📚 pH Calculations: Key Points for IB & OCR Chemistry | pH计算 考点精讲
Understanding pH calculations is essential for both IB Chemistry and OCR A-Level Chemistry. This guide covers the most important concepts, formulas, and problem-solving strategies you need to master. We will go through strong and weak acids/bases, buffers, salt hydrolysis, and titration curves, with step-by-step methods and common pitfalls. Whether you are preparing for Paper 2 or the multiple-choice exam, these high-yield points will boost your confidence.
理解pH计算对于IB化学和OCR A-Level化学都至关重要。本指南涵盖你需要掌握的最重要概念、公式和解题策略。我们将逐步讲解强/弱酸、强/弱碱、缓冲溶液、盐类水解和滴定曲线，并提供分步方法与常见误区。无论你是在准备论述题还是选择题，这些高频考点都能提升你的信心。
1. What is pH? – The Core Definition | 什么是pH？核心定义
pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration: pH = –log₁₀[H⁺]. This means that every one-unit decrease in pH represents a tenfold increase in [H⁺]. At 25°C, pure water has [H⁺] = 1.0 × 10⁻⁷ mol dm⁻³, giving a pH of 7.00.
pH定义为氢离子浓度的负对数（以10为底）：pH = –log₁₀[H⁺]。这意味着pH每降低1，[H⁺]就增大到原来的10倍。25°C时，纯水的[H⁺] = 1.0 × 10⁻⁷ mol dm⁻³，对应的pH为7.00。
Because the scale is logarithmic, it compresses a huge range of concentrations. For any aqueous solution at 25°C, the product [H⁺][OH⁻] = Kw = 1.0 × 10⁻¹⁴. The corresponding pOH = –log₁₀[OH⁻] and pH + pOH = 14.00.
由于pH标度是对数的，它可以将极大范围的氢离子浓度压缩表示。对于任何25°C的水溶液，[H⁺][OH⁻] = Kw = 1.0 × 10⁻¹⁴。相应有pOH = –log₁₀[OH⁻]，且pH + pOH = 14.00。
2. Water Autoionization and Temperature Effects | 水的自离解与温度效应
Water autoionizes very slightly: 2H₂O ⇌ H₃O⁺ + OH⁻. The equilibrium constant Kw = 1.0 × 10⁻¹⁴ at 298 K. This is an endothermic process, so Kw increases as temperature rises. At 40°C, Kw ≈ 3.8 × 10⁻¹⁴ and the pH of pure water is about 6.77, but the solution remains neutral because [H⁺] = [OH⁻].
水发生非常微弱的自离解：2H₂O ⇌ H₃O⁺ + OH⁻。在298 K时平衡常数 Kw = 1.0 × 10⁻¹⁴。这是吸热过程，因此温度升高时Kw变大。在40°C时，Kw ≈ 3.8 × 10⁻¹⁴，纯水的pH约为6.77，但溶液仍为中性，因为[H⁺] = [OH⁻]。
In exam questions, always check the temperature stated. If T ≠ 25°C, pH = 7 is not the neutral point; instead use the given Kw to find [H⁺] for neutral solutions and adapt any formula that relies on Kw.
在考试题中，务必检查题目给出的温度。如果T ≠ 25°C，则pH = 7不再是中性点；应使用给定的Kw求出中性溶液的[H⁺]，并调整所有依赖Kw的公式。
3. Strong Acids and Bases – Full Dissociation | 强酸与强碱 – 完全解离
Strong acids (e.g. HCl, HNO₃, H₂SO₄ in its first dissociation) fully dissociate in water, so [H⁺] equals the initial concentration of the acid times the number of protons released per formula unit. For HCl: [H⁺] = Cacid; for H₂SO₄: approximately [H⁺] ≈ 2 × Cacid if you treat both protons as fully dissociated (though the second dissociation is weak, IB/OCR often simplify it as complete).
强酸（如HCl、HNO₃、H₂SO₄的第一步解离）在水中完全解离，因此[H⁺]等于酸的初始浓度乘以每个分子释放的质子数。对于HCl：[H⁺] = C酸；对于H₂SO₄：若将两个质子都视为完全解离（尽管第二步是弱解离，但IB/OCR常简化为完全），则近似有[H⁺] ≈ 2 × C酸。
Strong bases such as NaOH, KOH, and Ba(OH)₂ fully dissociate to give OH⁻. For NaOH: [OH⁻] = Cbase; for Ba(OH)₂: [OH⁻] = 2 × Cbase. Then pH = 14.00 – pOH, where pOH = –log₁₀[OH⁻].
强碱如NaOH、KOH和Ba(OH)₂完全解离产生OH⁻。对于NaOH：[OH⁻] = C碱；对于Ba(OH)₂：[OH⁻] = 2 × C碱。然后pH = 14.00 – pOH，其中pOH = –log₁₀[OH⁻]。
When strong acids or bases are extremely dilute (C < 10⁻⁶ mol dm⁻³), the autoionization of water becomes significant. You must include the [H⁺] from water to avoid pH values that are physically impossible (e.g. a strong acid giving a pH > 7). Use the charge balance equation to solve exactly.
