Problem 100 - Chemistry: The Central Science

Question

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{\mathrm{s}}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\), and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the "Chemistry and Life" box on page 755, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}_{\text {k }}\) whose \(K_{p}=1.0 \times 10^{-40}\), (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Step-by-Step Solution

Answer

(a) The solubility-constant expressions are: 1. Hydroxyapatite: \(K_{sp}=[\mathrm{Ca}^{2+}]^5[\mathrm{PO}_4^{3-}]^3[\mathrm{OH}^-]\) 2. Fluoroapatite: \(K_{sp}=[\mathrm{Ca}^{2+}]^5[\mathrm{PO}_4^{3-}]^3[\mathrm{F}^-]\) (b) The molar solubility of each compound is: 1. Hydroxyapatite: \(1.08\times10^{-6}\,\text{mol/L}\) 2. Fluoroapatite: \(2.05\times10^{-8}\,\text{mol/L}\)

Step 1: (a) Write the expression for the solubility-constant for hydroxyapatite and fluoroapatite

: Let's first write the balanced chemical equations for the dissolution of each compound in water: 1. Hydroxyapatite: \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH}\rightleftharpoons5\mathrm{Ca}^{2+}+3\mathrm{PO}_4^{3-}+\mathrm{OH}^-\) 2. Fluoroapatite: \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{F}\rightleftharpoons5\mathrm{Ca}^{2+}+3\mathrm{PO}_4^{3-}+\mathrm{F}^-\) Now let's write the solubility constant expressions for each compound using their respective balanced chemical equations: 1. Hydroxyapatite: \(K_{sp}=[\mathrm{Ca}^{2+}]^5[\mathrm{PO}_4^{3-}]^3[\mathrm{OH}^-]\) 2. Fluoroapatite: \(K_{sp}=[\mathrm{Ca}^{2+}]^5[\mathrm{PO}_4^{3-}]^3[\mathrm{F}^-]\) We have now written the expressions for the solubility constants of hydroxyapatite and fluoroapatite.

Step 2: (b) Calculate the molar solubility of each compound

: 1. Hydroxyapatite: Let the common molar solubility of hydroxyapatite be \(s\) mol/L. Then, the concentrations of each ion are: \([\mathrm{Ca}^{2+}]=5s\) \([\mathrm{PO}_4^{3-}]=3s\) \([\mathrm{OH}^-]=s\) Substituting these values into the solubility constant expression: \(K_{sp} = [5s]^5[3s]^3[s]\) We are given that \(K_{sp}=6.8\times10^{-27}\), so we can solve for \(s\): \(6.8\times10^{-27}=(5s)^5(3s)^3(s)\) Solving for s, we get: \(s\approx1.08\times10^{-6}\,\text{mol/L}\) The molar solubility of hydroxyapatite is approximately \(1.08\times10^{-6}\) mol/L. 2. Fluoroapatite: Let the common molar solubility of fluoroapatite be \(s'\) mol/L. Then, the concentrations of each ion are: \([\mathrm{Ca}^{2+}]=5s'\) \([\mathrm{PO}_4^{3-}]=3s'\) \([\mathrm{F}^-]=s'\) Substituting these values into the solubility constant expression: \(K_{sp}=[5s']^5[3s']^3[s']\) We are given that \(K_{sp}=1.0\times10^{-40}\), so we can solve for \(s'\): \(1.0\times10^{-40}=(5s')^5(3s')^3(s')\) Solving for s', we get: \(s'\approx2.05\times10^{-8}\,\text{mol/L}\) The molar solubility of fluoroapatite is approximately \(2.05\times10^{-8}\) mol/L.

