Acids and Bases | 酸与碱

📚 Acids and Bases | 酸与碱

Acids and bases form one of the most fundamental topics in A-Level chemistry. This revision guide covers the key definitions, pH calculations, ionic equilibria, titration curves, and buffer systems required by the AQA specification, with a focus on the quantitative skills examiners expect you to demonstrate.

酸与碱是A-Level化学中最基础的主题之一。本复习指南涵盖AQA考纲要求的关键定义、pH计算、离子平衡、滴定曲线和缓冲溶液体系，重点聚焦考官期望你展现的定量计算技能。

1. Bronsted-Lowry Theory | 布朗斯特-劳里理论

The Bronsted-Lowry model defines an acid as a proton donor and a base as a proton acceptor. A proton is a hydrogen ion, H⁺. This theory extends beyond aqueous solutions and explains acid-base behaviour in terms of proton transfer between species.

布朗斯特-劳里模型将酸定义为质子供体，碱定义为质子受体。质子即氢离子H⁺。该理论超越了水溶液范围，从物种间质子转移的角度解释酸碱行为。

When an acid donates a proton, it forms its conjugate base. When a base accepts a proton, it forms its conjugate acid. An acid-base pair that differs by a single proton is called a conjugate acid-base pair.

当酸给出质子时，形成其共轭碱；当碱接受质子时，形成其共轭酸。相差一个质子的酸碱对称为共轭酸碱对。

HA ⇌ H⁺ + A⁻

Here, HA is the acid and A⁻ is its conjugate base. In the reverse direction, A⁻ acts as a base accepting a proton to reform HA.

在此，HA是酸，A⁻是其共轭碱。在逆反应方向上，A⁻作为碱接受质子重新形成HA。

Common examples include: HCl ⇌ H⁺ + Cl⁻, where Cl⁻ is the conjugate base; and NH₃ + H⁺ ⇌ NH₄⁺, where NH₄⁺ is the conjugate acid of the base NH₃.

常见例子包括：HCl ⇌ H⁺ + Cl⁻，其中Cl⁻是共轭碱；以及NH₃ + H⁺ ⇌ NH₄⁺，其中NH₄⁺是碱NH₃的共轭酸。

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

Water undergoes a very slight self-ionisation: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). The equilibrium constant for this process is the ionic product of water, denoted Kw.

水发生极微弱的自电离：H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)。该过程的平衡常数称为水的离子积，记作Kw。

At 25 °C, Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶. This is a temperature-dependent value: as temperature increases, Kw increases because the ionisation of water is endothermic.

在25 °C时，Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶。该值随温度变化：温度升高时Kw增大，因为水的电离是吸热过程。

In pure water, [H⁺] = [OH⁻] = √Kw = 1.00 × 10⁻⁷ mol dm⁻³, giving a neutral pH of 7 at 25 °C. Importantly, a solution is only neutral when [H⁺] = [OH⁻], regardless of pH. At 50 °C, Kw = 5.48 × 10⁻¹⁴, so neutral pH = ½pKw ≈ 6.63.

在纯水中，[H⁺] = [OH⁻] = √Kw = 1.00 × 10⁻⁷ mol dm⁻³，在25 °C时pH = 7呈中性。重要的是，溶液只有在[H⁺] = [OH⁻]时才为中性，与pH值无关。在50 °C时，Kw = 5.48 × 10⁻¹⁴，因此中性pH = ½pKw ≈ 6.63。

Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C

Kw enables conversion between [H⁺] and [OH⁻] in any aqueous solution, which is essential for calculating pH of strong bases and for understanding buffer systems.