当强酸或强碱极度稀释时（C < 10⁻⁶ mol dm⁻³），水的自离解变得不可忽略。必须将水产生的[H⁺]一并计入，才能避免得出不符合实际的pH（如强酸算出的pH > 7）。可使用电荷守恒方程精确求解。
4. Weak Acids and Ka – The Equilibrium Approach | 弱酸与Ka – 平衡处理法
Weak acids partially dissociate: HA ⇌ H⁺ + A⁻. The acid dissociation constant Ka = [H⁺][A⁻]/[HA]. A common approximation is [H⁺] = √(Ka × CHA), valid when CHA/Ka > 100 (or degree of dissociation α < 5%). This shortcut saves time in multiple-choice questions.
弱酸部分解离：HA ⇌ H⁺ + A⁻。酸解离常数 Ka = [H⁺][A⁻]/[HA]。常用近似公式为 [H⁺] = √(Ka × CHA)，当 CHA/Ka > 100（或解离度 α < 5%）时适用。这一简化在选择题中能节省不少时间。
If the approximation is invalid, use an ICE table and solve the quadratic equation: Ka = x²/(C – x), where x = [H⁺]. Always state your assumption and check it. The pH is then –log₁₀(x).
若近似条件不成立，则需要使用ICE表格并求解二次方程：Ka = x²/(C – x)，其中x = [H⁺]。务必陈述你的假设并检验其有效性。然后pH = –log₁₀(x)。
pKa = –log₁₀ Ka. A smaller pKa indicates a stronger weak acid. Understand that Ka is temperature-dependent.
pKa = –log₁₀ Ka。pKa越小，说明该弱酸的酸性相对越强。注意 Ka 的大小受温度影响。
5. Weak Bases and Kb | 弱碱与Kb
Weak bases such as NH₃ or amines react with water: B + H₂O ⇌ BH⁺ + OH⁻. The base dissociation constant Kb = [BH⁺][OH⁻]/[B]. The same approximation applies: [OH⁻] = √(Kb × CB) provided CB/Kb > 100.
弱碱如NH₃或胺类与水反应：B + H₂O ⇌ BH⁺ + OH⁻。碱解离常数 Kb = [BH⁺][OH⁻]/[B]。同样的近似条件适用：[OH⁻] = √(Kb × CB)，前提是 CB/Kb > 100。
Calculate pOH from [OH⁻] first, then convert to pH using pH = 14.00 – pOH. Remember to check whether the temperature is 25°C; otherwise use the correct Kw.
先由[OH⁻]求出pOH，再通过pH = 14.00 – pOH换算。切记检查温度是否为25°C；若不是，需用相应Kw进行换算。
Many weak base problems provide Ka of the conjugate acid instead. Convert using Kb = Kw / Ka (for the conjugate pair).
许多弱碱问题给出的是其共轭酸的Ka值。此时可用 Kb = Kw / Ka（共轭酸碱对）进行转换。
6. pKa and pKb – Linking Conjugate Pairs | pKa与pKb – 共轭对的关系
For a conjugate acid–base pair at 25°C: Ka × Kb = Kw. Taking negative logs gives pKa + pKb = 14.00. This relationship is particularly useful when only one constant is known.
对于25°C下的共轭酸碱对：Ka × Kb = Kw。两边取负对数可得 pKa + pKb = 14.00。当只知道其中一个常数时，这一关系尤为有用。
Example: the pKa of ammonium ion NH₄⁺ is 9.25; therefore the pKb of ammonia NH₃ is 14.00 – 9.25 = 4.75. Understanding this link avoids memorizing both sets of data.
举例：铵根离子NH₄⁺的pKa为9.25；因此氨NH₃的pKb为14.00 – 9.25 = 4.75。理解这一联系就不需要同时记忆两种数据了。
7. Buffer Solutions – The Henderson–Hasselbalch Equation | 缓冲溶液 – 亨德森-哈塞尔巴尔赫方程
A buffer is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid) that resists pH changes. For an acidic buffer, the pH is given by the famous Henderson–Hasselbalch equation:
pH = pKa + log₁₀([A⁻]/[HA])
缓冲溶液是由弱酸与其共轭碱（或弱碱与其共轭酸）组成的混合物，能抵抗pH变化。对于酸性缓冲溶液，pH可由亨德森-哈塞尔巴尔赫方程计算：
pH = pKa + log₁₀([A⁻]/[HA])
When [A⁻] = [HA], pH = pKa. The buffering capacity is best when the ratio [A⁻]/[HA] is between 0.1 and 10, i.e. pH within pKa ± 1. Dilution does not change the pH of a buffer (provided the ratio stays constant) because both concentrations scale by the same factor.
当[A⁻] = [HA]时，pH = pKa。当[A⁻]/[HA]比值在0.1到10之间（即pH在pKa ± 1范围内）时，缓冲能力最强。稀释不会改变缓冲溶液的pH（假设比值不变），因为两组分的浓度按相同比例变化。
For basic buffers, you can either use pOH = pKb + log₁₀([conjugate acid]/[base]) or convert to the acidic version using pKa of the conjugate acid. Both methods are acceptable; choose whichever is faster.
