Kw at 25°C equals:

$K_w$ is the **ionic product of water**. Water undergoes self-ionization to a very small extent according to the equilibrium equation: $$H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$$ The equilibrium constant for this process, known as the ionic product constant, is calculated as: $$K_w = [H^+][OH^-]$$ At the standard temperature of $25^\circ C$ ($298\text{ K}$), experimental measurements show that the concentration of hydrogen ions in pure water is: $$[H^+] = 1.0 \times 10^{-7}\text{ mol/L}$$ Since pure water is neutral, the concentration of hydroxide ions is equal to the concentration of hydrogen ions: $$[OH^-] = 1.0 \times 10^{-7}\text{ mol/L}$$ Substituting these values into the $K_w$ expression: $$K_w = (10^{-7}) \times (10^{-7}) = 10^{-14}$$ Thus, at $25^\circ C$, $K_w = 10^{-14}$.