A-Level Chemistry: Applications of the Acid Dissociation Constant Ka | A-Level化学：弱酸的电离常数Ka应用

📚 A-Level Chemistry: Applications of the Acid Dissociation Constant Ka | A-Level化学：弱酸的电离常数Ka应用

The acid dissociation constant, Ka, is a quantitative measure of the strength of a weak acid in solution. It is one of the most important equilibrium constants in A-Level chemistry, appearing in topics ranging from pH calculations to buffer systems and titration curves. This article explores the core applications of Ka, with step-by-step reasoning and worked examples tailored to the CIE A-Level syllabus.

酸电离常数 Ka 是衡量弱酸在溶液中酸性强度的定量指标。它是A-Level化学中最重要的平衡常数之一，涉及pH计算、缓冲溶液、滴定曲线等多个主题。本文将围绕CIE A-Level考纲，深入探讨Ka的核心应用，并配逐步推理与例题解析。

1. Defining Ka and Its Expression | 定义Ka及其表达式

For a weak acid HA that partially dissociates in water: HA(aq) ⇌ H⁺(aq) + A⁻(aq). The equilibrium constant is written as Ka = [H⁺][A⁻] / [HA], where square brackets denote concentrations in mol dm⁻³ at equilibrium.

对于在水中部分电离的弱酸HA：HA(aq) ⇌ H⁺(aq) + A⁻(aq)。其平衡常数写作 Ka = [H⁺][A⁻] / [HA]，方括号表示平衡时的浓度，单位为 mol dm⁻³。

Because weak acids ionise only slightly, [HA] at equilibrium is often approximated as the initial acid concentration, c₀. This approximation is valid when Ka is very small (typically Ka < 10⁻⁴) and the degree of dissociation is negligible compared to c₀.

由于弱酸仅发生微弱电离，平衡时的[HA]通常近似等于初始酸浓度 c₀。该近似在Ka很小（通常Ka < 10⁻⁴）且电离程度相较于c₀可忽略时成立。

Key point: the units of Ka are always mol dm⁻³ because the concentration terms cancel to leave one concentration unit. For example, ethanoic acid has Ka = 1.74 × 10⁻⁵ mol dm⁻³.

关键点：Ka的单位始终为 mol dm⁻³，因为浓度项相消后保留一个浓度单位。例如，乙酸的 Ka = 1.74 × 10⁻⁵ mol dm⁻³。

2. Using Ka to Calculate the pH of a Weak Acid | 用Ka计算弱酸溶液的pH

The most common application of Ka is finding the pH of a weak acid solution. For a monoprotic acid HA, if x mol dm⁻³ of HA dissociates, then at equilibrium [H⁺] = [A⁻] = x and [HA] = c₀ − x. Substituting into the Ka expression gives Ka = x² / (c₀ − x).

Ka最常见的应用是求弱酸溶液的pH。对于一元酸HA，若 x mol dm⁻³ 的HA发生电离，则平衡时 [H⁺] = [A⁻] = x，[HA] = c₀ − x。代入Ka表达式得 Ka = x² / (c₀ − x)。

When x is much smaller than c₀ (usually if c₀/Ka > 1000), the equation simplifies to Ka ≈ x² / c₀. Then x = √(Ka × c₀). Since x = [H⁺], pH = −log₁₀[H⁺] = −log₁₀√(Ka × c₀).

当x远小于c₀时（通常 c₀/Ka > 1000），方程简化为 Ka ≈ x² / c₀。于是 x = √(Ka × c₀)。由于 x = [H⁺]，所以 pH = −log₁₀[H⁺] = −log₁₀√(Ka × c₀)。

pH = ½(pKa − log₁₀c₀)

Worked example: Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, Ka = 1.74 × 10⁻⁵ mol dm⁻³. Since 0.100 / 1.74 × 10⁻⁵ ≈ 5747 > 1000, use the approximation: x = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³. Thus pH = −log₁₀(1.32 × 10⁻³) = 2.88.

例题：计算0.100 mol dm⁻³乙酸的pH，已知Ka = 1.74 × 10⁻⁵ mol dm⁻³。因0.100 / 1.74 × 10⁻⁵ ≈ 5747 > 1000，可用近似：x = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³。故 pH = −log₁₀(1.32 × 10⁻³) = 2.88。

3. Ka and the Quadratic Equation: When Approximations Fail | Ka与二次方程：何时近似失效

For very dilute weak acids or those with relatively large Ka, the approximation c₀ − x ≈ c₀ is not valid. In such cases you must solve the quadratic equation x² + Ka x − Ka c₀ = 0, using the positive root.

