IGCSE WJEC Chemistry: Mastering pH Calculations | IGCSE WJEC 化学：pH计算考点精讲

📚 IGCSE WJEC Chemistry: Mastering pH Calculations | IGCSE WJEC 化学：pH计算考点精讲

The pH scale is a cornerstone of acid-base chemistry, and at IGCSE WJEC level you are expected to perform straightforward calculations involving strong acids and strong bases. Understanding how to interconvert between hydrogen ion concentration and pH, using the ionic product of water K_w, and appreciating the logarithmic nature of the scale are all essential skills. This revision guide walks you through every calculation type you might encounter, with clear steps, examples, and tips to avoid common pitfalls.

pH标度是酸碱化学的基础，在IGCSE WJEC层次，你需要掌握强酸和强碱的简单计算。理解氢离子浓度与pH值之间的相互换算，运用水的离子积K_w，以及领会pH标度的对数本质，这些都是必备技能。本复习指南将带你遍历你可能遇到的所有计算类型，提供清晰的步骤、示例和避免常见错误的技巧。

1. Introduction to pH and the pH Scale | pH与pH标度介绍

The term pH stands for ‘power of hydrogen’ and is a measure of the acidity or alkalinity of an aqueous solution. The scale typically runs from 0 to 14, with acidic solutions having a pH less than 7, neutral solutions a pH of 7, and alkaline solutions a pH greater than 7. It is important to remember that pH is a logarithmic scale: a change of one pH unit represents a ten‑fold change in hydrogen ion concentration.

pH一词代表“氢离子活度”，用于衡量水溶液的酸碱度。标度通常从0到14，酸性溶液的pH小于7，中性溶液pH等于7，碱性溶液pH大于7。务必记住pH是对数标度：pH值每变化一个单位，氢离子浓度就变化十倍。

The hydrogen ion concentration [H⁺] is measured in mol dm⁻³. For pure water at 25 °C, [H⁺] = 1 × 10⁻⁷ mol dm⁻³, giving a neutral pH of 7. The pH scale is derived from the negative logarithm (base 10) of [H⁺]. This compact notation avoids dealing with very small numbers, e.g. 0.0000001 mol dm⁻³ becoming simply pH 7.

氢离子浓度[H⁺]以mol dm⁻³为单位。在25 °C时，纯水的[H⁺] = 1 × 10⁻⁷ mol dm⁻³，对应的中性pH为7。pH标度由[H⁺]的负对数（以10为底）得来。这种简洁的表示法避免处理极小的数字，例如0.0000001 mol dm⁻³直接记作pH 7。

2. The pH Formula and Logarithm Basics | pH公式与对数基础

The fundamental equation for pH calculation is:

pH计算的基本公式是：

pH = –log₁₀[H⁺(aq)]

Here, [H⁺(aq)] is the concentration of hydrogen ions in mol dm⁻³. The negative sign ensures that pH values are generally positive. On your scientific calculator you will use the ‘log’ button (which is base 10). To find pH, enter the [H⁺] value, press log, then change the sign. For example, if [H⁺] = 0.01 mol dm⁻³, log(0.01) = –2, so pH = –(–2) = 2.

其中[H⁺(aq)]是氢离子浓度，单位为mol dm⁻³。负号使得pH值通常为正。在科学计算器上，你会用到“log”键（以10为底）。求pH值时，输入[H⁺]数值，按log键，再变号。例如，若[H⁺] = 0.01 mol dm⁻³，log(0.01) = –2，因此pH = –(–2) = 2。

If you are given the pH and need to find [H⁺], you rearrange the formula:

如果已知pH值，需要求[H⁺]，则把公式变形：

[H⁺] = 10⁻⁽pH⁾

For a pH of 3, [H⁺] = 10⁻³ = 0.001 mol dm⁻³. This inverse relationship is crucial for many exam questions. Practise using the 10^x or antilog function (often ‘shift’ or ‘2nd’ + ‘log’) on your calculator.

