Acids and Bases | Key Points | 酸与碱考点精讲

📚 Acids and Bases | Key Points | 酸与碱考点精讲

Acids and bases are fundamental concepts in chemistry, central to understanding numerous reactions from laboratory syntheses to biochemical pathways. In the AQA A-Level Science specification, the topic is treated both qualitatively and quantitatively, requiring a firm grasp of proton transfer, equilibrium constants, pH calculations, and buffer systems. This revision article distils the essential learning points, linking theory to common examination applications.

酸和碱是化学中的基本概念，对于理解从实验室合成到生化路径的众多反应至关重要。在 AQA A-Level 科学课程中，该主题不仅涉及定性理解，更强调定量处理，要求牢固掌握质子转移、平衡常数、pH 计算以及缓冲体系。本篇复习文章提炼了核心考点，将理论与常见考试应用紧密结合。

1. Brønsted–Lowry Theory of Acids and Bases | 布朗斯特-劳里酸碱理论

A Brønsted–Lowry acid is a proton (H⁺) donor, and a Brønsted–Lowry base is a proton acceptor. This definition expands acid–base chemistry beyond aqueous solutions and focuses on the transfer of H⁺ between species. For example, when hydrogen chloride gas dissolves in water, HCl donates a proton to H₂O, forming H₃O⁺ and Cl⁻. Here HCl acts as the acid and H₂O acts as the base.

布朗斯特-劳里酸是质子（H⁺）供体，而布朗斯特-劳里碱是质子受体。这一定义将酸碱化学拓展到水溶液以外，聚焦于物种间的 H⁺ 转移。例如，氯化氢气体溶于水时，HCl 向 H₂O 提供一个质子，生成 H₃O⁺ 和 Cl⁻。在此，HCl 充当酸，H₂O 充当碱。

Water is amphoteric, meaning it can behave as either an acid or a base depending on the reaction partner. In the reverse reaction, H₃O⁺ can donate a proton to Cl⁻, so H₃O⁺ is the acid and Cl⁻ is the base. The AQA specification expects you to identify acids and bases in forward and reverse reactions and recognise that an acid–base reaction involves two conjugate pairs.

水是两性的，这意味着它既可以作为酸，也可以作为碱，取决于反应伙伴。在逆反应中，H₃O⁺ 可以向 Cl⁻ 提供一个质子，因此 H₃O⁺ 是酸，Cl⁻ 是碱。AQA 大纲要求你能够辨识正反应和逆反应中的酸和碱，并认识到酸碱反应涉及两对共轭对。

2. Conjugate Acid-Base Pairs | 共轭酸碱对

A conjugate acid–base pair consists of two species that differ by a single proton. The acid has one more H⁺ than its conjugate base. For the general weak acid HA, the pair is HA/A⁻; for the hydronium ion, the pair is H₃O⁺/H₂O. Another classic pair is NH₄⁺/NH₃, where the ammonium ion donates a proton to become ammonia.

共轭酸碱对由仅相差一个质子的两个物种组成。酸比其共轭碱多一个 H⁺。对于一般弱酸 HA，共轭对为 HA/A⁻；对于水合氢离子，共轭对为 H₃O⁺/H₂O。另一经典共轭对是 NH₄⁺/NH₃，铵离子失去一个质子后变为氨。

The strength of an acid is inversely related to the strength of its conjugate base. A strong acid, such as HCl, has an extremely weak conjugate base (Cl⁻) that has negligible tendency to accept a proton. Conversely, a weak acid like ethanoic acid, CH₃COOH, has a relatively strong conjugate base, CH₃COO⁻, which readily combines with H⁺. This relationship is crucial when predicting the direction of acid–base equilibria.

酸的强度与其共轭碱的强度成反比。强酸（如 HCl）具有极弱的共轭碱（Cl⁻），几乎无接受质子的倾向。相反，弱酸如乙酸（CH₃COOH）具有相对较强的共轭碱 CH₃COO⁻，它容易与 H⁺ 结合。这一关系在预测酸碱平衡方向时至关重要。

3. Strong and Weak Acids | 强酸与弱酸

Strong acids, including HCl, HNO₃ and H₂SO₄ (first dissociation), undergo complete dissociation in aqueous solution. This means the concentration of H⁺ equals the initial concentration of the acid (for monoprotic acids). Weak acids, such as carboxylic acids and carbonic acid, only partially dissociate, establishing a dynamic equilibrium described by an acid dissociation constant, Ka.

