Answer
$[H_3O^+] = 4 \times 10^{- 8}M$
$[OH^-] = 2.5 \times 10^{- 7}M$
Work Step by Step
We use the pH to find:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 7.4}$
$[H_3O^+] = 4 \times 10^{- 8}M$
Using the concentration of $H_3O^+$, we find:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 4 \times 10^{- 8} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 4 \times 10^{- 8}}$
$[OH^-] = 2.5 \times 10^{- 7}M$
