Question

Tooth enamel is composed of the mineral hydroxyapatite. 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, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F},\) 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 approximately \(1.59 \times 10^{-8}\) moles per liter. Adding acid increases the solubility of hydroxyapatite. Fluorapatite's solubility in pure water is approximately \(8.26 \times 10^{-13}\) moles per liter, much lower than hydroxyapatite's solubility. Fluoridation of drinking water forms fluorapatite, which protects tooth enamel by making it less susceptible to dissolution in acidic environments, thereby reducing the dissolution rate of tooth enamel and helping to prevent tooth decay.

Step-by-step solution

Calculate the solubility of hydroxyapatite

Hydroxyapatite has the chemical formula \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}\mathrm{OH}\), and its dissolution in water can be represented as: \[ \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}\mathrm{OH}(s) \rightleftharpoons 5\mathrm{Ca^{2+}}(aq) + 3\mathrm{PO_{4}^{3-}}(aq) + \mathrm{OH^-}(aq) \] Let \(s\) be the solubility of hydroxyapatite in moles per liter. Then, the concentrations of the ions in the solution will be: - \([\mathrm{Ca^{2+}}] = 5s\) - \([\mathrm{PO_{4}^{3-}}] = 3s\) - \([\mathrm{OH^-}] = s\) Using the \(K_{\mathrm{sp}}\) value, we have: \[K_{\mathrm{sp}} = 6.8 \times 10^{-37} = [\mathrm{Ca^{2+}}]^{5}[\mathrm{PO_{4}^{3-}}]^3[\mathrm{OH^-}]\] Substituting the concentrations, we get: \[6.8 \times 10^{-37} = (5s)^5 (3s)^3 s\] Now, we need to solve for \(s\) (the solubility of hydroxyapatite).

Solve for s (hydroxyapatite solubility)

To find the value of s, first simplify the equation: \[6.8 \times 10^{-37} = 5^5 \times 3^3 \times s^9\] Divide both sides by \(5^5 \times 3^3\): \[s^9 = \frac{6.8 \times 10^{-37}}{5^5 \times 3^3}\] Now, take the ninth root to obtain s: \[s = \sqrt[9]{\frac{6.8 \times 10^{-37}}{5^5 \times 3^3}}\] After calculating the ninth root, we get: \[s \approx 1.59 \times 10^{-8}\] Thus, the solubility of hydroxyapatite in pure water is approximately \(1.59 \times 10^{-8}\) moles per liter.

Effect of adding acid on hydroxyapatite's solubility

Adding an acid to the solution will increase the concentration of H+ ions, which will react with OH- ions to form water. This will lead to a decrease in the concentration of OH- ions. According to Le Chatelier's principle, the equilibrium will shift to the right to produce more OH- ions, leading to an increased dissolution of hydroxyapatite. Therefore, the solubility of hydroxyapatite will increase in the presence of an acid.

Calculate the solubility of fluorapatite

For fluorapatite (\(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}\mathrm{F}\)), we have: \[ \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}\mathrm{F}(s) \rightleftharpoons 5\mathrm{Ca^{2+}}(aq) + 3\mathrm{PO_{4}^{3-}}(aq) + \mathrm{F^-}(aq) \] Let \(s'\) be the solubility of fluorapatite in moles per liter. Then, the concentrations of the ions in the solution will be: - \([\mathrm{Ca^{2+}}] = 5s'\) - \([\mathrm{PO_{4}^{3-}}] = 3s'\) - \([\mathrm{F^-}] = s'\) Using the \(K_{\mathrm{sp}}\) value for fluorapatite, we have: \[K_{\mathrm{sp}} = 1 \times 10^{-60} = [\mathrm{Ca^{2+}}]^{5}[\mathrm{PO_{4}^{3-}}]^3[\mathrm{F^-}]\] Substituting the concentrations, we get: \[1 \times 10^{-60} = (5s')^5 (3s')^3 s'\] Now, we need to solve for \(s'\) (the solubility of fluorapatite).

