📚 A-Level Chemistry Year 2: Acids, Bases, Buffers and pH Calculations | A-Level 化学第二年：酸、碱、缓冲溶液与 pH 计算
In Year 2 chemistry, the study of acids, bases, buffers and pH calculations forms a central part of physical chemistry. These ideas connect equilibrium theory with practical laboratory skills and are essential for understanding many reactions in aqueous solution. Exam questions in this topic are usually calculation-heavy, so accuracy with units, significant figures and approximations matters as much as understanding the theory.
在化学第二年，酸、碱、缓冲溶液和 pH 计算的学习构成物理化学的核心部分。这些概念将平衡理论与实验技能联系起来，对于理解水溶液中许多反应至关重要。本专题的考题往往计算量较大，因此单位、有效数字与近似处理是否准确，与理论理解同样重要。
1. The Brønsted-Lowry Theory of Acids and Bases | 布朗斯特-劳里酸碱理论
According to the Brønsted-Lowry theory, an acid is a proton donor and a base is a proton acceptor. A proton is represented by H⁺. This definition is broader than the older idea that acids simply produce H⁺ in water, because it works in non-aqueous solvents and explains why species such as NH₃ behave as bases.
根据布朗斯特-劳里理论，酸是质子给体，碱是质子受体。质子用 H⁺ 表示。这一定义比"酸在水中产生 H⁺"的旧观点更为宽泛，因为它同样适用于非水溶剂，并能解释 NH₃ 等物种为何表现为碱。
When an acid donates a proton, the species left behind is its conjugate base. Similarly, when a base accepts a proton, it forms its conjugate acid. Each acid-base reaction involves two conjugate acid-base pairs. Note that a proton is a bare hydrogen nucleus and never exists free in aqueous solution; it is always transferred to a base, commonly water.
当酸给出质子后，剩下的物种就是其共轭碱。类似地，当碱接受质子后，就形成其共轭酸。每个酸碱反应都包含两对共轭酸碱对。注意质子是裸露的氢核，在水溶液中不会自由存在，它总是转移给某个碱，通常就是水。
Water is amphiprotic: it can act as an acid or as a base depending on its partner. This dual behaviour is the key to understanding Kw, buffers and salt hydrolysis later in the topic.
水是两性的：根据其反应对象不同，它可以充当酸或碱。这种双重行为是理解后续 Kw、缓冲溶液和盐类水解的关键。
- HCl + H₂O → H₃O⁺ + Cl⁻; HCl is an acid and Cl⁻ is its conjugate base.
- HCl + H₂O → H₃O⁺ + Cl⁻；HCl 是酸，Cl⁻ 是其共轭碱。
- NH₃ + H₂O ⇌ NH₄⁺ + OH⁻; NH₃ is a base and NH₄⁺ is its conjugate acid.
- NH₃ + H₂O ⇌ NH₄⁺ + OH⁻；NH₃ 是碱，NH₄⁺ 是其共轭酸。
2. Conjugate Pairs, Strong and Weak Acids | 共轭酸碱对与强酸弱酸
The strength of an acid is linked to the strength of its conjugate base in an inverse way. A strong acid such as HCl dissociates almost completely in water, so its conjugate base Cl⁻ has virtually no tendency to accept a proton back. A weak acid such as CH₃COOH dissociates only partially, so its conjugate base CH₃COO⁻ is a noticeably stronger proton acceptor.
酸的强度与其共轭碱的强度呈反比关系。像 HCl 这样的强酸在水中几乎完全电离，因此其共轭碱 Cl⁻ 几乎没有再接受质子的倾向。而像 CH₃COOH 这样的弱酸只部分电离，因此其共轭碱 CH₃COO⁻ 是明显更强的质子受体。
Strong acids you should recall include HCl, HBr, HI, HNO₃, H₂SO₄ (first dissociation) and HClO₄. Strong bases include NaOH, KOH, Ba(OH)₂ and Ca(OH)₂. Everything else in the A-Level specification, including all carboxylic acids and ammonia, is treated as weak.
需要记住的强酸包括 HCl、HBr、HI、HNO₃、H₂SO₄（第一步电离）和 HClO₄。强碱包括 NaOH、KOH、Ba(OH)₂ 和 Ca(OH)₂。考纲中其他物质，包括所有羧酸与氨，都按弱酸弱碱处理。
(→) is used for near-complete dissociation of a strong acid, while an equilibrium arrow (⇌ dissociation of a weak acid. Examiners frequently award marks for using the correct arrow.
