Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the Chemistry and Life box on page 790 , fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\), whose $K_{s p}=1.0 \times 10^{-60}$ (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Hydroxyapatite and fluoroapatite dissolve in water according to the following reactions: Hydroxyapatite: \(Ca_{5}\left(PO_{4}\right)_{3}OH_{(s)} \rightarrow 5Ca^{2+}_{(aq)} + 3PO_{4}^{3-}_{(aq)} + OH^-_{(aq)}\) Fluoroapatite: \(Ca_{5}\left(PO_{4}\right)_{3}F_{(s)} \rightarrow 5Ca^{2+}_{(aq)} + 3PO_{4}^{3-}_{(aq)} + F^-_{(aq)}\) Now let's write the expressions for the solubility constants (Ksp) for each compound.

For hydroxyapatite, Ksp is expressed as: \[K_{sp,hydroxyapatite} = [Ca^{2+}]^5[PO_4^{3-}]^3[OH^-]\] For fluoroapatite, Ksp is expressed as: \[K_{sp,fluoroapatite} = [Ca^{2+}]^5[PO_4^{3-}]^3[F^-]\]

Let the molar solubility of hydroxyapatite be x, then for every mole of hydroxyapatite dissolved: \([Ca^{2+}] = 5x\), \([PO_4^{3-}] = 3x\), \([OH^-] = x\) Substituting these values in the expression for Ksp of hydroxyapatite: \(K_{sp,hydroxyapatite} = (5x)^5 \cdot (3x)^3 \cdot x\) Plug the given value of Ksp, \(6.8 \times 10^{-27}\): \(6.8 \times 10^{-27} = (5x)^5 \cdot (3x)^3 \cdot x\) Now, let the molar solubility of fluoroapatite be y, then for every mole of fluoroapatite dissolved: \([Ca^{2+}] = 5y\), \([PO_4^{3-}] = 3y\), \([F^-] = y)\) Substituting these values in the expression for Ksp of fluoroapatite: \(K_{sp,fluoroapatite} = (5y)^5 \cdot (3y)^3 \cdot y\) Plug the given value of \(Ksp, 1.0 \times 10^{-60}\): \(1.0 \times 10^{-60} = (5y)^5 \cdot (3y)^3 \cdot y\)

Now, we have two equations: \(6.8 \times 10^{-27} = (5x)^5 \cdot (3x)^3 \cdot x\) for hydroxyapatite \(1.0 \times 10^{-60} = (5y)^5 \cdot (3y)^3 \cdot y\) for fluoroapatite We will now solve for x (the molar solubility of hydroxyapatite) and y (the molar solubility of fluoroapatite). We can solve these equations using a calculator: For hydroxyapatite: \(x \approx 1.18 \times 10^{-5}\) mol/L For fluoroapatite: \(y \approx 1.03 \times 10^{-12}\) mol/L So, the molar solubility of hydroxyapatite is approximately \(1.18 \times 10^{-5}\) mol/L and the molar solubility of fluoroapatite is approximately \(1.03 \times 10^{-12}\) mol/L.