what is the equilibrium constant of pure water at 25°c?\no 10^14\no 10^(-14)\no 10^(-7)\no 10^7

Explanation:

Step1: Recall water - autoionization

The auto - ionization of water is given by the equation $H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$. The equilibrium constant expression for this reaction is $K_w = [H^+][OH^-]$.

Step2: Know the ion concentrations at 25°C

At 25°C, in pure water, $[H^+]=[OH^-]=1.0\times 10^{- 7}\ mol/L$.

Step3: Calculate $K_w$

Substitute the values of $[H^+]$ and $[OH^-]$ into the $K_w$ expression: $K_w=(1.0\times 10^{-7})\times(1.0\times 10^{-7}) = 1.0\times 10^{-14}$.

Answer:

$10^{-14}$ (Second option)