强酸，包括 HCl、HNO₃ 和 H₂SO₄（第一步解离），在水溶液中完全解离。这意味着对于一元强酸，H⁺ 浓度等于酸的初始浓度。弱酸，如羧酸和碳酸，仅部分解离，建立起由酸解离常数 Ka 描述的动态平衡。

It is a common exam task to differentiate between strong and weak acids of the same concentration by comparing their pH, electrical conductivity, or rate of reaction with a metal such as magnesium. A 0.1 mol dm⁻³ solution of HCl has a pH of 1.0, whereas 0.1 mol dm⁻³ CH₃COOH has a pH of about 2.9, reflecting the much lower [H⁺] due to partial dissociation.

考试中常见的任务是，通过比较相同浓度强酸和弱酸的 pH、电导率或与金属（如镁）的反应速率来区分二者。0.1 mol dm⁻³ 的 HCl 溶液 pH 为 1.0，而 0.1 mol dm⁻³ CH₃COOH 的 pH 约为 2.9，这反映出由于部分解离，[H⁺] 低得多。

4. pH and [H⁺] Calculations | pH 与 [H⁺] 计算

pH is defined as the negative logarithm to base 10 of the hydrogen ion concentration:

pH = –log₁₀[H⁺]

Conversely, [H⁺] can be recovered from pH using [H⁺] = 10⁻^pH (Unicode: 10⁻ᵖᴴ, but using 10⁻ᴾᴴ is tricky; I’ll use [H⁺] = 10⁻ᵖᴴ with superscript tag?). The specification says use Unicode, but HTML ^{is acceptable? The instruction says “数学/化学符号一律用 Unicode（上标 ⁻³ ² ⁺、下标 ₐ ₙ ₂、→ ⇌ × ÷ ± ½ √ ∫ Δ Σ π θ ° 等）；【严禁使用 LaTeX，如 、\[ \]、$$、\ce{}、\frac 一律不用】。”. It doesn’t forbid HTML sup/sub tags, but it says use Unicode. I’ll use Unicode superscripts for simple ones: [H⁺] = 10⁻ᵖᴴ however pH is a single symbol. I’ll write [H⁺] = 10⁻ᵖᴴ? Not standard. Better to use: [H⁺] = 10^–pH, but they want Unicode. So I can write: [H⁺] = 10⁻ᴾᴴ? Not good. I’ll describe in words: [H⁺] = 10 raised to the power –pH. Or use: [H⁺] = 10⁻⁽ᵖᴴ⁾? Too messy. I’ll use [H⁺] = 10⁻^pH with the sup tag, which is standard HTML and allowed because “严禁使用 LaTeX” but HTML is okay. The instruction says “公式/方程用}

…

居中加粗；数学/化学符号一律用 Unicode”. It’s fine to mix HTML tags for superscript if needed. I’ll use ^{for pH as power. For example, [H⁺] = 10^–pH. I’ll do that. For simple superscripts like H⁺, I can use Unicode directly. So I’ll use ^{for complex exponents. This is acceptable.}}

Thus, for a strong monoprotic acid of concentration c, pH = –log₁₀(c). Diluting a strong acid by a factor of 10 increases the pH by 1. For strong bases, first calculate [OH⁻] from the base concentration (taking into account stoichiometry, e.g., Ba(OH)₂ gives 2 OH⁻ per formula unit), then use Kw to find [H⁺], and finally pH.

因此，对于浓度为 c 的一元强酸，pH = –log₁₀(c)。将强酸稀释 10 倍，pH 增加 1。对于强碱，首先根据碱的浓度和化学计量比（例如 Ba(OH)₂ 每个单元产生 2 个 OH⁻）计算 [OH⁻]，然后利用 Kw 求 [H⁺]，最后计算 pH。

At 25 °C, pOH = –log₁₀[OH⁻] and pH = 14 – pOH. Remember that the ion product of water changes with temperature, so the familiar ‘pH + pOH = 14’ only holds at 25 °C.

