Question

What is the solubility of Ca₃(PO₄)₂ (s) in mg/L, in a solution buffered at pH 7.2 using a phosphate buffer such that TOTPO₄ = 10^-2M?

Answer 1

pH = 7.2

pH = -log[H+]

[H+] = 6.31 x 10^-8 M

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

Ksp = [Ca2+]^3.[PO4^2-]^2 = 1.3 x 10^-32

with s amount of salt in solution,

Ksp = (3s)^3.(2s)^2

2HPO4^2- <==> 2H+ + 2PO4^3-

2s = [HPO4^2-] + [PO4^3-]

2HPO4^2- <==> 2H+ + 2PO4^3-     Ka3 = 1/4.8 x 10^-13

[HPO4^2-] = (6.31 x 10^-8)^2.[PO4^3-]^2/4.8 x 10^-13 = 8.3 x 10^-3[PO4^2-]

2s = [PO4^3-] + 8.3 x 10^-3[PO4^3-]

[PO4^3-] = 2s

Feed in Ksp equation,

1.3 x 10^-32 = (3s)^3.(2s)^2

solubility of Ca3(PO4)2 = s = 1.64 x 10^-7 M

                                            = 1.64 x 10^-7 x 1000 x 310.18

                                            = 0.051 mg/L