Answer
There will be precipitation.
Work Step by Step
1. Find $[OH^-]$:
$pOH = 14 - pH = 14 -8.5 = 5.5$
$[OH^-] = 10^{-pOH} = 10^{-5.5}M$
2. Find $[Co^{2+}]:$
$[Co^{2+}] = [Co(NO_3)_2] = 0.02M$
3. Find the product of the concentrations fo $(Co(OH)_2)$:
$P = [Co^{2+}][OH^-]^2$
$P = 0.02 * (10^{-5.5})^2$
$P = 2 \times 10^{-2} * 10^{-11}$
$P = 2 \times 10^{-13}$
4. Compare that product with the Ksp:
$Ksp = 1.3 \times 10^{-15}$
$P = 2 \times 10^{-13}$
* Since P has a higher value, there will be precipitation.
