📚 A-Level Edexcel Chemistry: pH Calculations – Key Concepts and Exam Tips | A-Level Edexcel 化学：pH 计算考点精讲
pH calculations are a cornerstone of A-Level Edexcel Chemistry, appearing in both Paper 1 and Paper 2 as well as in the practical skills questions. Being able to calculate the pH of strong and weak acids, strong and weak bases, and buffer solutions is essential for tackling quantitative problems. This guide walks through every key concept, formula and exam technique in a logical order, from the definition of pH of pH to complex buffer calculations and the ionic product of water, through Ka and Kb calculations, to buffer systems, titration curves and indicator choice. Worked examples, common errors and mark-scheme language are included so you can write answers that actually score the marks.
pH 计算是 A-Level Edexcel 化学的基石，在试卷一、试卷二以及实验技能题中都会出现。掌握强酸、弱酸、强碱、弱碱以及缓冲溶液 pH 的计算方法，对于解决定量问题至关重要。本指南按逻辑顺序梳理所有核心概念、公式与考试技巧：从 pH 的定义和水的离子积，到 Ka、Kb 计算，再到缓冲体系、滴定曲线与指示剂选择。文中配有例题、常见错误分析和得分点用语，帮助你写出真正能拿分的答案。
1. The pH Scale and Definition | pH 标度与定义
hydrogen ion concentration in aqueous solution. The definition required by Edexcel is pH = – on concentration. Pure water at 298 K has [H⁺] = 1.0 × 10⁻⁷ mol dm⁻³, which gives a neutral pH of exactly 7.00. Solutions with pH below 7 are acidic, and solutions with pH above 7 are alkaline at this temperature.
pH 是水溶液中氢离子浓度的对数度量。Edexcel 要求掌握的定义为 pH = –log₁₀[H⁺]，其中 [H⁺] 的单位是 mol dm⁻³。由于该标度是对数形式，pH 每变化 1 个单位，氢离子浓度就变化 10 倍。在 298 K 下，纯水中 [H⁺] = 1.0 × 10⁻⁷ mol dm⁻³，因此中性 pH 正好是 7.00。在此温度下，pH 小于 7 为酸性，大于 7 为碱性。
pH = –log₁₀[H⁺] ⇔ [H⁺] = 10⁻ᵖᴴ mol dm⁻³
Always give pH values to two decimal places, because the number before the decimal point comes from the power of ten and is not a significant figure. Remember also that the value 7.00 for neutrality is only true at 298 K, so never quote it as a universal constant.
pH 值一律保留两位小数，因为小数点前的数字来自 10 的幂次，不计入有效数字。还要记住，7.00 这个中性值只在 298 K 成立，不要把 7 当成普适常数。
2. The Ionic Product of Water, Kw | 水的离子积 Kw
Water self-ionises to a very small extent according to the equilibrium H₂O ⇌ m⁻⁶. At 298 K, Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶, which gives pKw = 14.00. This single expression links [H⁺] and [OH⁻] and is the bridge between acid calculations and base calculations.
水会发生极微弱的自电离，平衡为 H₂O ⇌ H⁺ + OH⁻。以浓度表示的该平衡常数就是水的离子积 Kw，单位为 mol² dm⁻⁶。在 298 K 时，Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶，即 pKw = 14.00。这个表达式把 [H⁺] 与 [OH⁻] 联系起来，是酸计算与碱计算之间的桥梁。
or example, 310 K, Kw is larger than 1.0 × 10⁻¹⁴ and the pH of pure water falls below 7.00, even though the water is still neutral because [H⁺] = [OH⁻]. Exam questions often test this idea directly, so always check the temperature given in the question before assuming pKw = 14.
