Hydroxylapatite, Ca10(PO4)6(OH)2, found in teeth and bones within the human body

Question

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)?

Explanation / Answer

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

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