Kw允许在任何水溶液中相互转换[H⁺]和[OH⁻]，这对于计算强碱的pH和理解缓冲体系至关重要。

3. The pH Scale | pH标度

The pH scale is a logarithmic measure of hydrogen ion concentration. It is defined as pH = -log₁₀[H⁺]. This definition applies to dilute aqueous solutions at any temperature, although the interpretation of acidic, neutral, and basic depends on the value of Kw at that temperature.

pH标度是氢离子浓度的对数度量。定义为pH = -log₁₀[H⁺]。该定义适用于任意温度下的稀水溶液，但酸性、中性、碱性的判断取决于该温度下的Kw值。

High [H⁺] corresponds to a low pH value. A change of one pH unit represents a tenfold change in hydrogen ion concentration. For example, a solution with pH 3 has [H⁺] = 1 × 10⁻³ mol dm⁻³, which has ten times more H⁺ than a solution with pH 4, where [H⁺] = 1 × 10⁻⁴ mol dm⁻³.

高[H⁺]对应低pH值。pH每变化一个单位代表氢离子浓度变化十倍。例如，pH 3的溶液[H⁺] = 1 × 10⁻³ mol dm⁻³，其H⁺浓度是pH 4溶液（[H⁺] = 1 × 10⁻⁴ mol dm⁻³）的十倍。

Conversely, [H⁺] = 10⁻ᵖᴴ mol dm⁻³. This inverse relationship must be used carefully when converting from pH to concentration in titration calculations.

反之，[H⁺] = 10⁻ᵖᴴ mol dm⁻³。在滴定计算中，从pH反推浓度时必须谨慎使用这一逆关系。

For acidic solutions at 25 °C: pH < 7. For alkaline solutions: pH > 7. For neutral solutions: pH = 7.

在25 °C时：酸性溶液pH < 7；碱性溶液pH > 7；中性溶液pH = 7。

To measure pH: a pH meter gives the most accurate reading; universal indicator gives a colour comparison; and litmus paper only distinguishes acid from base.

测量pH的方法：pH计给出最准确的读数；万能指示剂通过颜色对比；石蕊试纸仅区分酸和碱。

4. Strong Acids: pH Calculation | 强酸的pH计算

A strong acid is one that fully dissociates in aqueous solution. Common examples include hydrochloric acid (HCl), nitric acid (HNO₃), and sulfuric acid (H₂SO₄) for the first ionisation step.

强酸在水溶液中完全解离。常见例子包括盐酸（HCl）、硝酸（HNO₃），以及第一步电离的硫酸（H₂SO₄）。

For a monoprotic strong acid HA: HA(aq) → H⁺(aq) + A⁻(aq). Since dissociation is complete, [H⁺] = [HA]₀, the initial concentration of the acid.

对于一元强酸HA：HA(aq) → H⁺(aq) + A⁻(aq)。由于完全解离，[H⁺] = [HA]₀，即酸的初始浓度。

Worked example: Calculate the pH of a 0.0250 mol dm⁻³ solution of nitric acid.

例题：计算0.0250 mol dm⁻³硝酸溶液的pH。

[H⁺] = 0.0250 mol dm⁻³
pH = -log₁₀(0.0250) = 1.60

For sulfuric acid, the first proton dissociates fully (H₂SO₄ → H⁺ + HSO₄⁻), but the second dissociation (HSO₄⁻ ⇌ H⁺ + SO₄²⁻) is only partial. In AQA calculations, you may assume full dissociation of both protons for dilute solutions as a simplification, giving [H⁺] = 2 × [H₂SO₄]₀, but be prepared for questions that specify only the first dissociation is complete.

对于硫酸，第一个质子完全解离（H₂SO₄ → H⁺ + HSO₄⁻），但第二个解离（HSO₄⁻ ⇌ H⁺ + SO₄²⁻）仅是部分的。在AQA计算中，对于稀溶液可以简化假设两个质子均完全解离，得到[H⁺] = 2 × [H₂SO₄]₀，但需注意题目是否指定仅第一步完全解离。

5. Strong Bases: pH Calculation | 强碱的pH计算

A strong base fully dissociates in water. Examples include sodium hydroxide (NaOH) and potassium hydroxide (KOH). For these monoprotic bases: NaOH(aq) → Na⁺(aq) + OH⁻(aq), so [OH⁻] = [NaOH]₀.