对于pH = 3，[H⁺] = 10⁻³ = 0.001 mol dm⁻³。这种互逆关系对许多考题至关重要。请练习使用计算器上的10^x或反对数功能（通常是“shift”或“2nd” + “log”键）。

3. Calculating pH of Strong Acids | 强酸pH计算

Strong acids, such as hydrochloric acid (HCl), sulfuric acid (H₂SO₄) and nitric acid (HNO₃), fully dissociate in water. This means the concentration of hydrogen ions is directly related to the acid concentration and its basicity (the number of H⁺ produced per molecule).

强酸，如盐酸(HCl)、硫酸(H₂SO₄)和硝酸(HNO₃)，在水中完全解离。这意味着氢离子浓度与酸的浓度及其碱度（每个分子产生的H⁺数）直接相关。

For monobasic acids (e.g. HCl, HNO₃): one mole of acid releases one mole of H⁺. So [H⁺] = concentration of the acid. To find the pH of 0.05 mol dm⁻³ HCl: [H⁺] = 0.05 mol dm⁻³, pH = –log(0.05) = 1.30 (to 2 d.p.). Note that WJEC expects answers to be given to two decimal places unless instructed otherwise.

对于一元酸（如HCl、HNO₃）：一摩尔酸释放一摩尔H⁺。因此[H⁺] = 酸的浓度。求0.05 mol dm⁻³ HCl的pH：[H⁺] = 0.05 mol dm⁻³，pH = –log(0.05) = 1.30（保留两位小数）。注意WJEC要求答案保留两位小数，除非另有说明。

For dibasic strong acids (e.g. H₂SO₄): each mole releases two moles of H⁺. Therefore [H⁺] = 2 × acid concentration. For 0.01 mol dm⁻³ H₂SO₄, [H⁺] = 0.02 mol dm⁻³, pH = –log(0.02) = 1.70. Common mistake: forgetting to multiply by the basicity. Always check whether the acid is monobasic or dibasic.

对于二元强酸（如H₂SO₄）：每摩尔释放两摩尔H⁺。因此[H⁺] = 2 × 酸浓度。对于0.01 mol dm⁻³ H₂SO₄，[H⁺] = 0.02 mol dm⁻³，pH = –log(0.02) = 1.70。常见错误：忘记乘以碱度。一定要检查酸是一元还是二元。

4. The Ionic Product of Water, K_w | 水的离子积 K_w

Water undergoes slight self-ionisation: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq). The equilibrium constant for this process is called the ionic product of water, K_w. At 25 °C, K_w = 1.0 × 10⁻¹⁴ mol² dm⁻⁶.

水会发生微弱的自偶电离：2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)。这个过程的平衡常数称为水的离子积K_w。在25 °C时，K_w = 1.0 × 10⁻¹⁴ mol² dm⁻⁶。

K_w = [H⁺][OH⁻]

This simple expression is vital because it links the concentrations of H⁺ and OH⁻ in any aqueous solution. In pure water, [H⁺] = [OH⁻] = 1 × 10⁻⁷ mol dm⁻³, so pH = 7. When an acid is added, [H⁺] increases, thus [OH⁻] decreases to keep K_w constant. For a base, the reverse is true.

这个简单的表达式至关重要，因为它将任何水溶液中的H⁺和OH⁻浓度联系起来。在纯水中，[H⁺] = [OH⁻] = 1 × 10⁻⁷ mol dm⁻³，因此pH = 7。当加入酸时，[H⁺]增大，于是[OH⁻]减小以保持K_w恒定。对于碱，情况相反。

You will use K_w to calculate the pH of a strong base, or to find [OH⁻] given a pH. While the value of K_w changes with temperature, at IGCSE WJEC you are only required to use 1.0 × 10⁻¹⁴ at 25 °C.

你会用K_w计算强碱的pH，或根据pH求[OH⁻]。虽然K_w的值随温度变化，但在IGCSE WJEC考试中，你只需在25 °C时使用1.0 × 10⁻¹⁴。

5. Calculating pH of Strong Bases | 强碱pH计算

Strong bases such as sodium hydroxide (NaOH) and potassium hydroxide (KOH) fully dissociate to give OH⁻ ions. For a monoacidic base (one OH⁻ per formula unit): [OH⁻] = base concentration.

