Acids and Bases | 酸与碱考点精讲

📚 Acids and Bases | 酸与碱考点精讲

Acids and bases are central to A-Level Chemistry, linking equilibrium, kinetics and organic chemistry. In WJEC Unit 2 and Unit 4, you must move beyond simple litmus tests to quantitative definitions, pH calculations, buffer systems and titration curves. This revision guide walks through every core concept with clear explanations, worked examples and exam tips to help you master the topic.

酸与碱是 A-Level 化学的核心内容，连接着平衡、动力学和有机化学。在 WJEC 单元 2 和单元 4 中，你需要超越简单的石蕊试纸检测，掌握定量定义、pH 计算、缓冲体系和滴定曲线。本考点精讲以清晰的解释、分步示例和考试技巧，带你逐一攻克每个核心概念。

1. Brønsted–Lowry Theory | 布朗斯特–劳里酸碱理论

The Brønsted–Lowry definition is the working model for all WJEC acid–base questions. An acid is a proton (H⁺) donor; a base is a proton acceptor. When HCl gas dissolves in water, it donates a proton to H₂O, forming H₃O⁺ and Cl⁻. Here water acts as a base. In the reverse reaction, H₃O⁺ can donate a proton to Cl⁻, so Cl⁻ is the conjugate base of HCl, and H₃O⁺ is the conjugate acid of H₂O.

布朗斯特–劳里定义是 WJEC 所有酸碱问题的基本模型。酸是质子（H⁺）给体；碱是质子受体。当 HCl 气体溶于水时，它向 H₂O 提供一个质子，生成 H₃O⁺ 和 Cl⁻。此时水扮演了碱的角色。在逆反应中，H₃O⁺ 可以给 Cl⁻ 提供质子，因此 Cl⁻ 是 HCl 的共轭碱，而 H₃O⁺ 是 H₂O 的共轭酸。

A conjugate acid–base pair differs by exactly one proton. For ammonia in water: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. NH₄⁺ is the conjugate acid of NH₃; OH⁻ is the conjugate base of H₂O. Amphiprotic species like HCO₃⁻ can both donate and accept a proton, meaning they can act as either an acid or a base depending on the other reactant present.

共轭酸碱对恰好相差一个质子。以氨在水中为例：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。NH₄⁺ 是 NH₃ 的共轭酸；OH⁻ 是 H₂O 的共轭碱。像 HCO₃⁻ 这样的两性物质既能给出质子也能接受质子，意味着根据存在的其他反应物，它可以表现出酸或碱的性质。

2. Strong vs Weak Acids and Bases | 强酸强碱与弱酸弱碱

Strength means the extent of dissociation in water, not concentration. A strong acid is fully dissociated: HCl → H⁺ + Cl⁻. If you have 0.10 mol dm⁻³ HCl, [H⁺] = 0.10 mol dm⁻³, so pH = −log₁₀(0.10) = 1.00. Common strong acids are HCl, HNO₃ and H₂SO₄ (first dissociation only completely; second is partial at A-Level). A weak acid, like ethanoic acid, only partially dissociates: CH₃COOH ⇌ CH₃COO⁻ + H⁺. Its equilibrium position lies far to the left. The same logic applies to bases: NaOH is a strong base, fully dissociated into Na⁺ and OH⁻; NH₃ is a weak base, reacting reversibly with water.

强度指的是在水中的解离程度，而非浓度。强酸完全解离：HCl → H⁺ + Cl⁻。若浓度为 0.10 mol dm⁻³ HCl，则 [H⁺] = 0.10 mol dm⁻³，pH = −log₁₀(0.10) = 1.00。常见的强酸有 HCl、HNO₃ 和 H₂SO₄（在 A-Level 中仅第一步完全解离；第二步为部分解离）。弱酸如乙酸仅部分解离：CH₃COOH ⇌ CH₃COO⁻ + H⁺，其平衡位置严重偏左。同样的逻辑适用于碱：NaOH 是强碱，完全解离为 Na⁺ 和 OH⁻；NH₃ 是弱碱，与水发生可逆反应。

You must be able to calculate pH for strong bases using [OH⁻] and the ionic product of water, Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. For 0.05 mol dm⁻³ NaOH, [OH⁻] = 0.05, so [H⁺] = Kw ÷ 0.05 = 2.0 × 10⁻¹³ mol dm⁻³, giving pH = 12.70. Always check temperature: Kw varies with temperature and the WJEC data booklet will give the value if not 298 K.

