We address the solubility equilibrium...
#Ca_3(PO_4)_2(s) rightleftharpoons3Ca^(2+) + 2PO_4^(3-)#...
...now this expression is a little unrealistic...and I will expand on this later. However, for the nonce, we will assume this reactivity.
And now...#K_"sp"=1.08xx10^-23-=[Ca^(2+)]^3[PO_4^(3-)]^2#
And if we let #S="solubility of calcium phosphate"#...then clearly #[Ca^(2+)]=3S#, and #[PO_4^(3-)]=2S#...and if we substitute these values back into the solubility expression....
#K_"sp"=1.08xx10^-23-=[Ca^(2+)]^3[PO_4^(3-)]^2=(3S)^3(2S)^2=108S^5#...
And so #S=""^5sqrt(K_"sp"/(108))=""^5sqrt((1.08xx10^-23)/(108))=1.0xx10^-5*mol*L^-1# (if I have pressed on my calculator buttons properly).
And this gives a gram solubility of #1xx10^-5*mol*L^-1xx310.3*g*mol^-1=3.1*mg*L^-1#...i.e. approx. #"3 ppm"#...
And note that in aqueous solution, the phosphate anion would probably be present as #HPO_4^(2-)#...obviously, there would be a competing equilibrium, but this requires a bit more work....