对于极稀的弱酸或Ka较大的弱酸，近似 c₀ − x ≈ c₀ 不再成立。此时必须求解二次方程 x² + Ka x − Ka c₀ = 0，取正根。

Example: Calculate [H⁺] in 0.0100 mol dm⁻³ chloroethanoic acid, Ka = 1.30 × 10⁻³ mol dm⁻³. Ratio c₀/Ka = 0.0100 / 1.30 × 10⁻³ ≈ 7.7, far below 1000. Set up: 1.30 × 10⁻³ = x² / (0.0100 − x). Rearranging: x² + 1.30 × 10⁻³ x − 1.30 × 10⁻⁵ = 0. Solving gives x = 3.03 × 10⁻³ mol dm⁻³. Notice that x is about 30% of c₀, so ignoring it would cause significant error.

例：计算0.0100 mol dm⁻³氯乙酸的[H⁺]，已知Ka = 1.30 × 10⁻³ mol dm⁻³。比值c₀/Ka = 0.0100 / 1.30 × 10⁻³ ≈ 7.7，远小于1000。列式：1.30 × 10⁻³ = x² / (0.0100 − x)，整理得 x² + 1.30 × 10⁻³ x − 1.30 × 10⁻⁵ = 0。解得 x = 3.03 × 10⁻³ mol dm⁻³。注意x约为c₀的30%，忽略它会造成显著误差。

In the CIE exam, if the approximation is not valid, the question often requires solving the quadratic. Always check the condition c₀/Ka > 1000 before simplifying.

在CIE考试中，若近似不成立，题目往往要求解二次方程。简化前务必检查条件 c₀/Ka > 1000。

4. Determining Ka from pH Measurement | 通过pH测量确定Ka

Conversely, if the pH of a weak acid solution is known experimentally, Ka can be calculated. From the measured pH you obtain [H⁺] = 10⁻ᵖᴴ. For a monoprotic acid, [A⁻] = [H⁺] (from stoichiometry), and the remaining acid concentration is [HA] = c₀ − [H⁺]. Then substitute into Ka = [H⁺]² / (c₀ − [H⁺]).

反过来，若实验测得弱酸溶液的pH，则可计算Ka。由pH得到 [H⁺] = 10⁻ᵖᴴ。对于一元酸，[A⁻] = [H⁺]（按化学计量），剩余酸浓度为 [HA] = c₀ − [H⁺]。然后代入 Ka = [H⁺]² / (c₀ − [H⁺])。

Worked example: A 0.0500 mol dm⁻³ solution of a weak acid HA has pH 3.45. Then [H⁺] = 10⁻³·⁴⁵ = 3.55 × 10⁻⁴ mol dm⁻³. Assuming [HA] ≈ 0.0500 − 3.55 × 10⁻⁴ ≈ 0.0496, Ka = (3.55 × 10⁻⁴)² / 0.0496 = 2.54 × 10⁻⁶ mol dm⁻³.

例题：某弱酸HA的0.0500 mol dm⁻³溶液pH为3.45。则 [H⁺] = 10⁻³·⁴⁵ = 3.55 × 10⁻⁴ mol dm⁻³。设[HA] ≈ 0.0500 − 3.55 × 10⁻⁴ ≈ 0.0496，Ka = (3.55 × 10⁻⁴)² / 0.0496 = 2.54 × 10⁻⁶ mol dm⁻³。

This method is often used in practical-based exam questions. It assumes that all H⁺ comes from the acid and that activity coefficients are unity, which is acceptable for dilute solutions.

该方法常用于基于实验的考题。它假设所有H⁺均来自酸，且活度系数为1，这在稀溶液中可接受。

5. Ka, pKa and Acid Strength Comparison | Ka、pKa与酸强度比较

Acid strength is measured by the position of equilibrium of the dissociation. A larger Ka means the equilibrium lies further to the right, so a stronger weak acid. Because Ka values can span many orders of magnitude, pKa = −log₁₀Ka is often used. A smaller pKa corresponds to a stronger acid.

酸强度由电离平衡的位置衡量。Ka越大，平衡越偏右，酸越强。由于Ka跨度大，通常使用 pKa = −log₁₀Ka。pKa越小，酸性越强。

Acid	Ka / mol dm⁻³	pKa	Relative strength
Ethanoic acid	1.74 × 10⁻⁵	4.76	weakest
Chloroethanoic acid	1.30 × 10⁻³	2.89	medium
Dichloroethanoic acid	5.01 × 10⁻²	1.30	strong

Electron-withdrawing groups such as −Cl stabilise the conjugate base by spreading negative charge, increasing Ka. Therefore chloroethanoic acid is stronger than ethanoic acid.

