Question

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

Ca10(PO4)6OH2 <======> 10Ca(2+) + 6PO4(3-) + 2OH-

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 6.90 × 10-6 M?

Answer 1

if we fix [OH-] = 6.9*10^-6

then

Ca10(PO4)6OH2 <======> 10Ca(2+) + 6PO4(3-) + 2OH-

assume "S" as [Ca10(PO4)6OH2]

so

[Ca+2] = 10S

[PO4-3] = 6S

the equilibrium:

Ksp = [Ca+2]^10[PO4-3]^6[OH-]^2

substitute "S" and Ksp

2.34*10^-59= (10S)^10 (6S)^6 * (6.9*10^-6)^2

solve for S

(10^10)(S^10)(6^6)(S^6) = (2.34*10^-59) /  ((6.9*10^-6)^2)

(S^16) = (2.34*10^-59) /  ((6.9*10^-6)^2) / ( (10^10)(6^6) ) = 1.05344*10^-63

S = ( 1.05344*10^-63)^(1/16) = 0.000115

Solubility will be = 0.000115 mol/L