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GCSE Edexcel Chemistry: pH Calculations - Key Points | GCSE Edexcel 化学:pH计算 考点精讲

2026-07-15 00:49 · 本文部分内容由AI辅助创作

📚 GCSE Edexcel Chemistry: pH Calculations - Key Points | GCSE Edexcel 化学:pH计算 考点精讲

Mastering pH calculations is essential for Edexcel GCSE Chemistry. This guide breaks down every concept you need: from the definition of pH and the logarithmic scale to working with strong monoprotic and diprotic acids, dilution effects, and interpreting pH curves. Step-by-step explanations, worked examples, and common pitfalls will help you tackle any exam question with confidence.

掌握pH计算对于Edexcel GCSE化学至关重要。本指南将解析你所需的每一个概念:从pH的定义与对数标度,到处理强的一元酸和二元酸、稀释效应以及解读pH曲线。逐步说明、计算示例和常见误区将帮助你自信应对任何考题。

1. The pH Scale and Its Meaning | pH标度及其含义

The pH scale is a measure of the acidity or alkalinity of an aqueous solution. It typically ranges from 0 (very acidic) to 14 (very alkaline), with 7 being neutral at 25°C.

pH标度是衡量水溶液酸碱性的指标。其范围通常为0(强酸性)至14(强碱性),在25°C时7为中性。

A low pH (0–6) indicates a higher concentration of hydrogen ions, H⁺. A high pH (8–14) indicates a higher concentration of hydroxide ions, OH⁻, and thus a lower H⁺ concentration.

低pH(0–6)表示氢离子(H⁺)浓度更高。高pH(8–14)表示氢氧根离子(OH⁻)浓度更高,因此H⁺浓度较低。


2. pH and Hydrogen Ion Concentration | pH与氢离子浓度

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⁺]

Where [H⁺] is the concentration of hydrogen ions in mol/dm³. This relationship means that as [H⁺] increases, pH decreases.

其中[H⁺]是氢离子浓度,单位为mol/dm³。此关系意味着[H⁺]增大时,pH下降。

Because the scale is logarithmic, a change of one pH unit corresponds to a tenfold change in [H⁺]. For example, a solution with pH 3 has 10 times more H⁺ than one with pH 4, and 100 times more than pH 5.

由于标度是对数的,一个pH单位的变化对应着[H⁺]的十倍变化。例如,pH 3的溶液中H⁺浓度是pH 4的10倍,是pH 5的100倍。


3. Calculating pH from Hydrogen Ion Concentration | 由氢离子浓度计算pH

To find the pH of a strongly acidic solution where you know [H⁺], simply apply the formula. Most GCSE questions give [H⁺] directly or require you to deduce it from the acid concentration.

若已知强酸溶液中的[H⁺],直接使用公式即可。大多数GCSE题目会直接给出[H⁺]或要求你从酸浓度推断。

Worked example 1: Find the pH of a solution with [H⁺] = 0.001 mol/dm³.

示例1:计算[H⁺] = 0.001 mol/dm³溶液的pH。

pH = –log₁₀(0.001) = –log₁₀(10⁻³) = 3

imply the exponent (without the negative sign) when the concentration is written as a power of 10.

注意0.001 = 10⁻³,因此当浓度写为10的幂时,pH即为指数(去掉负号)。

Worked example 2: [H⁺] = 2.5 × 10⁻⁵ mol/dm³. Find the pH to 2 decimal places.

示例2:[H⁺] = 2.5 × 10⁻⁵ mol/dm³,计算pH至两位小数。

pH = –5 × 10⁻⁵) = 4.60 (using calculator). You will be expected to use a calculator for such non‑integer values.

pH = –log₁₀(2.5 × 10⁻⁵) = 4.60(使用计算器)。此类非整数计算需要使用计算器。

[H⁺] / mol dm⁻³pH酸性程度
1.0 (10⁰)0极强酸
0.1 (10⁻¹)1强酸
0.01 (10⁻²)2较强酸
0.001 (10⁻³)3弱酸

4. Calculating Hydrogen Ion Concentration from pH | 由pH计算氢离子浓度

Rearranging the pH equation gives:

重排pH公式可得:

[H⁺] = 10⁻pH mol/dm³

This is essential when a question provides the pH and asks for the H⁺ concentration.

