posted by Darcy .
Ca(OH)2 has a Ksp of 6.5×10−6.
If 0.370 g of Ca(OH)2 is added to 500 mL of water and the mixture is allowed to come to equilibrium, will the solution be saturated?
You can do this two ways.
1. Determine the solubility of Ca(OH)2 from the Ksp and compare with 0.379 g.
2. Determine Qsp, using the 0.370 g, and compare with Ksp. If Qsp>or=Ksp, it is saturated. I'll show you how to do 1.
mols Ca(OH)2 in 0.370 g = 0.370/molar mass Ca(OH)2 = estimated 0.005 M.
........Ca(OH)2 ==> Ca^2+ + 2OH^-
Ksp = 6.5E-6 = (Ca^2+)(OH^-)^2
6.5E-6 = (x)(2x) = 4x^3 and I get approx
x^3 = approx 1.6E-6 or x = approx 0.012 M
Convert to mols by mols = M x L = approx 0.012M x 0.5L = 0.006 mols. Then convert to grams. g = mols x molar mass = approx 0.006 x 74 = approx 0.4 g but the solution has only 0.370 so it isn't saturated.
Post your work if you get confused.
The other method is quicker.