强碱在水中完全解离。例如氢氧化钠（NaOH）和氢氧化钾（KOH）。对于这些一元碱：NaOH(aq) → Na⁺(aq) + OH⁻(aq)，因此[OH⁻] = [NaOH]₀。

Group 2 hydroxides such as calcium hydroxide Ca(OH)₂ and barium hydroxide Ba(OH)₂ release two hydroxide ions per formula unit: [OH⁻] = 2 × [Ca(OH)₂]₀.

第二主族氢氧化物如氢氧化钙Ca(OH)₂和氢氧化钡Ba(OH)₂，每个化学式单元释放两个氢氧根离子：[OH⁻] = 2 × [Ca(OH)₂]₀。

To find the pH of a strong base at 25 °C, use the Kw expression: [H⁺] = Kw / [OH⁻], then pH = -log₁₀[H⁺].

在25 °C下计算强碱的pH，需使用Kw表达式：[H⁺] = Kw / [OH⁻]，然后pH = -log₁₀[H⁺]。

Worked example: Calculate the pH of a 0.100 mol dm⁻³ solution of barium hydroxide at 25 °C.

例题：计算25 °C时0.100 mol dm⁻³氢氧化钡溶液的pH。

[OH⁻] = 2 × 0.100 = 0.200 mol dm⁻³
[H⁺] = 1.00 × 10⁻¹⁴ / 0.200 = 5.00 × 10⁻¹⁴
pH = -log₁₀(5.00 × 10⁻¹⁴) = 13.30

An alternative approach for bases is to calculate pOH = -log₁₀[OH⁻], then pH = 14 – pOH at 25 °C. This works because pH + pOH = pKw = 14 at 25 °C.

另一种方法：先算pOH = -log₁₀[OH⁻]，然后用pH = 14 – pOH（仅25 °C下成立）。这是因为pH + pOH = pKw = 14（25 °C时）。

6. Weak Acids and Ka | 弱酸与Ka

A weak acid only partially dissociates in water. Examples include ethanoic acid (CH₃COOH), carbonic acid (H₂CO₃), and hydrofluoric acid (HF). The equilibrium lies well to the left, with most of the acid present as undissociated molecules.

弱酸在水中仅部分解离。例如乙酸（CH₃COOH）、碳酸（H₂CO₃）和氢氟酸（HF）。平衡强烈偏向左侧，大部分酸以未解离分子形式存在。

For a weak monoprotic acid HA: HA(aq) ⇌ H⁺(aq) + A⁻(aq). The acid dissociation constant Ka is defined as:

对于一元弱酸HA：HA(aq) ⇌ H⁺(aq) + A⁻(aq)。酸解离常数Ka定义为：

Ka = [H⁺][A⁻] / [HA] (units: mol dm⁻³)

Larger Ka values indicate stronger acids. For example, Ka(CH₃COOH) = 1.74 × 10⁻⁵ mol dm⁻³ at 25 °C. The relationship pKa = -log₁₀Ka is commonly used, and weaker acids have larger pKa values.

Ka值越大表明酸越强。例如，25 °C时Ka(CH₃COOH) = 1.74 × 10⁻⁵ mol dm⁻³。常用关系pKa = -log₁₀Ka，酸越弱pKa值越大。

Worked example: A 0.500 mol dm⁻³ solution of a weak acid HA has pH 2.45. Calculate Ka for HA at 25 °C.

例题：0.500 mol dm⁻³弱酸HA溶液的pH = 2.45。计算25 °C时HA的Ka。

[H⁺] = 10⁻²·⁴⁵ = 3.55 × 10⁻³ mol dm⁻³
Since HA ⇌ H⁺ + A⁻, [A⁻] = [H⁺] = 3.55 × 10⁻³
Ka = (3.55 × 10⁻³)² / (0.500 – 3.55 × 10⁻³) = 2.54 × 10⁻⁵ mol dm⁻³

For weak acids where dissociation is very small (typically less than 5%), the approximation [HA] at equilibrium ≈ [HA]₀ can be used. If the extent of dissociation exceeds 5%, the quadratic equation must be solved for full accuracy.

