Answer
$ K_{sp} (Ca(OH)_2) = (5.47 \times 10^{-5})$
Work Step by Step
1. Calculate the hydroxide ion concentration:
pH + pOH = 14
12.68 + pOH = 14
pOH = 1.32
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 1.32}$
$[OH^-] = 4.79 \times 10^{- 2}$
2. Write the $K_{sp}$ expression:
$ Ca(OH)_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2OH^{-}(aq)$
$ K_{sp} = [Ca^{2+}]^ 1[OH^{-}]^ 2$
3. Determine the ion concentrations:
$[OH^{-}] = [Ca(OH)_2] * 2 = 0.0479$
$[Ca(OH)_2] = \frac{0.0479}{2} = 0.0239$
$[Ca^{2+}] = [Ca(OH)_2] * 1 = [0.0239] * 1 = 0.0239$
4. Calculate the $K_{sp}$:
$ K_{sp} = (0.0239)^ 1 \times (0.0479)^ 2$
$ K_{sp} = (0.0239) \times (2.29 \times 10^{-3})$
$ K_{sp} = (5.47 \times 10^{-5})$
