Question

what is the solubility of calcium phosphate in 0.1M calcium chloride?

Answer 1

given the concentration of CaCl2 = 0.1 M

Since CaCl2 is completely soluble in water, [Ca²⁺(aq)] = 0.1 M

The dissociation reaction of Ca₃PO₄(s) is

--------- Ca₃(PO₄)₂(s) ---------- > 3 Ca²⁺(aq) + 2PO₄^3-(aq) ; Ksp = 1x10^-26

eqm.conc(M):------------------------- 0.1 M, ------- 2S

Ksp = 1x10^-26 = [Ca²⁺(aq)]³x[PO₄^3-(aq)]²

=> Ksp = 1x10^-26 = (0.1)³x(2S)²  = 4x10^-3xS²  

=> S = squareroot(2.5x10^-24) = 1.58x10^-12 M (answer)

Hence solubility of calcium phosphate in 0.1M calcium chloride is 1.58x10^-12 M