在方程式中，单箭头（→）用于表示强酸的近乎完全电离，而平衡箭头（⇌）用于表示弱酸的部分电离。考试中常因箭头使用是否正确而给分。
3. The Ionic Product of Water, Kw | 水的离子积 Kw
Pure water undergoes slight self-ionisation: H₂O ⇌ H⁺ + OH⁻. The equilibrium constant for this process is called the ionic product of water, Kw. Because the concentration of water is effectively constant, it is incorporated into the constant itself.
纯水会发生轻微的自电离：H₂O ⇌ H⁺ + OH⁻。该过程的平衡常数称为水的离子积 Kw。由于水的浓度实际上可视为常数，它已被并入常数之中。
Kw = [H⁺][OH⁻]
ause Kw is an equilibrium constant, its value increases when temperature increases. The self-ionisation of water is endothermic (ΔH is positive), so heating shifts the equilibrium to the right, producing more H⁺ and more OH⁻.
在 298 K 时，Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶。由于 Kw 是平衡常数，温度升高时其数值增大。水的自电离是吸热过程（ΔH 为正值），因此加热使平衡右移，产生更多 H⁺ 和更多 OH⁻。
ater still contains equal concentrations of H⁺ and OH⁻ and is still neutral, but the pH is about 6.63. This is a common exam trap: neutral does not always mean pH 7.
在 298 K 的纯水中，[H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³，pH 为 7.00。在更高温度下，例如 323 K 时 Kw = 5.5 × 10⁻¹⁴ mol² dm⁻⁶，纯水中 H⁺ 与 OH⁻ 浓度依然相等，仍为中性，但 pH 约为 6.63。这是常见考点陷阱：中性并不总意味着 pH 等于 7。
4. The pH Scale, pOH and Logarithmic Thinking | pH 标度、pOH 与对数思维
The pH of a solution is defined as the negative logarithm to base 10 of the hydrogen ion concentration in mol dm⁻³. Because the scale is logarithmic, a change of one pH unit corresponds to a tenfold change in [H⁺].
溶液的 pH 定义为氢离子浓度（单位 mol dm⁻³）以 10 为底的负对数。由于该标度是对数的，pH 每变化 1 个单位，相当于 [H⁺] 变化 10 倍。
pH = −log₁₀[H⁺] [H⁺] = 10⁻ᵖᴴ
The same convention is applied to hydroxide ions and to equilibrium constants. The pOH of a solution is defined as −log₁₀[OH⁻], and at 298 K the following relationship holds.
同样的约定也适用于氢氧根离子与平衡常数。溶液的 pOH 定义为 −log₁₀[OH⁻]，在 298 K 时存在如下关系。
pOH = −log₁₀[OH⁻] pH + pOH = 14.00 (at 298 K)
and are not significant figures. For example, [H⁺] = 2.5 × 10⁻³ mol dm⁻³ gives pH = 2.60.
计算对数时，pH 应保留两位小数，因为小数点前的数字来自 10 的幂次，不属于有效数字。例如 [H⁺] = 2.5 × 10⁻³ mol dm⁻³ 时，pH = 2.60。
5. Calculating the pH of Strong Acids and Strong Bases | 强酸与强碱的 pH 计算
For a strong monoprotic acid, dissociation is complete, so the hydrogen ion concentration equals the acid concentration. For a strong diprotic acid such as H₂SO₄, the first ionisation is complete and the second is strong enough in dilute solution that [H⁺] is approximately twice the acid concentration.
对于强一元酸，由于完全电离，氢离子浓度等于酸的浓度。对于 H₂SO₄ 这样的强二元酸，第一步电离完全，在稀溶液中第二步电离程度也很高，因此 [H⁺] 约等于酸浓度的两倍。
For a strong base, the calculation must start from [OH⁻]. Convert to pOH first, then subtract from 14 to obtain the pH. Remember that Ba(OH)₂ and Ca(OH)₂ produce two OH⁻ ions per formula unit.
