What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1×10−6).
CaSO4(S)↔Ca2+(aq)+SO2−4(aq)Ksp=[Ca2+][SO2−4]
Let the solubility of CaSO4 be s.
Then, kSP=s29.1×10−6=s2s=3.02×10−3mol/L
Molecular mass of CaSO4=136 g/mol
Solubility of in gram/L
=3.02×10−3×136=0.41g/L
This means that we need 1L of water to dissolve 0.41g of CaSO4
Therefore, to dissolve 1g of CaSO4 we require =10.41L=2.44L of water.