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

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

In GCSE CIE Chemistry, pH calculations form a core part of the Acids, Bases and Salts topic. You need to understand the pH scale, the mathematical relationship between hydrogen ion concentration and pH, and be able to perform calculations for strong acids and strong bases. This article covers all the essential points, common pitfalls, and exam-style tips to help you score full marks.

在GCSE CIE化学中，pH计算是酸、碱和盐这一主题的核心内容。你需要理解pH标度、氢离子浓度与pH之间的数学关系，并能够对强酸和强碱进行计算。本文涵盖所有基本要点、常见错误以及考试风格技巧，帮助你拿到满分。

1. The pH Scale – Measuring Acidity and Alkalinity | pH标度——衡量酸碱性强弱

The pH scale is a numerical scale from 0 to 14 used to indicate how acidic or alkaline a solution is. A pH less than 7 indicates an acidic solution, pH = 7 is neutral, and pH greater than 7 indicates an alkaline solution. The lower the pH, the higher the concentration of hydrogen ions, H⁺.

pH标度是一个从0到14的数值范围，用于表示溶液的酸性或碱性程度。pH小于7表示酸性溶液，pH=7为中性，pH大于7表示碱性溶液。pH值越低，氢离子(H⁺)的浓度越高。

Acids produce H⁺ ions in water; alkalis produce OH⁻ ions.
酸在水中产生H⁺离子；碱产生OH⁻离子。
The scale is logarithmic: a change of one pH unit means a tenfold change in H⁺ concentration.
这个标度是对数标度：每变化一个pH单位，H⁺浓度变化十倍。

In the CIE exam, you may be asked to compare the acidity of two solutions based on their pH values or to predict pH change upon dilution.

在CIE考试中，你可能会被要求根据pH值比较两种溶液的酸性，或预测稀释后的pH变化。

2. Definition of pH – The Logarithmic Formula | pH的定义——对数公式

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⁺]

Here [H⁺] is the concentration of hydrogen ions in mol/dm³. This formula is given in the CIE data sheet, but you must understand how to use it in both directions.

这里[H⁺]是氢离子浓度，单位为 mol/dm³。该公式在CIE数据表中给出，但你必须理解如何正反两个方向使用它。

You also need the inverse relationship:

你还需要逆运算关系式：

[H⁺] = 10⁻ᵖᴴ

Logarithm basics: log₁₀(10⁻³) = –3, so –log₁₀(10⁻³) = 3, giving pH = 3 for 0.001 mol/dm³ H⁺.

对数基础：log₁₀(10⁻³) = –3，因此 –log₁₀(10⁻³) = 3，所以0.001 mol/dm³ H⁺溶液的pH为3。

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

For a strong monoprotic acid like HCl, which fully dissociates, [H⁺] equals the acid concentration. Example: 0.005 mol/dm³ HCl has [H⁺] = 0.005 mol/dm³. So pH = –log₁₀(0.005) ≈ 2.3. (Your calculator gives 2.3010; round to 2.3 in answers unless told otherwise.)

对于像HCl这样的强一元酸，完全电离，[H⁺]等于酸的浓度。例如：0.005 mol/dm³ HCl的[H⁺] = 0.005 mol/dm³。因此 pH = –log₁₀(0.005) ≈ 2.3。（计算器显示2.3010；除非另有要求，答案常取一位小数2.3。）

For diprotic strong acids such as H₂SO₄, each mole of acid produces two moles of H⁺. If the acid concentration is 0.01 mol/dm³, [H⁺] = 0.02 mol/dm³. Then pH = –log₁₀(0.02) ≈ 1.7.

对于二元强酸如H₂SO₄，每摩尔酸产生两摩尔H⁺。如果酸的浓度为0.01 mol/dm³，则[H⁺] = 0.02 mol/dm³，pH = –log₁₀(0.02) ≈ 1.7。

Exam tip: Always check whether the acid is monoprotic or diprotic before using the concentration directly.

考试提示：在直接使用浓度前，务必检查酸是一元酸还是二元酸。

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

To find [H⁺] when pH is known, use the equation [H⁺] = 10⁻ᵖᴴ. For instance, if a solution has pH = 4.5, then [H⁺] = 10⁻⁴·⁵ = 3.16 × 10⁻⁵ mol/dm³ (often given in standard form).

