mol H⁺ 来处理。
• 未验证弱酸近似条件的有效性：当酸极稀或 Ka 较大时，[HA] ≈ c 的近似可能失效，需要求助于二次方程。
• 有效数字使用不当：pH的小数位数应等于浓度的有效数字位数。常见错误是保留过多的小数位。
• 配制缓冲液时忽略体积变化：溶液混合后，必须根据总体积重新计算浓度。
• 混淆Ka与Kb：对于共轭酸碱对，务必使用正确的常数。若题目给出 Kb 而要求计算含共轭酸的酸性缓冲液的pH，须先将 Kb 转换为 Ka。

---

11. Summary of Key Formulas | 核心公式汇总

The following table summarises the key equations needed for pH calculations in CIE A-Level Chemistry.

下表汇总了CIE A-Level化学中pH计算所需的核心公式。

Quantity | 量	Formula | 公式
pH | pH值	pH = −log₁₀[H⁺]
pOH | pOH值	pOH = −log₁₀[OH⁻]
Ionic product of water | 水的离子积	Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ (25°C)
pH + pOH | pH与pOH之和	pH + pOH = 14 (at 25°C)
Acid dissociation constant | 酸解离常数	Ka = [H⁺][A⁻] / [HA]
Base dissociation constant | 碱解离常数	Kb = [BH⁺][OH⁻] / [B]
Weak acid [H⁺] | 弱酸氢离子浓度	[H⁺] = √(Ka × c)
Weak base [OH⁻] | 弱碱氢氧根浓度	[OH⁻] = √(Kb × c)
Conjugate pair relationship | 共轭对关系	Ka × Kb = Kw
Henderson-Hasselbalch (acidic buffer) | 酸性缓冲液	pH = pKa + log₁₀([A⁻] / [HA])
Henderson-Hasselbalch (basic buffer) | 碱性缓冲液	pOH = pKb + log₁₀([BH⁺] / [B])

---

12. Concluding Remarks | 结语

Mastering pH calculations requires a clear understanding of the underlying equilibrium principles and disciplined attention to assumptions and significant figures. For strong acids and bases, complete dissociation allows direct calculation. For weak acids and bases, the equilibrium constant approach with simplifying assumptions is essential. Buffer solutions and temperature effects add further depth to the topic. By systematically practising each type of calculation and being mindful of the common pitfalls highlighted in this article, students can approach pH-related examination questions with confidence.

Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）