当题目给出pH并要求H⁺浓度时,此公式至关重要。

Worked example: A solution has a pH of 2.4. Calculate [H⁺].

示例:某溶液的pH为2.4,计算[H⁺]。

[H⁺] = 10⁻²·⁴ = 3.98 × 10⁻³ mol/dm³ (≈ 4.0 × 10⁻³). Use the 10x or inverse log button on your calculator.

[H⁺] = 10⁻²·⁴ = 3.98 × 10⁻³ mol/dm³(约4.0 × 10⁻³)。在计算器上使用10x或逆对数功能。

You can check your answer by placing this [H⁺] back into the pH formula: –log₁₀(3.98×10⁻³) ≈ 2.4.

你可以将[H⁺]代入pH公式来验证答案:–log₁₀(3.98×10⁻³) ≈ 2.4。


5. Strong Acids and Complete Ionisation | 强酸与完全电离

A strong acid is one that completely dissociates (ionises) in water, releasing all its hydrogen ions. Common strong acids at GCSE are:

强酸是指在水中完全解离(电离),释放出所有氢离子的酸。GCSE常见的强酸有:

Because ionisation is complete, the concentration of H⁺ equals the acid concentration for a monoprotic acid (see next section). Weak acids, such as ethanoic acid, only partially ionise, so [H⁺] is much smaller than the acid concentration and pH calculations are not required at GCSE.

由于电离是完全的,对于一元酸(见下一节),H⁺浓度等于酸的浓度。弱酸(如乙酸)仅部分电离,因此[H⁺]远小于酸的浓度,GCSE不要求进行弱酸pH计算。


6. Monoprotic vs Diprotic Strong Acids | 一元强酸与二元强酸

e. HCl and HNO₃ are monoprotic. For a monoprotic strong acid of concentration c, [H⁺] = c.

一元酸每个分子释放一个质子(H⁺)。HCl和HNO₃是一元酸。对于浓度为c的一元强酸,[H⁺] = c。

(assuming both protons are fully ionised, which is true for sulfuric acid at GCSE level).

二元酸每个分子释放两个质子。H₂SO₄是二元酸。对于二元强酸,[H⁺] = 2 × c(在GCSE阶段假定两个质子均完全电离)。

AcidType[H⁺] from concentration cExample: c = 0.05 mol/dm³
HClMonoprotic[H⁺] = c[H⁺] = 0.05, pH = 1.30
HNO₃Monoprotic[H⁺] = c[H⁺] = 0.05, pH = 1.30
H₂SO₄Diprotic[H⁺] = 2c[H⁺] = 0.10, pH = 1.00

Always check how many ionisable hydrogens are present before calculating pH.

在计算pH之前,务必检查酸分子中有多少个可电离的氢。


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7. Effect of Dilution on pH | 稀释对pH的影响

Diluting an acid with water reduces the concentration of H⁺, so the pH increases (moves towards 7). Diluting an alkali reduces the concentration of OH⁻, which means [H⁺] increases slightly and pH decreases (moves towards 7).

用水稀释酸会降低H⁺浓度,因此pH升高(向7靠近)。稀释碱会降低OH⁻浓度,意味着[H⁺]略微增加,pH降低(向7靠近)。

Dilution calculation for strong acid: If you add water to a known volume of strong acid, use the dilution formula:

强酸稀释计算:若向已知体积的强酸中加水,使用稀释公式:

c₁V₁ = c₂V₂

= original volume, c₂ = new [H⁺], V₂ = final total volume. Then calculate pH from new c₂.

其中c₁ = 初始[H⁺],V₁ = 原体积,c₂ = 新[H⁺],V₂ = 最终总体积。然后由c₂计算pH。

Worked example: 25 cm³ of 0.10 mol/dm³ HCl is diluted to 250 cm³. Find the new pH.

示例:25 cm³ 0.10 mol/dm³ HCl稀释至250 cm³,求新的pH。

[H⁺]₁ = 0.10, V₁=25, V₂=250. c₂ = (0.10 × 25) / 250 = 0.010 mol/dm³. pH = –log(0.010) = 2.0 (from 1.0 before dilution).

