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

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

Water is often thought of as a neutral liquid, but at the molecular level it undergoes a very slight self-ionization. This produces low concentrations of hydrogen ions and hydroxide ions even in pure water.

水通常被认为是中性液体，但在分子水平上，水会发生极轻微的自电离。即使在纯水中，也会产生低浓度的氢离子和氢氧根离子。

The equilibrium constant for this self-ionization is called the ionic product of water, K_w. It is a fundamental concept in acid-base chemistry and links [H⁺] and [OH⁻] in all aqueous solutions.

水自电离的平衡常数称为水的离子积 K_w。它是酸碱化学中的基本概念，联系着所有水溶液中的 [H⁺] 和 [OH⁻]。

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1. Water's Autoionization | 水的自电离

Pure water contains very small concentrations of hydrogen ions and hydroxide ions because water molecules can transfer a proton from one molecule to another.

纯水中含有极少量的氢离子和氢氧根离子，因为水分子之间可以发生质子转移。

The process can be represented as:

该过程可以表示为：

H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

In A-level Chemistry, this is often simplified to:

在 A-level 化学中，通常简化为：

H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)

The equilibrium lies very far to the left, so only about one water molecule in every 555 million molecules is ionized at 25 °C.

平衡非常偏向左边，因此在 25 °C 时，大约每 5.55 亿个水分子中只有一个水分子电离。

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2. Defining the Ionic Product of Water, K_w | 定义水的离子积 K_w

For the self-ionization of water, the equilibrium constant can be written as:

对于水的自电离，平衡常数可以写成：

K_c = [H⁺][OH⁻] / [H₂O]

Since the concentration of water in pure water is effectively constant, it is incorporated into the equilibrium constant. This gives the ionic product of water:

由于纯水中水的浓度实际上是恒定的，因此将其并入平衡常数，得到水的离子积：

K_w = [H⁺][OH⁻]

K_w has units of concentration squared, mol² dm⁻⁶, because it is the product of two concentration terms.

K_w 的单位是浓度的平方，即 mol² dm⁻⁶，因为它是两个浓度项的乘积。

This expression applies not only to pure water but to all dilute aqueous solutions at a given temperature.

该表达式不仅适用于纯水，也适用于给定温度下所有稀的水溶液。

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3. The Value of K_w at 25 °C | 25 °C 时 K_w 的值

°C, the concentrations of H⁺ and OH⁻ in pure water are each equal to 1.0 × 10⁻⁷ mol dm⁻³.

在 25 °C 时，纯水中 H⁺ 和 OH⁻ 的浓度均等于 1.0 × 10⁻⁷ mol dm⁻³。

Therefore, the ionic product of water at 25 °C is:

因此，25 °C 时水的离子积为：

K_w = (1.0 × 10⁻⁷) × (1.0 × 10⁻⁷) = 1.0 × 10⁻¹⁴ mol² dm⁻⁶

This value is given on most exam data sheets and is valid only when the temperature is 25 °C.

该数值在大多数考试数据表中都会给出，并且仅在温度为 25 °C 时有效。

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

The self-ionization of water is an endothermic process because energy is required to break O-H bonds.

水的自电离是一个吸热过程，因为断裂 O-H 键需要能量。

According to Le Chatelier's principle, increasing the temperature shifts the equilibrium to the right, so K_w increases with temperature.

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。

The table below shows how K_w and the pH of pure water change with temperature:

下表显示了 K_w 和纯水的 pH 如何随温度变化：

Temperature / °C	Kw / mol² dm⁻⁶	pH of pure water
0	1.14 × 10⁻¹⁵	7.47
25	1.00 × 10⁻¹⁴	7.00
50	5.48 × 10⁻¹⁴	6.63
100	5.50 × 10⁻¹³	6.13

Even though the pH of pure water falls below 7 at high temperatures, the water is still neutral because [H⁺] still equals [OH⁻].

尽管纯水的 pH 在高温下会低于 7，但水仍然是中性的，因为 [H⁺] 仍然等于 [OH⁻]。

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5. Relationship between K_w, pH and pOH | K_w、pH 和 pOH 之间的关系

The pH of a solution is defined as:

溶液的 pH 定义为：

pH = -log₁₀[H⁺]

Similarly, pOH is defined as:

类似地，pOH 定义为：

pOH = -log₁₀[OH⁻]

Taking the negative logarithm of the K_w expression gives:

对 K_w 表达式取负对数，得到：

-log₁₀K_w = -log₁₀[H⁺] + (-log₁₀[OH⁻])

This gives the very useful relationship:

由此得到非常有用的关系式：

pK_w = pH + pOH

At 25 °C, pK_w = -log₁₀(1.0 × 10⁻¹⁴) = 14.00, so:

在 25 °C 时，pK_w = -log₁₀(1.0 × 10⁻¹⁴) = 14.00，因此：

pH + pOH = 14.00

This equation is only valid at 25 °C. At other temperatures, pH + pOH still equals pK_w, but pK_w is not 14.00.

