Answer
$[H_3O^+] = 2.818 \times 10^{- 9}M$
$[OH^-] = 3.548 \times 10^{- 6}M$
Work Step by Step
1. Find the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 8.55}$
$[H_3O^+] = 2.818 \times 10^{- 9}M$
2. Now, use $K_w$ to find $[OH^-]$
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 2.818 \times 10^{- 9} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 2.818 \times 10^{- 9}}$
$[OH^-] = 3.548 \times 10^{- 6}M$
