What is [H_3O^+] in an aqueous solution whose pOH=8.26?

May 6, 2017

By definition $p O H = - {\log}_{10} \left[H {O}^{-}\right]$..........and

Explanation:

And also, we KNOW that in aqueous solution,

$p H + p O H = 14$

$a .$ And thus, for $p O H = 8.26$, $\left[H {O}^{-}\right] = {10}^{- 8.26} \cdot m o l \cdot {L}^{-} 1$

$= 5 \times {10}^{-} 9 \cdot m o l \cdot {L}^{-} 1$

And thus for $p H = 5.74$,............................................

$\left[{H}_{3} {O}^{+}\right] = {10}^{- 5.74} \cdot m o l \cdot {L}^{-} 1 = 1.82 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$.

I will let you do $b .$, and $c .$. Post the values back in thread if you want them checked.