Question

The molar solubility of Mg3(PO4)2 is 6.26×10-6 mol/L at 25°C.

1. Calculate the solubility of Mg3(PO4)2 at 25°C. (in g/L)

2. Calculate Ksp at 25°C.

Answer 1

Mg₃(PO₄)₂         3Mg²⁺ + 2PO₄^3-

Molar solubility = 6.26×10^-6 mol/L

1) Solubility of Mg₃(PO₄)₂ = Molar solubility x Molar mass = (6.26×10^-6 mol/L) x 262.9 g/mol

                                                                                                = 1.64 x 10^-3 g/L

Solubility of Mg₃(PO₄)₂ = 1.64 x 10^-3 g/L

2) Ksp = (3s)³ x (2s)² = 108s⁵

Ksp = 108 x (6.26×10^-6)⁵ = 1.038 x 10^-24

Ksp = 1.038 x 10^-24