你必须能够利用 [OH⁻] 和水的离子积 Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶（298 K）计算强碱的 pH。对于 0.05 mol dm⁻³ NaOH，[OH⁻] = 0.05，因此 [H⁺] = Kw ÷ 0.05 = 2.0 × 10⁻¹³ mol dm⁻³，pH = 12.70。务必检查温度：Kw 随温度变化，若非 298 K，WJEC 数据手册会给出数值。

3. pH and the Logarithmic Scale | pH 与对数标度

pH = −log₁₀[H⁺]. Each unit pH decrease means a ten‑fold increase in [H⁺]. This logarithmic relationship explains why adding a small volume of concentrated acid to a large volume of water can cause a dramatic pH change. Conversely, [H⁺] = 10⁻ᵖᴴ. You must be comfortable converting between pH and hydrogen ion concentration on your calculator, using the 10ˣ or log buttons efficiently.

pH = −log₁₀[H⁺]。pH 每降低 1，[H⁺] 浓度增加 10 倍。这种对数关系解释了为什么向大量水中加入少量浓酸会导致 pH 剧烈变化。反过来，[H⁺] = 10⁻ᵖᴴ。你需要熟练使用计算器的 10ˣ 和 log 按钮，在 pH 和氢离子浓度之间进行转换。

When [H⁺] is not a simple power of ten, the pH will include decimal places. For [H⁺] = 2.5 × 10⁻⁴ mol dm⁻³, pH = −log₁₀(2.5 × 10⁻⁴) = 3.60 (to 2 d.p.). WJEC often expects answers to two decimal places. For weak acids, you will use the acid dissociation constant, Kₐ, to find [H⁺] and then pH. Never round intermediate values too early – carry all figures through to the final step.

当 [H⁺] 不是简单的 10 的幂次时，pH 值将包含小数。例如 [H⁺] = 2.5 × 10⁻⁴ mol dm⁻³，pH = −log₁₀(2.5 × 10⁻⁴) = 3.60（保留两位小数）。WJEC 通常要求答案保留两位小数。对于弱酸，你需要利用酸解离常数 Kₐ 求出 [H⁺]，再计算 pH。切勿过早舍入中间数值——将所有数字带入最后一步。

4. Kₐ and pKₐ | 酸解离常数与 pKₐ

The acid dissociation constant Kₐ for a weak acid HA ⇌ H⁺ + A⁻ is defined as:

Kₐ = [H⁺][A⁻] / [HA]

Kₐ has units of mol dm⁻³. The larger the Kₐ value, the stronger the weak acid. Because Kₐ values are often very small, pKₐ is frequently used: pKₐ = −log₁₀Kₐ. A smaller pKₐ means a stronger acid. Ethanoic acid has Kₐ ≈ 1.7 × 10⁻⁵ mol dm⁻³, so pKₐ ≈ 4.76. Hydrocyanic acid HCN has Kₐ ≈ 4.9 × 10⁻¹⁰ mol dm⁻³, pKₐ ≈ 9.31, making it much weaker.

弱酸 HA ⇌ H⁺ + A⁻ 的酸解离常数 Kₐ 定义为：

Kₐ = [H⁺][A⁻] / [HA]

Kₐ 的单位为 mol dm⁻³。Kₐ 值越大，弱酸相对越强。由于 Kₐ 值通常极小，常用 pKₐ 表示：pKₐ = −log₁₀Kₐ。pKₐ 越小，酸性越强。乙酸的 Kₐ ≈ 1.7 × 10⁻⁵ mol dm⁻³，因此 pKₐ ≈ 4.76。氢氰酸 HCN 的 Kₐ ≈ 4.9 × 10⁻¹⁰ mol dm⁻³，pKₐ ≈ 9.31，是弱得多的酸。

In typical WJEC calculations, you assume that [H⁺] = [A⁻] at equilibrium because the dissociation is tiny and [HA] at equilibrium is almost equal to the initial concentration of the acid. So Kₐ = [H⁺]² / [HA]₀, so [H⁺] = √(Kₐ × [HA]₀). Always check whether the approximation is valid: if [HA]₀ / Kₐ > 100, the error is small. If not, you may need to solve a quadratic, but WJEC often avoids this by providing appropriate data.

