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

📚 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。

pH < 7: acidic, more H⁺ than OH⁻ / 酸性，H⁺多于OH⁻
pH = 7: neutral, [H⁺] = [OH⁻] / 中性，[H⁺] = [OH⁻]
pH > 7: alkaline, more OH⁻ than H⁺ / 碱性，OH⁻多于H⁺

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

Notice that 0.001 = 10⁻³, so the pH is simply 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 = –log₁₀(2.5 × 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⁻³	pH	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 10^x 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10^x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：

Hydrochloric acid, HCl / 盐酸，HCl
Sulfuric acid, H₂SO₄ / 硫酸，H₂SO₄
Nitric acid, HNO₃ / 硝酸，HNO₃

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

A monoprotic acid releases one proton (H⁺) per molecule. 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。

A diprotic acid releases two protons per molecule. H₂SO₄ is diprotic. For a diprotic strong acid, [H⁺] = 2 × 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）。

Acid	Type	[H⁺] from concentration c	Example: c = 0.05 mol/dm³
HCl	Monoprotic	[H⁺] = c	[H⁺] = 0.05, pH = 1.30
HNO₃	Monoprotic	[H⁺] = c	[H⁺] = 0.05, pH = 1.30
H₂SO₄	Diprotic	[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。

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

where c₁ = original [H⁺], V₁ = 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。

Concentrated strong acid: many moles, fully ionised, very low pH. / 浓强酸：摩尔数多，完全电离，pH很低。
Dilute strong acid: few moles, still fully ionised, moderate pH. / 稀强酸：摩尔数少，仍完全电离，pH适中。
Concentrated weak acid: many moles, but only a tiny fraction ionised; pH is not as low as expected from its concentration. / 浓弱酸：摩尔数多，但仅极少部分电离；pH不如从其浓度预期的那么低。

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⁻³	pH
1.0	0
0.1	1
0.01	2
0.001	3

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。

The curve starts at a low pH (acidic).
It rises gradually at first, then very steeply near the equivalence point (pH 7 for strong acid–strong base).
After the steep rise, the curve levels off at a high pH (alkaline).

曲线起始于低pH（酸性）。
最初缓慢上升，在接近终点（强酸强碱为pH 7）时非常陡峭。
陡升之后，曲线在较高pH（碱性）趋于平缓。

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

Confusing [H⁺] with acid concentration for diprotic acids. Always multiply by 2 for H₂SO₄. / 混淆二元酸的[H⁺]与酸浓度。对于H₂SO₄，始终要乘以2。
Forgetting to convert volumes to dm³. If volumes are given in cm³, remember 1 dm³ = 1000 cm³ when using c = n/V. / 忘记将体积换算为dm³。若题目以cm³给出体积，在使用c = n/V时请记住1 dm³ = 1000 cm³。
Using the wrong key on the calculator. For 10⁻pH use the “10ˣ” or “shift log” button, not “eˣ”. / 按错计算器按键。计算10⁻pH时使用”10ˣ”或”shift log”键，而不是”eˣ”。
Rounding errors. Keep intermediate values in your calculator; only round the final pH to the required decimal places (usually 2). / 舍入误差。将中间结果保留在计算器中；只将最终pH值按要求的小数位数（通常为2位）进行舍入。
Misreading pH scale direction. Lower pH = higher acidity = higher [H⁺]. Visualise the scale. / 误读pH标度方向。pH越低 = 酸性越强 = [H⁺]越高。心中想象标度。
Assuming all acids are strong. Only HCl, HNO₃, H₂SO₄ are strong; others like ethanoic, citric, carbonic are weak. You won’t calculate pH for weak acids. / 假定所有酸都是强酸。只有HCl、HNO₃、H₂SO₄是强酸；其他如乙酸、柠檬酸、碳酸是弱酸。弱酸不要求计算pH。

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