在 25 °C 时，pOH = –log₁₀[OH⁻]，且 pH = 14 – pOH。切记水的离子积随温度变化，因此熟悉的“pH + pOH = 14”仅在 25 °C 下成立。

5. The Ionic Product of Water, Kw | 水的离子积 Kw

Water undergoes slight self-ionisation: 2H₂O ⇌ H₃O⁺ + OH⁻, often simplified to H₂O ⇌ H⁺ + OH⁻. The equilibrium constant for this process is Kw:

Kw = [H⁺][OH⁻]

At 25 °C, Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶. Because the dissociation is endothermic, Kw increases with rising temperature. Consequently, the pH of pure water is 7.0 only at 25 °C; at higher temperatures, pure water has a pH less than 7 but is still neutral because [H⁺] = [OH⁻].

水经历微弱的自解离：2H₂O ⇌ H₃O⁺ + OH⁻，常简化为 H₂O ⇌ H⁺ + OH⁻。该过程的平衡常数为 Kw：

Kw = [H⁺][OH⁻]

在 25 °C 时，Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶。由于解离过程吸热，Kw 随温度升高而增大。因此，纯水的 pH 仅在 25 °C 下为 7.0；在更高温度下，纯水的 pH 低于 7，但仍为中性，因为 [H⁺] = [OH⁻]。

Kw is indispensable for calculating the pH of strong base solutions and for linking pH and pOH. A common error is to forget that Kw remains constant (at a given temperature) even when acid or base is added; the product [H⁺][OH⁻] always equals Kw. This relationship is used to find [H⁺] in an alkaline solution given its [OH⁻].

Kw 对于强碱溶液 pH 的计算以及连接 pH 与 pOH 不可或缺。常见错误是忘记即使加入酸或碱，Kw（在给定温度下）仍保持恒定；乘积 [H⁺][OH⁻] 始终等于 Kw。这一关系被用于根据碱性溶液的 [OH⁻] 求 [H⁺]。

6. Weak Acid Dissociation Constant, Ka | 弱酸解离常数 Ka

For a weak acid HA that partially dissociates: HA ⇌ H⁺ + A⁻, the acid dissociation constant Ka is given by:

Ka = [H⁺][A⁻] / [HA]

The larger the Ka, the stronger the weak acid. Because only a tiny fraction of HA dissociates, we often assume that [HA]ₑq ≈ [HA]ᵢₙᵢₜᵢₐₗ and that [H⁺] = [A⁻]. This yields the simplified expression:

[H⁺] = √(Ka × c)

where c is the initial concentration of the weak acid. This approximation is valid when c / Ka > 100, a condition that the AQA exam may ask you to check.

对于部分解离的弱酸 HA：HA ⇌ H⁺ + A⁻，酸解离常数 Ka 表示为：

Ka = [H⁺][A⁻] / [HA]

Ka 值越大，弱酸相对越强。由于只有极少部分的 HA 解离，我们通常假设 [HA]平衡 ≈ [HA]初始，且 [H⁺] = [A⁻]。由此得到简化表达式：

[H⁺] = √(Ka × c)

其中 c 为弱酸的初始浓度。当 c / Ka > 100 时，该近似成立，AQA 考试可能会要求你验证此条件。

Understanding Ka also explains why dilution of a weak acid increases its degree of dissociation (α). Although [H⁺] decreases upon dilution, the proportion of molecules that dissociate becomes larger as the equilibrium shifts to the right to counteract the reduction in concentration.

理解 Ka 还可解释为何稀释弱酸会增加其解离度 (α)。虽然稀释后 [H⁺] 降低，但为抵消浓度减小的影响，平衡向右移动，解离的分子比例反而增大。

7. pKa and Acid Strength | pKa 与酸强度

Analogous to pH, pKa is defined as the negative logarithm of Ka:

pKa = –log₁₀(Ka)

A smaller pKa indicates a stronger weak acid (e.g., methanoic acid, pKa ≈ 3.75, is stronger than ethanoic acid, pKa ≈ 4.76). The pKa value is extremely useful during titrations and buffer calculations. When exactly half of the weak acid has been neutralised, [HA] = [A⁻], and therefore pH = pKa. This point is known as the half-equivalence point and is used to determine the pKa of an unknown weak acid from its titration curve.