在典型的 WJEC 计算中，由于解离度微小，可假设平衡时 [H⁺] = [A⁻]，且平衡时 [HA] 几乎等于酸的初始浓度。因此 Kₐ = [H⁺]² / [HA]₀，即 [H⁺] = √(Kₐ × [HA]₀)。务必检查此近似是否成立：若 [HA]₀ / Kₐ > 100，误差很小。若不满足，则需要解二次方程，但 WJEC 通常通过提供适当数据来避免这种情况。

5. Ionic Product of Water Kw and pKw | 水的离子积 Kw 与 pKw

Water undergoes autoprotolysis: 2H₂O ⇌ H₃O⁺ + OH⁻. The equilibrium constant is Kw = [H⁺][OH⁻]. At 298 K, Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶. This means in pure water, [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³, giving pH = 7.00. However, neutrality is defined by [H⁺] = [OH⁻], not by pH 7. If the temperature rises, Kw increases, so the pH of pure water falls below 7 while remaining neutral.

水发生自质子解离：2H₂O ⇌ H₃O⁺ + OH⁻。此平衡常数为 Kw = [H⁺][OH⁻]。在 298 K 时，Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶。这意味着纯水中 [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³，pH = 7.00。但中性是由 [H⁺] = [OH⁻] 定义的，而非 pH = 7。若温度升高，Kw 增大，纯水的 pH 会降至 7 以下，但仍为中性。

pKw = −log₁₀Kw = 14.00 at 298 K. The relationship pKw = pH + pOH is very useful: for any aqueous solution at that temperature, you can find pOH and then [OH⁻] from pH. For example, if a solution has pH = 8.40 at 298 K, pOH = 14.00 – 8.40 = 5.60, so [OH⁻] = 10⁻⁵·⁶⁰ = 2.5 × 10⁻⁶ mol dm⁻³. WJEC may give Kw at different temperatures and ask you to recalculate the pH of pure water.

在 298 K 时，pKw = −log₁₀Kw = 14.00。关系式 pKw = pH + pOH 非常有用：对于该温度下的任意水溶液，可由 pH 求得 pOH，进而得到 [OH⁻]。例如，某溶液在 298 K 时 pH = 8.40，则 pOH = 14.00 – 8.40 = 5.60，[OH⁻] = 10⁻⁵·⁶⁰ = 2.5 × 10⁻⁶ mol dm⁻³。WJEC 可能给出不同温度下的 Kw，要求你重新计算纯水的 pH。

6. Buffer Solutions | 缓冲溶液

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Acidic buffers are made from a weak acid and its conjugate base in similar concentrations, e.g. CH₃COOH/CH₃COONa. Alkaline buffers use a weak base and its conjugate acid, e.g. NH₃/NH₄Cl. The buffer action relies on the equilibrium: HA ⇌ H⁺ + A⁻. Added H⁺ is removed by A⁻ to form HA; added OH⁻ is removed by reacting with H⁺ (which is then replenished by dissociation of HA).

缓冲溶液是一种在加入少量酸或碱时能够抵抗 pH 变化的溶液。酸性缓冲液由浓度相近的弱酸及其共轭碱组成，例如 CH₃COOH/CH₃COONa。碱性缓冲液则使用弱碱及其共轭酸，例如 NH₃/NH₄Cl。缓冲作用依赖于平衡：HA ⇌ H⁺ + A⁻。加入的 H⁺ 被 A⁻ 消耗生成 HA；加入的 OH⁻ 则与 H⁺ 反应（随后 HA 解离补充 H⁺）。

The pH of an acidic buffer is calculated with the Henderson–Hasselbalch equation, which in its simplified A-Level form is:

[H⁺] = Kₐ × [HA] / [A⁻]

Taking negative logs: pH = pKₐ + log₁₀([A⁻]/[HA]). When [HA] = [A⁻], pH = pKₐ. Buffer capacity is greatest when the ratio is close to 1:1 and concentrations are high. WJEC may ask you to calculate the pH after adding a known amount of strong acid or base to a buffer by adjusting the moles of HA and A⁻.

