Question

Tooth enamel is composed of hydroxyapatite, whose simplest formula is Ca5(PO4)3OH, and whose corresponding Ksp = 6.8 × 10-27.As discussed in the “Chemistry and Life” box on page 730, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite,

Ca5(PO4)3F,whose Ksp = 1.0 × 10-60.

a)Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite.

(b) Calculate the molar solubility of each of these compounds.

Answer 1

a)

Ca5(PO4)3OH(s) ↔ [Ca5(PO4)3]+(aq) + OH-(aq)

Ksp = [ [Ca5(PO4)3]+ ] [ OH-] = 6.28×10-27

Ca5(PO4)3F(s) ↔ [Ca5(PO4)3]+(aq) + F-(aq)

Ksp = [[ Ca5(PO4)3]+] [ F-] = 1.0 × 10-60

 

b)

At saturated solution of Ca5(PO4)3OH

 

Solubility of Ca5(PO4)3OH = √6.28 × 10-27

                                                = 7.92 × 10-14mol/L

 

At saturated solution of Ca5(PO4)3F

Solubility of Ca5(PO4)3F = √1.0 × 10-60

                                        = 1.0 ×10-30 mol/L

a)

Ca5(PO4)3OH(s) ↔ [Ca5(PO4)3]+(aq) + OH-(aq)

Ksp = [ [Ca5(PO4)3]+ ] [ OH-] = 6.28×10-27