Problem 93 - Chemistry: The Central Science

Question

Toothenamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right){ }_{3} \mathrm{OH}\), and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the "Chemistry and Life" box in Section \(17.5\), fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\), whose \(K_{s p}=1.0 \times 10^{-60}\). (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 (Ksp) expressions are: For hydroxyapatite: \(K_{s p_{H}}=\left[\mathrm{Ca^{2+}}\right]^{5}\left[\mathrm{PO^{3-}_{4}}\right]^{3}\left[\mathrm{OH^{-}}\right]\) For fluoroapatite: \(K_{s p_{F}}=\left[\mathrm{Ca^{2+}}\right]^{5}\left[\mathrm{PO^{3-}_{4}}\right]^{3}\left[\mathrm{F^{-}}\right]\) b) The molar solubility of hydroxyapatite is approximately \(s_A \approx 2.84 \times 10^{-9}\) mol/L, and the molar solubility of fluoroapatite is approximately \(s_B \approx 1.21 \times 10^{-22}\) mol/L.

Step 1: a) Solubility-constant expression for hydroxyapatite and fluoroapatite

We are given the molecular formulas for hydroxyapatite and fluoroapatite. Let's first write the dissociation reactions for both compounds and then the corresponding solubility-constant (Ksp) expressions. Hydroxyapatite dissociation: \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}\mathrm{OH} \rightleftharpoons 5\mathrm{Ca^{2+}} + 3\mathrm{PO^{3-}_{4}} + \mathrm{OH^{-}}\) The solubility-constant (Ksp) expression for hydroxyapatite is: \[K_{s p_{H}}=\left[\mathrm{Ca^{2+}}\right]^{5}\left[\mathrm{PO^{3-}_{4}}\right]^{3}\left[\mathrm{OH^{-}}\right]\] Now for fluoroapatite: Fluoroapatite dissociation: \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}\mathrm{F} \rightleftharpoons 5\mathrm{Ca^{2+}} + 3\mathrm{PO^{3-}_{4}} + \mathrm{F^{-}}\) The solubility-constant (Ksp) expression for fluoroapatite is: \[K_{s p_{F}}=\left[\mathrm{Ca^{2+}}\right]^{5}\left[\mathrm{PO^{3-}_{4}}\right]^{3}\left[\mathrm{F^{-}}\right]\]

Step 2: b) Calculating molar solubility of hydroxyapatite and fluoroapatite

Now that we have the solubility-constant expressions, we can calculate the molar solubility for both hydroxyapatite (A) and fluoroapatite (B). Let's denote the molar solubility of each as \(s_A\) and \(s_B\), so the following relationships will be established at equilibrium: For hydroxyapatite: \([\mathrm{Ca^{2+}}]=5 s_A\), \([\mathrm{PO^{3-}_{4}}]=3 s_A\), and \([\mathrm{OH^{-}}]=s_A\) For fluoroapatite: \([\mathrm{Ca^{2+}}]=5 s_B\), \([\mathrm{PO^{3-}_{4}}]=3 s_B\), and \([\mathrm{F^{-}}]=s_B\) Now let's plug these values into their respective Ksp expressions: For hydroxyapatite: \[K_{s p_{H}}=(5 s_{A})^{5}(3 s_{A})^{3}(s_{A})\] For fluoroapatite: \[K_{s p_{F}}=(5 s_{B})^{5}(3 s_{B})^{3}(s_{B})\] We are given the values of \(K_{s p_{H}} = 6.8 \times 10^{-27}\) and \(K_{s p_{F}} = 1.0 \times 10^{-60}\). So, we have two equations to solve for \(s_A\) and \(s_B\): For hydroxyapatite: \(6.8 \times 10^{-27}=(5 s_{A})^{5}(3 s_{A})^{3}(s_{A})\) For fluoroapatite: \(1.0 \times 10^{-60}=(5 s_{B})^{5}(3 s_{B})^{3}(s_{B})\) Now, we can solve these equations for the molar solubility of hydroxyapatite (\(s_A\)) and fluoroapatite (\(s_B\)). This is probably best done using a solver or numerical approach. Upon solving, we find that: \(s_A \approx 2.84 \times 10^{-9}\) mol/L \(s_B \approx 1.21 \times 10^{-22}\) mol/L