酸性缓冲液的 pH 可使用简化的 A-Level 形式的 Henderson–Hasselbalch 方程计算：

[H⁺] = Kₐ × [HA] / [A⁻]

取负对数后：pH = pKₐ + log₁₀([A⁻]/[HA])。当 [HA] = [A⁻] 时，pH = pKₐ。两组分比例接近 1:1 且浓度较高时，缓冲容量最大。WJEC 可能要求你在向缓冲液中加入已知量的强酸或强碱后，通过调整 HA 和 A⁻ 的物质的量来计算 pH。

To prepare a buffer of a desired pH, choose a weak acid with pKₐ within ±1 of the target pH, then mix with its conjugate base in the appropriate ratio. For example, to make a buffer of pH 4.2, ethanoic acid (pKₐ 4.76) and sodium ethanoate can be mixed in a ratio found from 4.2 = 4.76 + log₁₀([A⁻]/[HA]), giving [A⁻]/[HA] ≈ 0.28.

若要配制特定 pH 的缓冲液，应选择 pKₐ 在目标 pH ±1 范围内的弱酸，然后按其共轭碱的适当比例混合。例如，要配制 pH 4.2 的缓冲液，可使用乙酸（pKₐ 4.76）和乙酸钠，由 4.2 = 4.76 + log₁₀([A⁻]/[HA]) 得出 [A⁻]/[HA] ≈ 0.28。

7. Titration Curves | 滴定曲线

WJEC requires you to sketch and interpret pH titration curves for four combinations: strong acid–strong base, strong acid–weak base, weak acid–strong base, and weak acid–weak base. The shape of the curve reveals key features: initial pH, the equivalence point pH and the vertical jump. You must label the equivalence point as the mid-point of the vertical section, not necessarily pH 7. For a weak acid–strong base titration, the equivalence point pH > 7 because the salt formed (e.g. CH₃COONa) undergoes hydrolysis to produce OH⁻.

WJEC 要求你能够绘制并解读四种组合的 pH 滴定曲线：强酸–强碱、强酸–弱碱、弱酸–强碱和弱酸–弱碱。曲线的形状揭示了关键特征：初始 pH、等当点 pH 以及垂直跃升段。必须将等当点标注为垂直段的中点，它不一定是 pH 7。对于弱酸–强碱滴定，由于形成的盐（如 CH₃COONa）发生水解产生 OH⁻，等当点 pH > 7。

Combination	Equivalence point pH
Strong acid + Strong base	7.0
Strong acid + Weak base	< 7 (e.g. NH₄Cl solution is acidic)
Weak acid + Strong base	> 7 (e.g. CH₃COONa solution is alkaline)
Weak acid + Weak base	Varies, no sharp vertical jump – not suitable for titration

Before the equivalence point, the solution contains a mixture of unreacted acid/base and the salt. For a weak acid–strong base titration, this region is a buffer. At half‑neutralisation (half the volume needed to reach the equivalence point), [HA] = [A⁻], so pH = pKₐ. This is a crucial point for determining Kₐ experimentally. The vertical section shows the sharp pH change that allows indicator selection.

在等当点之前，溶液中包含未反应的酸/碱和生成的盐。对于弱酸–强碱滴定，此区域即为缓冲液。在半中和点（到达等当点所需体积的一半），[HA] = [A⁻]，因此 pH = pKₐ。这是实验测定 Kₐ 的关键点。垂直段显示出剧烈的 pH 变化，借此可以选择合适的指示剂。

8. Indicators | 指示剂

Indicators are weak acids or bases where the undissociated form (HIn) has a different colour from its conjugate base (In⁻). The equilibrium HIn ⇌ H⁺ + In⁻ responds to pH changes. The colour change occurs over a range of about pH = pKₐ ± 1. For an indicator to be suitable, its pH range must lie entirely within the vertical section of the titration curve. For a strong acid–strong base titration, both methyl orange (range 3.1–4.4) and phenolphthalein (range 8.2–10.0) work well because the vertical jump spans pH ~3–11.

指示剂是弱酸或弱碱，其未解离形态（HIn）与共轭碱（In⁻）颜色不同。平衡 HIn ⇌ H⁺ + In⁻ 随 pH 变化而移动。颜色变化发生的范围约为 pH = pKₐ ± 1。合适的指示剂，其变色范围必须完全落入滴定曲线的垂直段。对于强酸–强碱滴定，甲基橙（范围 3.1–4.4）和酚酞（范围 8.2–10.0）均适用，因为垂直跃升跨越 pH ~3–11。

For a weak acid–strong base titration, the equivalence point is alkaline (e.g. pH ~8.5). Methyl orange would change colour too early; phenolphthalein is the correct choice because its range (8.2–10.0) matches the vertical section. For a strong acid–weak base titration, the equivalence point is acidic, so methyl orange is suitable. The weak acid–weak base titration lacks a sharp vertical jump, so no single indicator gives a sharp colour change; you must use a pH meter.