对于碱性缓冲溶液，可以使用pOH = pKb + log₁₀([共轭酸]/[碱])，也可以转化为共轭酸的pKa再用亨德森方程。两种方法均可，哪个更快就用哪个。
8. pH of Salt Solutions – Hydrolysis | 盐溶液的pH – 水解
Salts containing the conjugate base of a weak acid (e.g. CH₃COONa) produce basic solutions because the anion hydrolyses water: A⁻ + H₂O ⇌ HA + OH⁻. The hydrolysis constant Kh = Kw/Ka. The [OH⁻] can be approximated by √(Kh × Csalt) if the salt is not too dilute.
含有弱酸共轭碱的盐（如CH₃COONa）会水解产生碱性溶液：A⁻ + H₂O ⇌ HA + OH⁻。水解常数 Kh = Kw/Ka。若盐浓度不太低，可用 [OH⁻] ≈ √(Kh × C盐) 进行估算。
Salts containing the conjugate acid of a weak base (e.g. NH₄Cl) give acidic solutions: BH⁺ + H₂O ⇌ B + H₃O⁺. Here Kh = Kw/Kb, and [H⁺] ≈ √(Kh × Csalt). For amphoteric salts, you may need to compare Ka and Kb of the ions.
含有弱碱共轭酸的盐（如NH₄Cl）形成酸性溶液：BH⁺ + H₂O ⇌ B + H₃O⁺。此时 Kh = Kw/Kb，[H⁺] ≈ √(Kh × C盐)。对于两性盐，可能需要比较离子的Ka和Kb。
9. Titration Curves and the pH at Key Points | 滴定曲线与关键点pH
Understanding the shape of pH titration curves allows you to calculate pH at four important stages: (i) initial point – pure acid/base; (ii) before equivalence – buffer region; (iii) at equivalence – salt hydrolysis; (iv) after equivalence – excess titrant.
理解pH滴定曲线形状能帮助你计算四个关键阶段的pH：(i) 初始点 – 纯酸/碱；(ii) 等当点前 – 缓冲区域；(iii) 等当点 – 盐的水解；(iv) 等当点后 – 过量滴定剂。
- Initial point: For a weak acid, use [H⁺] = √(Ka × C); for strong acid, pH = –logC. 初始点：弱酸用[H⁺] = √(Ka × C)；强酸直接用pH = –logC。
- Buffer region: Apply Henderson–Hasselbalch with the remaining weak acid and the salt formed. 缓冲区域：用剩余弱酸和生成盐的浓度代入亨德森-哈塞尔巴尔赫方程。
- Equivalence point: Strong acid–strong base gives pH = 7; weak acid–strong base gives pH > 7 because the conjugate base hydrolyses; weak base–strong acid gives pH < 7. Calculate the concentration of the hydrolysis product and use Kh. 等当点：强酸强碱pH=7；弱酸强碱pH>7（因共轭碱水解）；弱碱强酸pH<7。计算水解产物的浓度后使用Kh。
- After equivalence: Calculate excess [H⁺] or [OH⁻] and convert to pH. 等当点后：计算过量的[H⁺]或[OH⁻]后换算。
10. Dilution and Mixing – Keeping Track of Moles | 稀释与混合 – 追踪物质的量
When an acidic solution is diluted with water, the number of moles of H⁺ remains the same (for strong acids) or adjusts slightly through equilibrium shifts (for weak acids). For a strong acid, simply use the new total volume: [H⁺] = initial moles / new volume. But if the resulting [H⁺] falls below ~10⁻⁶ mol dm⁻³, you must include the contribution from water autoionization.
酸性溶液用水稀释时，H⁺的物质的量（对强酸而言）保持不变，或通过平衡移动略有调整（对弱酸）。对于强酸，直接使用新的总体积：[H⁺] = 初始物质的量 / 新体积。但如果计算出的[H⁺]低于约10⁻⁶ mol dm⁻³，则必须计入水的自离解贡献。
When mixing two solutions, determine whether one reactant is in excess. Calculate the excess [H⁺] or [OH⁻] after neutralization, then find the pH. If the mixtures are buffered, use the Henderson–Hasselbalch equation with the new total volume.
当混合两种溶液时，先判断哪种反应物过量。计算中和后剩余的[H⁺]或[OH⁻]，然后再求pH。如果混合物成为缓冲溶液，则在新总体积下使用亨德森-哈塞尔巴尔赫方程。
11. Common Pitfalls and Smart Tips | 常见陷阱与实用技巧
- Always confirm the temperature (298 K or other). Kw = 1.0 × 10⁻¹⁴ is only valid at 25°C. 务必确认温度（298 K或其他）。Kw = 1.0 × 10⁻¹⁴ 仅在25°C时成立。
- Do not forget to check the assumption C/Ka > 100 before using the square-root approximation; otherwise solve the quadratic. 使用平方根
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