# Since pure water is neutral in pH, what does it contain?

Sep 21, 2016

Shouldn't you tell us what this signifies?

#### Explanation:

Water undergoes autoprotolysis: 2H_2OrightleftharpoonsH_3O^(+) + ""^(-)OH.

This equilibrium has been carefully measured, and under standard conditions, ${K}_{w}$ $=$ [H_3O^+][""^(-)OH] $=$ ${10}^{-} 14$.

For distilled water, [H_3O^(+)] = [""^(-)OH]=10^-7*mol*L^-1.

For the relationship,[H_3O^+][""^(-)OH] $=$ ${10}^{-} 14$, we can take ${\log}_{10}$ of both sides to give:

log_10[H_3O^+]+log_10[""^(-)OH] $=$ ${\log}_{10} \left({10}^{-} 14\right)$, i.e.

log_10[H_3O^+]+log_10[""^(-)OH] $=$ $- 14$, equivalently,

-log_10[H_3O^+]-log_10[""^(-)OH] $=$ $+ 14$,

If we define $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$, then:

$p H + p O H = 14.$

And of course neutrality requires that $\left[{H}_{3} {O}^{+}\right] = \left[H {O}^{-}\right]$, and $p H = \left(p O H\right) = 7.$