[H⁺]₁ = 0.10, V₁=25, V₂=250。c₂ = (0.10 × 25) / 250 = 0.010 mol/dm³。pH = –log(0.010) = 2.0(稀释前为1.0)。

Remember: each tenfold dilution increases the pH by 1 unit for a strong acid.

记住:对于强酸,每稀释10倍,pH增加1个单位。


8. Concentration vs Strength of an Acid | 酸的浓度与强度

These two terms are often confused. Concentration refers to how many moles of acid are dissolved in 1 dm³ of water. Strength refers to the degree of ionisation.

这两个术语常被混淆。浓度是指1 dm³水中溶解了多少摩尔酸。强度是指电离的程度。

At GCSE, you only calculate pH for strong acids, where concentration directly gives [H⁺].

在GCSE阶段,只要求计算强酸的pH,此时酸的浓度直接给出[H⁺]。


9. The Logarithmic Nature – Every 10‑Fold Change | 对数性质——每10倍变化

A core statement in the specification: as hydrogen ion concentration increases by a factor of 10, the pH decreases by 1. This relationship alone can solve many multiple‑choice questions without a calculator.

考纲中的核心表述:当氢离子浓度增加10倍时,pH下降1。仅凭此关系即可解决许多选择题,无需计算器。

Example sequence:

[H⁺] / mol dm⁻³pH
1.00
0.11
0.012
0.0013

Conversely, if pH increases by 1, [H⁺] decreases by a factor of 10. This is very useful when comparing two solutions.

反之,如果pH增加1,[H⁺]降低为原来的1/10。这在比较两种溶液时非常有用。


10. pH Curves and Neutralisation | pH曲线与中和反应

Edexcel GCSE Core Practical 3.6 involves measuring the change in pH as a base (e.g., calcium hydroxide) is added to a fixed volume of acid. Plotting pH against volume of base added gives a pH curve.

Edexcel GCSE核心实践3.6涉及测量在向固定体积的酸中加入碱(如氢氧化钙)时pH的变化。绘制pH对加入碱体积的曲线,得到pH曲线。

You may be asked to interpret such graphs, estimate the volume needed for neutralisation, or describe the trend. The steep part indicates that a small addition of base causes a large pH change around the equivalence point.

你可能需要解读此类图线、估算中和所需体积或描述趋势。陡峭部分表明在等当点附近,少量碱的加入会引起pH的巨变。


11. Common Pitfalls and Exam Tips | 常见误区与考试技巧


12. Practice Questions and Model Answers | 练习题与标准答案

Q1: Calculate the pH of 0.025 mol/dm³ nitric acid.

问题1:计算0.025 mol/dm³硝酸的pH。

HNO₃ is monoprotic strong acid, so [H⁺] = 0.025.
pH = –log(0.025) = 1.60 (2 d.p.).

HNO₃是一元强酸,因此[H⁺] = 0.025。
pH = –log(0.025) = 1.60(保留两位小数)。

Q2: A solution of sulfuric acid has pH = 1.40. Find its concentration.

问题2:某硫酸溶液的pH为1.40,求其浓度。

[H⁺] = 10⁻¹·⁴⁰ = 0.0398 mol/dm³.
Since H₂SO₄ is diprotic, [H₂SO₄] = [H⁺]/2 = 0.0199 mol/dm³ ≈ 0.020 mol/dm³.

[H⁺] = 10⁻¹·⁴⁰ = 0.0398 mol/dm³。
由于H₂SO₄是二元酸,[H₂SO₄] = [H⁺]/2 = 0.0199 mol/dm³ ≈ 0.020 mol/dm³。

Q3: 50 cm³ of 0.20 mol/dm³ HCl is mixed with 150 cm³ of water. Find the pH of the resulting solution.

问题3:将50 cm³ 0.20 mol/dm³ HCl与150 cm³水混合,求所得溶液的pH。

Total V = 200 cm³. New [H⁺] = (0.20 × 50)/200 = 0.050 mol/dm³. pH = –log(0.050) = 1.30.

总体积V = 200 cm³。新[H⁺] = (0.20 × 50)/200 = 0.050 mol/dm³。pH = –log(0.050) = 1.30。


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