Answer
$[H_3O^+] = 5.888 \times 10^{- 12}$
$[OH^-] = 1.698 \times 10^{- 3}$
Work Step by Step
1. Find the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 11.23}$
$[H_3O^+] = 5.888 \times 10^{- 12}$
2. Use the $K_w$ to calculate $OH^-$:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 5.888 \times 10^{- 12} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 5.888 \times 10^{- 12}}$
$[OH^-] = 1.698 \times 10^{- 3}$