吸电子基团（如−Cl）通过分散负电荷稳定共轭碱，使Ka增大。因此氯乙酸比乙酸更强。

6. Relating Ka to Kb: The Ionic Product of Water | Ka与Kb的关系：水的离子积

For a conjugate acid-base pair, Ka × Kb = Kw, where Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C. This relation allows the strength of a weak base to be determined from the Ka of its conjugate acid.

对于共轭酸碱对，Ka × Kb = Kw，其中 Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶（25 °C）。该关系允许通过共轭酸的Ka确定弱碱的强度。

Example: The Ka of the ammonium ion, NH₄⁺, is 5.62 × 10⁻¹⁰ mol dm⁻³. Thus the Kb of ammonia, NH₃, is Kb = Kw / Ka = 1.00 × 10⁻¹⁴ / 5.62 × 10⁻¹⁰ = 1.78 × 10⁻⁵ mol dm⁻³.

例：铵离子NH₄⁺的Ka为 5.62 × 10⁻¹⁰ mol dm⁻³。因此氨NH₃的Kb = Kw / Ka = 1.00 × 10⁻¹⁴ / 5.62 × 10⁻¹⁰ = 1.78 × 10⁻⁵ mol dm⁻³。

This reciprocal relationship explains why a stronger acid has a weaker conjugate base. It is essential when calculating the pH of salt solutions such as sodium ethanoate.

这种倒数关系解释了为何酸越强其共轭碱越弱。它在计算盐溶液（如乙酸钠）的pH时至关重要。

7. Ka in Buffer Solutions: The Henderson-Hasselbalch Equation | 缓冲溶液中的Ka：Henderson-Hasselbalch方程

A buffer solution contains a weak acid and its conjugate base. Its pH can be derived from the Ka expression. Rearranging Ka = [H⁺][A⁻] / [HA] gives [H⁺] = Ka × [HA] / [A⁻]. Taking negative logs gives the Henderson-Hasselbalch equation:

缓冲溶液含有弱酸及其共轭碱。其pH可由Ka表达式推导。由 Ka = [H⁺][A⁻] / [HA] 得 [H⁺] = Ka × [HA] / [A⁻]。取负对数得Henderson-Hasselbalch方程：

pH = pKa + log₁₀([A⁻] / [HA])

When the concentrations of acid and salt are equal, [A⁻] = [HA], so pH = pKa. This is the most effective buffer region. For example, a buffer made with 0.20 mol dm⁻³ ethanoic acid (pKa = 4.76) and 0.30 mol dm⁻³ sodium ethanoate has pH = 4.76 + log₁₀(0.30 / 0.20) = 4.76 + 0.18 = 4.94.

当酸与盐浓度相等时，[A⁻] = [HA]，因此pH = pKa。这是缓冲能力最强的区域。例如，由0.20 mol dm⁻³乙酸（pKa = 4.76）和0.30 mol dm⁻³乙酸钠组成的缓冲溶液，pH = 4.76 + log₁₀(0.30 / 0.20) = 4.76 + 0.18 = 4.94。

The buffer capacity is greatest when pH ≈ pKa. Adding small amounts of strong acid or base changes the ratio [A⁻]/[HA] only slightly, so pH remains nearly constant.

当pH ≈ pKa时缓冲容量最大。加入少量强酸或强碱只会轻微改变比值[A⁻]/[HA]，因此pH几乎保持不变。

8. Ka and Titration Curves | Ka与滴定曲线

In a titration of a weak acid with a strong base, the shape of the curve is governed by Ka. At the half-equivalence point, exactly half of the acid has been neutralised, so [HA] = [A⁻]. From the Henderson-Hasselbalch equation, pH = pKa. Hence the pKa of a weak acid can be read directly from the titration curve.

在弱酸与强碱的滴定中，曲线形状由Ka决定。在半中和点，一半酸被中和，此时 [HA] = [A⁻]。由Henderson-Hasselbalch方程，pH = pKa。因此可直接从滴定曲线读取弱酸的pKa。

The equivalence point of a weak acid–strong base titration is above 7 because the conjugate base A⁻ hydrolyses to produce OH⁻. The pH at equivalence depends on the Kb of A⁻, which is related to Ka by Kb = Kw / Ka.

弱酸-强碱滴定的等当点高于7，因为共轭碱A⁻水解产生OH⁻。等当点pH取决于A⁻的Kb，而Kb = Kw / Ka。

Choose an indicator whose colour-change interval contains the pH at the equivalence point.

选择变色范围包含等当点pH的指示剂。
For ethanoic acid (Ka ≈ 10⁻⁵), the equivalence point pH is around 8–9, so phenolphthalein (range 8.3–10.0) is suitable.

