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

📚 The Ionic Product of Water, K_w | 水的离子积，K_w

Water is not just a passive solvent. A tiny fraction of water molecules undergo auto-ionisation (self-ionisation), transferring a proton from one water molecule to another. This equilibrium produces hydroxonium ions, H₃O⁺, and hydroxide ions, OH^-. For A-Level Chemistry, we simplify this to H⁺ and OH^-. The equilibrium constant for this reaction, with the concentration of water incorporated into the constant, is called the ionic product of water, K_w. Understanding K_w is essential because it links [H⁺] and [OH^-] in any aqueous solution and underpins calculations of pH, pOH, and the behaviour of acids and bases across different temperatures.

水不仅仅是惰性溶剂。极少部分的水分子会发生自电离（自身质子转移），产生水合氢离子 H₃O⁺ 和氢氧根离子 OH^-。在 A-Level 化学中我们常简写为 H⁺ 和 OH^-。这个反应的平衡常数，在将水的浓度并入常数后，就被称为水的离子积常数 K_w。掌握 K_w 至关重要，因为它把水溶液中 [H⁺] 与 [OH^-] 联系起来，也是不同温度下 pH、pOH 计算以及酸、碱行为分析的基础。

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1. Auto-ionisation of Water | 水的自电离

Pure water conducts electricity to a very small extent, indicating the presence of ions. The auto-ionisation reaction can be written as: 2H₂O(l) ⇌ H₃O⁺(aq) + OH^-(aq). For simplicity, we often write H₂O(l) ⇌ H⁺(aq) + OH^-(aq). This process is endothermic, meaning the forward reaction absorbs heat.

纯水有极微弱的导电性，说明存在离子。自电离反应可写为：2H₂O(l) ⇌ H₃O⁺(aq) + OH^-(aq)。简化常写作 H₂O(l) ⇌ H⁺(aq) + OH^-(aq)。该过程吸热，正向反应吸收热量。

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2. Definition and Expression of K_w | K_w 的定义与表达式

Applying the equilibrium law, K_c = [H⁺][OH^-] / [H₂O]. Since the concentration of liquid water is essentially constant (approximately 55.5 mol dm^-3), it can be combined with K_c to give a new constant — the ionic product of water, K_w. Thus, K_w = [H⁺][OH^-]. The units are mol² dm^-6 because each concentration term has units mol dm^-3.

根据平衡定律，K_c = [H⁺][OH^-] / [H₂O]。由于液态水的浓度基本恒定（约 55.5 mol dm^-3），它可以与 K_c 合并成一个新常数——水的离子积 K_w。因此，K_w = [H⁺][OH^-]。每项浓度单位都是 mol dm^-3，所以 K_w 的单位是 mol² dm^-6。

K_w is an equilibrium constant that applies to all dilute aqueous solutions, not just pure water. At a given temperature, the product of [H⁺] and [OH^-] must remain constant, even if one of the ion concentrations is very high and the other very low.

K_w 是一个适用于所有稀水溶液的平衡常数，而不仅仅针对纯水。在给定温度下，[H⁺] 与 [OH^-] 的乘积必须保持不变，哪怕其中一种离子浓度极高而另一种极低。

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3. Value of K_w at 25°C | 25°C 时的 K_w 值

At 298 K (25 °C), the ionic product of water has been determined experimentally to be K_w = 1.0 × 10^-14 mol² dm^-6. You are expected to know this value and use it in calculations. In pure water, because [H⁺] = [OH^-], we have [H⁺] = √K_w = 1.0 × 10^-7 mol dm^-3, leading to a neutral pH of 7.00.

在 298 K (25°C) 时，实验测得水的离子积 K_w = 1.0 × 10^-14 mol² dm^-6。你应当熟记这个数值并用于计算。在纯水中，由于 [H⁺] = [OH^-]，可得 [H⁺] = √K_w = 1.0 × 10^-7 mol dm^-3，对应中性 pH = 7.00。

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4. Temperature Dependence of K_w | K_w 的温度依赖性

Since the auto-ionisation of water is endothermic, increasing the temperature shifts the equilibrium to the right according to Le Chatelier's principle. As a result, K_w increases with temperature. The value of K_w is only 1.0 × 10^-14 at 25°C. At 40°C, K_w ≈ 2.92 × 10^-14; at 60°C, K_w ≈ 9.61 × 10^-14. This means pure water remains neutral (since [H⁺] = [OH^-]), but the pH drops below 7. Neutrality does not mean pH = 7; it means [H⁺] = [OH^-].