强碱如氢氧化钠(NaOH)和氢氧化钾(KOH)完全解离产生OH⁻离子。对于一元碱（每个单元一个OH⁻）：[OH⁻] = 碱的浓度。

To find the pH of a strong base, you must first calculate [OH⁻], then use K_w to find [H⁺], and finally convert to pH. Step‑by‑step for 0.1 mol dm⁻³ NaOH:

要求强碱的pH，你必须先计算[OH⁻]，然后用K_w求[H⁺]，最后转化为pH。以0.1 mol dm⁻³ NaOH为例的步骤：

Step 1: [OH⁻] = 0.1 mol dm⁻³
第1步：[OH⁻] = 0.1 mol dm⁻³
Step 2: [H⁺] = K_w / [OH⁻] = 1.0 × 10⁻¹⁴ / 0.1 = 1.0 × 10⁻¹³ mol dm⁻³
第2步：[H⁺] = K_w / [OH⁻] = 1.0 × 10⁻¹⁴ / 0.1 = 1.0 × 10⁻¹³ mol dm⁻³
Step 3: pH = –log(1.0 × 10⁻¹³) = 13.00
第3步：pH = –log(1.0 × 10⁻¹³) = 13.00

For a dibasic strong base like Ba(OH)₂, [OH⁻] = 2 × base concentration. Always check the formula of the base for the number of OH⁻ ions.

对于二元强碱如Ba(OH)₂，[OH⁻] = 2 × 碱的浓度。务必检查碱的化学式中OH⁻离子的数目。

An alternative shortcut: pOH = –log[OH⁻], and pH + pOH = 14 (at 25 °C). In the above example, pOH = –log(0.1) = 1, so pH = 14 – 1 = 13. This is acceptable in WJEC answers if clearly explained.

另一种快捷方式：pOH = –log[OH⁻]，且pH + pOH = 14（25 °C时）。在上例中，pOH = –log(0.1) = 1，所以pH = 14 – 1 = 13。如果能清楚解释，WJEC答题时可以接受。

6. Dilution and pH Changes | 稀释与pH变化

Diluting an acid with water decreases [H⁺] and therefore increases the pH. Because the relationship is logarithmic, a ten‑fold dilution (i.e. adding water until the volume is ten times larger) increases the pH by 1 unit – but only for strong acids. For example, 10 cm³ of 0.1 mol dm⁻³ HCl (pH = 1) diluted to 100 cm³ gives 0.01 mol dm⁻³ HCl (pH = 2).

用水稀释酸会降低[H⁺]，从而提高pH。由于是对数关系，十倍稀释（即加水至体积为原来的十倍）会使pH增加1个单位——但仅对强酸而言。例如，10 cm³的0.1 mol dm⁻³ HCl (pH = 1) 稀释至100 cm³得0.01 mol dm⁻³ HCl (pH = 2)。

For strong bases, a ten‑fold dilution reduces [OH⁻] by a factor of 10, which increases [H⁺] by a factor of 10 (since K_w is constant), causing the pH to fall by 1 unit. Diluting 0.1 mol dm⁻³ NaOH from 10 cm³ to 100 cm³ changes pH from 13 to 12.

对于强碱，十倍稀释使[OH⁻]降低至十分之一，由于K_w恒定，[H⁺]增大十倍，导致pH下降1个单位。将10 cm³的0.1 mol dm⁻³ NaOH稀释至100 cm³，pH从13变为12。

When the concentration of an acid or base becomes extremely low (e.g. 10⁻⁸ mol dm⁻³ HCl), the contribution of H⁺ from the self-ionisation of water becomes significant, and the pH approaches 7. At IGCSE WJEC, you are not expected to perform such calculations, but you should know that a highly dilute strong acid is still acidic and its pH will be just below 7, never reaching or exceeding 7 by dilution alone.

当酸或碱的浓度极低时（如10⁻⁸ mol dm⁻³ HCl），水自偶电离产生的H⁺贡献变得显著，pH会趋近于7。在IGCSE WJEC中，不要求此类计算，但你应该知道，极稀的强酸仍为酸性，其pH会略低于7，仅靠稀释不会达到或超过7。

7. Relationship Between pH and [H⁺] – Reverse Calculations | pH与[H⁺]的关系——反向计算

Examination questions frequently ask you to find the hydrogen ion concentration from a given pH. The conversion formula, as introduced earlier, is [H⁺] = 10⁻⁽pH⁾. You must be comfortable converting a pH like 3.60 into a concentration.

