Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The pH and pOH Scales - Page 743: 20

$[H_3O^+] = 3.548 \times 10^{- 8}$ $[OH^-] = 2.818 \times 10^{- 7}$

1. Calculate $[H_3O^+]$ $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 7.45}$ $[H_3O^+] = 3.548 \times 10^{- 8}$ 2. Find $[OH^-]$ $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $ 3.548 \times 10^{- 8} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 3.548 \times 10^{- 8}}$ $[OH^-] = 2.818 \times 10^{- 7}$