What is the pOH of an aqueous solution at 25.0 "^oC that contains 3.98 x 10^-9 M hydronium ion?

Nov 6, 2016

pOH=-log_(10)[""^(-)OH]=5.60
We know or should know that under standard conditions (the which operate here): $p H + p O H = 14$.
$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $- {\log}_{10} \left[3.98 \times {10}^{-} 9\right] = 8.40$.
And thus $p O H = 14 - 8.60 = 5.60$