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
Explanation:
Step1: Recall water - autoionization
The auto - ionization of water is given by the equation $H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$, and the equilibrium constant expression $K_w=[H^+][OH^-]$.
Step2: Know the ion concentrations in pure water at 25°C
In pure water at 25°C, $[H^+]=[OH^-]=1.0\times10^{- 7}\ mol/L$.
Step3: Calculate the equilibrium constant
$K_w=(1.0\times10^{-7})\times(1.0\times10^{-7}) = 1.0\times10^{-14}$.
Answer:
$10^{-14}$