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 69.0 L of pure water?

Given:

Ksp of CaSO4 = 7.10 * 10⁻⁵

Molar mass of CaSO4 = 136.1 g/mol

Volume of the solvent = 69.0 L

To determine:

The grams of CaSO4 that would dissolve in 69.0 L of solvent

Explanation:

CaSO4 (s) ↔ Ca2 + (aq) + SO42 - (aq)

Ksp = [Ca2+][SO42-]

Let 's' be the solubility i. e. moles of CaSO4 dissolved in a given volume of water

Therefore,

Ksp = (s) (s)

s = √Ksp = √7.10 * 10⁻⁵ = 8.43 * 10⁻³ moles/lit

i. e 0.00843 moles of CaSO4 can dissolve in 1 L of water

Therefore, # moles that can dissolve in 69.0 L = 0.00843 * 69.0 = 0.582 moles of CaSO4

The corresponding mass of CaSO4 = 0.582 moles * 136.1 g/mole = 79.21 g

Ans: 79.2 g of CaSO4 can dissolve in 69.0 L of water