类似于 pH，pKa 被定义为 Ka 的负对数：

pKa = –log₁₀(Ka)

pKa 越小，弱酸相对越强（例如甲酸的 pKa ≈ 3.75，强于乙酸的 pKa ≈ 4.76）。pKa 值在滴定和缓冲液计算中非常有用。当弱酸恰好被中和一半时，[HA] = [A⁻]，因此 pH = pKa。这一点称为半当量点，可用于从滴定曲线确定未知弱酸的 pKa。

You should be able to interconvert between Ka and pKa using the log relationship and recognise that a one-unit change in pKa corresponds to a tenfold change in Ka. This logarithmic scale helps compare acid strengths quickly.

你应当能够利用对数关系在 Ka 与 pKa 之间相互转换，并理解 pKa 每变化 1 单位对应 Ka 变化 10 倍。这种对数标度有助于快速比较酸的强度。

8. Buffer Solutions | 缓冲溶液

A buffer solution resists changes in pH upon the addition of small amounts of acid or base. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in appreciable concentrations. A classic example is the ethanoic acid / sodium ethanoate mixture: CH₃COOH and CH₃COO⁻.

缓冲溶液能在加入少量酸或碱时抵抗 pH 变化。它由相当浓度的弱酸及其共轭碱（或弱碱及其共轭酸）组成。经典例子是乙酸/乙酸钠混合物：CH₃COOH 与 CH₃COO⁻。

When H⁺ is added, the conjugate base A⁻ removes it: A⁻ + H⁺ → HA. When OH⁻ is added, the weak acid HA neutralises it: HA + OH⁻ → A⁻ + H₂O. The pH of a buffer can be calculated using the Henderson–Hasselbalch equation:

pH = pKa + log₁₀([A⁻]/[HA])

This equation assumes that the concentrations of A⁻ and HA at equilibrium are approximately equal to the initial concentrations used to prepare the buffer.

当加入 H⁺ 时，共轭碱 A⁻ 将其消除：A⁻ + H⁺ → HA。当加入 OH⁻ 时，弱酸 HA 将其中和：HA + OH⁻ → A⁻ + H₂O。缓冲液的 pH 可用 Henderson–Hasselbalch 方程计算：

pH = pKa + log₁₀([A⁻]/[HA])

该方程假设平衡时 A⁻ 和 HA 的浓度近似等于配制缓冲液时使用的初始浓度。

Buffer capacity is greatest when [A⁻] = [HA], i.e., when pH = pKa. The effective buffering range is typically pKa ± 1. In the lab, buffers can be prepared by partially neutralising a weak acid with a strong base, or by mixing a weak acid with a salt of its conjugate base.

当 [A⁻] = [HA]，即 pH = pKa 时，缓冲容量最大。有效缓冲范围通常为 pKa ± 1。在实验室中，缓冲液可通过用强碱部分中和弱酸，或将弱酸与其共轭碱的盐混合来制备。

9. Titration Curves and Equivalence Point | 滴定曲线与等当点

A pH titration curve plots pH against the volume of titrant added. Its shape depends on the strengths of the acid and base involved. Four key combinations are examined at A-Level:

Strong acid–strong base: equivalence point pH = 7; steep vertical region from pH 3 to 11.
Weak acid–strong base: equivalence point pH > 7 (due to the formation of the conjugate base, which hydrolyses); buffer region before the endpoint where pH = pKa at half-equivalence.
Strong acid–weak base: equivalence point pH < 7; buffer region before endpoint.
Weak acid–weak base: no sharp vertical jump; not suitable for simple indicator titration.

绘制 pH 随滴定剂加入体积变化的曲线即为 pH 滴定曲线。其形状取决于所涉及酸和碱的强度。A-Level 考试中考察四种关键组合：

强酸–强碱：等当点 pH = 7；垂直突跃区从 pH 3 到 11。
弱酸–强碱：等当点 pH > 7（因形成共轭碱并水解）；终点前存在缓冲区域，半当量点处 pH = pKa。
强酸–弱碱：等当点 pH < 7；终点前有缓冲区域。
弱酸–弱碱：无垂直突跃；不适合使用单一指示剂进行滴定。

The equivalence point is where stoichiometric amounts of acid and base have reacted. It is not necessarily pH 7. The vertical section of the curve is used to select an appropriate indicator, whose colour change range must lie within the sharp pH change.