考试中经常要求根据已知pH求氢离子浓度。如前所述，转换公式为[H⁺] = 10⁻⁽pH⁾。你必须能熟练地将3.60这样的pH值转换为浓度。

pH	[H⁺] / mol dm⁻³
2	1 × 10⁻² = 0.01
4.5	10⁻⁴·⁵ = 3.16 × 10⁻⁵
7	1 × 10⁻⁷
10.3	10⁻¹⁰·³ = 5.01 × 10⁻¹¹

Practice using the 10^x key. For a pH of 3.60, you should obtain [H⁺] = 2.51 × 10⁻⁴ mol dm⁻³. The ability to express answers in standard form is often expected.

练习使用10^x键。对于pH = 3.60，你应得出[H⁺] = 2.51 × 10⁻⁴ mol dm⁻³。通常需要用科学记数法表示答案。

When finding [OH⁻] from pH, first determine [H⁺] using 10⁻⁽pH⁾, then apply [OH⁻] = K_w / [H⁺]. For a solution with pH = 10.0, [H⁺] = 10⁻¹⁰, so [OH⁻] = 1.0 × 10⁻¹⁴ / 1.0 × 10⁻¹⁰ = 1.0 × 10⁻⁴ mol dm⁻³.

当从pH求[OH⁻]时，先用10⁻⁽pH⁾算出[H⁺]，再套用[OH⁻] = K_w / [H⁺]。对于pH = 10.0的溶液，[H⁺] = 10⁻¹⁰，则[OH⁻] = 1.0 × 10⁻¹⁴ / 1.0 × 10⁻¹⁰ = 1.0 × 10⁻⁴ mol dm⁻³。

8. pH Curves and Indicators (Brief Summary) | pH曲线与指示剂（简单总结）

Although the core focus here is calculation, WJEC IGCSE expects you to interpret pH curves from titrations. A pH curve plots pH against the volume of titrant added. For a strong acid–strong base titration, the curve has a steep vertical portion around pH 7. From the curve you can deduce the equivalence point and select an appropriate indicator.

虽然这里的重点是计算，但WJEC IGCSE要求你能解读酸碱滴定中的pH曲线。pH曲线描绘的是pH随滴定剂加入体积的变化。对于强酸-强碱滴定，曲线在pH 7附近有一段陡直的垂直部分。从曲线可以判断等当点并选择合适的指示剂。

Indicators such as litmus, phenolphthalein and methyl orange change colour over different pH ranges. At WJEC level, you do not need to calculate the pH at the equivalence point for weak acid or weak base titrations, but you may be asked to state why a particular indicator is suitable – e.g. phenolphthalein (range pH 8.2–10.0) is ideal for strong acid–strong base because its change coincides with the steep portion of the curve.

指示剂如石蕊、酚酞和甲基橙在不同的pH范围内变色。在WJEC层次，你不需要计算弱酸或弱碱滴定中等当点的pH，但可能被问到为什么某种指示剂适用——例如酚酞（变色范围pH 8.2–10.0）非常适合强酸-强碱滴定，因为其变色范围与曲线陡直部分吻合。

Remember: a pH meter provides a direct pH reading, while an indicator gives a rough endpoint. Calculation of pH strengthens your ability to predict the shape of these curves.

记住：pH计直接给出pH读数，而指示剂给出一个粗略的终点。pH计算能增强你预测这些曲线形状的能力。

9. Common Mistakes and Tips | 常见错误与技巧

Mistake 1: Forgetting to account for dibasic acids or bases. Always write the dissociation equation first: H₂SO₄ → 2H⁺ + SO₄²⁻. Then [H⁺] = 2 × [H₂SO₄].

错误1：忘记考虑二元酸或二元碱。一定要先写解离方程式：H₂SO₄ → 2H⁺ + SO₄²⁻，然后[H⁺] = 2 × [H₂SO₄]。

Mistake 2: Confusing the use of the negative sign. pH = –log[H⁺]; missing the minus sign gives a negative pH for acids, which is rarely correct except for very concentrated strong acids.