Key Concepts

Hydroxyapatite

Hydroxyapatite is a naturally occurring mineral form of calcium apatite with the formula \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH}\). It is the main component of tooth enamel and bone, providing both rigidity and structure. The solubility product constant (\(K_{sp}\)) for hydroxyapatite is incredibly low, \(6.8 \times 10^{-27}\), indicating its lack of solubility in water.
This low solubility is crucial for structural integrity in biological materials, such as our bones and teeth. In water, hydroxyapatite dissolves into calcium ions, phosphate ions, and hydroxide ions:

• \([\mathrm{Ca}^{2+}] = 5s\)
• \([\mathrm{PO}_4^{3-}] = 3s\)
• \([\mathrm{OH}^-] = s\)

The equilibrium expression for its solubility product is given by:
\[ K_{sp} = [\mathrm{Ca}^{2+}]^5 [\mathrm{PO}_4^{3-}]^3 [\mathrm{OH}^-] \]
This expression helps in determining the molar solubility, \(s\), of hydroxyapatite in various conditions.

Fluoroapatite

Fluoroapatite is another form of calcium apatite but with fluoride replacing the hydroxide group. Its chemical formula is \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{F}\). Unlike hydroxyapatite, fluoroapatite is even less soluble, with a solubility product constant (\(K_{sp}\)) of \(1.0 \times 10^{-40}\).
This extremely low solubility means fluoroapatite is highly resistant to dissolution, which makes it a preferred mineral for strengthening tooth enamel. This mineral forms when fluoride reacts with hydroxyapatite, contributing to enhanced dental health by making teeth more resistant to acidic conditions and decay.
Like hydroxyapatite, fluoroapatite dissolves into calcium ions, phosphate ions, and fluoride ions. These concentrations in solution are:

• \([\mathrm{Ca}^{2+}] = 5s'\)
• \([\mathrm{PO}_4^{3-}] = 3s'\)
• \([\mathrm{F}^-] = s'\)

The solubility constant expression for fluoroapatite is:
\[ K_{sp} = [\mathrm{Ca}^{2+}]^5 [\mathrm{PO}_4^{3-}]^3 [\mathrm{F}^-] \]
This helps in calculating the molar solubility, \(s'\), and understanding the conditions affecting fluoride incorporation in dental prophylaxis.

Molar Solubility

Molar solubility refers to the number of moles of a compound that can dissolve per liter of solution to reach saturation. For slightly soluble salts like hydroxyapatite and fluoroapatite, the concept of molar solubility is pivotal in predicting how much of these minerals can dissolve in biological systems.
The molar solubility \(s\) for hydroxyapatite, calculated from its solubility product \(K_{sp} = 6.8 \times 10^{-27}\), is approximately \(1.08 \times 10^{-6}\) mol/L. This reflects the minimal dissolution required for maintaining the integrity of tooth enamel and bone.
In contrast, fluoroapatite has a much lower molar solubility of \(2.05 \times 10^{-8}\) mol/L, derived from its \(K_{sp} = 1.0 \times 10^{-40}\). A lower molar solubility suggests a stronger and more resistant structure, explaining why fluoride treatments are beneficial for dental health.
These values highlight how a small change in solubility can significantly affect biological properties and the chemical stability of these minerals.

Dissolution Reactions

Dissolution reactions describe the process where a solid dissolves in a solvent, dissociating into its constituent ions. For hydroxyapatite and fluoroapatite, these reactions are crucial in understanding their roles in biological systems.
The dissolution of hydroxyapatite in water can be expressed as:
\[ \mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH} \rightleftharpoons 5\mathrm{Ca}^{2+} + 3\mathrm{PO}_4^{3-} + \mathrm{OH}^- \]
This equilibrium shows how hydroxyapatite dissolves into calcium, phosphate, and hydroxide ions, albeit minimally due to its low solubility.
Comparatively, the dissolution of fluoroapatite is:
\[ \mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{F} \rightleftharpoons 5\mathrm{Ca}^{2+} + 3\mathrm{PO}_4^{3-} + \mathrm{F}^- \]
Given the very low \(K_{sp}\) of fluoroapatite, this reaction proceeds to an even lesser extent, resulting in very few ions in solution.
Understanding these reactions helps in predicting the stability and effects of these minerals in applications like toothpaste, where the dissolution of fluoride ions can enhance enamel strength and resistance to decay.