对于强碱，计算必须从 [OH⁻] 出发：先算 pOH，再用 14 相减得到 pH。注意 Ba(OH)₂ 和 Ca(OH)₂ 每摩尔可产生两摩尔 OH⁻。
| Solution | Working | Answer |
| 0.010 mol dm⁻³ HCl | pH = −log(0.010) | pH = 2.00 |
| 0.050 mol dm⁻³ NaOH | pOH = −log(0.050) = 1.30 | pH = 12.70 |
| 0.0050 mol dm⁻³ Ba(OH)₂ | [OH⁻] = 0.010, pOH = 2.00 | pH = 12.00 |
6. Weak Acids and the Acid Dissociation Constant Ka | 弱酸与酸解离常数 Ka
A weak acid HA dissociates only partially in water, setting up an equilibrium. The acid dissociation constant Ka expresses the position of this equilibrium, and a larger Ka means a stronger acid.
弱酸 HA 在水中只部分电离，形成平衡。酸解离常数 Ka 反映该平衡的位置，Ka 越大表示酸性越强。
HA ⇌ H⁺ + A⁻ Ka = [H⁺][A⁻] ÷ [HA]
. We also assume that all the H⁺ present comes from the weak acid, so [H⁺] = [A⁻]. Combining these two approximations gives a very useful result for exam calculations.
由于电离程度很小，可以假设 HA 的平衡浓度基本等于初始酸浓度 c；同时假设溶液中的 H⁺ 全部来自该弱酸，因此 [H⁺] = [A⁻]。把这两个近似结合，就得到考试中极常用的结果。
[H⁺] = √Ka × c
ethanoic acid has Ka = 1.8 × 10⁻⁵ mol dm⁻³ at 298 K. For a 0.100 mol dm⁻³ solution, [H⁺] = √1.8 × 10⁻⁵ × 0.100 = √1.8 × 10⁻⁶ mol dm⁻³, so pH = 2.87. The percentage dissociation is (1.34 × 10⁻³ ÷ 0.100) × 100 = 1.34 %, confirming that the approximation is valid.
例如，在 298 K 时乙酸 Ka = 1.8 × 10⁻⁵ mol dm⁻³。对 0.100 mol dm⁻³ 溶液，[H⁺] = √1.8 × 10⁻⁵ × 0.100 = √1.8 × 10⁻⁶ = 1.34 × 10⁻³ mol dm⁻³，故 pH = 2.87。电离度为 (1.34 × 10⁻³ ÷ 0.100) × 100 = 1.34 %，说明近似是合理的。
7. pKa and the Link Between Ka and pH | pKa 与 Ka、pH 之间的关系
Because Ka values are often very small, chemists use pKa, defined in the same way as pH but applied to Ka. A lower pKa corresponds to a stronger acid, and pKa values are quoted to two decimal places.
由于 Ka 数值往往很小，化学家引入 pKa，其定义方式与 pH 类似，只是对象为 Ka。pKa 越小，酸性越强；pKa 通常保留两位小数。
pKa = −log₁₀Ka Ka = 10⁻ᵖᴷᵃ
The most useful form of the relationship links pH, pKa and the ratio of the conjugate base concentration to the acid concentration. This is often called the Henderson-Hasselbalch equation and is written below.
最实用的关系式把 pH、pKa 以及共轭碱浓度与酸浓度之比联系起来，通常称为亨德森-哈塞尔巴赫方程，形式如下。
pH = pKa + log₁₀([A⁻] ÷ [HA])
ve this occurs at the half-equivalence point, which is why the half-equivalence pH can be read directly as the pKa of the weak acid.
由此可得一个非常重要的结论：当 [A⁻] = [HA] 时，1 的对数为零，故 pH = pKa。在滴定曲线上，这出现在半当量点，因此半当量点的 pH 可直接读作该弱酸的 pKa。
8. Weak Bases and the Base Dissociation Constant Kb | 弱碱与碱解离常数 Kb
A weak base B accepts a proton from water, generating hydroxide ions and the conjugate acid BH⁺. The base dissociation constant Kb is defined in an exactly parallel way to Ka.
弱碱 B 从水中接受质子，生成氢氧根离子与共轭酸 BH⁺。碱解离常数 Kb 的定义与 Ka 完全对应。
B + H₂O ⇌ BH⁺ + OH⁻ Kb = [BH⁺][OH⁻] ÷ [B]
For a conjugate acid-base pair, the two constants are related through the ionic product of water. Taking negative logarithms of both sides gives a second, equally useful relationship.