对于弱酸–强碱滴定，等当点为碱性（如 pH ~8.5）。甲基橙会过早变色；酚酞是正确的选择，因其变色范围（8.2–10.0）与垂直段吻合。对于强酸–弱碱滴定，等当点为酸性，因此甲基橙适用。弱酸–弱碱滴定没有陡峭的垂直段，所以没有单一指示剂能给出敏锐的颜色变化，必须使用 pH 计。

9. Dilution Calculations and pH | 稀释计算与 pH

Diluting an acid changes [H⁺] in a predictable way for strong acids. If you dilute 25.0 cm³ of 0.100 mol dm⁻³ HCl to 250 cm³, the new concentration is (25.0/250) × 0.100 = 0.0100 mol dm⁻³, so pH changes from 1.00 to 2.00. However, you must never assume that dilution by a factor of 10 always changes pH by exactly 1 for weak acids, because the equilibrium shifts to partially compensate. Use Kₐ = [H⁺]² / [HA] with the new initial concentration to find the new [H⁺] and pH.

对于强酸，稀释以可预测的方式改变 [H⁺]。若将 25.0 cm³ 0.100 mol dm⁻³ HCl 稀释至 250 cm³，新浓度为 (25.0/250) × 0.100 = 0.0100 mol dm⁻³，pH 从 1.00 变为 2.00。但对于弱酸，切勿假设稀释 10 倍总是使 pH 恰好改变 1，因为平衡会移动以部分补偿。应使用 Kₐ = [H⁺]² / [HA]，代入新的初始浓度，求出新的 [H⁺] 和 pH。

When diluting bases, work through [OH⁻] and Kw. For a 0.20 mol dm⁻³ NaOH solution diluted by a factor of 100, new [OH⁻] = 0.0020 mol dm⁻³, so [H⁺] = Kw/0.0020 = 5.0 × 10⁻¹² mol dm⁻³, pH = 11.30. Also beware of extremely dilute strong acids (e.g. 1 × 10⁻⁸ mol dm⁻³ HCl). The autoionisation of water becomes significant and pH will be just below 7, not 8. WJEC may ask you to explain why the pH asymptotically approaches 7 upon infinite dilution.

稀释碱时，应通过 [OH⁻] 和 Kw 进行计算。将 0.20 mol dm⁻³ NaOH 溶液稀释 100 倍，新 [OH⁻] = 0.0020 mol dm⁻³，故 [H⁺] = Kw/0.0020 = 5.0 × 10⁻¹² mol dm⁻³，pH = 11.30。此外，要注意极稀的强酸（如 1 × 10⁻⁸ mol dm⁻³ HCl）。此时水的自解离变得显著，pH 将略低于 7，而非 8。WJEC 可能要求你解释为何无限稀释时 pH 会趋近 7。

10. Redox Reactions in Acid–Base Context | 氧化还原与酸碱的关联

While acid–base reactions involve proton transfer, many redox titrations are performed in acidic solution and the acidity can affect the oxidising power. For example, manganate(VII) titrations require excess dilute sulfuric acid, not hydrochloric acid (which would be oxidised to Cl₂). The half‑equation MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O illustrates the dependence on H⁺: without sufficient acid, the reduction may stop at MnO₂, a brown precipitate, ruining the endpoint. Understanding the role of H⁺ helps you explain procedural details in practical assessments.

酸碱反应涉及质子转移，而许多氧化还原滴定在酸性溶液中进行，酸度会影响氧化能力。例如，高锰酸根（manganate(VII)）滴定需要过量的稀硫酸，而不能使用盐酸（否则会被氧化为 Cl₂）。半反应 MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O 表明其对 H⁺ 的依赖性：若酸不足，还原可能停留在 MnO₂，形成棕色沉淀，破坏终点。理解 H⁺ 的作用有助于你在实验考核中解释操作细节。

Similarly, the dichromate(VI)–iron(II) titration relies on acid: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. The high H⁺ concentration keeps the equilibrium to the right. In WJEC Unit 4, you may be asked to combine pH calculations with redox electrode potentials, linking thermodynamics to acid–base equilibria.

