Answer
$ K_{sp} (Ca(OH)_2)= (5.52 \times 10^{-5})$
Work Step by Step
1. Calculate the molar mass $(Ca(OH)_2)$:
40.08* 1 + 2 * ( 16* 1 + 1.008* 1 ) = 74.096g/mol
2. Calculate the number of moles $(Ca(OH)_2)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 1.78}{ 74.096}$
$n(moles) = 0.0240$
3. Find the concentration in mol/L $(Ca(OH)_2)$:
$0.024$ mol in 1L: $0.024 M (Ca(OH)_2)$
4. Write the $K_{sp}$ expression:
$ Ca(OH)_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2OH^{-}(aq)$
$ K_{sp} = [Ca^{2+}]^ 1[OH^{-}]^ 2$
5. Determine the ions concentrations:
$[Ca^{2+}] = [Ca(OH)_2] * 1 = [0.024] * 1 = 0.024$
$[OH^{-}] = [Ca(OH)_2] * 2 = 0.048$
6. Calculate the $K_{sp}$:
$ K_{sp} = (0.024)^ 1 \times (0.048)^ 2$
$ K_{sp} = (0.024) \times (2.3 \times 10^{-3})$
$ K_{sp} = (5.52 \times 10^{-5})$
