Weak acids – using the acid dissociation constant, Kₐ | 弱酸 – 使用酸解离常数 Kₐ

📚 Weak acids – using the acid dissociation constant, Kₐ | 弱酸 – 使用酸解离常数 Kₐ

Weak acids are a fundamental concept in A-Level Chemistry. Unlike strong acids that undergo complete ionisation, weak acids only partially dissociate in aqueous solution, establishing a dynamic equilibrium between the unionised acid molecules and the ions formed. The acid dissociation constant, Kₐ, provides a quantitative measure of the extent of this dissociation. Understanding how to use Kₐ to calculate pH, assess acid strength, and predict the behaviour of weak acid solutions is essential for success in the Cambridge A-Level examination. This article will guide you through the definition of Kₐ, its relationship with pKₐ, the assumptions behind pH calculations, the effect of dilution on dissociation, and common applications.

弱酸是 A-Level 化学中的一个基本概念。与完全电离的强酸不同，弱酸在水溶液中仅部分解离，在未电离的酸分子与生成的离子之间建立起动态平衡。酸解离常数 Kₐ 定量描述了这种解离的程度。理解如何使用 Kₐ 计算 pH、评估酸的强度以及预测弱酸溶液的行为，对于在剑桥 A-Level 考试中取得成功至关重要。本文将带你了解 Kₐ 的定义、它与 pKₐ 的关系、pH 计算背后的假设、稀释对解离的影响以及常见应用。

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1. What are Weak Acids? | 什么是弱酸？

A weak acid is a proton donor that partially ionises in water. At equilibrium, a significant proportion of the acid molecules remain undissociated. Typical examples include ethanoic acid (CH₃COOH), which gives vinegar its sour taste, carbonic acid (H₂CO₃) found in fizzy drinks, and hydrofluoric acid (HF). In contrast, strong acids such as hydrochloric acid (HCl) or sulfuric acid (H₂SO₄) dissociate completely under the same conditions. The partial dissociation of a weak acid HA can be represented by the reversible reaction: HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq), often simplified to HA(aq) ⇌ H⁺(aq) + A⁻(aq).

弱酸是一种在水中部分电离的质子给予体。达到平衡时，相当大比例的酸分子仍然未解离。典型的例子包括使醋变酸的乙酸（CH₃COOH）、汽水中存在的碳酸（H₂CO₃）以及氢氟酸（HF）。相比之下，像盐酸（HCl）或硫酸（H₂SO₄）这样的强酸在相同条件下完全解离。弱酸 HA 的部分解离可用可逆反应表示：HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)，通常简写为 HA(aq) ⇌ H⁺(aq) + A⁻(aq)。

Because only a fraction of the acid releases H⁺ ions, a 0.1 mol dm⁻³ solution of a weak acid will have a much lower hydrogen ion concentration – and consequently a higher pH – than a 0.1 mol dm⁻³ solution of a strong acid. The equilibrium position lies far to the left, reflecting limited proton transfer.

由于只有一小部分酸释放出 H⁺ 离子，0.1 mol dm⁻³ 的弱酸溶液中氢离子浓度远低于同浓度的强酸溶液，因而 pH 值更高。平衡位置强烈偏向左侧，反映出有限的质子转移。

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

For a general weak acid HA dissociating in water according to HA(aq) ⇌ H⁺(aq) + A⁻(aq), the equilibrium constant, known as the acid dissociation constant Kₐ, is given by:

对于一般弱酸 HA 在水中按 HA(aq) ⇌ H⁺(aq) + A⁻(aq) 解离，其平衡常数称为酸解离常数 Kₐ，表达式为：

Kₐ = [H⁺][A⁻] / [HA]

The square brackets denote equilibrium concentrations in mol dm⁻³. Note that water is not included in the expression because its concentration is effectively constant in dilute aqueous solutions. Kₐ is a measure of the extent of acid dissociation: the larger the value of Kₐ, the stronger the weak acid, and the higher the proportion of molecules that have donated a proton.

方括号表示以 mol dm⁻³ 为单位的平衡浓度。注意水的浓度没有出现在表达式中，因为在稀水溶液中它实际上是常数。Kₐ 是酸解离程度的量度：Kₐ 值越大，弱酸越强，提供质子的分子比例就越高。

l dm⁻³ at 25°C. It is essential to remember that Kₐ, like all equilibrium constants, depends on temperature. At a given temperature, no matter how the equilibrium mixture is prepared, the ratio [H⁺][A⁻]/[HA] will adjust to equal Kₐ.