对于一对共轭酸碱，两个常数通过水的离子积相互联系。对两边同时取负对数，可得到另一个同样有用的关系式。
Ka × Kb = Kw pKa + pKb = 14.00 (at 298 K)
For ammonia, Kb = 1.8 × 10⁻⁵ mol dm⁻³. In a 0.100 mol dm⁻³ solution, [OH⁻] = √1.8 × 10⁻⁵ × 0.100 t approximation used for weak acids applies, provided the degree of dissociation is small.
对于氨，Kb = 1.8 × 10⁻⁵ mol dm⁻³。在 0.100 mol dm⁻³ 溶液中，[OH⁻] = √1.8 × 10⁻⁵ × 0.100 = 1.34 × 10⁻³ mol dm⁻³，故 pOH = 2.87，pH = 11.13。只要电离度很小，弱酸所用的平方根近似同样适用。
9. How Buffer Solutions Work | 缓冲溶液的作用原理
A buffer solution resists changes in pH when small amounts of acid or alkali are added, or when the solution is diluted. It consists of a weak acid and its conjugate base, or a weak base and its conjugate acid, both present in significant and comparable concentrations.
缓冲溶液在加入少量酸或碱、或进行稀释时能抵抗 pH 的变化。它由弱酸与其共轭碱、或弱碱与其共轭酸组成，两者浓度都较大且数量级相当。
An acidic buffer is typically made from a weak acid and the salt of that weak acid, for example CH₃COOH with CH₃COONa. The weak acid supplies a reservoir of HA and the salt supplies A⁻. Added H⁺ is removed by A⁻, and added OH⁻ is removed by HA, so the ratio [A⁻] ÷ [HA] and therefore the pH stay almost constant.
酸性缓冲溶液通常由弱酸与其盐配制，例如 CH₃COOH 与 CH₃COONa。弱酸提供 HA 储备，盐提供 A⁻。加入的 H⁺ 被 A⁻ 消耗，加入的 OH⁻ 被 HA 消耗，因此 [A⁻] ÷ [HA] 之比、进而 pH 几乎保持不变。
A basic buffer is made from a weak base and its salt, for example NH₃ with NH₄Cl. Because the working equilibrium involves OH⁻, this type of buffer is most effective in the alkaline region.
碱性缓冲溶液由弱碱与其盐配制，例如 NH₃ 与 NH₄Cl。由于其工作平衡涉及 OH⁻，这类缓冲液在碱性区域最为有效。
- Buffer capacity is greatest when [A⁻] ≈ [HA], that is when pH ≈ pKa.
- 缓冲容量在 [A⁻] ≈ [HA]、即 pH ≈ pKa 时最大。
- A buffer is useful over an approximate range of pH = pKa ± 1.
- 缓冲液的有效范围大致为 pH = pKa ± 1。
- Dilution does not change the ratio [A⁻] ÷ [HA], so the pH is nearly unchanged.
- 稀释不改变 [A⁻] ÷ [HA] 之比，因此 pH 几乎不变。
10. Calculating the pH of a Buffer Solution | 缓冲溶液 pH 的计算
Buffer calculations are straightforward once you recognise the pattern: the weak acid provides the HA term, the salt provides the A⁻ term, and the equation below links them to pH. It is simply the Henderson-Hasselbalch equation written with salt and acid concentrations.
只要识别出模式，缓冲溶液计算就很简单：弱酸提供 HA 项，盐提供 A⁻ 项，下面的方程把它们与 pH 联系起来。它实际上就是写成盐浓度与酸浓度形式的亨德森-哈塞尔巴赫方程。
pH = pKa + log₁₀([salt] ÷ [acid])
l dm⁻³ CH₃COONa. Substituting gives pH = 4.76 + log₁₀(0.200 ÷ 0.100) = 4.76 + 0.30 = 5.06.
考虑由 0.100 mol dm⁻³ CH₃COOH（pKa = 4.76）与 0.200 mol dm⁻³ CH₃COONa 配成的缓冲液。代入得 pH = 4.76 + log₁₀(0.200 ÷ 0.100) = 4.76 + 0.30 = 5.06。
If a small amount of HCl is added, the added H⁺ reacts with CH₃COO⁻ to form CH₃COOH. The salt concentration falls slightly and the acid concentration rises slightly, but the ratio changes very little, so the pH change is small. If the same amount of HCl were added to pure water, the pH would fall dramatically. This comparison is a standard exam question and should be answered with an equation and a full explanation.
若加入少量 HCl，加入的 H⁺ 与 CH₃COO⁻ 反应生成 CH₃COOH。
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