由于水的自电离是吸热过程，根据勒夏特列原理，升高温度会使平衡向右移动。因此 K_w 随温度升高而增大。K_w = 1.0 × 10^-14 仅在 25°C 时成立。在 40°C 时，K_w ≈ 2.92 × 10^-14；60°C 时约为 9.61 × 10^-14。这意味着纯水仍保持中性（因为 [H⁺] = [OH^-]），但 pH 会降至 7 以下。中性不等于 pH=7，而是 [H⁺] = [OH^-]。

A common exam question provides K_w at a different temperature and asks for the pH of pure water. Remember to take the square root of the given K_w to find [H⁺], then calculate pH = -log₁₀[H⁺].

常见的考试题会给出某个温度下的 K_w，要求计算纯水的 pH。切记将给定的 K_w 开平方得到 [H⁺]，再计算 pH = -log₁₀[H⁺]。

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5. Relationship between [H⁺] and [OH^-] in Pure Water | 纯水中 [H⁺] 与 [OH^-] 的关系

In pure water, the only source of H⁺ and OH^- is the auto-ionisation equilibrium. Consequently, the concentrations must be equal: [H⁺] = [OH^-]. At 25°C, this gives each ion a concentration of 1.0 × 10^-7 mol dm^-3. Even a tiny amount of dissolved CO₂ can disturb this equality by producing carbonic acid, but for pure water in theory this relationship holds.

在纯水中，H⁺ 和 OH^- 的唯一来源是自电离平衡。因此二者浓度必然相等：[H⁺] = [OH^-]。25°C 时两种离子浓度均为 1.0 × 10^-7 mol dm^-3。即使微量溶入的 CO₂ 生成碳酸也可打破这种相等，但理论上纯水严格满足此关系。

Using K_w, you can always find one ion concentration if the other is known. For example, if [H⁺] = 3.2 × 10^-4 mol dm^-3, then [OH^-] = K_w / [H⁺] = 1.0 × 10^-14 / 3.2 × 10^-4 ≈ 3.1 × 10^-11 mol dm^-3.

借助 K_w，只要已知一种离子的浓度，总能求出另一种。例如，若 [H⁺] = 3.2 × 10^-4 mol dm^-3，则 [OH^-] = K_w / [H⁺] = 1.0 × 10^-14 / 3.2 × 10^-4 ≈ 3.1 × 10^-11 mol dm^-3。

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6. [H⁺] and [OH^-] in Acidic and Basic Solutions | 酸碱溶液中的 [H⁺] 与 [OH^-]

In an acidic solution, the added acid provides an excess of H⁺. The equilibrium H₂O ⇌ H⁺ + OH^- shifts to the left, suppressing the OH^- concentration. The product [H⁺][OH^-] still equals K_w. Similarly, in an alkaline solution, extra OH^- causes a drastic decrease in [H⁺]. The equations [H⁺] = K_w / [OH^-] and [OH^-] = K_w / [H⁺] always apply.

在酸性溶液中，加入的酸提供了过量的 H⁺。平衡 H₂O ⇌ H⁺ + OH^- 左移，抑制了 OH^- 浓度。乘积 [H⁺][OH^-] 仍等于 K_w。类似地，在碱性溶液中，额外的 OH^- 会使 [H⁺] 急剧下降。公式 [H⁺] = K_w / [OH^-] 和 [OH^-] = K_w / [H⁺] 始终成立。

A typical exam task is to find the pH of a very dilute acid, where the contribution of H⁺ from water is no longer negligible. For instance, 1.0 × 10^-8 mol dm^-3 HCl would give an apparent pH of 8 by simply taking -log (1.0 × 10^-8), but this ignores water's auto-ionisation. The correct approach considers total [H⁺] = 1.0 × 10^-8 + [OH^-], then uses K_w to solve the resulting quadratic.

常见考题是求极稀酸的 pH，此时水电离产生的 H⁺ 已不可忽略。例如 1.0 × 10^-8 mol dm^-3 HCl，若简单地用 -log (1.0 × 10^-8) 会得出 pH = 8，但忽略了水的自电离。正确做法是让总 [H⁺] = 1.0 × 10^-8 + [OH^-]，再利用 K_w 解二次方程。

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7. pK_w and its Relationship with pH and pOH | pK_w 及其与 pH、pOH 的关系

Analogous to pH, we define pK_w = -log₁₀K_w. At 25°C, pK_w = 14.00. Since K_w = [H⁺][OH^-], taking negative logarithms gives pK_w en pH and pOH. For any aqueous solution at 25°C, if pH = 5.3, then pOH = 14.0 - 5.3 = 8.7.

