What is the [H_3O^+] in a solution with [OH^-] = 2.3 * 10^-12 M?

Jun 4, 2016

The ion product $\left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right] = {10}^{- 14}$ at $1$ $a t m$ and $298 \cdot K$.

Explanation:

Thus $\left[{H}_{3} {O}^{+}\right] = \frac{\left[{10}^{-} 14\right]}{\left[H {O}^{-}\right]}$

Chemists typically use $p H$ and $p O H$, where $p H = - {\log}_{10} \left(\left[{H}_{3} {O}^{+}\right]\right)$ and $p O H = - {\log}_{10} \left(\left[H {O}^{-}\right]\right)$.

The use of logarithms allows us to use the relationship:

$p H + p O H = 14$ under standard conditions.

Under non-standard conditions, for instance at $373 \cdot K$, how do you think ${K}_{w}$ would evolve?