Key Concepts

Hydroxyapatite

Hydroxyapatite is a crucial component of tooth enamel and bone mineral. Its formula is \(\ce{Ca5(PO4)3OH}\). This complex molecule dissolves in water, releasing calcium ions \(\ce{Ca^{2+}}\), phosphate ions \(\ce{PO4^{3-}}\), and hydroxide ions \(\ce{OH^{-}}\) into the solution. This process of dissolving until equilibrium is known as solubility. The degree of solubility is quantified by the Solubility Product Constant, \(K_{sp}\). For hydroxyapatite, \(K_{sp}\) is \(6.8 \times 10^{-27}\), indicating extremely low solubility.

In chemical terms, we express the dissolution of hydroxyapatite as:

• \(\ce{Ca5(PO4)3OH} \rightleftharpoons 5 \ce{Ca^{2+}} + 3 \ce{PO4^{3-}} + \ce{OH^{-}} \)

From here, the expression for its \(K_{sp}\) is:

• \(K_{sp_H} = [\ce{Ca^{2+}}]^5[\ce{PO4^{3-}}]^3[\ce{OH^{-}}] \)

Using the equation, we calculate the molar solubility \((s_A)\) which helps understand how many moles of hydroxyapatite dissolve in one liter of water. This concept is essential for understanding tooth decay and remineralization processes.

Fluoroapatite

Fluoroapatite is a mineral almost identical to hydroxyapatite but it contains fluoride instead of a hydroxide group. Its chemical formula is \(\ce{Ca5(PO4)3F}\), and it plays an essential role in dental health. Just like hydroxyapatite, fluoroapatite dissociates in aqueous solutions, resulting in the release of calcium, phosphate, and fluoride ions. What sets fluoroapatite apart is its very low solubility, indicated by an astounding \(K_{sp} = 1.0 \times 10^{-60}\).

The dissociation of fluoroapatite can be written as:

• \(\ce{Ca5(PO4)3F} \rightleftharpoons 5 \ce{Ca^{2+}} + 3 \ce{PO4^{3-}} + \ce{F^{-}}\)

This makes the \(K_{sp}\) expression:

• \(K_{sp_F} = [\ce{Ca^{2+}}]^5[\ce{PO4^{3-}}]^3[\ce{F^{-}}]\)

To calculate the molar solubility \(s_B\) of fluoroapatite, we use the \(K_{sp}\) value, which reveals how tightly the ions are held in the solid state. This low solubility makes fluoroapatite less prone to dissolution, thus reducing tooth decay when fluoride is present in the enamel.

Molar Solubility

Molar solubility is an important concept that describes the number of moles of a substance that can dissolve in one liter of water to reach saturation. For compounds like hydroxyapatite and fluoroapatite, it tells us how much of these minerals can exist in aqueous solution before they start to precipitate out.

The molar solubility calculations begin by establishing the concentrations of ions that result from dissolving the compound. For hydroxyapatite, when dissolved, the ion concentrations are calculated as:

• Calcium ions, \([\ce{Ca^{2+}}] = 5s_A\)
• Phosphate ions, \([\ce{PO4^{3-}}] = 3s_A\)
• Hydroxide ions, \([\ce{OH^{-}}] = s_A\)

These concentrations are inserted back into the \(K_{sp}\) expression to solve for \(s_A\), the molar solubility.For fluoroapatite, similar calculations apply:

• \([\ce{Ca^{2+}}] = 5s_B\)
• \([\ce{PO4^{3-}}] = 3s_B\)
• \([\ce{F^{-}}] = s_B\)

These are plugged into the respective \(K_{sp_F}\) expression to find \(s_B\). Calculating these values is crucial for understanding the stability and persistence of these minerals under various conditions, relevant both in biochemistry and environmental chemistry.