Weak Acids – Using the Acid Dissociation Constant Kₐ | 弱酸——酸解离常数 Kₐ 的应用

📚 Weak Acids – Using the Acid Dissociation Constant Kₐ | 弱酸——酸解离常数 Kₐ 的应用

Weak acids appear throughout A-level chemistry: in buffer calculations, in titration curves, in the action of everyday substances such as ethanoic acid in vinegar, and in the behaviour of indicators. Unlike strong acids, they do not ionise completely in water, so their behaviour has to be described with an equilibrium constant rather than with simple stoichiometry. This article explains what the acid dissociation constant Kₐ is, how it is defined and used, and how to carry out the pH, concentration and pKₐ calculations that examiners most often ask for.

弱酸贯穿整个 A-level 化学课程：缓冲溶液的计算、滴定曲线、醋中乙酸这类日常物质的作用，以及指示剂的行为都离不开它。与强酸不同，弱酸在水中不能完全电离，因此描述它们的性质必须借助平衡常数，而不能只用简单的化学计量关系。本文将说明酸解离常数 Kₐ 的含义、定义方式和应用方法，并系统讲解考试中最常出现的 pH、浓度与 pKₐ 计算。

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1. What Is a Weak Acid? | 什么是弱酸？

A weak acid is a proton donor that only partially dissociates in water. Instead of going to completion, the dissociation reaches a dynamic equilibrium between the undissociated acid molecules and the ions formed. At equilibrium the forward and reverse reactions are still happening, but at equal rates, so the concentrations of all species remain constant.

弱酸是一种在水中只部分解离的质子给体。解离反应不会进行到底，而是在未解离的酸分子和生成的离子之间达到动态平衡。平衡时正逆反应仍在进行，但速率相等，因此各物种的浓度保持不变。

HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)

In this equation HA is the weak acid, H₂O acts as a base accepting a proton, H₃O⁺ is the oxonium (hydronium) ion and A⁻ is the conjugate base of the acid. Because only a small fraction of HA molecules donate a proton, the position of equilibrium lies well to the left.

在这个方程式中，HA 是弱酸，H₂O 作为碱接受质子，H₃O⁺ 是水合氢离子，A⁻ 是该酸的共轭碱。由于只有很小一部分 HA 分子给出质子，平衡位置明显偏左。

For A-level calculations, the equilibrium is usually simplified to:

在 A-level 计算中，通常将平衡简化为：

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

This means an equilibrium mixture contains undissociated HA molecules together with H⁺ (strictly H₃O⁺) ions and A⁻ ions, all present at the same time. The concentration of H⁺ is therefore much smaller than the concentration of the acid you started with, which is exactly why a weak acid has a higher pH and reacts more slowly than a strong acid of the same concentration.

这意味着平衡混合物中同时含有未解离的 HA 分子、H⁺（严格来说是 H₃O⁺）离子和 A⁻ 离子。因此 H⁺ 的浓度远小于起始酸的浓度，这正是相同浓度下弱酸的 pH 更高、反应更慢的原因。

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2. Defining the Acid Dissociation Constant Kₐ | 酸解离常数 Kₐ 的定义

The position of this equilibrium is quantified by the acid dissociation constant, Kₐ. It is the equilibrium constant for the dissociation of the acid in water, and for the simplified equation above it takes the form shown below.

该平衡的位置由酸解离常数 Kₐ 定量描述。它是酸在水中解离反应的平衡常数，对应上面的简化方程式，其表达式如下。

Kₐ = [H⁺][A⁻] ÷ [HA]

Only the aqueous species appear in the expression: water is the solvent and its concentration is effectively constant, so it is not included. The square brackets mean equilibrium concentration in mol dm⁻³, which is why Kₐ has units of mol dm⁻³ in A-level work and why those units must be quoted when the question asks for them.