错误2：弄混负号的使用。pH = –log[H⁺]；漏掉负号会使酸得到负pH值，除了极浓强酸外一般不正确。

Mistake 3: Applying dilution rules to weak acids or mixing up the direction of pH change. Adding water to an acid increases pH; adding water to an alkali decreases pH.

错误3：将稀释规律用于弱酸，或搞混pH变化的方向。加水稀释酸，pH增大；加水稀释碱，pH减小。

Mistake 4: Not using standard form correctly. When [H⁺] = 2.5 × 10⁻³, your calculator may display 0.0025. WJEC often wants the answer in standard form, so practise the conversion.

错误4：未能正确使用科学记数法。当[H⁺] = 2.5 × 10⁻³时，计算器可能显示0.0025。WJEC常要求用科学记数法作答，因此要练习转换。

Top tip: Always carry out a quick sanity check. Strong acids typically have pH 0–2 for concentrations ≈ 0.1–1 mol dm⁻³; strong bases have pH 12–14. If you calculate a pH of 8 for 0.1 mol dm⁻³ HCl, you have made an error.

首要技巧：始终进行快速合理性检查。浓度在0.1–1 mol dm⁻³左右的强酸pH通常为0–2；强碱pH为12–14。如果你计算出0.1 mol dm⁻³ HCl的pH为8，那你就出错了。

10. Worked Examples | 典型例题

Example 1: Calculate the pH of 0.002 mol dm⁻³ HNO₃.

例题1：计算0.002 mol dm⁻³ HNO₃的pH。

HNO₃ is monobasic, so [H⁺] = 0.002 mol dm⁻³. pH = –log(0.002) = 2.70 (2 d.p.).

HNO₃是一元酸，所以[H⁺] = 0.002 mol dm⁻³。pH = –log(0.002) = 2.70（两位小数）。

Example 2: A solution of NaOH has a concentration of 0.05 mol dm⁻³. Find its pH at 25 °C.

例题2：某NaOH溶液的浓度为0.05 mol dm⁻³。求其在25 °C时的pH。

[OH⁻] = 0.05 mol dm⁻³. [H⁺] = K_w / 0.05 = 1.0×10⁻¹⁴ / 5.0×10⁻² = 2.0×10⁻¹³ mol dm⁻³. pH = –log(2.0×10⁻¹³) = 12.70.

[OH⁻] = 0.05 mol dm⁻³。[H⁺] = K_w / 0.05 = 1.0×10⁻¹⁴ / 5.0×10⁻² = 2.0×10⁻¹³ mol dm⁻³。pH = –log(2.0×10⁻¹³) = 12.70。

Example 3: The pH of a KOH solution is 11.30. Determine the concentration of the alkali.

例题3：某KOH溶液的pH为11.30，求该碱的浓度。

[H⁺] = 10⁻¹¹·³⁰ = 5.01×10⁻¹² mol dm⁻³. Then [OH⁻] = 1.0×10⁻¹⁴ / 5.01×10⁻¹² = 2.00×10⁻³ mol dm⁻³. Since KOH is monoacidic, [KOH] = 2.00×10⁻³ mol dm⁻³.

[H⁺] = 10⁻¹¹·³⁰ = 5.01×10⁻¹² mol dm⁻³。然后[OH⁻] = 1.0×10⁻¹⁴ / 5.01×10⁻¹² = 2.00×10⁻³ mol dm⁻³。由于KOH为一元碱，[KOH] = 2.00×10⁻³ mol dm⁻³。

Example 4: What is the pH of the solution formed when 25 cm³ of 0.10 mol dm⁻³ HCl is mixed with 25 cm³ of water?

例题4：将25 cm³ 0.10 mol dm⁻³ HCl与25 cm³水混合，所得溶液的pH是多少？

Total volume = 50 cm³. The acid is diluted by a factor of 2, so new [HCl] = 0.05 mol dm⁻³. [H⁺] = 0.05 mol dm⁻³. pH = –log(0.05) = 1.30.

总体积 = 50 cm³。酸被稀释了2倍，所以新[HCl] = 0.05 mol dm⁻³。[H⁺] = 0.05 mol dm⁻³。pH = –log(0.05) = 1.30。

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IGCSE WJEC Chemistry: Mastering pH Calculations | IGCSE WJEC 化学：pH计算 考点精讲