What is the pOH of a 2.1*10^-6 M solution?

Jul 2, 2016

We assume that you mean a $2.1 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$ solution of $K O H$.
$p O H = - {\log}_{10} \left[H {O}^{-}\right]$
$=$ $- {\log}_{10} \left(2.1 \times {10}^{-} 6\right)$ $=$ $- \left(- 5.68\right)$ $=$ $5.68$.
What is $p H$ of this solution? You should be able to answer this easily.