Calculate the solublity for Ca_3(PO_4 )_2 given a Ksp = 0.0000120

Question

Calculate the solublity for Ca_3(PO_4 )_2 given a Ksp = 0.0000120

Answer 1

Literature values for Ksp for Ca3(PO4)2 are significantly less than that given in this problem. Usually they are on the order of 10^-29 not 10^-5. With that caveat in mind, and using the assigned Ksp value, we approach the problem as follows:

Ca₃(PO₄)₂(s) <==> 3Ca²⁺(aq) + 2PO₄^3-(aq)

Ksp = 1.20x10^-5 = [Ca²⁺]³[PO₄^3-]²

Let x = solubility of Ca₃(PO₄)₂, then [Ca²⁺] = 3x and [PO₄^3-]= 2x

Ksp = 1.20x10^-5 = [3x]³[2x]² = 108x⁵

x⁵ = 1.1x10^-7

x = 0.0407 M = solubility of Ca₃(PO₄)₂

You could also solve as follows:

Let x = [PO₄^3-] and the [Ca²⁺] = 1.5 x

Ksp = 1.20x10^-5 = (1.5x)³(x)² = 3.375x⁵

x = 0.0813 M = [PO₄^3-], therefore [Ca₂(PO₄)₃] = 1/2 x 0.0813 = 0.0407 M