## Chemistry: The Molecular Science (5th Edition)

$pH = 2.886$ $pOH = 11.114$
1. Since $HNO_3$ is a strong acid: $[H_3O^+] = [HNO_3] = 0.0013M$ 2. Calculate the pH value: $pH = -log[H_3O^+]$ $pH = -log( 1.3 \times 10^{- 3})$ $pH = 2.886$ 3. Use the pH value to calculate the pOH. $pOH + pH = 14$ $pOH = 14 - pH = 14 - 2.886$ $pOH = 11.114$