由于 Kₐ 值跨越许多数量级，化学家通常用科学记数法表示它们，例如，乙酸的 Kₐ 在 25°C 时为 1.8 × 10⁻⁵ mol dm⁻³。务必记住，Kₐ 和所有平衡常数一样与温度有关。在给定温度下，无论平衡混合物如何配制，比值 [H⁺][A⁻]/[HA] 都会调整到等于 Kₐ。

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3. Kₐ Expression and Units | Kₐ 表达式与单位

entration term has units mol dm⁻³, so the units of Kₐ are (mol dm⁻³ × mol dm⁻³) / (mol dm⁻³) = mol dm⁻³. In many A-Level problems, units are omitted for convenience, but you may be required to state them, especially when comparing Kₐ values or discussing the equilibrium position.

当我们写出 Kₐ 表达式 Kₐ = [H⁺][A⁻] / [HA] 时，单位取决于解离的化学计量比。对于一元酸 HA，每个浓度项的单位是 mol dm⁻³，因此 Kₐ 的单位为 (mol dm⁻³ × mol dm⁻³) / (mol dm⁻³) = mol dm⁻³。在许多 A-Level 题目中，为了方便常省略单位，但你可能需要指明它们，特别是在比较 Kₐ 值或讨论平衡位置时。

Strictly speaking, the thermodynamic equilibrium constant is dimensionless because each concentration is divided by a standard concentration (1 mol dm⁻³). However, the Cambridge syllabus typically accepts Kₐ expressed with units of mol dm⁻³. Always check the phrasing of the question and mark schemes.

严格来说，热力学平衡常数是无单位的，因为每个浓度要除以标准浓度（1 mol dm⁻³）。然而，剑桥课程大纲通常接受用 mol dm⁻³ 作为单位的 Kₐ。务必留意问题的措辞和评分标准。

For a weak dibasic acid, such as carbonic acid, dissociation occurs in two steps with two distinct Kₐ values: Kₐ₁ and Kₐ₂. At A-Level, calculations focus almost exclusively on the first dissociation of monoprotic weak acids.

对于像碳酸这样的弱二元酸，解离分两步进行，有两个不同的 Kₐ 值：Kₐ₁ 和 Kₐ₂。在 A-Level 层次，计算几乎只关注一元弱酸的第一步解离。

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4. From Kₐ to pKₐ | 从 Kₐ 到 pKₐ

To simplify the discussion of acid strength, the pKₐ scale is used. The pKₐ value is defined as the negative logarithm to base 10 of Kₐ:

为了简化对酸强度的讨论，我们使用 pKₐ 标度。pKₐ 值定义为 Kₐ 的以 10 为底的负对数：

pKₐ = −log₁₀(Kₐ)

1.8 × 10⁻⁵, so pKₐ = −log₁₀(1.8 × 10⁻⁵) = 4.74. Hydrocyanic acid, HCN, with Kₐ = 4.9 × 10⁻¹⁰, has pKₐ = 9.31, indicating it is an extremely weak acid.

因此，Kₐ 较大的较强弱酸具有较小的 pKₐ，而极弱的弱酸则具有较大的 pKₐ。例如，乙酸的 Kₐ = 1.8 × 10⁻⁵，因此 pKₐ = −log₁₀(1.8 × 10⁻⁵) = 4.74。氢氰酸 HCN 的 Kₐ = 4.9 × 10⁻¹⁰，pKₐ = 9.31，表明它是极弱的酸。

Kₐ expression simplifies to [H⁺] = Kₐ, and consequently pH = pKₐ. This is a key principle used in estimating the pKₐ of an unknown weak acid from a pH titration curve.

在处理缓冲溶液和滴定问题时，pKₐ 标度尤其有用。半中和的弱酸溶液中 [HA] = [A⁻]，因此 Kₐ 表达式简化为 [H⁺] = Kₐ，从而 pH = pKₐ。这是利用 pH 滴定曲线估算未知弱酸 pKₐ 的关键原理。

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5. Calculating the pH of a Weak Acid Solution | 计算弱酸溶液的 pH

Calculating the pH of a weak acid is one of the most common types of problems in A-Level chemistry. Consider an aqueous solution of a weak monoprotic acid HA with initial concentration c. The dissociation can be analysed using an ICE (Initial, Change, Equilibrium) table:

计算弱酸的 pH 是 A-Level 化学中最常见的问题类型之一。考虑初始浓度为 c 的一元弱酸 HA 水溶液。可借助 ICE（初始、变化、平衡）表分析其解离：

Species	Initial	Change	Equilibrium
HA	c	−x	c − x
H⁺	~0	+x	x
A⁻	0	+x	x

Here x is the equilibrium concentration of H⁺ (and also A⁻). Plugging into the Kₐ expression gives:

在此 x 是 H⁺（以及 A⁻）的平衡浓度。代入 Kₐ 表达式得到：

Kₐ = (x)(x) / (c − x) = x² / (c − x)

tic equation, but for weak acids, the degree of dissociation is typically very small, so x ≪ c. Therefore, we can use the approximation c − x ≈ c. This simplifies the expression to:

这是一个二次方程，但对弱酸而言，解离度通常很小，因此 x ≪ c。于是我们可以采用近似 c − x ≈ c。这样表达式简化为：

Kₐ ≈ x² / c

Solving for x (which equals [H⁺]) yields:

解出 x（等于 [H⁺]）得到：

[H⁺] ≈ √Kₐ × c

Finally, the pH is given by pH = −log₁₀[H⁺].

最后，pH 由 pH = −log₁₀[H⁺] 给出。

pH of a 0.100 mol dm⁻³ solution of ethanoic acid at 25°C, given Kₐ = 1.8 × 10⁻⁵ mol dm⁻³.

我们用一个数字例子来演练。计算 25°C 下 0.100 mol dm⁻³ 乙酸溶液的 pH，已知 Kₐ = 1.8 × 10⁻⁵ mol dm⁻³。

[H⁺] ≈ √1.8 × 10⁻⁵ × 0.100 = √1.8 × 10⁻⁶ = 1.34 × 10⁻³ mol dm⁻³

pH = −log₁₀(1.34 × 10⁻³) = 2.87

Always check the validity of the approximation by calculating the percentage dissociation: % dissociation = (x / c) × 100% = (1.34 × 10⁻³ / 0.100) × 100% ≈ 1.34%. Because this is less than 5%, the approximation is acceptable. If the percentage exceeds 5%, the quadratic equation must be solved accurately.

务必通过计算百分解离度来检查近似的合理性：解离度 % = (x / c) × 100% = (1.34 × 10⁻³ / 0.100) × 100% ≈ 1.34%。由于该值小于 5%，近似是可行的。若百分比超过 5%，则必须精确求解二次方程。

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6. The Approximation Assumptions | 近似假设

When we apply the simplified formula [H⁺] ≈ √Kₐ × c he most dilute solutions (c < 1 × 10⁻⁶ mol dm⁻³) of weak acids with Kₐ > 1 × 10⁻¹², the H⁺ from water can be safely ignored.

当我们应用简化公式 [H⁺] ≈ √Kₐ × c 时，我们做了两个重要假设。首先，假设水的自解离对 H⁺ 的贡献可以忽略。除极稀溶液（c < 1 × 10⁻⁶ mol dm⁻³）外，对于 Kₐ > 1 × 10⁻¹² 的弱酸，来自水的 H⁺ 可以安全地忽略不计。

to the initial concentration that c − x ≈ c. This assumption is justified by the so-called "5% rule" – if the percent dissociation is less than about 5% n within the typical precision of A-Level data. When using the formula, always check the 5% t fails, solve the quadratic equation: Kₐ = x² / (c − x) rearranges to x² + Kₐx − Kₐc = 0.

其次，我们假设解离的 HA 的量相对初始浓度很小，以致于 c − x ≈ c。这个假设由所谓的“5% 规则”来检验——如果百分解离度小于约 5%，近似答案与精确解在 A-Level 数据的典型精度范围内是一致的。使用该公式时，一定要检查 5% 规则。如果不满足，则求解二次方程：Kₐ = x² / (c − x) 整理为 x² + Kₐx − Kₐc = 0。

f dissociation increases. For example, a 1.0 × 10⁻⁴ mol dm⁻³ solution of ethanoic acid has ~12% tolysis may be needed, but Cambridge exams will usually warn you or provide suitable Kₐ and concentration values that keep the approximation valid.

