Answer
- $K_{sp} = 5.0 \times 10^{-4}$
The $K_{sp}$ differs from the value of the appendix, because this constant value depends on the temperature of the solution, so, this solution was probably at a different temperature.
Work Step by Step
1. Write the $K_{sp}$ expression for $Ca(OH)_2$:
- $Ca(OH)_2(aq) \lt -- \gt Ca^{2+}(aq) + 2OH^-(aq)$
$K_{sp} = [Ca^{2+}][OH^-]^2$
2. As we calculated in 122a, the molarity of this saturated solution is: $0.050M$.
$x = $ molar solubility.
$K_{sp} = x * (2x)^2$
$K_{sp} = x * 4x^2 = 4x^3$
$K_{sp} = 4 * (0.050)^3 = 4 * 1.25 \times 10^{-4} = 5.0 \times 10^{-4}$