对于解离度很小的弱酸（通常小于5%），可近似取平衡时[HA] ≈ [HA]₀。若解离度超过5%，则需解二次方程以获得完全准确的答案。

7. pH of Weak Acid Solutions | 弱酸溶液的pH

Calculating the pH of a weak acid requires Ka and the initial concentration, using the ICE table approach. The key assumption: since HA dissociates minimally, [H⁺] ≈ [A⁻] and [HA] at equilibrium ≈ initial [HA].

计算弱酸溶液的pH需要Ka和初始浓度，采用ICE表格法。关键假设：由于HA解离极小，[H⁺] ≈ [A⁻]，且平衡时[HA] ≈ 初始[HA]。

Worked example: Ethanoic acid has Ka = 1.74 × 10⁻⁵ mol dm⁻³. Calculate the pH of a 0.100 mol dm⁻³ solution of ethanoic acid.

例题：乙酸的Ka = 1.74 × 10⁻⁵ mol dm⁻³。计算0.100 mol dm⁻³乙酸溶液的pH。

Ka = [H⁺]² / [HA]₀ → [H⁺]² = Ka × [HA]₀
[H⁺]² = 1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶
[H⁺] = 1.32 × 10⁻³ mol dm⁻³
pH = -log₁₀(1.32 × 10⁻³) = 2.88

This gives the general formula for dilute solutions of weak monoprotic acids: [H⁺] = √(Ka × [HA]₀).

这给出了稀一元弱酸溶液的通用公式：[H⁺] = √(Ka × [HA]₀)。

As the acid is diluted, the pH increases, but [H⁺] from water dissociation becomes significant in extremely dilute solutions. For concentrations below about 1 × 10⁻⁶ mol dm⁻³, the contribution from water’s autoionisation must be considered.

随着酸被稀释，pH升高，但在极稀溶液中水的自电离对[H⁺]的贡献变得显著。当浓度低于约1 × 10⁻⁶ mol dm⁻³时，必须考虑水自电离的贡献。

8. pH Changes During Titrations | 滴定过程中的pH变化

Strong acid-strong base titrations show a sharp vertical region around the equivalence point. The equivalence point of a strong acid-strong base titration is exactly 7 at 25 °C, because both ions fully dissociate and the salt formed is neutral.

强酸-强碱滴定在等当点附近有陡峭的垂直区间。强酸-强碱滴定的等当点恰好为7（25 °C），因为酸和碱完全解离，生成的盐为中性。

Weak acid-strong base titrations have an equivalence point above 7 (alkaline), because the conjugate base A⁻ hydrolyses: A⁻ + H₂O ⇌ HA + OH⁻. This produces excess OH⁻ in solution.

弱酸-强碱滴定的等当点高于7（碱性），因为共轭碱A⁻发生水解：A⁻ + H₂O ⇌ HA + OH⁻，使溶液中OH⁻过量。

Strong acid-weak base titrations have an equivalence point below 7 (acidic), because the conjugate acid BH⁺ hydrolyses: BH⁺ + H₂O ⇌ B + H₃O⁺.

强酸-弱碱滴定的等当点低于7（酸性），因为共轭酸BH⁺发生水解：BH⁺ + H₂O ⇌ B + H₃O⁺。

At the half-equivalence point of a weak acid-strong base titration, [HA] = [A⁻], and therefore pH = pKa. This is a useful feature for determining Ka experimentally.

在弱酸-强碱滴定的半等当点处，[HA] = [A⁻]，因此pH = pKa。这是实验测定Ka的重要特征。

When selecting an indicator, the indicator’s pKa should lie within the vertical region of the titration curve, and the colour change should bracket the equivalence point pH.

