# How is the pH of a solution related to the [H_3O^+]?

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$
So given a neutral solution, $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$, and since, by definition, [H_3O^+]=[""^(-)OH], i,e,. $\left[{H}_{3} {O}^{+}\right]$ $=$ ${10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$
And thus $p H$ $=$ $- {\log}_{10} \left({10}^{-} 7\right)$ $=$ $7$ (because ${\log}_{a} {a}^{b}$ $=$ $b$ by definition). You might need to review the logarithmic function.