what is the equilibrium constant of pure water at 25°c?\n10^{14}\n10^{-14}\n10^{-7}\n10^{7}
what is the equilibrium constant of pure water at 25°c?\n10^{14}\n10^{-14}\n10^{-7}\n10^{7}
Answer
Brief Explanations:
The auto - ionization of water is represented as $H_2O(l)+H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)$. At 25°C, the equilibrium constant for this reaction, $K_w = [H_3O^+][OH^-]=1.0\times10^{- 14}$.
Answer:
$10^{-14}$