# What is the "pOH" of a solution if the ["OH"^-] = 33.5 xx 10^-2 "M"?

$\text{pOH} = 0.475$
The chemical definition of $\text{pOH}$ or in fact, the $\text{p}$ of any value/constant is given as the cologarithm of that thing. That is the negative (or reciprocal) logarithm of that thing.
$\therefore$ "pOH"=-log["OH"^-]=log(1/(["OH"^-]))
$\text{pOH} = - \log \left(3.35 \times {10}^{-} 1\right) = 0.475$