Question

The Ksp of (hydroxyapatite ) Ca5(PO4)3OH is 6.8 * 10^-37 , and its molar concentration is 2.7 * 10^-5 mol/L ,, if this (hydroxyapatite) reacts with fluoride , then the F- ions replace OH- ions and the product is Ca5(PO4)3F ,, if we know the Ksp of ( fluoroapatite ) is 1.0 * 10^-60 , then what is solubility in mol/L of fluoroapatite in water ?

thank you a lot

Answer 1

Ca5(PO4)3F--> 5Ca^2+ + 3PO4^3- + F^-
I..solid.......0.........0.......0
C..solid.......5x......3x........x
E..solid.......5x......3x........s

Ksp = (Ca^2+)^5*(PO4^3-)^3*(F^-)
1E-60 = (5x)^5*(3x)^3*(x)
Solve for x = solubility in mols/L.