已知pH求[H⁺]时，使用公式[H⁺] = 10⁻ᵖᴴ。例如，若溶液pH = 4.5，则[H⁺] = 10⁻⁴·⁵ = 3.16 × 10⁻⁵ mol/dm³（通常以科学记数法表示）。

CIE questions sometimes ask you to compare two solutions: “Solution A has pH 3, solution B has pH 5. How many times more concentrated is the H⁺ in A than in B?” The answer: 10⁽⁵⁻³⁾ = 10² = 100 times.

CIE考题有时要求比较两种溶液：“溶液A的pH为3，溶液B的pH为5。A中H⁺浓度是B的多少倍？”答案为：10⁽⁵⁻³⁾ = 10² = 100倍。

5. pH of Strong Bases – Using the Ionic Product of Water | 强碱的pH——利用水的离子积

For alkaline solutions, you cannot directly use [OH⁻] to find pH. You need the ionic product of water, K_w = [H⁺][OH⁻] = 1 × 10⁻¹⁴ mol²/dm⁶ at 25°C (provided in data sheet).

对于碱性溶液，不能直接用[OH⁻]求pH。需要用到水的离子积K_w = [H⁺][OH⁻] = 1 × 10⁻¹⁴ mol²/dm⁶（25°C时，数据表提供）。

Steps for a strong base like NaOH: first find [OH⁻] (equal to base concentration for a monobasic base), then calculate [H⁺] = K_w / [OH⁻], and finally pH = –log₁₀[H⁺].

对于像NaOH这样的强碱，步骤为：先求出[OH⁻]（对于一元碱等于碱的浓度），然后计算[H⁺] = K_w / [OH⁻]，最后pH = –log₁₀[H⁺]。

Example: 0.1 mol/dm³ NaOH gives [OH⁻] = 0.1. Then [H⁺] = 1×10⁻¹⁴ / 0.1 = 1×10⁻¹³ mol/dm³. pH = –log₁₀(10⁻¹³) = 13.

示例：0.1 mol/dm³ NaOH溶液中[OH⁻] = 0.1。则[H⁺] = 1×10⁻¹⁴ / 0.1 = 1×10⁻¹³ mol/dm³。pH = –log₁₀(10⁻¹³) = 13。

For bases like Ba(OH)₂, which produces two OH⁻ per formula unit, [OH⁻] = 2 × base concentration. Always follow this two-step method.

对于像Ba(OH)₂这样的碱，每摩尔产生两摩尔OH⁻，故[OH⁻] = 2 × 碱的浓度。始终遵循这种两步法。

6. Dilution and pH Changes | 稀释与pH变化

Diluting an acid with water decreases [H⁺] and therefore increases the pH, moving it closer to 7 but never above 7 (unless you add an alkali). For a strong acid, each tenfold dilution raises the pH by 1 unit. Diluting a base lowers the pH towards 7.

用水稀释酸会降低[H⁺]，因此pH升高，向7靠近但不会超过7（除非加入碱）。对于强酸，每稀释10倍，pH升高1个单位。稀释碱会使pH降低，向7靠近。

Example: 10 cm³ of 0.1 mol/dm³ HCl (pH=1) diluted to 100 cm³ gives concentration 0.01 mol/dm³, [H⁺]=0.01, pH=2.

示例：10 cm³ 0.1 mol/dm³ HCl（pH=1）稀释到100 cm³，浓度变为0.01 mol/dm³，[H⁺]=0.01，pH=2。

Be careful: adding water to an acid increases the volume, so use the dilution formula C₁V₁ = C₂V₂ to find the new concentration before calculating pH.

注意：向酸中加水会增加体积，因此在计算pH前要用稀释公式C₁V₁ = C₂V₂求出新的浓度。

7. pH of Neutralisation Reactions | 中和反应的pH

When an acid and a base react, the pH of the resulting solution depends on the relative amounts. If exactly enough acid is added to neutralise the base, the salt formed and water give a neutral solution with pH=7, assuming both are strong.

当酸和碱反应时，所得溶液的pH取决于两者的相对量。如果恰好加入足够量的酸中和碱，生成的盐和水形成中性溶液，pH=7，前提是两者都是强电解质。

If there is excess acid, the solution will be acidic, and the pH is determined by the concentration of leftover H⁺. Calculate moles of H⁺ from acid, moles of OH⁻ from base, find excess moles, then divide by total volume to get [H⁺] (or [OH⁻]), and finally compute pH.