该方程仅在 25 °C 时成立。在其他温度下，pH + pOH 仍等于 pK_w，但 pK_w 不再是 14.00。

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6. Calculating pH of Strong Bases Using K_w | 使用 K_w 计算强碱的 pH

For a strong base such as NaOH, dissociation in water is complete:

对于 NaOH 等强碱，它在水中完全解离：

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

This means the hydroxide concentration is usually known directly from the concentration of the base. The hydrogen ion concentration is then found using K_w.

这意味着氢氧根离子浓度通常可直接由碱的浓度得出。然后利用 K_w 求出氢离子浓度。

For example, calculate the pH of 0.10 mol dm⁻³ NaOH at 25 °C.

例如，计算 25 °C 时 0.10 mol dm⁻³ NaOH 的 pH。

Step 1: [OH⁻] = 0.10 mol dm⁻³

步骤 1：[OH⁻] = 0.10 mol dm⁻³

Step 2: [H⁺] = K_w / [OH⁻] = 1.0 × 10⁻¹⁴ / 0.10 = 1.0 × 10⁻¹³ mol dm⁻³

步骤 2：[H⁺] = K_w / [OH⁻] = 1.0 × 10⁻¹⁴ / 0.10 = 1.0 × 10⁻¹³ mol dm⁻³

Step 3: pH = -log₁₀(1.0 × 10⁻¹³) = 13.00

步骤 3：pH = -log₁₀(1.0 × 10⁻¹³) = 13.00

Remember that for bases with two hydroxide ions, such as Ba(OH)₂, the [OH⁻] is twice the concentration of the base.

请记住，对于含有两个氢氧根离子的碱，例如 Ba(OH)₂，[OH⁻] 是碱浓度的两倍。

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7. Calculating pH of Dilute Acids and Bases | 计算稀酸和稀碱的 pH

⁻⁶ mol dm⁻³, the hydrogen ions from the self-ionization of water can no longer be ignored.

当强酸非常稀时，例如浓度低于 1 × 10⁻⁶ mol dm⁻³，来自水自电离的氢离子就不能再忽略。

In such cases, the total [H⁺] comes from both the acid and the autoionization of water. A similar situation occurs for very dilute strong bases.

在这种情况下，总 [H⁺] 来自酸和水的自电离。非常稀的强碱也会出现类似情况。

To solve these problems accurately, you can set up an equation using K_w and the known concentration of acid or base. For example, for a very dilute HCl solution of concentration c:

为了准确求解这类问题，可以利用 K_w 和已知的酸或碱浓度建立方程。例如，对于浓度为 c 的极稀 HCl 溶液：

Let [H⁺] from water = x, so total [H⁺] = c + x and [OH⁻] = x

K_w = (c + x)(x)

Solving for x then gives the true [H⁺] and hence the true pH. At A-level, you may be asked to explain why the pH of a very dilute acid approaches but never exceeds 7 at 25 °C.

解出 x 即可得到真实的 [H⁺]，从而得到真实的 pH。在 A-level 考试中，你可能会被要求解释为什么在 25 °C 时极稀酸的 pH 接近但永远不会超过 7。

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8. pK_w and its Significance | pK_w 及其意义

The quantity pK_w is defined as:

pK_w 定义为：

pK_w = -log₁₀K_w

At 25 °C, pK_w = 14.00. At higher temperatures, because K_w increases, pK_w decreases.

在 25 °C 时，pK_w = 14.00。在较高温度下，由于 K_w 增大，pK_w 减小。

For example, at 50 °C, K_w = 5.48 × 10⁻¹⁴, so pK_w = 13.26. This means the neutral pH at 50 °C is 6.63, not 7.00.

例如，在 50 °C 时，K_w = 5.48 × 10⁻¹⁴，因此 pK_w = 13.26。这意味着 50 °C 时的中性 pH 是 6.63，而不是 7.00。

to test whether you understand that neutrality is defined by [H⁺] = [OH⁻], not by pH = 7.

