Question

Why are the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{OH}^{-}\) equal in pure water?

Short answer

In pure water, \(\text{H}_3\text{O}^+\) and \(\text{OH}^-\) concentrations are equal due to the autoionization reaching equilibrium, resulting in equal formation and recombination rates of these ions.

Step-by-step solution

Understanding Water Autoionization

Pure water undergoes autoionization where two water molecules react to form \(\text{H}_3\text{O}^+\) and \(\text{OH}^-\) ions. The reaction can be represented as: \[\text{H}_2\text{O} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^- \]

Le Chatelier's Principle

According to Le Chatelier's Principle, the water autoionization reaction reaches a point of dynamic equilibrium, meaning the rate at which water ionizes is equal to the rate at which \(\text{H}_3\text{O}^+\) and \(\text{OH}^-\) recombine to form water.

Equilibrium Expression

At equilibrium, the concentrations of \(\text{H}_3\text{O}^+\) and \(\text{OH}^-\) are equal because there is no external source or sink for \(\text{H}_3\text{O}^+\) or \(\text{OH}^-\) that would disturb this balance in pure water. Therefore, we have \[\text{[H}_3\text{O}^+] = \text{[OH}^- \]

Key concepts

Dynamic Equilibrium

In pure water, a fascinating phenomenon occurs known as autoionization. This is when two water molecules react to form hydronium ions \((\text{H}_3\text{O}^+)\) and hydroxide ions \((\text{OH}^-)\). The key reaction can be represented as: \[ \text{H}_2\text{O} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^- \]This reaction does not go to completion but instead reaches a state called dynamic equilibrium. In dynamic equilibrium, the forward reaction rate (water molecules forming ions) equals the reverse reaction rate (ions recombining to form water). This means the concentration of the reactants and products remains constant over time. A dynamic equilibrium is a hallmark of reversible reactions, indicating no net change in the concentrations of the involved species even though both forward and reverse reactions are continuously occurring.

Le Chatelier's Principle

According to Le Chatelier's Principle, when a system at equilibrium is subjected to an external change, the system will adjust to partially counteract that change, re-establishing equilibrium. In the case of water autoionization, the equilibrium can be disturbed by changes such as adding acids or bases. For instance, adding an acid increases the concentration of \(\text{H}_3\text{O}^+\) ions.
Le Chatelier's Principle predicts that the system will respond by partially reducing this disturbance. This can be done by the increased \(\text{H}_3\text{O}^+\) ions recombining with \(\text{OH}^-\) ions, thus reducing their concentrations and moving the equilibrium back toward the reactants. Conversely, adding a base increases the \(\text{OH}^-\) concentration, pushing the equilibrium to produce more \(\text{H}_3\text{O}^+\) ions and reducing \(\text{OH}^-\) via combining back to form water.

Ion Concentrations in Pure Water

In pure water, without any external influencing factors like additional acids or bases, the concentrations of \(\text{H}_3\text{O}^+\) and \(\text{OH}^-\) are inherently equal. This is because the water autoionization reaction reaches a natural state of dynamic equilibrium. In this state, there is no external source or sink of \(\text{H}_3\text{O}^+\) or \(\text{OH}^-\), so the ions are produced and consumed at the same rate. This balance is mathematically represented by the equation: \[ \text{[H}_3\text{O}^+] = \text{[OH}^-] \]
At 25°C, the product of these ion concentrations is a constant known as the ion-product constant for water \(Kw\), which equals \(1.0 \times 10^{-14}\). Thus, in pure water, \(\text{[H}_3\text{O}^+]\) and \(\text{[OH}^-]\) are both \(1.0 \times 10^{-7} \text{M}\).