对于乙酸（Ka ≈ 10⁻⁵），等当点pH约为8–9，适合用酚酞（变色范围8.3–10.0）。

9. Ka and Degree of Dissociation | Ka与电离度

The degree of dissociation, α, is the fraction of acid molecules that have ionised: α = [H⁺] / c₀. For a weak acid, Ka = c₀α² / (1 − α). For very weak acids, α is small, so Ka ≈ c₀α², and α = √(Ka / c₀).

电离度α是已电离的酸分子比例：α = [H⁺] / c₀。对于弱酸，Ka = c₀α² / (1 − α)。对极弱酸，α很小，故 Ka ≈ c₀α²，α = √(Ka / c₀)。

This shows that α increases as the acid is diluted. Hence pH increases with dilution, but the acid becomes more dissociated. This is a common exam question.

这说明稀释会使α增大。因此pH随稀释而升高，但酸的电离度增大了。这是常见考点。

Example: For ethanoic acid with Ka = 1.74 × 10⁻⁵, at c₀ = 0.10 mol dm⁻³, α = √(1.74 × 10⁻⁵ / 0.10) = 0.0132 (1.32%). At c₀ = 0.0010 mol dm⁻³, α = √(1.74 × 10⁻⁵ / 0.0010) = 0.132 (13.2%).

例：对于Ka = 1.74 × 10⁻⁵的乙酸，当c₀ = 0.10 mol dm⁻³时，α = √(1.74 × 10⁻⁵ / 0.10) = 0.0132（1.32%）。当c₀ = 0.0010 mol dm⁻³时，α = √(1.74 × 10⁻⁵ / 0.0010) = 0.132（13.2%）。

10. Factors Affecting Ka: Temperature and Solvent | 影响Ka的因素：温度与溶剂

Ka is a constant at a given temperature. Like all equilibrium constants, it changes with temperature. For most weak acids, ionisation is endothermic, so Ka increases with increasing temperature, meaning pH decreases.

Ka在给定温度下是常数。与所有平衡常数一样，它随温度变化。大多数弱酸的电离是吸热的，因此升温会使Ka增大，pH降低。

Changing concentration, adding a common ion, or adding a catalyst does not change Ka. Only temperature changes the value of Ka. Addition of a strong acid suppresses dissociation by the common ion effect, but the Ka value remains the same.

改变浓度、加入共同离子或加催化剂不会改变Ka。只有温度会改变Ka。加入强酸通过同离子效应抑制电离，但Ka不变。

In exam questions, always state: “Ka is constant at constant temperature.”

在考试题中，务必说明：“在恒定温度下Ka为常数。”

11. Common Mistakes and Exam Tips | 常见错误与考试技巧

Forgetting to convert pH to [H⁺] before using Ka — always calculate [H⁺] = 10⁻ᵖᴴ.

使用Ka前忘记将pH换算为[H⁺]——务必计算 [H⁺] = 10⁻ᵖᴴ。
Assuming the approximation holds without checking c₀/Ka > 1000.

未检查c₀/Ka > 1000就作近似。
Using [HA] ≠ c₀ when the acid is significantly dissociated.

当酸明显电离时，仍认为[HA] = c₀。
Confusing Ka and Kb. Remember: strong acid → large Ka; strong conjugate base → large Kb.

混淆Ka与Kb。记住：强酸→Ka大；强共轭碱→Kb大。
Wrong units — Ka is in mol dm⁻³, but pKa is unitless.

单位错误——Ka单位是 mol dm⁻³，但pKa无单位。

Always show your working and state whether you have used the approximation. This gains method marks even if the numerical answer is wrong.

始终写出过程，并说明是否使用了近似。即使数值答案错误也能获得步骤分。

12. Summary: The Power of Ka | 总结：Ka的威力

Ka links together pH, acid strength, buffer action, and titration behaviour. Mastering its applications allows you to solve a wide range of equilibrium problems with confidence.

Ka将pH、酸强度、缓冲作用和滴定行为联系起来。掌握其应用，你就能自信地解决各类平衡问题。

Key equation to remember: Ka = [H⁺][A⁻] / [HA], and its logarithmic form pH = pKa + log₁₀([A⁻]/[HA]). Always apply the small-dissociation approximation only when justified, and remember that Ka is constant at constant temperature.

需牢记的关键方程：Ka = [H⁺][A⁻] / [HA]，及其对数形式 pH = pKa + log₁₀([A⁻]/[HA])。只有在合理时才使用小电离近似，并记住在恒温下Ka为常数。

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A-Level Chemistry: Applications of the Acid Dissociation Constant Ka | A-Level化学：弱酸的电离常数Ka应用