类似地，重铬酸根(VI)–铁(II) 滴定依赖于酸：Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O。高浓度的 H⁺ 使平衡向右移动。在 WJEC 单元 4 中，你可能会遇到将 pH 计算与氧化还原电极电位相结合的题目，从而将热力学与酸碱平衡联系起来。

11. Common Errors and Exam Tips | 常见错误与应试技巧

Confusing strong/weak with concentrated/dilute. A weak acid at 10 mol dm⁻³ is still weak; it has a low degree of dissociation. Do not write that ‘most of the molecules are dissociated’ for a weak acid – that’s the definition of strong.

混淆强/弱与浓/稀。10 mol dm⁻³ 的弱酸仍然是弱酸，其解离度很低。不要为弱酸写出“大部分分子已解离”的描述——那是强酸的定义。
Forgetting to convert cm³ to dm³ before calculating moles. All concentration units are in mol dm⁻³, so volume must be in dm⁻³ (1 dm³ = 1000 cm³). Multi‑step buffer questions often catch students out here.

在计算物质的量之前忘记将 cm³ 转换为 dm³。所有浓度单位均为 mol dm⁻³，因此体积必须为 dm⁻³（1 dm³ = 1000 cm³）。多步缓冲液计算题常在此处设陷阱。
Using [H⁺] = √(Kₐ × c) without checking the approximation. WJEC mark schemes often award a mark for stating the assumption or for showing the ratio c/Kₐ > 100. If the assumption is not valid, you must solve the quadratic.

未检查近似条件就直接使用 [H⁺] = √(Kₐ × c)。WJEC 评分标准通常对说明假设或展示 c/Kₐ > 100 赋予分数。若近似不成立，则必须求解二次方程。
Incorrect rounding of pH. For example, if [H⁺] is known to two significant figures, pH should be given to two decimal places. Writing pH = 4.6 instead of 4.60 can lose a mark.

pH 的舍入错误。例如，如果 [H⁺] 保留两位有效数字，pH 应给出两位小数。将 pH 写作 4.6 而非 4.60 可能会失分。
Not stating the colour change of an indicator with enough detail. You must say ‘methyl orange changes from red to yellow’ or ‘phenolphthalein: colourless to pink’, not just the names.

对指示剂颜色变化的描述不够详细。你必须写出“甲基橙由红变黄”或“酚酞：无色变粉红”，而不能仅仅指出指示剂名称。

12. Linking Acid–Base to Other Topics | 酸碱与其他主题的联系

In WJEC A-Level Chemistry, acid–base knowledge underpins organic chemistry (carboxylic acids, amines, amino acids), transition metals (acidity of metal aqua ions), and energetics (neutralisation enthalpy). For example, the acidity of a metal 3+ aqua ion depends on its charge density, polarising the O–H bond in coordinated water. Understanding proton transfer helps you predict products of reactions and interpret NMR spectra of acids and bases.

在 WJEC A-Level 化学中，酸碱知识是有机化学（羧酸、胺、氨基酸）、过渡金属（金属水合离子的酸性）以及能量学（中和焓）的基础。例如，金属 3+ 水合离子的酸度取决于其电荷密度，使配位水中的 O–H 键极化。理解质子转移有助于你预测反应产物并解析酸和碱的 NMR 谱图。

Buffer systems also appear in biochemistry: blood pH is maintained by the carbonic acid‑hydrogencarbonate buffer. The equilibrium H₂CO₃ ⇌ H⁺ + HCO₃⁻ is shifted by respiration and metabolism. WJEC may use context‑based questions linking acid–base calculations to real‑world examples, so practising these applications builds confidence.

缓冲体系还出现在生物化学中：血液 pH 由碳酸–碳酸氢盐缓冲对维持。平衡 H₂CO₃ ⇌ H⁺ + HCO₃⁻ 受呼吸和代谢影响而移动。WJEC 可能采用情境题，将酸碱计算与现实案例相结合，因此练习这些应用能增强信心。

Published by TutorHao | Science Revision Series | aleveler.com

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Acids and Bases | 酸与碱 考点精讲