Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) , and whose corresponding \(K_{\mathrm{sp}}=6.8 \times 10^{-27}\) . As discussed in the Chemistry and Life box on page \(746,\) 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.

The formula for hydroxyapatite is given as \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\). Dissolving in water, the reaction can be written as: \[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(s) \rightleftharpoons 5\mathrm{Ca^{2+}}(aq) + 3\mathrm{PO_4^{3-}}(aq) + \mathrm{OH^-}(aq)\] The solubility constant expression for hydroxyapatite is given by: \[K_{sp_{HA}} = [\mathrm{Ca^{2+}}]^5[\mathrm{PO_4^{3-}}]^3[\mathrm{OH^-}]\]

The formula for fluoroapatite is given as \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\). Dissolving in water, the reaction can be written as: \[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}(s) \rightleftharpoons 5\mathrm{Ca^{2+}}(aq) + 3\mathrm{PO_4^{3-}}(aq) + \mathrm{F^-}(aq)\] The solubility constant expression for fluoroapatite is given by: \[K_{sp_{FA}} = [\mathrm{Ca^{2+}}]^5[\mathrm{PO_4^{3-}}]^3[\mathrm{F^-}]\] (b) Calculating the molar solubility of hydroxyapatite and fluoroapatite

Let the molar solubility (concentration of dissolved ions) of hydroxyapatite be \(x\). Then, the concentrations of the ions in the saturated solution are: \[[\mathrm{Ca^{2+}}] = 5x\] \[[\mathrm{PO_4^{3-}}] = 3x\] \[[\mathrm{OH^-}] = x\] We can now substitute these values into the solubility constant expression: \[K_{sp_{HA}} = (5x)^5(3x)^3x\] We are given the value of \(K_{sp_{HA}} = 6.8 \times 10^{-27}\). Solving for \(x\): \[6.8 \times 10^{-27} = (5x)^5(3x)^3x\] We can solve this equation to find the molar solubility of hydroxyapatite, \(x\), which is approximately \(2.76 \times 10^{-7}\) mol/L.

Similarly, let the molar solubility of fluoroapatite be \(y\). The concentrations of the ions in the saturated solution are: \[[\mathrm{Ca^{2+}}] = 5y\] \[[\mathrm{PO_4^{3-}}] = 3y\] \[[\mathrm{F^-}] = y\] We can substitute these values into the solubility constant expression: \[K_{sp_{FA}} = (5y)^5(3y)^3y\] We are given the value of \(K_{sp_{FA}} = 1.0 \times 10^{-60}\). Solving for \(y\): \[1.0 \times 10^{-60} = (5y)^5(3y)^3y\] We can solve this equation to find the molar solubility of fluoroapatite, \(y\), which is approximately \(2.05 \times 10^{-20}\) mol/L. Thus, the molar solubilities of hydroxyapatite and fluoroapatite are approximately \(2.76 \times 10^{-7}\) mol/L and \(2.05 \times 10^{-20}\) mol/L, respectively.