Question

What is the minimum volume of water required to dissolve 1g of calcium sulfate at 298 K? (For calcium sulfate, Ksp is 9.1 × 10–6).

Answer 1

CaSO4(s) ↔ Ca2+(aq) + SO2-4(aq)
Ksp = [Ca2+][SO2-4]
Let the solubility of CaSO4 be s.
Then, Ksp = s2 = 9.1 × 10-6 = s2
s = 3.02 × 10-3 mol / L
Molecular mass of CaSO4 = 136 g/mol
Solubility of CaSO4 in gram/L = 3.02 × 10-3 × 136 = 0.41 g/L
This means that we need 1L of water to dissolve 0.41g of CaSO4 Therefore, to dissolve 1g of CaSO4 we require = $\frac{1}{0.41}$ L = 2.44L of water.