Solubility Constant Product?

Question

if 0,5g Ca(OH)2 was put into 1 liter pure water, calculate how much Ca(OH)2 can dissolve?Ksp Ca(OH)2=3,2x10^-5

Answer 1

If s is the solubility of Ca(OH)2 (in mol/L) then according to the equilibrium:

Ca(OH)2 <==> Ca2+ + 2OH-

the concentrations of Ca2+ and OH- are: [Ca2+] = s M and [OH-] = 2s M. Now

Ksp = [Ca2+]*[OH-]^2, so

3,2*10^-5 = s * (2s)^2 or 4s^3 = 3.2*10^-5 => s^3 = 8*10^-6 =>
s = 2*10^-2 M.

Find the moles of Ca(OH)2 that dissolved:

n = s*V, n = 0.02*1, n = 0.02 mol

And find the mass of Ca(OH)2 that dissolved (molar mass of Ca(OH)2 = 74):

m = n*Mr, m = 0.02*74, m = 1.48 g.

But that's greater than 0.5 g. So all Ca(OH)2 is dissolved.