Answer
$[Ca(OH)_2] = 5.61\times 10^{- 5}M$
Work Step by Step
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]$
