In: Chemistry
Hydroxylapatite, Ca10(PO4)6(OH)2, found in teeth and bones within the human body, has a solubility constant of Ksp = 2.34 × 10–59, and dissociates according to Ca10(PO4)6(OH)2 (s) ⇄10Ca2+ (aq) + 6 PO4 3– (aq) + 2 OH– (aq) Solid hydroxylapatite is dissolved in water to form a saturated solution.
(a) What is the pH of this saturated solution?
(b) What is the solubility of Hydroxylapatitein this solutionif [OH– ] is fixed at 5.0 × 10–8 M (optimal pH of saliva)?
(c) What is the concentration of Ca2+in the solution with pH of 2.55 (pH of common drinks)?
a)
pH from OH-
so..
Ksp = [Ca+2]^10 * [PO4-3]^6 *[OH-]^2
If saturated, assume 1 mol of Ca10(PO4)6(OH)2 = S
[Ca+2] = 10*S
[PO4-3] = 6S
[OH-] = 2S
substitute
2.34*10^-59 = (10S)^10 * (6S)^6 * (2S)^2
(10^10)(6^6)(2^2) * S^18 = 2.34*10^-59
S^18 = (2.34*10^-59 ) / ((10^10)(6^6)(2^2))
S^18 = 1.2538*10^-17
S = (1.2538*10^-17)^(1/18) = 0.115083M
since
[OH-] = 2S = 2*0.115083 = 0.230166 M
pOH = -log(OH)= -log(0.230166) = 0.637
pH = 14-0.637 = 13.363
b)
solubility if w efix OH = 5*10^-8
Ksp = [Ca+2]^10 * [PO4-3]^6 *[OH-]^2
2.34*10^-59 = (10S)^10 * (6S)^6 * (5*10^-8)^2
substitute for OH
(2.34*10^-59) /( (5*10^-8)^2) = (10^10)(6^6)*S^16
(9.36*10^-45)/ ( (10^10)(6^6) ) = S^16
2.006*10^-59 = S^16
S = (2.006*10^-59)^(1/16)
S = 0.0002144 M
c)
find Ca+2 for pH = 2.55
[OH-] = 10^-(14-pH) = 10^-(14-2.55) =3.5481*10^-12
substitute similar to b
Ksp = [Ca+2]^10 * [PO4-3]^6 *[OH-]^2
2.34*10^-59 = (10S)^10 * (6S)^6 * (3.5481*10^-12)^2
(2.34*10^-59) / ((3.5481*10^-12)^2) = (10^10)(6^6) * (S^16)
(1.8587*10^-36 ) / ( (10^10)(6^6)) = (S^16)
3.9838*10^-51 = S^16
S = (3.9838*10^-51) ^(1/16)
S = 0.0007079 M