选择指示剂时，指示剂的pKa应位于滴定曲线的垂直区间内，且颜色变化应覆盖等当点pH。

Suitable indicators: strong acid-strong base use methyl orange (3.1-4.4) or phenolphthalein (8.2-10.0). Weak acid-strong base use phenolphthalein. Strong acid-weak base use methyl orange.

指示剂选择：强酸-强碱可用甲基橙（3.1-4.4）或酚酞（8.2-10.0）。弱酸-强碱用酚酞。强酸-弱碱用甲基橙。

9. Buffer Solutions | 缓冲溶液

A buffer solution resists changes in pH when small amounts of acid or base are added, or when diluted. It maintains a nearly constant pH. There are two types: acidic buffers and basic buffers.

缓冲溶液在加入少量酸或碱或稀释时能抵抗pH变化，保持pH基本恒定。分为两类：酸性缓冲液和碱性缓冲液。

An acidic buffer is a mixture of a weak acid and its conjugate base salt, typically a sodium or potassium salt. Example: CH₃COOH / CH₃COONa. Basic buffers are a mixture of a weak base and its conjugate acid salt, e.g., NH₃ / NH₄Cl.

酸性缓冲液是弱酸及其共轭碱盐（通常为钠盐或钾盐）的混合物。例如：CH₃COOH / CH₃COONa。碱性缓冲液是弱碱及其共轭酸盐的混合物，如NH₃ / NH₄Cl。

How does an acidic buffer work? If H⁺ is added, it combines with the conjugate base A⁻ to form HA, removing the excess protons. If OH⁻ is added, it reacts with HA to form A⁻ and water. The pH remains nearly constant because the ratio [A⁻]/[HA] changes only slightly.

酸性缓冲液如何工作？加入H⁺时，其与共轭碱A⁻结合形成HA，清除过量质子。加入OH⁻时，OH⁻与HA反应生成A⁻和水。由于[A⁻]/[HA]的比值仅微小变化，pH保持基本恒定。

For buffer calculations, the Henderson-Hasselbalch equation is used at 25 °C:

缓冲液计算中，25 °C时使用亨德森-哈塞尔巴尔赫方程：

pH = pKa + log₁₀([salt] / [acid])

Worked example: A buffer is prepared by mixing 25.0 cm³ of 0.200 mol dm⁻³ ethanoic acid with 25.0 cm³ of 0.200 mol dm⁻³ sodium ethanoate. Calculate the pH if Ka = 1.74 × 10⁻⁵ mol dm⁻³.

例题：将25.0 cm³的0.200 mol dm⁻³乙酸与25.0 cm³的0.200 mol dm⁻³乙酸钠混合制备缓冲液。已知Ka = 1.74 × 10⁻⁵ mol dm⁻³，计算pH。

Total volume = 50.0 cm³
[CH₃COOH] = (0.200 × 25.0) / 50.0 = 0.100 mol dm⁻³
[CH₃COO⁻] = (0.200 × 25.0) / 50.0 = 0.100 mol dm⁻³
pH = pKa + log₁₀(0.100/0.100) = -log₁₀(1.74 × 10⁻⁵) + 0 = 4.76

Buffers resist pH change best when [salt] : [acid] ratio is near 1:1. The buffer capacity refers to the amount of acid or base it can neutralise before the pH changes significantly; maximum capacity occurs when this ratio equal 1.

当[盐] : [酸]的比值接近1:1时，缓冲液抵抗pH变化的能力最强。缓冲容量指在pH显著变化前能中和的酸或碱的量；当比值为1时容量最大。

10. Applications of Buffers | 缓冲体系的应用

Biological buffers play essential roles in living organisms. Blood maintains pH between 7.35 and 7.45 using the carbonic acid-hydrogencarbonate buffer system. The equilibrium is: H₂CO₃ ⇌ H⁺ + HCO₃⁻. The enzyme catalase and haemoglobin function optimally within this narrow pH range.