与 pH 类似，我们定义 pK_w = -log₁₀K_w。25°C 时 pK_w = 14.00。因为 K_w = [H⁺][OH^-]，取负对数后得到 pK_w = pH + pOH。这个非常实用的公式可在 pH 与 pOH 间互相转换。对于 25°C 下的任何水溶液，若 pH = 5.3，那么 pOH = 14.0 - 5.3 = 8.7。

At higher temperatures, pK_w becomes smaller, so the sum pH + pOH is less than 14. Nevertheless, in pure water, pH always equals pOH, meaning each is half of pK_w. Thus, a shift in pK_w directly shifts the neutral point on the pH scale.

在更高温度下，pK_w 变小，因此 pH + pOH 之和小于 14。然而在纯水中，pH 始终等于 pOH，各占 pK_w 的一半。因此 pK_w 的改变直接推动 pH 标度上的中性点发生移动。

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8. Calculating pH of Strong Bases Using K_w | 利用 K_w 计算强碱的 pH

To find the pH of a solution of a strong base such as NaOH, first calculate [OH^-] from the concentration of the base, remembering that strong bases dissociate completely. Then use [H⁺] = K_w / [OH^-], and finally pH = -log₁₀[H⁺]. For example, 0.010 mol dm^-3 NaOH gives [OH^-] = 1.0 × 10^-2 mol dm^-3. At 25°C, [H⁺] = 1.0 × 10^-14 / 1.0 × 10^-2 = 1.0 × 10^-12 mol dm^-3, hence pH = 12.00.

计算强碱（如 NaOH）溶液的 pH 时，先由碱的浓度求出 [OH^-]，牢记强碱完全电离。然后用 [H⁺] = K_w / [OH^-]，最后 pH = -log₁₀[H⁺]。例如 0.010 mol dm^-3 NaOH 中 [OH^-] = 1.0 × 10^-2 mol dm^-3。25°C 时 [H⁺] = 1.0 × 10^-14 / 1.0 × 10^-2 = 1.0 × 10^-12 mol dm^-3，故 pH = 12.00。

If the same NaOH solution were at 40°C where K_w = 2.92 × 10^-14, then [H⁺] = 2.92 × 10^-14 / 1.0 × 10^-2 = 2.92 × 10^-12 mol dm^-3, giving pH ≈ 11.53. Always check the temperature provided in examination questions.

若同一 NaOH 溶液处于 40°C，K_w = 2.92 × 10^-14，那么 [H⁺] = 2.92 × 10^-14 / 1.0 × 10^-2 = 2.92 × 10^-12 mol dm^-3，pH ≈ 11.53。务必留意考题中给定的温度。

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9. Dilution of Acids and Bases in Relation to K_w | 稀释酸碱时与 K_w 的关系

Diluting an acidic solution with water reduces [H⁺] and therefore increases pH. However, the product [H⁺][OH^-] must always equal K_w. As the acid is diluted, the auto-ionisation of water contributes a larger fraction of the total H⁺. When an acid is diluted to extremely low concentrations (below about 4.0 × 10^-7 mol dm^-3), the pH tends towards the neutral value of the water at that temperature, rather than becoming alkaline.

用水稀释酸性溶液会降低 [H⁺]，pH 因而上升。但乘积 [H⁺][OH^-] 必须始终等于 K_w。随着酸被稀释，水的自电离在总 H⁺ 中所占比例越来越大。当酸稀释至极低浓度（约低于 4.0 × 10^-7 mol dm^-3）时，pH 会趋近于该温度下纯水的中性值，而不会变成碱性。

Similarly, diluting a strong base shifts pH towards the neutral point. For a 0.001 mol dm^-3 NaOH solution, pH ≈ 11 at 25°C; if you dilute it 1000-fold, the [OH^-] is about 1 × 10^-6 mol dm^-3. The [H⁺] from water can no longer be neglected. The resulting pH is close to 7, reflecting the influence of K_w.

同样，稀释强碱会使 pH 向中性点靠拢。对于 0.001 mol dm^-3 NaOH 溶液，25°C 时 pH ≈ 11；若稀释 1000 倍，[OH^-] 约为 1 × 10^-6 mol dm^-3。此时水的自电离产生的 H⁺ 已不可忽略，最终 pH 接近 7，体现 K_w 的制约。

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10. Comparison of K_w Values at Different Temperatures | 不同温度下 K_w 值的对比

The table below illustrates how K_w and the neutral pH change with temperature. Remember that the solution remains electrically neutral; it is the pH of neutrality that shifts.

下表展示 K_w 和中性 pH 如何随温度变化。请记住溶液本身仍为电中性，移动的是“中性”的 pH 值。

Temperature / °C	Kw / mol2 dm-6	Neutral pH
0	1.14 × 10-15	7.47
10	2.92 × 10-15	7.27
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