A 75.0-L solution is 0.0251 M in Ca2+. If Na2SO4 is added to the solution in order to precipitate the calcium, what minimum mass of Na2SO4 is required to get a precipitate?
Precipitation occurs when the ionic product exceeds the solubility product constant.
CaSO4 → Ca2+ + SO42-
Solubility product for CaSO4 = 7.10 x 10-5
Concentration of 75.0 L Ca2+ is 0.0251 M
Ksp = [Ca2+][SO42-]
[SO42-] = Ksp/ Ca2+]
= 7.10 x 10-5/ 0.0251 = 2.83 x 10-3 M
Thus, minimum concentration of SO42- 2.83 x 10-3 M is required to get a precipitate of CaSO4
Moles of SO42- = 2.83 x 10-3 mol/L x 75.0 L = 0.212 mol