Question

Teeth can be protected from decay by chemical treatment with a dilute solution of fluoride ion, which makes the enamel more resistant to attack by acid. Fluoride functions both by increasing the rate at which enamel remineralizes and by causing the partial conversion of hydroxyapatite to fluorapatite through exchange of \(\mathrm{F}\) for \(\mathrm{OH}^{-}\) in healthy enamel. $$\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{~F})(s)+\mathrm{OH}^{-}(a q)$$ Use the \(K_{\mathrm{sp}}\) values provided to calculate the molar solubility of each of these compounds. $$ \begin{array}{r} \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s) \rightleftharpoons 5 \mathrm{Ca}^{2+}(a q)+3 \mathrm{PO}_{4}{ }^{3-}(a q)+\mathrm{OH}^{-}(a q) \\ K_{\mathrm{sp}}=2.3 \times 10^{-59} \\ \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{~F})(s) \rightleftharpoons 5 \mathrm{Ca}^{2+}(a q)+3 \mathrm{PO}_{4}^{3-}(a q)+\mathrm{F}^{-}(a q) \\ K_{\text {sp }}=3.2 \times 10^{-60} \end{array} $$

Short answer

The solubility of hydroxyapatite is higher than that of fluorapatite, indicating fluorapatite is less soluble and more stable.

Step-by-step solution

Understand Chemical Reaction

The chemical reaction given is the conversion of hydroxyapatite to fluorapatite by exchange of \( \mathrm{F}^{-} \) for \( \mathrm{OH}^{-} \). This reaction shows how \( \mathrm{F}^{-} \) helps in making the enamel more resistant to acids.

Write Dissolution Equation

For hydroxyapatite: \( \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s) \rightleftharpoons 5 \mathrm{Ca}^{2+}(aq) + 3 \mathrm{PO}_{4}^{3-}(aq) + \mathrm{OH}^{-}(aq) \). For fluorapatite: \( \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{~F})(s) \rightleftharpoons 5 \mathrm{Ca}^{2+}(aq) + 3 \mathrm{PO}_{4}^{3-}(aq) + \mathrm{F}^{-}(aq) \).

Establish Solubility Product Expression

The solubility product \( K_{\mathrm{sp}} \) is the product of the ion concentrations at equilibrium. For hydroxyapatite, \( K_{\mathrm{sp}} = [\mathrm{Ca}^{2+}]^5 [\mathrm{PO}_{4}^{3-}]^3 [\mathrm{OH}^{-}] \). For fluorapatite, \( K_{\mathrm{sp}} = [\mathrm{Ca}^{2+}]^5 [\mathrm{PO}_{4}^{3-}]^3 [\mathrm{F}^{-}] \).

Define Molar Solubility (s) in Terms of Ion Concentrations

Assume the molar solubility of hydroxyapatite is \( s \), then \( [\mathrm{Ca}^{2+}] = 5s, [\mathrm{PO}_{4}^{3-}] = 3s, [\mathrm{OH}^{-}] = s \). Similarly, for fluorapatite, \( s' \), \( [\mathrm{Ca}^{2+}] = 5s', [\mathrm{PO}_{4}^{3-}] = 3s', [\mathrm{F}^{-}] = s' \).

Calculate Hydroxyapatite Solubility (s)

Substitute into the \( K_{\mathrm{sp}} \) expression: \( 2.3 \times 10^{-59} = (5s)^5 (3s)^3 (s) \). Simplify and solve for \( s \): \( s = (2.3 \times 10^{-59} / (5^5 \times 3^3))^{1/9} \). Calculate \( s \).

Calculate Fluorapatite Solubility (s')

Substitute into the \( K_{\mathrm{sp}} \) expression: \( 3.2 \times 10^{-60} = (5s')^5 (3s')^3 (s') \). Simplify and solve for \( s' \): \( s' = (3.2 \times 10^{-60} / (5^5 \times 3^3))^{1/9} \). Calculate \( s' \).

Key concepts

Fluoride Treatment

Fluoride treatment is a popular method for strengthening tooth enamel and preventing decay. It works by incorporating fluoride ions into the tooth surface, making it more resistant to acid attacks from bacterial metabolism. Fluoride treatment can be found in various dental care products, including toothpaste and mouth rinses.

Using fluoride helps in two primary ways:

• It encourages the remineralization of enamel by attracting calcium and phosphate ions back to the tooth surface.
• It partially transforms hydroxyapatite, a natural mineral found in enamel, into fluorapatite. This new compound is less soluble in acid.

Applying fluoride regularly helps to maintain a healthy balance between mineral loss and gain, enhancing overall dental health.

Enamel Remineralization

Enamel remineralization is a process that repairs the early signs of tooth decay by restoring lost minerals to the enamel layer. This natural repair process is continuously occurring in the mouth as minerals dissolve and redeposit from saliva.

Fluoride plays a critical role in this process by:

• Forming a barrier over demineralized areas, making them less vulnerable to further acid attacks.
• Enhancing the deposition of calcium and phosphate, which are essential for rebuilding enamel.

When fluoride ions are present, they decrease the solubility of the tooth enamel. This results in stronger and more acid-resistant surfaces, helping preserve teeth and prevent cavities.

Hydroxyapatite and Fluorapatite Transformation

The transformation of hydroxyapatite to fluorapatite is an essential chemical reaction for strengthening teeth. Hydroxyapatite, the primary mineral in tooth enamel, can be partially converted to fluorapatite through the exchange of fluoride ions for hydroxide ions.

This reaction is expressed as:
\[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s)+\mathrm{F}^{-}(aq) \rightleftharpoons \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{~F})(s)+\mathrm{OH}^{-}(aq)\]

This transformation is beneficial because:

• Fluorapatite is more stable and less soluble than hydroxyapatite.
• It enhances the tooth's resistance to acidic conditions, which are common in the mouth due to diet and bacterial activity.

Overall, this conversion helps in maintaining healthy, decay-resistant teeth.