📚 Calculations Involving Acids and Bases (HL) | 酸和碱的计算（高阶）
Mastering calculations involving acids and bases is a central requirement of IB Chemistry Higher Level. Building on the stoichiometry and equilibrium principles from SL, you will quantify the strength of weak acids and bases, predict the pH of buffer solutions, and analyse titration curves mathematically. This article provides a structured guide to all the essential HL calculation types, with step‑by‑step methods and carefully explained approximations.
掌握酸碱计算是 IB 化学高阶课程的核心要求。在 SL 阶段的化学计量与平衡原理基础上，你将定量地研究弱酸和弱碱的强度、预测缓冲溶液的 pH，并用数学方法分析滴定曲线。本文提供了一份结构清晰的指南，涵盖所有重要的 HL 计算类型，并附有分步解法和详细解释的近似条件。
1. Introduction to Acid–Base Calculations | 酸碱计算简介
In Topic 8 (SL) and Topic 18 (HL) of the IB Chemistry syllabus, you move from qualitative descriptions of acids and bases to quantitative calculations. You will use the ionic product of water, dissociation constants, and the Henderson–Hasselbalch equation to determine pH, pOH, and concentrations of species in solution. A solid grasp of these calculations is indispensable for Paper 2 and the internal assessment.
在 IB 化学大纲的第 8 章（SL）和第 18 章（HL）中，你将从对酸和碱的定性描述过渡到定量计算。你将运用水的离子积、解离常数和亨德森–哈塞尔巴尔赫方程来求算溶液的 pH、pOH 以及各物种的浓度。扎实掌握这些计算对于试卷二和内部评估不可或缺。
2. The Ionic Product of Water, Kw | 水的离子积常数
Water undergoes autoprotolysis according to: 2H₂O(l) ⇌ reaction is Kw = [H₃O⁺][OH⁻]. At 298 K, Kw = 1.0 × 10⁻¹⁴. In pure water, [H₃O⁺] = [OH⁻] = √Kw pH of 7.00. Note that Kw increases with temperature, so the pH of neutrality shifts downward at higher temperatures.
水发生自耦电离：2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)。该反应的平衡常数是 Kw = [H₃O⁺][OH⁻]。在 298 K 时，Kw = 1.0 × 10⁻¹⁴。在纯水中，[H₃O⁺] = [OH⁻] = √Kw = 1.0 × 10⁻⁷ mol dm⁻³，因此中性 pH 为 7.00。需注意 Kw 随温度升高而增大，因此高温时中性的 pH 值会下降。
3. pH and pOH – The Basics | pH 和 pOH 基础
ample, if [OH⁻] = 2.5 × 10⁻³ mol dm⁻³, then pOH = −log(2.5 × 10⁻³) ≈ 2.60, and pH = 14.00 − 2.60 = 11.40.
溶液的 pH 定义为 pH = −log₁₀[H₃O⁺]。类似地，pOH = −log₁₀[OH⁻]。对 Kw 表达式取负对数可得到：pH + pOH = 14.00（298 K 时）。这些关系使你能够在氢离子浓度、氢氧根离子浓度和 pH 之间进行换算。例如，若 [OH⁻] = 2.5 × 10⁻³ mol dm⁻³，则 pOH = −log(2.5 × 10⁻³) ≈ 2.60，pH = 14.00 − 2.60 = 11.40。
4. Strong Acids and Strong Bases | 强酸和强碱
.2 × 10⁻², so the total [H₃O⁺] is slightly less than 2c at moderate concentrations. In typical HL problems, you may be told to assume complete dissociation for both protons unless instructed otherwise.
强酸（如 HCl 和 HNO₃）在水溶液中完全解离。因此，对于浓度为 c 的一元强酸，[H₃O⁺] = c，pH = −log c。二元强酸如 H₂SO₄ 分两步释放质子；第一步完全解离，第二步部分解离（Ka₂ ≈ 1.2 × 10⁻²），所以在中等浓度时总 [H₃O⁺] 略小于 2c。在典型的 HL 题目中，除非特别说明，通常可以假设两个质子均完全解离。
hydroxides such as Ba(OH)₂, [OH⁻] = 2 × c because each formula unit supplies two OH⁻ ions.
