# A solution has a [OH^-] of 1*10^-2. What is the pOH of this solution?

May 19, 2016

$p O H$ $=$ $2$.
$p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$ (i.e. just as $p H$ $=$ $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$.
Since know that when ${\log}_{a} b$ $= c$, then, by definition, ${a}^{c}$ $=$ $b$, thus $- {\log}_{10} \left(1 \times {10}^{-} 2\right)$ $=$ $- \left(- 2\right)$ $=$ $2$.
Given that in water at $298 \cdot K$, ${K}_{w}$ $=$ ${10}^{-} 14$, what is $p H$ of this solution.