生物缓冲体系在生命体中扮演重要角色。血液利用碳酸-碳酸氢根缓冲体系维持pH在7.35到7.45之间。平衡为：H₂CO₃ ⇌ H⁺ + HCO₃⁻。过氧化氢酶和血红蛋白在此狭窄pH范围内发挥最佳功能。

In agriculture, soil pH affects nutrient availability to plants. In medicine, buffer solutions are used to maintain the pH of pharmaceutical formulations and to calibrate pH meters.

在农业中，土壤pH影响植物对养分的吸收。在医学中，缓冲液用于维持药物制剂的pH以及校准pH计。

In industrial chemistry, fermentation processes require pH control for optimal enzyme activity, and buffer solutions are essential in many analytical procedures including biochemical assays and electrophoretic separations.

在工业化学中，发酵过程需要pH控制以维持最佳酶活性，缓冲液在生化分析和电泳分离等许多分析程序中不可或缺。

11. Summary of Key Equations | 核心公式汇总

Quantity	Equation	Notes
pH	pH = -log₁₀[H⁺]	Always check temperature
[H⁺]	[H⁺] = 10⁻ᵖᴴ	Used in reverse calculations
Kw	Kw = [H⁺][OH⁻]	1.00 × 10⁻¹⁴ at 25 °C
Ka	Ka = [H⁺][A⁻]/[HA]	For weak acid HA
pKa	pKa = -log₁₀Ka	Smaller pKa = stronger acid
Buffer pH	pH = pKa + log₁₀([salt]/[acid])	Valid when both species present

Note that pH + pOH = pKw is only equal to 14 at 25 °C. At other temperatures, use Kw directly to find [H⁺].

注意pH + pOH = pKw仅等于14（25 °C时）。在其他温度下，需直接用Kw求[H⁺]。

12. Common Exam Mistakes | 常见考试错误

When calculating [H⁺] from pH, remember the inverse logarithm: [H⁺] = 10⁻ᵖᴴ. Many students incorrectly use [H⁺] = -log₁₀pH, which reverses the operation.

从pH计算[H⁺]时，记得取反对数：[H⁺] = 10⁻ᵖᴴ。许多学生错误地使用[H⁺] = -log₁₀pH，将运算颠倒了。

Do not state that a solution is neutral solely because pH = 7. Neutrality requires [H⁺] = [OH⁻]. Since Kw is temperature-dependent, neutral pH is not always 7.

不要仅因pH = 7就断言溶液为中性。中性要求[H⁺] = [OH⁻]。由于Kw随温度变化，中性pH并不总是7。

For weak acid calculations, the approximation [HA] ≈ initial concentration is valid only if the acid is sufficiently weak and concentrated. If the degree of dissociation is greater than about 5%, solve the quadratic equation Ka = x²/([HA]₀ – x).

弱酸计算中，近似[HA] ≈ 初始浓度仅在酸足够弱且浓度不太稀时有效。若解离度大于约5%，需解二次方程Ka = x²/([HA]₀ – x)。

When diluting an acid or base, pH tends toward 7 but never crosses it due to dilution alone. Diluting a strong acid 10-fold raises pH by 1 unit only for concentrations above about 10⁻⁶ mol dm⁻³.

稀释酸或碱时，pH趋向于7但不会仅因稀释而越过7。强酸稀释10倍使pH升高1个单位，仅适用于浓度高于约10⁻⁶ mol dm⁻³的情况。

Finally, always include units in your Ka and Kw values. Ka is measured in mol dm⁻³, and Kw in mol² dm⁻⁶. Omitting units loses marks in written examinations.

最后，Ka和Kw的值务必包含单位。Ka的单位为mol dm⁻³，Kw的单位为mol² dm⁻⁶。漏写单位在笔试中会失分。

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Acids and Bases | 酸与碱