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

Ca3(PO4)2 <==> 3Ca^2+   + 2PO4^3-

                             3s             (2s+0.01)

(3s)^2 * (2s+0.01)^2 = 2.07*10^-33

as s is very small, 2s + 0.01 ~ 2s

so, the above equation becomes :

9s^2 * 2s^2 = 2.07*10^-33

or, 18s^4 = 2.07*10^-33

or, s = 1.072*10^-17 mol/L

          = (1.072*10^-17*310.18)g/L =3.33 *10^-15g/L = 3.33*10^-12mg/L