表达式中只出现水溶液中的物种：水是溶剂，其浓度实际上是常数，因此不写入表达式。方括号表示以 mol dm⁻³ 为单位的平衡浓度，所以 A-level 中 Kₐ 的单位是 mol dm⁻³；当题目要求写出单位时必须写出。

Kₐ is temperature dependent. Since dissociation of a weak acid is endothermic, raising the temperature shifts the equilibrium to the right and increases the value of Kₐ, making the acid slightly stronger.

Kₐ 随温度变化。由于弱酸解离是吸热过程，升高温度使平衡右移，Kₐ 值增大，酸的强度略有增强。

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3. pKₐ and the Conjugate Base | pKₐ 与共轭碱

Because Kₐ values are often very small numbers, chemists use the logarithmic form pKₐ, defined as the negative logarithm to base 10 of Kₐ. pKₐ is dimensionless, and the relationship runs in the opposite direction to Kₐ: a larger Kₐ means a smaller pKₐ.

由于 Kₐ 常常是非常小的数值，化学家使用对数形式 pKₐ，它定义为 Kₐ 以 10 为底的负对数。pKₐ 没有单位，其变化方向与 Kₐ 相反：Kₐ 越大，pKₐ 越小。

pKₐ = −log₁₀Kₐ   and   Kₐ = 10⁻ᵖᴷᵃ

= 9.31. The difference of nearly five pKₐ units corresponds to a ten-thousand-fold difference in acid strength.

需要记住的常见数值：乙酸的 Kₐ = 1.8 × 10⁻⁵ mol dm⁻³，因此 pKₐ = 4.76；而氢氰酸的 Kₐ = 4.9 × 10⁻¹⁰ mol dm⁻³，因此 pKₐ = 9.31。两者相差近五个 pKₐ 单位，相当于酸性强弱相差一万倍。

The conjugate base A⁻ is itself a base. The stronger the acid HA, the weaker its conjugate base A⁻, and vice versa. This is why the salt of a weak acid, such as sodium ethanoate, produces an alkaline solution when dissolved in water.

共轭碱 A⁻ 本身也是碱。酸 HA 越强，其共轭碱 A⁻ 越弱，反之亦然。这正是乙酸钠等弱酸盐溶于水后溶液显碱性的原因。

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4. Which Acid Is Stronger? Comparing Kₐ Values | 哪种酸更强？比较 Kₐ 数值

The simplest use of Kₐ is comparison. When comparing weak acids at the same temperature and the same concentration, the acid with the larger Kₐ (and the smaller pKₐ) is the stronger acid, because its equilibrium position lies further to the right and it produces more H⁺ ions.

Kₐ 最简单的用途是比较酸性强弱。在相同温度和相同浓度下，Kₐ 较大（pKₐ 较小）的酸更强，因为它的平衡位置更靠右，产生的 H⁺ 更多。

Acid 酸	Kₐ / mol dm⁻³	pKₐ	Relative strength 相对强度
HF 氢氟酸	6.8 × 10⁻⁴	3.17	strongest 最强
HCOOH 甲酸	1.8 × 10⁻⁴	3.75	strong 较强
C₆H₅COOH 苯甲酸	6.3 × 10⁻⁵	4.20	medium 中等
CH₃COOH 乙酸	1.8 × 10⁻⁵	4.76	weaker 较弱
HClO 次氯酸	2.9 × 10⁻⁸	7.54	very weak 很弱
HCN 氢氰酸	4.9 × 10⁻¹⁰	9.31	weakest 最弱

Remember that data tables are only valid at the stated temperature, normally 298 K. Comparing Kₐ values measured at different temperatures is a common source of lost marks.

要注意数据表只在标注温度（通常为 298 K）下有效。用不同温度下测得的 Kₐ 进行比较，是常见的失分点。

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5. Calculating pH from Kₐ and Concentration | 由 Kₐ 和浓度计算 pH

The standard exam calculation gives you the concentration of a weak acid and its Kₐ, and asks for the pH. Start by writing the equilibrium expression, then set up the concentration table.