对于 NaOH 或 KOH 等强碱，[OH⁻] 等于碱的浓度（假设完全解离）。先求出 pOH：pOH = −log[OH⁻]，然后 pH = 14.00 − pOH。对于 Ba(OH)₂ 等第 2 族氢氧化物，[OH⁻] = 2 × c，因为每个式单元提供两个 OH⁻ 离子。
5. Weak Acids and Ka | 弱酸与酸解离常数
eak acid only partially dissociates in water, establishing an equilibrium: HA(aq) + H₂O(l) ⇌ ₃O⁺][A⁻] / [HA]. The magnitude of Ka indicates the strength of the acid: a larger Ka means a stronger weak acid. Typical values range from about 10⁻⁵ for ethanoic acid to 10⁻¹⁰ for very weak acids like phenol.
弱酸在水中仅部分解离，建立平衡：HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)。酸解离常数 Ka 定义为：Ka = [H₃O⁺][A⁻] / [HA]。Ka 的大小反映了酸的强度：Ka 越大，弱酸越强。乙酸的 Ka 约为 10⁻⁵，而像苯酚这类极弱酸的 Ka 可低至 10⁻¹⁰。
The data booklet provides Ka and pKa values for a selection of weak acids. You are expected to write the Ka expression for any given weak acid and use it to calculate pH or concentration.
数据手册中提供了一系列弱酸的 Ka 和 pKa 值。你需要能写出任意给定弱酸的 Ka 表达式，并利用它来计算 pH 或浓度。
6. Weak Bases and Kb | 弱碱与碱解离常数
For a weak base B reacting with water: B(aq) + H₂O(l) ⇌ the most common weak base encountered, with Kb = 1.8 × 10⁻⁵ at 298 K. For a conjugate acid– seful when you are given Ka of the conjugate acid and need Kb for the base, or vice versa.
对于弱碱 B 与水的反应：B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)，碱解离常数 Kb = [BH⁺][OH⁻] / [B]。氨（NH₃）是最常见的弱碱，其 Kb = 1.8 × 10⁻⁵（298 K）。对于共轭酸碱对，始终有 Ka × Kb = Kw 的关系。当已知共轭酸的 Ka 而需要求碱的 Kb，或反过来时，这个关系非常有用。
7. pKa and pKb – A Logarithmic Scale | pKa 和 pKb – 对数尺度
the relationship Ka × Kb = Kw, we obtain pKa + pKb = 14.00 at 298 K for a conjugate pair.
就像用 pH 方便地表示 [H₃O⁺] 一样，pKa = −log₁₀Ka 和 pKb = −log₁₀Kb 被用来在易于处理的尺度上表示酸碱强度。pKa 越小，酸性越强。例如，乙酸的 pKa 为 4.76，而酸性更强的氯乙酸 pKa 为 2.86。由 Ka × Kb = Kw 的关系可推出共轭酸碱对在 298 K 时满足 pKa + pKb = 14.00。
8. Calculations Involving Weak Acids and Bases | 弱酸弱碱的计算
he approximation when the ratio c₀/Ka > 1000 or the percentage dissociation is less than 5%. The simplified expression becomes x = √Ka × c₀, and pH = −log x.
要计算初始浓度为 c₀ 的弱酸溶液的 pH，可建立 ICE（初始、变化、平衡）表格。设产生的 [H₃O⁺] 为 x，则 Ka = x² / (c₀ − x)。若酸浓度不太稀且 Ka 很小，则可近似为 c₀ − x ≈ c₀。IB 通常认可当 c₀/Ka > 1000 或解离度小于 5% 时使用该近似。简化后得 x = √Ka × c₀，pH = −log x。
a = 1.8 × 10⁻⁵). Since c₀/Ka = 0.100 / (1.8 × 10⁻⁵) ≈ 5556 > 1000, we can approximate. x = √1.8 × 10⁻⁵ × 0.100 = 1.34 × 10⁻³ mol dm⁻³. pH = −log(1.34 × 10⁻³) ≈ 2.87.
