How do I calculate for the solubility of Ca3(PO4)2 in Phosphoric Acid?

I know that the common ion effect will decrease the solubility of $$\ce{Ca3(PO4)2}$$ since Phosphoric acid has a same anion. However, phosphoric acid is a weak acid so I am really having a hard time even writing a net chemical equation.

• You should elaborate what you already did and SE members will help you where you have hit a snag. – Poutnik Jul 3 at 13:44
• You should consider the opposite effect, as phosphoric acid presence eliminates most of phosphate and hydrogenphosphate ions by association with hydronium ions. $$\ce{Ca3(PO4)2 + 4 H3PO4 -> 3 Ca(H2PO4)2}$$, resp. $$\ce{Ca3(PO4)2 + 4 H+ -> 3 Ca^2+ + 2 H2PO4-}$$ Solubility of calcium dihydrogenphosphate is 2g/100 mL. – Poutnik Jul 3 at 20:18

I'm quentin, I have your answer on the solubility of Ca3(PO4)2 :

Ca3(PO4)2 = 3 Ca^2+ + 2 PO4^3-

Ks = [Ca2+]3 [PO4^3-]2[Ca^2] = 3s and [PO4^2-] = 2s

Ks = s^5.3^3.2^2 = 108.s^5

10^-26 = 108.s^5 so

s = (10-26/108)1/5 = 2,5.10-6 mol.L^-1

the solubility of tricalcium phosphate is too low