如果酸过量，溶液呈酸性，pH由剩余H⁺的浓度决定。先计算酸的H⁺物质的量，碱的OH⁻物质的量，求出过量的物质的量，再除以总体积得到[H⁺]（或[OH⁻]），最后计算pH。

Example: 25 cm³ of 0.1 mol/dm³ HCl mixed with 20 cm³ of 0.1 mol/dm³ NaOH. Moles H⁺ = 0.0025, moles OH⁻ = 0.0020. Excess H⁺ = 0.0005 mol, total volume = 0.045 dm³. [H⁺] = 0.0005/0.045 ≈ 0.0111 mol/dm³. pH = –log₁₀(0.0111) ≈ 1.95.

示例：25 cm³ 0.1 mol/dm³ HCl与20 cm³ 0.1 mol/dm³ NaOH混合。H⁺物质的量=0.0025 mol，OH⁻物质的量=0.0020 mol。过量H⁺=0.0005 mol，总体积=0.045 dm³。[H⁺] = 0.0005/0.045 ≈ 0.0111 mol/dm³。pH ≈ 1.95。

8. Using Indicators to Estimate pH | 用指示剂估算pH

While pH meters give precise pH values, indicators provide a quick way to determine whether a solution is acidic, neutral, or alkaline, and can be used to find the approximate pH if you know their colour change ranges.

虽然pH计能给出精确的pH值，指示剂则能快速判断溶液是酸性、中性还是碱性，并且如果你知道其变色范围，可以用来估算近似的pH。

Indicator	Colour in acid	Colour in alkali	pH range
Litmus	Red	Blue	5 – 8
Methyl orange	Red	Yellow	3.1 – 4.4
Phenolphthalein	Colourless	Pink	8.2 – 10.0

For CIE, you do not need to memorise exact pH ranges for all indicators, but you should know that methyl orange is red in acid and yellow in alkali; phenolphthalein is colourless in acid and pink in alkali. Questions may ask which indicator is suitable for a given titration based on the pH at the equivalence point.

对于CIE，你不需要记住所有指示剂的精确pH范围，但应知道甲基橙在酸中呈红色、在碱中呈黄色；酚酞在酸中无色、在碱中呈粉红色。考题可能会问根据等当点的pH选择哪种指示剂合适。

9. Common Mistakes in pH Calculations | pH计算中的常见错误

Mistake 1: Forgetting that H₂SO₄ is diprotic. Using the acid concentration directly as [H⁺] will give a pH that is too high (less acidic). Always multiply by 2 for the first dissociation (both protons are strong for GCSE CIE).

错误1：忘记H₂SO₄是二元酸。直接将酸的浓度当作[H⁺]会使pH偏高（酸性偏弱）。对于CIE GCSE，第一步电离的两个质子都完全电离，因此要乘以2。

Mistake 2: Not converting volumes to dm³. If volumes are in cm³, divide by 1000 when calculating concentration in mol/dm³.

错误2：未将体积换算为dm³。如果体积单位是cm³，计算浓度（mol/dm³）时要除以1000。

Mistake 3: Confusing pH and pOH, or applying [H⁺] = 10⁻ᵖᴴ for bases directly. For bases, you must use K_w.

错误3：混淆pH和pOH，或对碱直接使用[H⁺] = 10⁻ᵖᴴ。对于碱，必须使用K_w。

Mistake 4: Not rounding answers correctly. pH values are usually given to one or two decimal places, and [H⁺] in standard form to 2 or 3 significant figures.

错误4：答案未正确四舍五入。pH值通常保留一位或两位小数，[H⁺]以科学记数法保留2至3位有效数字。

10. pH Calculation Worked Examples – Exam Style | pH计算例题——考试风格

Question 1: Calculate the pH of 0.0025 mol/dm³ nitric acid, HNO₃.

问题1：计算0.0025 mol/dm³硝酸(HNO₃)的pH值。

Solution: HNO₃ is monoprotic strong acid, so [H⁺] = 0.0025. pH = –log₁₀(0.0025) = 2.60 (to 2 d.p.).

解答：HNO₃是一元强酸，所以[H⁺] = 0.0025。pH = –log₁₀(0.0025) = 2.60（保留两位小数）。

Question 2: What is the pH of a solution made by dissolving 0.40 g of NaOH in water to make 100 cm³? (Mᵣ of NaOH = 40)

问题2：将0.40 g NaOH溶于水配成100 cm³溶液，该溶液的pH是多少？（NaOH的Mᵣ = 40）

Solution: Moles NaOH = mass/Mᵣ = 0.40/40 = 0.010 mol. Volume = 0.100 dm³. [OH⁻] = 0.010/0.100 = 0.10 mol/dm³. [H⁺] = K_w/[OH⁻] = 1×10⁻¹⁴/0.10 = 1×10⁻¹³. pH = 13.