例如，计算 0.100 mol dm⁻³ CH₃COOH（Ka = 1.8 × 10⁻⁵）的 pH。因 c₀/Ka ≈ 5556 > 1000，可使用近似。x = √1.8 × 10⁻⁵ × 0.100 = 1.34 × 10⁻³ mol dm⁻³，pH = −log(1.34 × 10⁻³) ≈ 2.87。
stified, you must solve the quadratic equation Ka = x² / (c₀ − x), which yields x = [−Ka + √Ka² + 4Ka c₀] / 2. Weak base calculations follow the same pattern: use Kb, set up ICE for [OH⁻], find pOH, then convert to pH.
当近似条件不满足时，必须求解二次方程 Ka = x² / (c₀ − x)，得到 x = [−Ka + √Ka² + 4Ka c₀] / 2。弱碱的计算遵循相同模式：利用 Kb 和 ICE 表格求 [OH⁻]，再算出 pOH 后转换为 pH。
9. Salt Hydrolysis | 盐的水解
ot neutral. For instance, sodium ethanoate (CH₃COONa) dissolves to give CH₃COO⁻ ions, which act as a weak base: CH₃COO⁻ + H₂O ⇌ H of the salt solution is then calculated as for a weak base using the salt concentration.
由弱酸强碱或强酸弱碱形成的盐，其水溶液不是中性的。例如，乙酸钠（CH₃COONa）溶解产生 CH₃COO⁻ 离子，它作为弱碱起作用：CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻。CH₃COO⁻ 的 Kb 可通过乙酸的 Ka 求得：Kb = Kw / Ka。此后将该盐溶液视作弱碱溶液，用盐的浓度计算 pH。
de (NH₄Cl) provides NH₄⁺ ions, the conjugate acid of NH₃. The acidic hydrolysis NH₄⁺ + H₂O ⇌ O⁺ has Ka = Kw / Kb(NH₃). The pH can be obtained using the weak acid calculation approach.
相反，氯化铵（NH₄Cl）提供 NH₄⁺ 离子，即 NH₃ 的共轭酸。水解反应 NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ 的 Ka = Kw / Kb(NH₃)。可用弱酸的计算方法求出 pH。
10. Buffer Solutions | 缓冲溶液
A buffer solution resists changes in pH upon addition of small amounts of acid or base. Acidic buffers consist of a weak acid and its conjugate base (e.g. CH₃COOH/CH₃COO⁻), while basic buffers use a weak base and its conjugate acid (e.g. NH₃/NH₄⁺). The buffer capacity is determined by the concentrations of the two components and their ratio.
缓冲溶液能够抵抗因加入少量酸或碱而引起的 pH 变化。酸性缓冲液由弱酸及其共轭碱组成（如 CH₃COOH/CH₃COO⁻），碱性缓冲液则由弱碱及其共轭酸组成（如 NH₃/NH₄⁺）。缓冲容量取决于两组分的浓度及其比值。
Buffers are vital in biological systems (e.g. blood pH control) and in laboratory work. From an equilibrium perspective, the weak acid provides a reservoir of HA that reacts with added OH⁻, while the conjugate base A⁻ reacts with added H₃O⁺.
缓冲液在生物体系（如血液 pH 调控）和实验室工作中至关重要。从平衡角度看，弱酸提供了可与外加 OH⁻ 反应的 HA 储层，而共轭碱 A⁻ 则与外加 H₃O⁺ 反应。
11. Henderson–Hasselbalch Equation | 亨德森–哈塞尔巴尔赫方程
The pH of an acidic buffer can be calculated directly using the Henderson–Hasselbalch equation:
pH = pKa + log([A⁻] / [HA])
the Ka expression. It shows that when [A⁻] = [HA], pH = pKa, which corresponds to the half‑ A⁻ and HA are within a factor of 10 of each other and are reasonably high relative to Ka.
