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 746, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, Ca5(PO4)3F, whose Ksp = 1.0 × 10^(-60). (b) Calculate the molar solubility of each of these compounds.
Verified step by step guidance
1
Identify the dissolution reactions for both hydroxyapatite and fluoroapatite. For hydroxyapatite, the reaction is: Ca_5(PO_4)_3OH (s) ⇌ 5Ca^{2+} (aq) + 3PO_4^{3-} (aq) + OH^{-} (aq). For fluoroapatite, the reaction is: Ca_5(PO_4)_3F (s) ⇌ 5Ca^{2+} (aq) + 3PO_4^{3-} (aq) + F^{-} (aq).
Write the expression for the solubility product constant (K_{sp}) for each compound. For hydroxyapatite: K_{sp} = [Ca^{2+}]^5 [PO_4^{3-}]^3 [OH^{-}]. For fluoroapatite: K_{sp} = [Ca^{2+}]^5 [PO_4^{3-}]^3 [F^{-}].
Assume the molar solubility of hydroxyapatite is 's'. Then, [Ca^{2+}] = 5s, [PO_4^{3-}] = 3s, and [OH^{-}] = s. Substitute these into the K_{sp} expression for hydroxyapatite and solve for 's'.
Similarly, assume the molar solubility of fluoroapatite is 's'. Then, [Ca^{2+}] = 5s, [PO_4^{3-}] = 3s, and [F^{-}] = s. Substitute these into the K_{sp} expression for fluoroapatite and solve for 's'.
Compare the molar solubilities of hydroxyapatite and fluoroapatite to understand the effect of fluoride on the solubility of tooth enamel.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Solubility Product Constant (Ksp)
The solubility product constant (Ksp) is an equilibrium constant that quantifies the solubility of a sparingly soluble ionic compound. It is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced equation. A lower Ksp value indicates a lower solubility in water, which is crucial for comparing the solubility of hydroxyapatite and fluoroapatite in this context.
Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution at equilibrium. It is directly related to Ksp, as the molar solubility can be calculated from the Ksp expression for a given compound. Understanding how to derive molar solubility from Ksp is essential for solving the problem regarding the solubility of hydroxyapatite and fluoroapatite.
Equilibrium in chemical reactions occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. In the context of solubility, this means that the dissolution of a solid and the precipitation of its ions reach a balance. Recognizing how equilibrium affects the solubility of compounds like hydroxyapatite and fluoroapatite is vital for calculating their molar solubility.
Verified step by step guidance
01:47
