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

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

Welcome to your focused revision guide for pH calculations – a key quantitative skill in the AQA GCSE Chemistry higher tier. This article explains the meaning of pH, shows you how to use the essential formula pH = –log₁₀[H⁺], and teaches you to work out pH for strong acids, strong bases, and dilutions. We will also tackle common exam traps and give you worked examples to build your confidence. Let’s turn those tricky log calculations into easy marks.

欢迎阅读 pH 计算精讲——这是 AQA GCSE 化学高阶内容中的核心定量技能。本文会解释 pH 的含义，教你如何运用关键公式 pH = –log₁₀[H⁺]，以及如何计算强酸、强碱和稀释溶液的 pH。我们还会剖析常见考试陷阱并提供例题，帮你将看似棘手的对数计算变成稳稳的得分点。

1. What is pH? | 什么是 pH？

pH is a number that tells us how acidic or alkaline a solution is. The letters stand for ‘power of hydrogen’ because the value is related to the concentration of hydrogen ions, H⁺, in the solution. A Danish chemist, Søren Sørensen, introduced the concept in 1909 to simplify working with very small concentrations. Mathematically, pH is defined as the negative logarithm to base 10 of the hydrogen ion concentration. This means every time the [H⁺] changes by a factor of 10, the pH shifts by 1 unit. The more H⁺ ions present, the lower the pH – so strong acids have low pH values, while strong alkalis have high pH values.

pH 是一个表示溶液酸碱度的数值。字母 p 代表“指数”(power)，H 代表氢离子，因此 pH 与溶液中氢离子 H⁺ 的浓度直接相关。1909 年丹麦化学家 Søren Sørensen 提出这一概念，以便处理极小的浓度值。数学上，pH 定义为氢离子浓度的以 10 为底的负对数。这意味着每当 [H⁺] 改变 10 倍，pH 就移动 1 个单位。H⁺ 离子越多，pH 越低——因此强酸 pH 低，强碱 pH 高。

2. The pH Scale and Hydrogen Ion Concentration | pH 标度与氢离子浓度

The pH scale typically runs from 0 to 14. A solution with pH 7 is neutral: [H⁺] equals [OH⁻] and both are 1.0 × 10⁻⁷ mol dm⁻³ at 25 °C. Solutions with pH less than 7 are acidic ([H⁺] > [OH⁻]), and those with pH greater than 7 are alkaline ([H⁺] < [OH⁻]). Because the scale is logarithmic, a solution of pH 3 has ten times the [H⁺] of a pH 4 solution, and a hundred times that of pH 5. This logarithmic nature is why a seemingly small change in pH can mean a huge change in acidity – a fact examiners love to test.

pH 标度通常为 0 到 14。pH = 7 的溶液为中性：此时 [H⁺] = [OH⁻]，在 25 °C 下两者均为 1.0 × 10⁻⁷ mol dm⁻³。pH 小于 7 的溶液呈酸性（[H⁺] > [OH⁻]），大于 7 的呈碱性（[H⁺] < [OH⁻]）。由于标度是对数性质的，pH = 3 的溶液其 [H⁺] 是 pH = 4 溶液的 10 倍，是 pH = 5 溶液的 100 倍。这种对数关系意味着 pH 的微小变化对应酸度的巨大改变——正是考官喜欢考察的知识点。

3. The pH Equation: pH = –log₁₀[H⁺] | pH 计算公式

The foundation of all GCSE pH calculations is the equation:

所有 GCSE pH 计算的基础都是这个公式：

pH = –log₁₀[H⁺]

Here [H⁺] is the hydrogen ion concentration in mol dm⁻³. The ‘log₁₀’ function is the common logarithm; your calculator’s ‘log’ button normally performs this. The negative sign flips the number so that higher [H⁺] gives a lower pH. For instance, if [H⁺] = 0.1 mol dm⁻³, log₁₀(0.1) = –1, so pH = –(–1) = 1. If you are not yet comfortable with logarithms, practise entering values such as 0.01 and 0.001 to see how log₁₀ and the negative sign interact.

式中 [H⁺] 是氢离子浓度，单位为 mol dm⁻³。log₁₀ 是常用对数，计算器上的 “log” 键通常直接使用。负号会使数值翻转，使得 [H⁺] 越高时 pH 越低。例如当 [H⁺] = 0.1 mol dm⁻³，log₁₀(0.1) = –1，因此 pH = –(–1) = 1。如果你对对数还不太熟悉，请多练习输入 0.01、0.001 等数值，观察 log₁₀ 与负号如何配合。

4. Calculating pH from [H⁺] | 由氢离子浓度计算 pH

Follow these steps to find the pH when you are given [H⁺]:

已知 [H⁺] 求 pH 时，按下列步骤操作：

1. Ensure the [H⁺] is in mol dm⁻³. 2. Press the ‘log’ button on your calculator. 3. Multiply by –1 (or simply change the sign). The result is the pH.

1. 确保 [H⁺] 的单位为 mol dm⁻³。2. 在计算器上按 “log” 键。3. 乘以 –1（或直接变号），所得即为 pH。

Example: Calculate the pH of a solution with [H⁺] = 2.5 × 10⁻³ mol dm⁻³.
log₁₀(2.5 × 10⁻³) ≈ –2.60, so pH = 2.60.

