Problem 100 - Chemistry The Central Science In Si Units

Question

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) and whose corresponding \(K_{\text {ip }}=6.8 \times 10^{-27}\). As discussed in the Chemistry and Life box on page 790 , 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 \(K_{y p}=1.0 \times 10^{-60}\) (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

Hydroxyapatite has a molar solubility of \(2.7 \times 10^{-6} \text{ M}\), and fluoroapatite has a molar solubility of \(9.32 \times 10^{-8} \text{ M}\).

Step 1: Write Solubility Product Expression for Hydroxyapatite

The solubility product expression for hydroxyapatite, \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH}\), is based on its dissociation in water: \[\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH} \rightleftharpoons 5\mathrm{Ca}^{2+} + 3\mathrm{PO}_4^{3-} + \mathrm{OH}^-\]. Thus, the \(K_{\text{ip}}\) expression is: \[K_{\text{ip}} = [\mathrm{Ca}^{2+}]^5 [\mathrm{PO}_4^{3-}]^3 [\mathrm{OH}^-]\] where \(K_{\text{ip}} = 6.8 \times 10^{-27}\).

Step 2: Write Solubility Product Expression for Fluoroapatite

The dissociation of fluoroapatite, \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{F}\), in water is: \[\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{F} \rightleftharpoons 5\mathrm{Ca}^{2+} + 3\mathrm{PO}_4^{3-} + \mathrm{F}^-\]. This results in the \(K_{yp}\) expression: \[K_{yp} = [\mathrm{Ca}^{2+}]^5 [\mathrm{PO}_4^{3-}]^3 [\mathrm{F}^-]\] where \(K_{yp} = 1.0 \times 10^{-60}\).

Step 3: Set up Molar Solubility for Hydroxyapatite

Let \(s\) be the molar solubility of hydroxyapatite. Therefore: \[[\mathrm{Ca}^{2+}] = 5s, \; [\mathrm{PO}_4^{3-}] = 3s, \; [\mathrm{OH}^-] = s\]. Substitute these into \(K_{\text{ip}}\) expression: \[K_{\text{ip}} = (5s)^5 (3s)^3 (s)\].

Step 4: Solve for Molar Solubility of Hydroxyapatite

Substituting into the expression gives: \[(5s)^5 \times (3s)^3 \times s = 6.8 \times 10^{-27}\]. Simplifying this gives: \[5^5 \times 3^3 \times s^{9} = 6.8 \times 10^{-27}\]. \[9375s^9 = 6.8 \times 10^{-27}\]. Solving for \(s\) gives: \[s = \left(\frac{6.8 \times 10^{-27}}{9375}\right)^{1/9} \approx 2.7 \times 10^{-6} \text{ M}\].

Step 5: Calculate Molar Solubility for Fluoroapatite

For fluoroapatite, the substitution is similar: \[[\mathrm{Ca}^{2+}] = 5s, \; [\mathrm{PO}_4^{3-}] = 3s, \; [\mathrm{F}^-] = s\]. The expression becomes: \[K_{yp} = 9375s^9 = 1.0 \times 10^{-60}\]. Solving for \(s\) gives: \[s = \left(\frac{1.0 \times 10^{-60}}{9375}\right)^{1/9} \approx 9.32 \times 10^{-8} \text{ M}\].

Key Concepts

Hydroxyapatite

Hydroxyapatite is a naturally occurring mineral form of calcium apatite. Its chemical formula is \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH} \).
It plays a crucial role as a component of bone and tooth enamel. Its structure allows it to support teeth and bones by providing strength and rigidity. Hydroxyapatite dissolves in water according to the following reaction: \[\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH} \rightleftharpoons 5\mathrm{Ca}^{2+} + 3\mathrm{PO}_4^{3-} + \mathrm{OH}^-\]To understand this, think about dissolving sugar in water, but in this case, we're dealing with much smaller molecules capable of forming ions. Solubility product (\(K_{\text{ip}}\)) describes how much of this compound can dissolve in water to form a saturated solution. The \(K_{\text{ip}}\) for hydroxyapatite is \(6.8 \times 10^{-27}\). This number is essentially a measure of the compound's tendency to dissolve: the smaller the number, the less soluble it is. Hence, hydroxyapatite is not very soluble, providing durable material for our teeth. Here's a simplification:

• 5 calcium ions (\(\mathrm{Ca}^{2+}\)) are released.
• 3 phosphate ions (\(\mathrm{PO}_4^{3-}\)) are freed.
• 1 hydroxide ion (\(\mathrm{OH}^-\)) is produced.

These contribute to the formation of the solubility product expression, which helps in calculating its molar solubility.

Fluoroapatite

Fluoroapatite, similar to hydroxyapatite, is a mineral form of calcium apatite. It is formed when fluoride (\(\mathrm{F}^-\)) ions replace the hydroxide ions in hydroxyapatite, resulting in the compound \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{F}\). This substitution results in a more stable and less soluble compound,
which contributes to the development of strong, decay-resistant teeth. The reaction for its dissolution in water is: \[\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{F} \rightleftharpoons 5\mathrm{Ca}^{2+} + 3\mathrm{PO}_4^{3-} + \mathrm{F}^-\]The solubility product (\(K_{yp}\)) for fluoroapatite is \(1.0 \times 10^{-60}\), which is significantly smaller than that of hydroxyapatite. This highlights that fluoroapatite is even less soluble, meaning that it dissolves less readily in water compared to hydroxyapatite.Here's what's happening with dissolving:

• 5 calcium ions (\(\mathrm{Ca}^{2+}\)) are created.
• 3 phosphate ions (\(\mathrm{PO}_4^{3-}\)) are released.
• 1 fluoride ion (\(\mathrm{F}^-\)) is produced.

This decreased solubility helps to fortify dental structure against decay, demonstrating why fluoride treatments fortify teeth.

Molar Solubility

Molar solubility is an important concept for understanding how much of a compound can dissolve in a solution to reach saturation. While hydroxyapatite and fluoroapatite have different solubility product constants (\(K_{\text{ip}}\) and \(K_{yp}\)), they both dissolve to produce ions in solution.To calculate molar solubility, we start by setting up expressions that represent the dissolution of each compound, using the previously formed equations based on the compound's dissociation in water:

• For hydroxyapatite: \((5s)^5 \times (3s)^3 \times s \). Here, \(s\) represents molar solubility.
• For fluoroapatite: \( (5s)^5 \times (3s)^3 \times s\).

Both these expressions can be simplified to become: 9375\(s^9\), representing the total dissolution process and reflecting the number of ions in the solution. By substituting the known \(K_{\text{ip}}\) or \(K_{yp}\) values into these equations, we can solve for \(s\), revealing the molar solubility:

• The calculated molar solubility for hydroxyapatite is approximately \(2.7 \times 10^{-6} \text{ M}\).
• For fluoroapatite, it is \(9.32 \times 10^{-8} \text{ M}\).

As a takeaway:- Hydroxyapatite has a higher molar solubility than fluoroapatite.- This means more hydroxyapatite can dissolve in solution compared to fluoroapatite. Understanding molar solubility helps assess how effective each compound will be when used in dental applications.