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

$[OH^-] = 3 \times 10^{-3}M$ $pH \approx 11.477$
1. Because $Sr(OH)_2$ is a strong base, and 1 $Sr(OH)_2$ have 2 $OH$ : $[OH^-] = [Sr(OH)_2] \times 2$ $[OH^-] = 1.5 \times 10^{-3} \times 2$ $[OH^-] = 3 \times 10^{-3}$M 2. Find the pOH: $pOH = -log[OH^-]$ $pOH = -log(3\times 10^{-3})$ $pOH \approx 2.523$ 3. Find the pH: $pH = 14 - pOH$ $pH \approx 14 - 2.523$ $pH \approx 11.477$