酸性缓冲液的 pH 可直接用亨德森–哈塞尔巴尔赫方程计算：
pH = pKa + log([A⁻] / [HA])
该方程由对 Ka 表达式取负对数推导而来。它表明当 [A⁻] = [HA] 时，pH = pKa，这正是滴定过程中半中和点的情况。当 A⁻ 与 HA 的浓度相差不超过 10 倍，且相对 Ka 足够大时，该方程最为可靠。
³ HA and 0.80 mol dm⁻³ A⁻, pH = 4.76 + log(0.80/0.10) = 4.76 + log 8 = 4.76 + 0.90 = 5.66.
示例：某缓冲液含 0.50 mol dm⁻³ CH₃COOH 和 0.50 mol dm⁻³ CH₃COONa。乙酸的 pKa = 4.76，因此 pH = 4.76 + log(0.50/0.50) = 4.76。若缓冲液含 0.10 mol dm⁻³ HA 和 0.80 mol dm⁻³ A⁻，则 pH = 4.76 + log(0.80/0.10) = 4.76 + log 8 = 4.76 + 0.90 = 5.66。
12. Acid–Base Titrations and pH Curves | 酸碱滴定与 pH 曲线
Titration calculations involve determining the pH at various stages: initial, before equivalence, at equivalence, and beyond equivalence. The shape of the pH curve depends on the strengths of the acid and base used.
滴定计算要求确定不同阶段的 pH：初始值、等价点前、等价点和等价点后。pH 曲线的形状取决于所用酸碱的强度。
For a strong acid–strong base titration, the equivalence point occurs at pH 7.00. The initial pH is simply that of the strong acid (or base), and after the equivalence, pH is dictated by the excess strong base (or acid). The Henderson–Hasselbalch equation is not needed; simple stoichiometry and dilution suffice.
对于强酸–强碱滴定，等价点位于 pH 7.00。初始 pH 就是强酸（或强碱）的 pH，等价点后 pH 则由过量的强碱（或强酸）决定。此处无需亨德森–哈塞尔巴尔赫方程，简单的化学计量和稀释计算即可。
For a weak acid– H is calculated using the weak acid Ka. In the buffer region (between initial and equivalence), apply the Henderson–Hasselbalch equation using the amounts of unreacted HA and produced A⁻. At the half‑ uivalence point, all HA has been converted to A⁻, so the solution is a salt that undergoes hydrolysis; calculate its pH as a weak base. After the equivalence point, the pH is governed by excess strong base.
对于弱酸–强碱滴定，初始 pH 用弱酸的 Ka 计算。在缓冲区域（初始至等价点之间），利用未反应 HA 和生成的 A⁻ 的量，使用亨德森–哈塞尔巴尔赫方程。在半等价点，pH = pKa。在等价点，所有 HA 已转化为 A⁻，此时溶液为发生水解的盐，需按弱碱计算其 pH。等价点后，pH 由过量的强碱决定。
are chosen so their pKa falls within the steep portion of the pH curve. For a strong acid– th methyl orange (pKa ≈ 3.5) and phenolphthalein (pKa ≈ 9.3) are suitable. For a weak acid–strong base titration, phenolphthalein is ideal because the equivalence pH > 7.
指示剂的选择应使其 pKa 落在 pH 曲线的陡升范围内。对于强酸–强碱滴定，甲基橙（pKa ≈ 3.5）和酚酞（pKa ≈ 9.3）均适用。对于弱酸–强碱滴定，酚酞是理想选择，因为等价点 pH > 7。
For a weak base–strong acid titration, the roles are reversed: the equivalence point occurs below pH 7, and methyl orange is a suitable indicator. Calculating all critical points follows the same logic using Kb, hydrolysis, and excess strong acid principles.
对于弱碱–强酸滴定，情况相反：等价点 pH 低于 7，甲基橙是合适的指示剂。计算所有关键点位时，遵循相同的逻辑，运用 Kb、水解以及过量强酸原理即可。
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