示例：计算 [H⁺] = 2.5 × 10⁻³ mol dm⁻³ 溶液的 pH。
log₁₀(2.5 × 10⁻³) ≈ –2.60，因此 pH = 2.60。

Always give pH to 1 or 2 decimal places as appropriate. Remember: an answer like pH = 2.6 could lose a mark if the question expects 2.60.

请根据题目要求将 pH 保留 1 位或 2 位小数。注意：如果要求保留两位小数，写 pH = 2.6 可能会丢分。

5. Finding [H⁺] from pH | 由 pH 求氢离子浓度

To work backwards from pH to [H⁺], rearrange the formula:

从 pH 反推 [H⁺] 时，需变换公式：

[H⁺] = 10⁻ᵖᴴ mol dm⁻³

Use the 10ˣ button (often ’10^x’ or ‘antilog’) on your calculator. Type the negative of the pH and apply the function.

使用计算器上的 10ˣ 键（常标为 “10^x” 或 “antilog”）。输入 pH 的负值并执行该功能。

Example: Find [H⁺] for a solution of pH 4.80.
[H⁺] = 10⁻⁴·⁸⁰ ≈ 1.58 × 10⁻⁵ mol dm⁻³.

示例：求 pH = 4.80 溶液的 [H⁺]。
[H⁺] = 10⁻⁴·⁸⁰ ≈ 1.58 × 10⁻⁵ mol dm⁻³。

Many candidates lose marks by forgetting to include the negative sign before the pH when using the 10ˣ key. Double-check your answer: the [H⁺] should be a small number if the pH is greater than 1.

很多考生在使用 10ˣ 键时会忘记在 pH 前加负号，导致丢分。检查答案是否合理：若 pH 大于 1，[H⁺] 应该是一个很小的数。

6. Strong Acids and Complete Dissociation | 强酸与完全解离

Strong acids such as hydrochloric acid (HCl), nitric acid (HNO₃) and sulfuric acid (H₂SO₄) completely dissociate in water. This means every molecule releases its H⁺ ions into the solution. For a strong monoprotic acid like HCl, the [H⁺] is simply equal to the concentration of the acid:

强酸（如盐酸 HCl、硝酸 HNO₃ 和硫酸 H₂SO₄）在水中完全解离，意味着每个分子都释放出其 H⁺ 离子。对于像 HCl 这样的一元强酸，[H⁺] 就等于酸的浓度：

0.10 mol dm⁻³ HCl → [H⁺] = 0.10 mol dm⁻³ → pH = –log₁₀(0.10) = 1.00

Sulfuric acid is a strong diprotic acid: each molecule produces two H⁺ ions upon complete dissociation. Therefore, for H₂SO₄, [H⁺] = 2 × acid concentration.

硫酸是二元强酸：完全解离时每个 H₂SO₄ 分子产生两个 H⁺。因此对于 H₂SO₄，[H⁺] = 2 × 酸浓度。

Example: 0.050 mol dm⁻³ H₂SO₄ → [H⁺] = 2 × 0.050 = 0.100 mol dm⁻³ → pH = 1.00.

示例：0.050 mol dm⁻³ H₂SO₄ → [H⁺] = 2 × 0.050 = 0.100 mol dm⁻³ → pH = 1.00。

Always check whether the acid is monoprotic or diprotic. This is a classic GCSE mistake.

务必辨别酸是一元还是二元——这是 GCSE 经典易错点。

7. pH of Strong Bases using Kw | 强碱的 pH 计算（使用水的离子积）

For strong alkalis such as sodium hydroxide (NaOH) or potassium hydroxide (KOH), dissociation is also complete, giving OH⁻ ions. To find the pH, you must first work out [OH⁻] and then use the ionic product of water, Kw:

对于强碱（如氢氧化钠 NaOH、氢氧化钾 KOH），解离也是完全的，产生 OH⁻ 离子。计算 pH 时需要先得到 [OH⁻]，再借助水的离子积 Kw：

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ (at 25 °C)

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶（25 °C 时）

Rearrange to find [H⁺]: [H⁺] = Kw ÷ [OH⁻]. Then calculate pH = –log₁₀[H⁺].

变换得 [H⁺] = Kw ÷ [OH⁻]，再计算 pH = –log₁₀[H⁺]。

Example: Calculate the pH of 0.20 mol dm⁻³ NaOH.
[OH⁻] = 0.20 mol dm⁻³.
[H⁺] = (1.0 × 10⁻¹⁴) / 0.20 = 5.0 × 10⁻¹⁴ mol dm⁻³.
pH = –log₁₀(5.0 × 10⁻¹⁴) ≈ 13.30.

示例：计算 0.20 mol dm⁻³ NaOH 的 pH。
[OH⁻] = 0.20 mol dm⁻³。
[H⁺] = (1.0 × 10⁻¹⁴) / 0.20 = 5.0 × 10⁻¹⁴ mol dm⁻³。
pH = –log₁₀(5.0 × 10⁻¹⁴) ≈ 13.30。

Note: the Kw value is only valid at 25 °C. Some questions may ask you to use a different Kw if the temperature changes, but at GCSE this is rare.

注意：Kw 值仅在 25 °C 时适用。若温度改变，某些题目可能给出不同的 Kw 值，但在 GCSE 中较少见。

8. Dilution and pH Change | 稀释与 pH 变化

When you dilute a strong acid, the concentration of H⁺ decreases, so the pH increases

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