What is the equilibrium constant of pure water at 25°C?
The equilibrium constant for the auto-ionization of water, denoted as \(K_w\), relates to the dissociation reaction of water into hydronium and hydroxide ions:
\[H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)\]
The equilibrium expression for this reaction is given by the product of the molar concentrations of the ions:
\[K_w = [H^+][OH^-]\]
At a standard temperature of \(25^{\circ}\text{C}\), pure water dissociates to a very small extent such that the concentration of both hydronium ions (\([H^+]\)) and hydroxide ions (\([OH^-]\)) is \(10^{-7} \text{ M}\).
To determine the value of \(K_w\) at \(25^{\circ}\text{C}\), we substitute the concentrations of the ions into the equilibrium expression:
\[K_w = [10^{-7} \text{ M}] \times [10^{-7} \text{ M}]\]
Using the laws of exponents, where \(a^m \times a^n = a^{m+n}\), we calculate:
\[K_w = 10^{-7 + (-7)} = 10^{-14}\]
Comparing this result to the given options, we find that the equilibrium constant of pure water at \(25^{\circ}\text{C}\) is \(10^{-14}\).
The equilibrium constant of water is essential in chemistry for calculating the pH of solutions. For example, if you know the concentration of an acid in water, you can use the relationship \(K_w = [H^+][OH^-]\) to determine the concentration of hydroxide ions present. This relationship acts as the foundation for acid-base equilibrium calculations in biological systems, environmental monitoring, and industrial chemical processing.