Question

Show calculations for Calcium phosphate molar solubility. Ca3(PO4)2. ksp = 2.08 x 10^-33. Can you please show me how you get from one step to the next? thanks!

Answer 1

-33 In aqueous solution, dissociation of cag (PO4)2 1,8 as, ca, (PO4), (5) = 3 caz tast 2 po - Carv) By defination, solubility Product constant (Ksp) of cag (P04), (POA) will be Ksp = [Ca2+][PO]? Griven, solubility Product constant (Ksp) of Cag (POA),= 2:08 x16 Therefore, Ksp = [caz+]3[PO]? = 2.08 x 1733 Let, molari solubil bility of caz (PO4)2 = 5 mol/L Then, [cat ] = = 35 mol/L and [POJ = 25 mol/L Now, [ Ca2+ ][ POB]= 2.08x1533 → (35) x (25)2 = 2.08 x1033 +2759 x 452 633 = 2:08 X10 * 108 5 5 = 2.08x 10 33 -33 2:08 X10 → 55 = 108 → 55 = 1.926 X 1035 + 5 = (1.926 x 1035) + 5 = 1:14x107 Therefore, molari solubility of Calcium phosphate, cag (PO4), molle, 18 1:14*107