Answer to Question #75679 in General Chemistry for Blake Bell

Question

The solubility product constant of calcium sulfate, CaSO4, is 7.10×10−5. Its molar mass is 136.1 g/mol. How many grams of calcium sulfate can dissolve in 69.5 L of pure water?

Answer 1

CaSO4(s) → Ca2+ + SO42-
So [Ca2+] = moles/L of CaSO4 that dissolves.
Ksp = [Ca2+]·[SO42-] = 7.10·10-5
Let [Ca2+] = x. Then [SO42-] = x, and (x)·(x) = 7.10·10-5.
x = 8.4·10-3 mol/L - [Ca2+]
69.5 L*(8.4·10-3 mol CaSO4 / 1 L)*136.1 g/mol CaSO4 = 79.46 g of calcium sulfate can dissolve.