Autoionization of water Knowledge-Based Study

Study notes

1. What is Autoionization?

   Even highly purified water conducts electricity to a tiny, measurable extent. This phenomenon occurs because water molecules undergo autoionization (also known as self-ionization), an equilibrium process where two water molecules react to produce hydronium and hydroxide ions.

   In this reaction, one water molecule donates a proton acting as a Brønsted–Lowry acid, while the second water molecule accepts the proton acting as a Brønsted–Lowry base:

   $$2\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$

   For simplicity, this equilibrium is frequently written in terms of a single dissociated water molecule:

   $$\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$$

   Because water acts as both an acid and a base simultaneously in this process, it is described as amphiprotic (or amphoteric).

   The Equilibrium Constant for Water

   Applying the law of mass action to the autoionization equilibrium gives an equilibrium constant expression. Because liquid water is a pure solvent with an activity close to 1, its concentration is omitted from the expression:

   $$K_w = [\text{H}_3\text{O}^+][\text{OH}^-]$$

   The constant $K_w$ is called the ion-product constant of water. At a standard temperature of 25 °C (298 K), experimental measurements establish:

   $$K_w = 1.0 \times 10^{-14}$$

   In pure water, stoichiometry dictates that every hydronium ion produced is accompanied by exactly one hydroxide ion:

   $$[\text{H}_3\text{O}^+] = [\text{OH}^-] = \sqrt{K_w} = 1.0 \times 10^{-7}\text{ M}$$

   This fundamental relationship holds true not only for pure water, but for any dilute aqueous solution at 25 °C.

   Is Autoionization Exothermic or Endothermic?

   Breaking covalent O–H bonds to produce separated ions requires energy. Consequently, the autoionization of water is an endothermic process ($\Delta H > 0$):

   $$\text{energy} + 2\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$

   According to Le Chatelier's principle, an increase in temperature shifts the equilibrium position toward the forward endothermic direction:

   • At temperatures above 25 °C: The forward shift increases ion formation, so $K_w > 1.0 \times 10^{-14}$. For example, at 50 °C, $K_w \approx 5.5 \times 10^{-14}$.
   • At temperatures below 25 °C: The reverse reaction is favored, so $K_w < 1.0 \times 10^{-14}$.

   Crucial Conceptual Point: As water is heated above 25 °C, $[\text{H}_3\text{O}^+]$ increases, causing the pH of pure water to drop below 7. However, the water remains strictly neutral because $[\text{H}_3\text{O}^+]$ remains exactly equal to $[\text{OH}^-]$.

   Defining Acidic, Basic, and Neutral Conditions

   Because the product $[\text{H}_3\text{O}^+][\text{OH}^-]$ remains fixed at a given temperature, changing the concentration of one ion inevitably changes the other inversely.

   Aqueous solutions are classified by the relative ratio of these two ions:

   • Neutral solution: $[\text{H}_3\text{O}^+] = [\text{OH}^-]$ (at 25 °C, both equal $1.0 \times 10^{-7}\text{ M}$).
   • Acidic solution: An acid adds excess hydronium ions, making $[\text{H}_3\text{O}^+] > [\text{OH}^-]$ (at 25 °C, $[\text{H}_3\text{O}^+] > 1.0 \times 10^{-7}\text{ M}$).
   • Basic solution: A base adds excess hydroxide ions, making $[\text{OH}^-] > [\text{H}_3\text{O}^+]$ (at 25 °C, $[\text{H}_3\text{O}^+] < 1.0 \times 10^{-7}\text{ M}$).

   If the concentration of either $[\text{H}_3\text{O}^+]$ or $[\text{OH}^-]$ is known, the other can always be determined using:

   $$[\text{OH}^-] = \frac{K_w}{[\text{H}_3\text{O}^+]}$$