Question

Hydroxylapatite, Ca10(PO4)6(OH)2, has a solubility constant of Ksp = 2.34 × 10-59, and dissociates according to

Ca10(PO4)6(OH)2(s) --> 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)

Solid hydroxylapatite is dissolved in water to form a saturated solution. What is the concentration of Ca2 in this solution if [OH–] is somehow fixed at 7.30 × 10-6 M?

I am confused as to how to solve for x with such large exponents

Answer 1

Ca10(PO4)6(OH)2 ==> 10Ca^2+ + 6PO4^3- + 2OH^-
I would substitute
(Ca^2+) = 10x
(PO4^3-) = 6x
(OH^-) = 2x + 7.3E-6
Ksp = (Ca^2+)^10(PO4^3-)^6(OH^-)^2
Ksp = (10x)^10(6x)^6(2x + 7.3E-6)^2
Solve for x. You want x = (Ca^2+).

Answer 2

Thank you. I think I understand how to set it up.