# What is the pH of a solution for which [H_3O^+]=6.31xx10^-4*mol*L^-1?

$p H = 3.20$
$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$
$= - {\log}_{10} \left(6.31 \times {10}^{-} 4\right) =$
$- {\log}_{10} \left\{6.31 \times {10}^{-} 4\right\} = - \left(- 3.20\right) = 3.20$