# If HO^-=2.7xx10^-11*mol*L^-1, what is pOH of this solution?

$p O H$ $=$ $10.57$
$p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$.
Thus $p O H$ $=$ $- {\log}_{10} \left(2.7 \times {10}^{-} 11\right)$.