标准考题会给出弱酸的浓度和 Kₐ，要求计算 pH。第一步写出平衡表达式，然后列出浓度变化表。

uilibrium [H⁺] = [A⁻] = x and [HA] = c − x. Substituting gives the exact expression below.

若 HA 的初始浓度为 c，有 x mol dm⁻³ 发生解离，则平衡时 [H⁺] = [A⁻] = x，[HA] = c − x。代入后得到下面的精确表达式。

Kₐ = x² ÷ (c − x)

e approximation c − x ≈ c can be used. This leads to the very useful working relationship:

由于弱酸解离程度很小，x 相对于 c 极小，因此可采用 c − x ≈ c 的近似。由此得到非常实用的计算关系式：

[H⁺] = √Kₐ × c   then   pH = −log₁₀[H⁺]

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Worked example: find the pH of 0.10 mol dm⁻³ ethanoic acid, Kₐ = 1.8 × 10⁻⁵ mol dm⁻³.

例题：求 0.10 mol dm⁻³ 乙酸溶液的 pH，已知 Kₐ = 1.8 × 10⁻⁵ mol dm⁻³。

[H⁺] = √1.8 × 10⁻⁵ × 0.10 = √1.8 × 10⁻⁶ = 1.34 × 10⁻³ mol dm⁻³

pH = −log₁₀(1.34 × 10⁻³) = 2.87

the weak acid is roughly seventy times less acidic in terms of hydrogen ion concentration.

注意 0.10 mol dm⁻³ 的强酸 pH = 1.00，因此从氢离子浓度看，该弱酸的酸性约弱七十倍。

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6. Approximations, and When They Fail | 近似处理及其适用条件

alid when the degree of dissociation is small, which is true when Kₐ is much smaller than the concentration. As a rough rule, if Kₐ ÷ c is less than about 10⁻³, the approximation is safe to within a few per cent.

当解离度很小时，[HA]ₑq ≈ [HA]₀ 的近似成立，即 Kₐ 远小于浓度时成立。经验规则是：若 Kₐ ÷ c 小于约 10⁻³，该近似带来的误差在百分之几以内，可以放心使用。

If the acid is relatively strong or very dilute, the exact quadratic must be solved instead. For 0.10 mol dm⁻³ ethanoic acid:

如果酸相对较强或溶液非常稀，则必须解精确的二次方程。对于 0.10 mol dm⁻³ 乙酸：

x² + 1.8 × 10⁻⁵x − 1.8 × 10⁻⁶ = 0  →  x = 1.33 × 10⁻³, pH = 2.87

The exact answer differs from the approximate one only in the third significant figure, which justifies using the shortcut in most exam questions. However, you should always state the assumption you have made, because marks are awarded for it.

精确结果与近似结果只在小数点后第三位有差异，这说明多数考题中使用快捷方法是合理的。不过，你应当始终写出所采用的假设，因为这一步是有分的。

Another assumption worth stating is that the ionisation of water contributes a negligible amount of H⁺. This holds for all but extremely dilute solutions, where [H⁺] from water becomes comparable to that from the acid.

另一个值得写出的假设是：水的电离所提供的 H⁺ 可以忽略。除极稀溶液外，这一假设均成立；只有在极稀溶液中，来自水的 [H⁺] 才与来自酸的 [H⁺] 相当。

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7. Calculating Kₐ from pH Data | 由 pH 数据计算 Kₐ

The calculation can also run in reverse. If you know the concentration of the acid and the pH of the solution, you can find Kₐ. This is how Kₐ values are determined experimentally, for example by measuring the pH of a series of solutions of known concentration.

计算也可以反向进行。若已知酸的浓度和溶液的 pH，就可以求出 Kₐ。这正是实验测定 Kₐ 的方法，例如测定一系列已知浓度溶液的 pH。

ed example: a 0.050 mol dm⁻³ solution of a monoprotic weak acid HA has pH = 3.20. Find Kₐ.

