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

$[Ca(OH)_2] = 5.61\times 10^{- 5}M$
1. Calculate [OH^-]: pH + pOH = 14 10.05 + pOH = 14 pOH = 3.95 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 3.95}$ $[OH^-] = 1.122 \times 10^{- 4}$ 2. Since $Ca(OH)_2$ is a strong base with 2 OH in each molecule: $[OH^-] = 2 * [Ca(OH)_2]$ $1.122\times 10^{- 4} = 2 * [Ca(OH)_2]$ $\frac{ 1.122\times 10^{- 4}}{ 2} = [Ca(OH)_2]$ $5.61\times 10^{- 5}M = [Ca(OH)_2]$