The Ksp of Hydroxyapatite, Ca5(PO4)3OH, is 6.8 x 10^-37. a) Show the dissociatio

Question

The Ksp of Hydroxyapatite, Ca5(PO4)3OH, is 6.8 x 10^-37.

a) Show the dissociation equation.

b) Derive the mathematical expression for Ksp.

c) Calculate the solubility of the calcium and phosphate ions in pure water in moles per liter.

d) Show by listing the three participating equations and calculating the equilibrium constant for the reaction that the addition of four equivalents of H+ heavily favors the dissolution of hydroxyapatite. (Hint: include in your calculation the Ksp, triple the inverse of Ka3, and the inverse of Kw.)

Explanation / Answer

a) Ca5(PO4)3OH <======> 5Ca (+2) + 3PO4 (-3) + OH-

b) Ksp = [Ca]^5*[PO4]^3*[OH]

See how the coefficients are used as exponents

Okay so if X amount of Hydroxyapatite
Then 5X Ca2+ forms, 3X Phosphate forms, and X OH forms

plug that into the Ksp

Ksp = [5X]^5[3X]^3[X]

Ksp = 84375 x^9

c)    Just plug and solve for X

X = 2.71E^-5

[Ca2+] = 5 x 2.71 E^-5 = 13.55 x E^-5

[PO43-] = 3 x 2.71 E^-5 = 8.13 x E^-5

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