Question

Kw, the equilibrium constant for the ionization of water by the equation below, is 1.0 x 10-14. what does that mean when we are considering pure water?

Answer 1

Answer: Concentration of $[H^+]\&[OH^-]$ are equal in pure water.

Explanation:

$H_2O\rightleftharpoons H^++OH^-$

$K_{eq}=\frac{[H^+][OH^-]}{[H_2O]}$

$K_w=K_{eq}[H_2O]=[H^+][OH^-]=1.0\times 10^{-14}$

Also, pure water has neutral with 7 pH value ,

$pH=7=-\log[H^+]$

$[H^+]=1\times 10^{-7}mol/L$

So, the$K_w$ of pure water is given

$K_w=[H^+][OH^-]=1\times 10^{-7}\times[OH^-]= 1\times 10^{-14}$

$[OH^-]=1\times 10^{-7}mol/L$

Hence, Concentration of $[H^+]\&[OH^-]$are equal in pure water