## General Chemistry (4th Edition)

$[H_3O^+] = 4 \times 10^{- 8}M$ $[OH^-] = 2.5 \times 10^{- 7}M$
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: $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $4 \times 10^{- 8} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 4 \times 10^{- 8}}$ $[OH^-] = 2.5 \times 10^{- 7}M$