# What is pOH for a solution that is 0.565*mol*L^-1 in HClO_4(aq)?

Apr 26, 2017

$p H + p O H = 14$

#### Explanation:

$p H + p O H = 14$ in aqueous solution under standard conditions.

$H C l {O}_{4} \left(a q\right) + {H}_{2} O \left(l\right) \rightarrow {H}_{3} {O}^{+} + C l {O}_{4}^{-}$

Now $\text{perchloric acid}$ is thus stoichiometric in ${H}_{3} {O}^{+}$, $\text{hydronium ion}$, and since $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, $p H = - {\log}_{10} \left(0.272\right) = 0.565$, and...................

$p O H = 14 - 0.565 = 13.4$