# Question #f3621

May 21, 2016

$p H$ and $p O H$ depend on the concentrations of $\left[{H}_{3} {O}^{+}\right]$ or $\left[H {O}^{-}\right]$ respectively.

#### Explanation:

$p H$ $=$ $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$

$p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$

For strong acids, for which protonation of the water molecule is almost quantitative, $p H$ $=$ $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$. Likewise for strong bases, $p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$.

For water at $298$ $K$, the following equilibrium applies:

${K}_{w}$ $=$ ${10}^{-} 14$ $=$ $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$. And, after taking negative ${\log}_{10}$ of both sides:

$- {\log}_{10} \left({10}^{-} 14\right)$ $=$ $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] - {\log}_{10} \left[H {O}^{-}\right]$.

And thus,

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