考试题目经常利用这一概念来检验你是否理解中性是由 [H⁺] = [OH⁻] 定义的，而不是由 pH = 7 定义的。

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9. Common Misconceptions about K_w | 关于 K_w 的常见误区

• Misconception 1: pH = 7 always means a solution is neutral.
• 误区 1： pH = 7 总是表示溶液为中性。

This is only true at 25 °C. At other temperatures, neutral pH equals ½ pK_w, which is not 7.00.

这仅在 25 °C 时成立。在其他温度下，中性 pH 等于 ½ pK_w，而不是 7.00。

• Misconception 2: K_w is always 1.0 × 10⁻¹⁴ mol² dm⁻⁶.
• 误区 2： K_w 总是 1.0 × 10⁻¹⁴ mol² dm⁻⁶。

K_w changes with temperature, so this value is only correct at 25 °C.

K_w 随温度变化，因此该数值仅在 25 °C 时正确。

• Misconception 3: Diluting an acid can eventually make it alkaline.
• 误区 3： 稀释酸最终会使其变成碱性。

Diluting an acid decreases [H⁺], but the pH approaches 7 from below and never becomes greater than 7 at 25 °C.

稀释酸会降低 [H⁺]，但在 25 °C 时 pH 从低于 7 的方向接近 7，永远不会大于 7。

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10. Worked Examples | 典型例题

Example 1: Calculate the pH of pure water at 40 °C if K_w = 2.92 × 10⁻¹⁴ mol² dm⁻⁶.

例题 1： 若 40 °C 时 K_w = 2.92 × 10⁻¹⁴ mol² dm⁻⁶，计算纯水的 pH。

In pure water, [H⁺] = [OH⁻], so:

在纯水中，[H⁺] = [OH⁻]，因此：

[H⁺]² = 2.92 × 10⁻¹⁴

[H⁺] = √2.92 × 10⁻¹⁴ = 1.71 × 10⁻⁷ mol dm⁻³

pH = -log₁₀(1.71 × 10⁻⁷) = 6.77

Example 2: Calculate the pH of 0.0050 mol dm⁻³ Ba(OH)₂ at 25 °C.

例题 2： 计算 25 °C 时 0.0050 mol dm⁻³ Ba(OH)₂ 的 pH。

[OH⁻] = 2 × 0.0050 = 0.010 mol dm⁻³

[OH⁻] = 2 × 0.0050 = 0.010 mol dm⁻³

[H⁺] = 1.0 × 10⁻¹⁴ / 0.010 = 1.0 × 10⁻¹² mol dm⁻³

[H⁺] = 1.0 × 10⁻¹⁴ / 0.010 = 1.0 × 10⁻¹² mol dm⁻³

pH = -log₁₀(1.0 × 10⁻¹²) = 12.00

pH = -log₁₀(1.0 × 10⁻¹²) = 12.00

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11. Exam Tips for K_w Questions | K_w 题目的考试技巧

• ck the temperature given in the question. If it is not 25 °C, do not use 1.0 × 10⁻¹⁴ for K_w or 14.00 for pK_w.
• 一定要检查题目中给出的温度。如果不是 25 °C，不要使用 1.0 × 10⁻¹⁴ 作为 K_w，也不要使用 14.00 作为 pK_w。
• Write out the K_w expression before substituting values.
• 在代入数值之前先写出 K_w 表达式。
• For strong bases, calculate [OH⁻] first, then use K_w to find [H⁺].
• 对于强碱，先计算 [OH⁻]，然后利用 K_w 求 [H⁺]。
• Remember that neutral does not always mean pH = 7; it means [H⁺] = [OH⁻].
• 记住中性并不总是意味着 pH = 7；它意味着 [H⁺] = [OH⁻]。
• If a base provides two OH⁻ ions per formula unit, multiply its concentration by 2.
• 如果每个化学式单元提供两个 OH⁻ 离子，则需要将其浓度乘以 2。

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12. Summary | 总结

The ionic product of water, K_w, is the equilibrium constant for the self-ionization of water and is given by K_w = [H⁺][OH⁻].

水的离子积 K_w 是水自电离的平衡常数，表达式为 K_w = [H⁺][OH⁻]。

At 25 °C, K_w = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ and pH + pOH = 14.00. However, K_w increases with temperature because water's self-ionization is endothermic.

在 25 °C 时，K_w = 1.0 × 10⁻¹⁴ mol² dm⁻⁶，pH + pOH = 14.00。然而，由于水的自电离是吸热过程，K

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