水的自电离是吸热过程，升温使平衡右移，Kw 增大。因此例如在 310 K 时，Kw 大于 1.0 × 10⁻¹⁴，纯水的 pH 会低于 7.00，但水依然是中性的，因为 [H⁺] = [OH⁻]。考试常直接考查这一点，所以在假设 pKw = 14 之前，务必先看题目给出的温度。
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ (298 K) pKw = –log₁₀Kw = 14.00
3. Strong Acids: Direct Calculation | 强酸：直接计算
A strong acid dissociates completely in water, so the concentration of hydrogen ions equals the concentration of the acid, provided the acid is monoprotic. Hydrochloric acid, nitric acid and the monoprotic organic acids behave in this way. For a diprotic strong acid such as sulfuric acid, Edexcel questions normally assume that both protons are released, so [H⁺] is twice the acid concentration. In that case you must multiply before taking the logarithm.
强酸在水中完全电离，因此对于一元酸，氢离子浓度等于酸的浓度。盐酸、硝酸以及一元有机强酸都符合这一规律。对于硫酸这样的二元强酸，Edexcel 的题目通常假设两个质子都释放出来，所以 [H⁺] 是酸浓度的两倍。此时必须先乘以 2，再取对数。
Worked example: ol dm⁻³ HCl. Since HCl is monoprotic and fully dissociated, [H⁺] = 0.050 mol dm⁻³, so pH = – assuming complete dissociation of both protons: [H⁺] = 2 × 0.050 = 0.100 mol dm⁻³, so pH = –log₁₀(0.100) = 1.00.
例题：计算 0.050 mol dm⁻³ HCl 的 pH。因 HCl 为一元强酸且完全电离，[H⁺] = 0.050 mol dm⁻³，故 pH = –log₁₀(0.050) = 1.30。再计算 0.050 mol dm⁻³ H₂SO₄ 的 pH（假设两个质子均完全电离）：[H⁺] = 2 × 0.050 = 0.100 mol dm⁻³，故 pH = –log₁₀(0.100) = 1.00。
A common exam trap is diluting a strong acid. Diluting by a factor of ten lowers [H⁺] by a factor of ten, so the pH increases by exactly 1.00 unit. Comparing the calculated pH of a strong acid with that of a weak acid of the same concentration is also a standard question: the weak acid always has the higher pH because it is only partially ionised.
常见陷阱是强酸的稀释。稀释 10 倍使 [H⁺] 降为原来的十分之一，pH 恰好增加 1.00 个单位。另一个标准题型是比较同浓度强酸与弱酸的 pH：弱酸只部分电离，因此 pH 一定更高。
4. Strong Bases and the pH + pOH = 14 Relationship | 强碱与 pH + pOH = 14 的关系
For bases it is easier to calculate [OH⁻] first, convert this to pOH using pOH = –log₁₀[OH⁻], and then use the relationship pH + pOH = pKw. At 298 K this becomes pH = 14.00 – ch formula unit releases two hydroxide ions, so [OH⁻] is twice the concentration of the base.
对于碱，先计算 [OH⁻] 更为方便：用 pOH = –log₁₀[OH⁻] 求出 pOH，再用关系式 pH + pOH = pKw。在 298 K 下即 pH = 14.00 – pOH。NaOH、KOH 等第一族氢氧化物完全电离，所以 [OH⁻] 等于碱的浓度。对于氢氧化钡 Ba(OH)₂，每个化学式单位释放两个氢氧根离子，因此 [OH⁻] 是碱浓度的两倍。
Worked example: Calculate the pH of 0.020 mol dm⁻³ NaOH. [OH⁻] = 0.020 mol dm⁻³, so pOH = –log₁₀(0.020) = 1.70, and pH = 14.00 – 0.020 mol dm⁻³, giving exactly the same pH of 12.30, which is a neat check on your method.
例题：计算 0.020 mol dm⁻³ NaOH 的 pH。[OH⁻] = 0.020 mol dm⁻³，pOH = –log₁₀(0.020) = 1.70，故 pH = 14.00 – 1.70 = 12.30。对于 0.010 mol dm⁻³ Ba(OH)₂，[OH⁻] = 0.020 mol dm⁻³，pH 同样为 12.30，这可以很好地检验你的解题思路。
etting to double the concentration for dibasic bases, or applying the relationship pH = 14 – ure other than 298 K. If the question gives a different value of Kw, you must use pH = pKw – pOH instead of the familiar 14.
