Question

The \(K_{\mathrm{sp}}\) of hydroxyapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\), is \(6.8 \times 10^{-37}\). Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\), forms. The \(K_{\mathrm{sp}}\) of this substance is \(1 \times\) \(10^{-60}\). Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

Short answer

The solubility of hydroxyapatite in pure water is \(2.14 \times 10^{-7} \, mol \, L^{-1}\), and the solubility of fluorapatite is \(1.50 \times 10^{-14} \, mol \, L^{-1}\). Adding acid increases the solubility of hydroxyapatite, while treating it with fluoride forms the less soluble fluorapatite, which helps prevent tooth decay by reducing the solubility of tooth enamel. This provides a rationale for the fluoridation of drinking water.

Step-by-step solution

Write the balanced dissolution equations for hydroxyapatite and fluorapatite

The dissociation of hydroxyapatite and fluorapatite into their constituent ions can be represented by the following balanced equations, respectively: Hydroxyapatite: \[Ca_5(PO_4)_3OH(s) \rightleftharpoons 5Ca^{2+}(aq) + 3PO_4^{3-}(aq) + OH^-(aq)\] Fluorapatite: \[Ca_5(PO_4)_3F(s) \rightleftharpoons 5Ca^{2+}(aq) + 3PO_4^{3-}(aq) + F^-(aq)\]

Set up the equilibrium expressions for hydroxyapatite and fluorapatite

The \(K_{sp}\) expressions for hydroxyapatite and fluorapatite can now be set up using the balanced equations: Hydroxyapatite: \[K_{sp}= [Ca^{2+}]^5 [PO_4^{3-}]^{3} [OH^-]\] Fluorapatite: \[K_{sp}= [Ca^{2+}]^5 [PO_4^{3-}]^{3} [F^-]\]

Calculate the solubility of hydroxyapatite

Let x be the solubility of hydroxyapatite in moles/liter. When hydroxyapatite dissolves, it produces 5 moles of \(Ca^{2+}\), 3 moles of \(PO_4^{3-}\), and 1 mole of \(OH^-\) ions for every mole of hydroxyapatite that dissolves in water. Therefore, we can set up the following relationships: \[Ca^{2+} = 5x\] \[PO_4^{3-} = 3x\] \[OH^- = x\] Now plug these into the equilibrium expression for hydroxyapatite: \[6.8 \times 10^{-37} = (5x)^5 (3x)^3(x)\] Solving for x, we get: \[x = 2.14 \times 10^{-7}\] So, the solubility of hydroxyapatite in pure water is about \(2.14 \times 10^{-7}\,mol\,L^{-1}\).

Calculate the solubility of fluorapatite

Similarly, let y be the solubility of fluorapatite in moles/liter. When fluorapatite dissolves, it produces 5 moles of \(Ca^{2+}\), 3 moles of \(PO_4^{3-}\), and 1 mole of \(F^-\) ions for every mole of fluorapatite that dissolves in water. We can set up the following relationships: \[Ca^{2+} = 5y\] \[PO_4^{3-} = 3y\] \[F^- = y\] Now plug these into the equilibrium expression for fluorapatite: \[1 \times 10^{-60} = (5y)^5 (3y)^3(y)\] Solving for y, we get: \[y = 1.50 \times 10^{-14}\] So, the solubility of fluorapatite in pure water is about \(1.50 \times 10^{-14}\,mol\,L^{-1}\).

Discuss the effect of acid on solubility and rationale for fluoridation of drinking water

Adding acid to a solution containing hydroxyapatite increases the solubility of hydroxyapatite, as it reacts with the \(OH^-\) ions formed during the dissolution, driving the reaction forward, and increasing the solubility. Treating hydroxyapatite with fluoride ions converts it into fluorapatite, making it less soluble, as seen by its significantly lower solubility than hydroxyapatite. This formation of insoluble fluorapatite helps in the prevention of tooth decay, providing a rationale for the fluoridation of drinking water, as it reduces the solubility of tooth enamel and the availability of ions to form plaque and cause dental caries.

Key concepts

Hydroxyapatite Solubility

Understanding the solubility of hydroxyapatite, which has a chemical formula of \( \mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3} \mathrm{OH} \), involves looking at its solubility product constant (\( K_{\mathrm{sp}} = 6.8 \times 10^{-37} \)). This measure indicates how much hydroxyapatite can dissolve in water. When it dissolves, it breaks down into calcium ions (\( Ca^{2+} \)), phosphate ions (\( PO_4^{3-} \)), and hydroxide ions (\( OH^- \)). The balance between these ions in solution defines its solubility at a given temperature.

On a microscopic level, dissolving hydroxyapatite involves the interplay between the solid phase and its ions in solution. The equilibrium is dynamic, with some individual hydroxyapatite particles dissolving while others precipitate out of the solution at any given time. The actual amount that dissolves in pure water is quite low, owing to the low value of its solubility product, leading it to be classified as sparingly soluble.

Fluorapatite Solubility

Fluorapatite \( \mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3} \mathrm{F} \) differs from hydroxyapatite due to the substitution of a fluoride ion (\( F^- \) for a hydroxide ion. This small change significantly affects its solubility. The solubility product constant (\( K_{\mathrm{sp}} \) of fluorapatite is even lower than that of hydroxyapatite, at \(1 \times 10^{-60}\). As a result, fluorapatite is less soluble in water.

Substituting the fluoride ion makes the crystal lattice of the compound more stable, and therefore, less likely to dissolve in water. This increased stability is what makes fluorapatite a critical material in dental health, serving to strengthen tooth enamel against decay.

Equilibrium Expression

The equilibrium expressions for solubility products involve the concentrations of the ions that are formed when a substance dissolves in water. They are derived based on the stoichiometry of the dissolving reaction. For hydroxyapatite, the expression is \( K_{\mathrm{sp}} = [Ca^{2+}]^5 [PO_4^{3-}]^{3} [OH^-] \), and for fluorapatite, it is \( K_{\mathrm{sp}} = [Ca^{2+}]^5 [PO_4^{3-}]^{3} [F^-] \). These expressions allow us to calculate the solubility of a substance when we know its \( K_{\mathrm{sp}} \), helping us predict whether a precipitate will form under certain conditions.

Effect of Acid on Solubility

Acids can significantly affect the solubility of substances like hydroxyapatite. When an acid is added to water, it increases the concentration of hydrogen ions (\( H^+ \)), which react with any hydroxide ions (\( OH^- \)) in the solution. According to Le Chatelier's principle, when the \( OH^- \)) ions are consumed, the equilibrium of the dissolving reaction shifts to replace them, thereby increasing the solubility of the hydroxyapatite. For instance, in an acidic environment, typical of cariogenic (cavity-causing) conditions in the mouth, the solubility of tooth enamel increases. By understanding this concept, we see why maintaining a neutral pH in the mouth is critical for dental health.

Fluoridation of Drinking Water

The fluoridation of drinking water is a public health measure aimed at reducing dental cavities. This process involves adding a controlled amount of fluoride to the public water supply. Fluoride helps in the remineralization process of teeth, converting hydroxyapatite in the enamel to the less soluble fluorapatite. This transformation enhances the resistance of teeth to acid attacks and reduces demineralization. Fluorapatite's lower solubility means that teeth are less likely to decay because they are prone to retaining fewer cavity-causing ions. Consequently, the practice of adding fluoride to water has been instrumental in improving dental health in communities around the world.