等当点是酸和碱按化学计量完全反应的时刻，不一定对应 pH 7。曲线的垂直部分用于选择合适的指示剂，其变色范围必须落在 pH 突跃范围内。

10. pH Indicators | pH 指示剂

Acid–base indicators are usually weak acids whose conjugate acid and base forms have distinctly different colours. The indicator equilibrium can be written as:

HIn ⇌ H⁺ + In⁻

where HIn and In⁻ have different colours. The colour perceived depends on the ratio [In⁻]/[HIn], which is governed by the pH of the solution and the indicator’s pKa (pKᵢₙ). The colour change occurs over a range of about pH = pKᵢₙ ± 1.

酸碱指示剂通常是一种弱酸，其共轭酸和共轭碱形式具有显著不同的颜色。指示剂平衡可写为：

HIn ⇌ H⁺ + In⁻

其中 HIn 和 In⁻ 颜色不同。观察到的颜色取决于 [In⁻]/[HIn] 的比值，而该比值由溶液 pH 和指示剂的 pKa (pKᵢₙ) 决定。颜色变化发生在约 pH = pKᵢₙ ± 1 的范围内。

For a strong acid–strong base titration, both methyl orange (range 3.1–4.4) and phenolphthalein (range 8.3–10.0) are suitable because the vertical pH jump spans their ranges. For a weak acid–strong base titration, phenolphthalein is suitable, but methyl orange is not, because its range lies below the equivalence point. Correct indicator choice is a frequent exam question.

对于强酸–强碱滴定，甲基橙（范围 3.1–4.4）和酚酞（范围 8.3–10.0）均适用，因为垂直 pH 突跃涵盖它们的变色范围。对于弱酸–强碱滴定，酚酞适用，但甲基橙不适用，因为其变色范围在等当点以下。正确选择指示剂是考试中的常见问题。

11. Standard Enthalpy of Neutralisation | 标准中和焓

For strong acid–strong base reactions, the standard enthalpy change of neutralisation (ΔH°ₙₑᵤₜ) is almost constant at about –57.5 kJ mol⁻¹. This is because the reaction is simply H⁺(aq) + OH⁻(aq) → H₂O(l). The spectator ions do not participate. However, if either the acid or base is weak, the enthalpy change is less exothermic because some energy is used to dissociate the weak acid or weak base.

对于强酸–强碱反应，标准中和焓变 (ΔH°中和) 几乎恒定为 –57.5 kJ mol⁻¹ 左右。这是因为反应本质上是 H⁺(aq) + OH⁻(aq) → H₂O(l)，旁观离子不参与反应。但如果酸或碱是弱电解质，焓变值会较小（放热较少），因为部分能量需用于弱酸或弱碱的解离。

This concept appears in thermochemical cycles and helps explain why weak acid–strong base reactions have a less negative ΔH° neutralisation. You may be asked to calculate the enthalpy of neutralisation from experimental data using q = mcΔT and to compare the values for different combinations.

这一概念出现在热化学循环中，并有助于解释为何弱酸–强碱反应的 ΔH°中和绝对值较小。你可能会被要求利用 q = mcΔT 从实验数据计算中和焓，并比较不同酸碱组合的值。

12. Applications of Buffer Systems | 缓冲体系的应用

Buffers are vital in biological systems and industrial processes. The pH of human blood is maintained at about 7.4 by a carbonic acid–hydrogencarbonate (H₂CO₃/HCO₃⁻) buffer. In cells, phosphate buffers (H₂PO₄⁻/HPO₄²⁻) are important. In the laboratory, buffers are used to calibrate pH meters and to maintain constant pH during biochemical assays.

缓冲液在生物体系和工业过程中至关重要。人体血液的 pH 通过碳酸–碳酸氢盐 (H₂CO₃/HCO₃⁻) 缓冲对维持在约 7.4。细胞中，磷酸盐缓冲液 (H₂PO₄⁻/HPO₄²⁻) 起重要作用。实验室内，缓冲液用于校准 pH 计以及在生物化学测定中维持恒定的 pH。

Understanding buffer action also underpins the action of many medicines and consumer products, such as antacid tablets and shampoos. Exam questions frequently require you to predict the effect of adding H⁺ or OH⁻ to a given buffer or to calculate the resulting pH shift using the Henderson–Hasselbalch equation.

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Acids and Bases | Key Points | 酸与碱 考点精讲