# If the pH=12 for a solution of potassium hydroxide, what is [HO^-]?

$\left[H {O}^{-}\right]$ $=$ $0.01 \cdot m o l \cdot {L}^{-} 1$.
We know, or should know that (i) $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$; and (ii), given that K_w=[H_3O^+][""^(-)OH]=10^-14 under standard conditions, $p {K}_{w} = p H + p O H = 14.$
We have $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = 12$; thus $\left[{H}_{3} {O}^{+}\right] = {10}^{-} 12 \cdot m o l \cdot {L}^{-} 1$, and $p O H = 2$, and thus $\left[H {O}^{-}\right] = {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$.
What is $\left[{K}^{+} \left(a q\right)\right]$?