例题：一元弱酸 HA 的 0.050 mol dm⁻³ 溶液 pH = 3.20，求 Kₐ。

[H⁺] = 10⁻³·²⁰ = 6.31 × 10⁻⁴ mol dm⁻³

[A⁻] = 6.31 × 10⁻⁴ ; [HA]ₑq = 0.050 − 0.000631 = 0.0494 mol dm⁻³

Kₐ = (6.31 × 10⁻⁴)² ÷ 0.0494 = 8.1 × 10⁻⁶ mol dm⁻³

Notice two things. First, [H⁺] is obtained from the pH using the inverse logarithm, not the pH itself. Second, the equilibrium concentration of HA is the initial concentration minus the amount that dissociated; for a precise answer this subtraction must be done, even though the difference is small.

注意两点。第一，[H⁺] 是由 pH 取反对数得到的，而不是直接使用 pH 数值。第二，HA 的平衡浓度等于初始浓度减去已解离的量；若要得到精确答案，这一步减法必须做，尽管差值很小。

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8. Degree of Ionisation and the Effect of Dilution | 电离度与稀释效应

The degree of ionisation, sometimes written α, is the fraction of acid molecules that have donated a proton. It is usually expressed as a percentage and is calculated as shown below.

电离度（有时记作 α）是已给出质子的酸分子所占的比例，通常用百分数表示，计算方法如下。

α = [H⁺] ÷ c × 100%

For 0.10 mol dm⁻³ ethanoic acid, α = 1.34 × 10⁻³ ÷ 0.10 × 100% = 1.34%. For a strong acid of the same concentration, α would be 100%.

对于 0.10 mol dm⁻³ 乙酸，α = 1.34 × 10⁻³ ÷ 0.10 × 100% = 1.34%。而同浓度的强酸，α 为 100%。

tion produces a classic exam result. If the same acid is diluted to 0.010 mol dm⁻³, [H⁺] = √1.8 × 10⁻⁵ × 0.010 = 4.24 × 10⁻⁴ mol dm⁻³, so α = 4.24% drogen ion concentration falls, and the pH rises, but the percentage ionisation increases.

稀释会产生一个经典考点。若将同一酸稀释至 0.010 mol dm⁻³，[H⁺] = √1.8 × 10⁻⁵ × 0.010 = 4.24 × 10⁻⁴ mol dm⁻³，因此 α = 4.24%。氢离子浓度下降、pH 上升，但电离百分比却增大了。

This is a direct consequence of Le Chatelier's principle. Adding water lowers the concentration of all species, so the equilibrium shifts towards the side with the greater number of dissolved particles, that is, towards the products, to oppose the change.

这是勒夏特列原理的直接结果。加入水降低了所有物种的浓度，平衡便向溶解微粒数较多的一侧（即生成物一侧）移动，以抵抗这一变化。

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9. Temperature, Equilibrium and Kₐ | 温度、平衡与 Kₐ

Kₐ is a true equilibrium constant, so it changes only with temperature. The dissociation of a weak acid is endothermic, with ΔH positive, so increasing the temperature shifts the equilibrium position to the right.

Kₐ 是真正的平衡常数，因此只随温度变化。弱酸的解离是吸热过程，ΔH 为正值，所以升高温度使平衡位置右移。

Consequently, at higher temperature the value of Kₐ increases, the acid becomes slightly stronger, and the pH of a solution of that acid decreases slightly. This is a good way to show that you understand the link between equilibrium position and the magnitude of Kₐ.

因此，温度升高时 Kₐ 增大，酸的强度略有增强，该酸溶液的 pH 略微降低。这是展示你理解“平衡位置与 Kₐ 大小之间联系”的好例子。

Note also that water itself ionises, with K𝓌 OH = 14 applies at this temperature, and pH is always calculated from [H⁺], never from [OH

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