## Chemistry: The Central Science (13th Edition)

$[H^+] = 3.98 \times 10^{-8}M$ $[OH^-] = 6.03 \times 10^{-7}M$ $pOH = 6.22$
$pH = 7.40$ $K_w = 2.4 \times 10^{-14}$ 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. $pOH = -log[OH^-]$ $pOH = -log(6.03 \times 10^{-7})$ $pOH = 6.22$