Answer
No precipitation will occur.
Work Step by Step
1. Find $[OH^-]$
$pOH = 14 -pH = 14 - 8 = 6$
$[OH^-] = 10^{-pOH} = 10^{-6}$
2. Find $[Ca^{2+}]$:
$[Ca^{2+}] = [CaCl_2] = 0.05M$
3. Calculate the product of the concentrations for $Ca(OH)_2$:
$P = [Ca^{2+}][OH^-]^2$
$P = 0.05 * (10^{-6})^2$
$P = 5 \times 10 ^{-2} * 10^{-12}$
$P = 5 \times 10^{-14}$
4. Compare this product with the Ksp:
$Ksp (Ca(OH)_2)= 6.5 \times 10^{-6} $
$P = 5 \times 10^{-14}$
Since the Ksp value is higher, there will be no precipitations.