常见错误包括：忘记二元碱浓度要乘以 2；以及在非 298 K 时仍套用 pH = 14 – pOH。若题目给出不同的 Kw 值，必须改用 pH = pKw – pOH。
5. Weak Acids and the Acid Dissociation Constant Ka | 弱酸与酸解离常数 Ka
A weak acid only partially dissociates, so an equilibrium is established: HA ⇌ anageable: the concentration of HA at equilibrium is approximately equal to the initial concentration, because so little has ionised, and [H⁺] = [A⁻] because each dissociation produces one of each ion.
弱酸只部分电离，因此存在平衡：HA ⇌ H⁺ + A⁻。酸解离常数定义为 Ka = [H⁺][A⁻] / [HA]，单位为 mol dm⁻³。两个简化假设让计算变得可行：一是由于电离量极小，平衡时 HA 的浓度近似等于初始浓度；二是由于每次电离同时产生一个 H⁺ 和一个 A⁻，所以 [H⁺] = [A⁻]。
Substituting these assumptions into the Ka expression gives the working equation [H⁺] = √Ka × c, where c is the initial concentration of the acid. You then take the negative logarithm to obtain the pH. Lower values of Ka correspond to weaker acids, because the equilibrium lies further to the left.
把这些假设代入 Ka 表达式，得到实用公式 [H⁺] = √Ka × c，其中 c 为酸的初始浓度。再取负对数即可得到 pH。Ka 越小，酸性越弱，因为平衡越趋向左侧。
Ka = [H⁺][A⁻] / [HA] ⇒ [H⁺] = √Ka × c
Worked example: Calculate the pH of 0.10 mol dm⁻³ ethanoic acid, Ka = 1.8 × 10⁻⁵ mol dm⁻³. [H⁺] = √1.8 × 10⁻⁵ × 0.10 = √1.8 × 10⁻⁶ = 1.34 × 10⁻³ mol dm⁻³. Therefore pH = – .00, which is about 74 times the hydrogen ion concentration of the ethanoic acid solution.
例题：计算 0.10 mol dm⁻³ 乙酸溶液的 pH，Ka = 1.8 × 10⁻⁵ mol dm⁻³。[H⁺] = √1.8 × 10⁻⁵ × 0.10 = √1.8 × 10⁻⁶ = 1.34 × 10⁻³ mol dm⁻³，故 pH = –log₁₀(1.34 × 10⁻³) = 2.87。作为对比，0.10 mol dm⁻³ HCl 的 pH 为 1.00，其氢离子浓度约为该乙酸溶液的 74 倍。
concentration and multiply by 100. In the example above, (1.34 × 10⁻³ / 0.10) × 100 = 1.34 %, which confirms that the acid is indeed only slightly ionised.
也可以计算电离度：用 [H⁺] 除以初始浓度再乘以 100。上例中 (1.34 × 10⁻³ / 0.10) × 100 = 1.34 %，说明该酸确实只轻微电离。
6. When You Can Ignore the Extent of Ionisation | 何时可以忽略电离量
mount that dissociates is small compared with the initial concentration. In practice, the shortcut is safe when the percentage dissociation is less than about 5 percent, which is roughly equivalent to c / Ka being greater than 100. For most A-Level questions involving ethanoic acid at concentrations of 0.01 mol dm⁻³ or above, the approximation holds comfortably.
近似式 [HA]平衡 ≈ [HA]初始 只有在电离量远小于初始浓度时才成立。实践中，当电离度低于约 5 % 时该简式是安全的，大致相当于 c / Ka 大于 100。对于大多数 A-Level 题目，只要乙酸的浓度在 0.01 mol dm⁻³ 或以上，该近似都完全成立。
ther than taking a square root. Writing x for [H⁺], the Ka expression becomes Ka = x² / (c – x), which rearranges to x² + Ka x – is solved with the quadratic formula. In an exam, always state the assumption explicitly: "assume the concentration of undissociated acid remains constant" is a standard mark-scheme phrase.
若该假设不成立，就不能直接开平方，而必须解一元二次方程。设 x = [H⁺]，则 Ka 表达式变为 Ka = x² / (c – x)，整理得 x² + Ka x –
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