Answer
$[H^+] = 3.98 \times 10^{-8}M$
$[OH^-] = 6.03 \times 10^{-7}M$
$pOH = 6.22$
Work Step by Step
$pH = 7.40$
$K_w = 2.4 \times 10^{-14}$
1. Find the [H+], using the pH.
$[H^+] = 10^{-pH}$
$[H^+] = 10^{-7.4} = 3.98 \times 10^{-8}$
2. Find the [OH-].
$[H^+] \times [OH^-] = K_w$
$3.98 \times 10^{-8} \times [OH^-] = 2.4 \times 10^{-14}$
$[OH^-] = \frac{2.4 \times 10^{-14}}{3.98 \times 10^{-8}}$
$[OH^-] = 6.03 \times 10^{-7}$
3. Find the pOH.
$pOH = -log[OH^-]$
$pOH = -log(6.03 \times 10^{-7})$
$pOH = 6.22 $