Solve for s' (fluorapatite solubility)

To find the value of s', first simplify the equation: \[1 \times 10^{-60} = 5^5 \times 3^3 \times s'^9\] Divide both sides by \(5^5 \times 3^3\): \[s'^9 = \frac{1 \times 10^{-60}}{5^5 \times 3^3}\] Now, take the ninth root to obtain s': \[s' = \sqrt[9]{\frac{1 \times 10^{-60}}{5^5 \times 3^3}}\] After calculating the ninth root, we get: \[s' \approx 8.26 \times 10^{-13}\] Thus, the solubility of fluorapatite in pure water is approximately \(8.26 \times 10^{-13}\) moles per liter.

Provide a rationale for fluoridation of drinking water

Comparing the solubilities of hydroxyapatite and fluorapatite, we can see that fluorapatite is much less soluble in water. This means that when fluoride is added to drinking water and forms fluorapatite, it renders the tooth enamel less susceptible to dissolution in an acidic environment. The low solubility of fluorapatite helps protect the teeth from decay by reducing the dissolution rate of tooth enamel in the presence of acids produced by bacteria in the mouth.

Key concepts

Hydroxyapatite

Hydroxyapatite, with the chemical formula \( \text{Ca}_5(\text{PO}_4)_3\text{OH} \), is a naturally occurring mineral that is a primary component of tooth enamel and bone. Its structure comprises calcium ions, \text{PO}_4 groups, and hydroxide ions. In water, hydroxyapatite can dissolve to a limited extent, determined by its solubility product constant, \( K_{\text{sp}} \).

When we consider the solubility calculation for hydroxyapatite, it provides us insight into the equilibrium between solid hydroxyapatite and its ions in aqueous solution. Since tooth enamel is mostly composed of this mineral, understanding its solubility can reveal how likely it is to demineralize in certain conditions, such as the presence of acids. Because the solubility of hydroxyapatite increases in acidic environments, it's clear why acids are a risk factor for tooth decay.

Such knowledge is not only vital for dentistry but also for creating treatments that can protect our teeth, implying a direct connection to public health measures like water fluoridation. By understanding the basic chemistry of hydroxyapatite, one can appreciate the strategies put in place to preserve our enamel strength.

Fluorapatite

Fluorapatite, similar to hydroxyapatite, is a variant mineral where the hydroxide group is replaced by a fluoride ion, resulting in the chemical formula \( \text{Ca}_5(\text{PO}_4)_3\text{F} \). Compared to hydroxyapatite, fluorapatite has a much lower solubility in water, as indicated by its lower \( K_{\text{sp}} \) value. This lower solubility signifies that fluorapatite, once formed in the tooth enamel, is less prone to being dissolved by the acids produced by oral bacteria.

Fluorapatite not only has a pivotal role in dental health but also serves as an example of how a small change in a molecule's structure can greatly affect its properties. The transition from hydroxyapatite to fluorapatite upon fluoride treatment strengthens teeth and makes them more resistant to decay. This characteristic underpins the rationale for the fluoridation of drinking water.

Ksp (Solubility Product Constant)

The Solubility Product Constant, abbreviated as \( K_{\text{sp}} \), is a fundamental concept in chemistry that quantifies the maximum amount of a compound that can dissolve in water at a given temperature. It is unique for every solid and is key in predicting the behavior of sparingly soluble salts.

The calculation of the \( K_{\text{sp}} \) involves the stoichiometric coefficients and concentration terms for the ions produced when the solid dissolves. For both hydroxyapatite and fluorapatite, this calculation requires considering the stoichiometry of the dissolution reaction to form the respective equations. Solving these equations provides the solubility in moles per liter, which can then be related to the physical context, such as dental health.

As students work through solubility calculations, they gain a deeper appreciation for \( K_{\text{sp}} \) as an indicator of solubility and its practical applications. For instance, the stark difference in the \( K_{\text{sp}} \) values for hydroxyapatite and fluorapatite is the scientific reason behind promoting fluoridation for dental protection.