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
