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 8.40 × 10-6 M?

Answer 1

Ca10(PO4)6(OH)2 >10Ca^2+ + 6PO4^3-+2OH^2
solid.............10x......6x......2x

Ksp = 2.34E-59 = (Ca^2+)^10(PO4^3-)^6*(OH^-)^2.

(Ca^2+)= 10x
(PO4^3-) = 6x
(OH^-) = 8.4E-6 + 2x

Substitute into Ksp expression and solve for x, then Ca^2+ = 10x.

Answer 2

Does this require the quadratic formula or is there a simpler way to solve for x?

Answer 3

I would assume 2x+8.4E-6 = 8.4E-6 and see what happens.

Answer 4

Is my step here correct mathematically?

2.34e-59 = (10x)^10(6x)^6(8.4e-6+2x)

and then I solve for x?

or is it

2.34e-59 = (10x)(6x)(8.4e-6+2x)

and then I solve for x?

Answer 5

nevermind i got it thanks for the help!

Answer 6

how did you solve it???