同样关键的是，当酸溶液变得非常稀时，该近似会变得更差，因为解离度增大。例如，1.0 × 10⁻⁴ mol dm⁻³ 的乙酸溶液的解离度约为 12%，使得近似公式不可靠。这种情况下，可能需要使用二次方程乃至更严谨的包含水的自递解的处理方法，但剑桥考试通常会给出提示或提供能使近似成立的合适 Kₐ 和浓度值。

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7. Degree of Dissociation and Dilution | 解离度与稀释

e dissociated at equilibrium: α = [H⁺] / c. For a weak acid obeying the approximation, α ≈ √Kₐ / c. This relationship reveals a fascinating behaviour known as Ostwald's dilution law: as the concentration of a weak acid solution decreases, its degree of dissociation increases.

解离度通常用 α 表示，定义为在平衡时已解离的酸分子所占的分数：α = [H⁺] / c。对于符合近似的弱酸，α ≈ √Kₐ / c。这一关系揭示了一种有趣的行为，即奥斯特瓦尔德稀释定律：随着弱酸溶液浓度降低，其解离度增大。

that the solution becomes more acidic. In fact, the hydrogen ion concentration [H⁺] = αc = √Kₐ × c decreases because c decreases. Hence, diluting a weak acid causes the pH to rise, moving it toward 7, but the pH never surpasses 7. For example, diluting 0.1 mol dm⁻³ ethanoic acid tenfold to 0.01 mol dm⁻³ changes the percentage dissociation from about 1.3% to around 4.2%, but the pH increases from 2.87 to 3.38. These counterintuitive trends make excellent exam questions.

尽管 α 会随稀释而增大，但若以为溶液变得更酸就错了。事实上，氢离子浓度 [H⁺] = αc = √Kₐ × c 随 c 减小而减小。因此，稀释弱酸会导致 pH 升高，趋近于 7，但 pH 绝不会超过 7。例如，将 0.1 mol dm⁻³ 乙酸稀释 10 倍至 0.01 mol dm⁻³，百分解离度从约 1.3% 变到约 4.2%，但 pH 却从 2.87 升至 3.38。这些反直觉的趋势是极好的考题素材。

Ostwald' qualitative effect and be able to use α = [H⁺]/c together with Kₐ to perform calculations.

奥斯特瓦尔德稀释定律可写成线性形式：1/α = 1 + c/(Kₐ α)，但在 A-Level 阶段你只需知道定性影响，并能将 α = [H⁺]/c 与 Kₐ 结合进行计算。

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8. Experimental Determination of Kₐ | Kₐ 的实验测定

on) is dissolved in water to make a solution of known concentration. The pH is measured with a calibrated pH meter. From the pH, [H⁺] is calculated as 10⁻ⁱpH. The equilibrium concentrations are [H⁺] = [A⁻] = 10⁻ⁱpH and [HA] ≈ c − [H⁺], assuming the acid is monoprotic and using the approximation that dissociation is small. Kₐ is then computed from Kₐ = [H⁺]² / (c − [H⁺]).

在实验室中，弱酸的 Kₐ 值可用 pH 计通过实验测定。将已知质量的纯酸（或标准溶液）溶于水制成已知浓度的溶液。用校准过的 pH 计测量 pH。根据 pH，可算得 [H⁺] = 10⁻ⁱpH。平衡浓度为 [H⁺] = [A⁻] = 10⁻ⁱpH，且 [HA] ≈ c − [H⁺]，假设酸是一元的并采用解离度很小的近似。然后由 Kₐ = [H⁺]² / (c − [H⁺]) 计算 Kₐ。

nt (where exactly half the acid has been neutralised) equals the pKₐ of the acid. At this point, [HA] = [A⁻], so the Kₐ expression gives [H⁺] = Kₐ and thus pH = pKₐ. Reading the pH from the titration curve at the half-neutralisation volume provides the pKₐ directly, and Kₐ = 10⁻ⁱpKₐ. This method is frequently featured in practical assessment papers.

一个更巧妙的方法是利用 pH 滴定曲线。用强碱（如 NaOH）滴定弱酸时，在半等量点（恰好一半的酸被中和）的 pH 等于该酸的 pKₐ。在这一点，[HA] = [A⁻]，因此 Kₐ 表达式给出 [H⁺] = Kₐ，从而 pH = pKₐ。在半中和体积处读取滴定曲线的 pH，就可以直接得到 pKₐ，进而 Kₐ = 10⁻ⁱpKₐ。这种方法经常出现在实验考核试卷中。

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9. Weak Acids and Buffer Solutions | 弱酸与缓冲溶液

A buffer solution resists changes in pH when small amounts of acid or base are added. An acidic buffer is typically composed of a weak acid and its conjugate base (often added as a soluble salt). The equilibrium HA

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