解答：NaOH物质的量=质量/Mᵣ = 0.40/40 = 0.010 mol。体积=0.100 dm³。[OH⁻] = 0.010/0.100 = 0.10 mol/dm³。[H⁺] = K_w/[OH⁻] = 1×10⁻¹⁴/0.10 = 1×10⁻¹³。pH = 13。

Question 3: 20 cm³ of 0.05 mol/dm³ H₂SO₄ is added to 30 cm³ of 0.10 mol/dm³ NaOH. Calculate the pH of the mixture.

问题3：将20 cm³ 0.05 mol/dm³ H₂SO₄加入30 cm³ 0.10 mol/dm³ NaOH中，计算混合物的pH。

Solution: Moles H⁺ from H₂SO₄ = 2 × (0.05 × 0.020) = 0.0020 mol. Moles OH⁻ from NaOH = 0.10 × 0.030 = 0.0030 mol. Excess OH⁻ = 0.0010 mol. Total volume = 0.050 dm³. [OH⁻] = 0.0010/0.050 = 0.020 mol/dm³. [H⁺] = 1×10⁻¹⁴/0.020 = 5×10⁻¹³ mol/dm³. pH = –log₁₀(5×10⁻¹³) ≈ 12.3.

解答：H₂SO₄提供的H⁺物质的量= 2 × (0.05 × 0.020) = 0.0020 mol。NaOH提供的OH⁻物质的量= 0.10 × 0.030 = 0.0030 mol。过量OH⁻ = 0.0010 mol。总体积= 0.050 dm³。[OH⁻] = 0.0010/0.050 = 0.020 mol/dm³。[H⁺] = 1×10⁻¹⁴/0.020 = 5×10⁻¹³ mol/dm³。pH ≈ 12.3。

11. Experimental Determination of pH | pH的实验测定

In the laboratory, you might use universal indicator solution or pH paper to estimate pH. The colour is compared with a chart. For more accurate measurements, a pH meter (with a probe) is used. You may be asked to describe how to use a pH meter: rinse the probe with distilled water, place it in the solution, stir gently, and record the reading when stable.

在实验室里，你可能会使用通用指示剂溶液或pH试纸来估计pH，颜色与色卡对比。如需更精确的测量，则使用pH计（带探头）。你可能会被要求描述如何使用pH计：用蒸馏水冲洗探头，放入溶液中，轻轻搅拌，待读数稳定后记录。

CIE practical-style questions can ask about the trend in pH when an acid is gradually added to an alkali, or the shape of a pH curve. For a strong acid–strong base titration, the pH jumps sharply from around 3 to 11 at the end point.

CIE的实验类题目可能会问，当酸逐渐加入碱中时pH的变化趋势，或pH曲线的形状。对于强酸–强碱滴定，在终点附近pH会从大约3急剧跳到11。

12. Quick Revision Checklist | 快速复习清单

pH = –log₁₀[H⁺] ; [H⁺] = 10⁻ᵖᴴ
For strong monoprotic acids, [H⁺] = acid concentration; for H₂SO₄, [H⁺] = 2 × concentration.
对于强一元酸，[H⁺] = 酸的浓度；对于H₂SO₄，[H⁺] = 2 × 浓度。
For bases, use K_w = 1 × 10⁻¹⁴ to find [H⁺] after determining [OH⁻].
对于碱，先确定[OH⁻]，再用K_w = 1 × 10⁻¹⁴求[H⁺]。
Dilution factor of 10 changes pH by 1 unit (for strong acids/bases).
稀释10倍，pH变化1个单位（对强酸/碱）。
In neutralisation, calculate excess moles of H⁺ or OH⁻, then concentration, then pH.
中和反应中，先计算过量的H⁺或OH⁻物质的量，再算浓度，最后求pH。
Always convert cm³ to dm³ (÷1000).
始终将cm³转换为dm³（除以1000）。
Round pH to 1 or 2 decimal places unless told otherwise.
除非另有要求，pH保留1或2位小数。

Mastering these pH calculation skills will also prepare you for more advanced topics such as buffer solutions at A Level, so it is worth understanding the underlying principles now.

掌握这些pH计算技能也会为你将来学习A Level的缓冲溶液等高级主题打下基础，因此现在就理解其根本原理是非常值得的。

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GCSE CIE Chemistry: pH Calculations | GCSE CIE 化学：pH计算 考点精讲