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 \(K_{s p}=6.8 \times 10^{-27} .\) As discussed in the "Chemistry and Life" box on page 730 , 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 \(K_{s p}=1.0 \times 10^{-60} \cdot(\mathrm{a})\) Write the expres- sion for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Short answer

The solubility-constant expressions for hydroxyapatite and fluoroapatite are: Hydroxyapatite: \(K_{sp} = [Ca^{2+}]^5 \cdot [PO_4^{3-}]^3 \cdot [OH^-] \) Fluoroapatite: \(K_{sp} = [Ca^{2+}]^5 \cdot [PO_4^{3-}]^3 \cdot [F^-] \) The molar solubility of hydroxyapatite is approximately \(1.18 \times 10^{-5} \, moles/L\) and the molar solubility of fluoroapatite is approximately \(1.32 \times 10^{-9} \, moles/L\).

Step-by-step solution

Write the solubility-constant expressions for hydroxyapatite and fluoroapatite

Let's start by writing the dissociation reactions for both compounds: Hydroxyapatite: \(Ca_5(PO_4)_3OH \longrightarrow 5Ca^{2+} + 3PO_4^{3-} + OH^-\) Fluoroapatite: \(Ca_5(PO_4)_3F \longrightarrow 5Ca^{2+} + 3PO_4^{3-} + F^-\) Now that we have the dissociation reactions, we can write the solubility-constant expressions. For hydroxyapatite: \(K_{sp} = [Ca^{2+}]^5 \cdot [PO_4^{3-}]^3 \cdot [OH^-] \) For fluoroapatite: \(K_{sp} = [Ca^{2+}]^5 \cdot [PO_4^{3-}]^3 \cdot [F^-] \)

Calculate the molar solubility of hydroxyapatite and fluoroapatite

Let the molar solubility of hydroxyapatite be \(x\) moles/L, so \([Ca^{2+}] = 5x, [PO_4^{3-}] = 3x, [OH^-] = x\) Substitute these concentrations into the \(K_{sp}\) expression for hydroxyapatite: \(6.8 \times 10^{-27} = (5x)^5 \cdot (3x)^3 \cdot x\) Now, we'll simplify the equation and solve for \(x\), the molar solubility of hydroxyapatite: \(6.8 \times 10^{-27} = (15^3)x^{14}\) \(x = 1.18 \times 10^{-5} \, moles/L\) Similarly, let the molar solubility of fluoroapatite be \(y\) moles/L, so \([Ca^{2+}] = 5y, [PO_4^{3-}] = 3y, [F^-] = y\) Substitute these concentrations into the \(K_{sp}\) expression for fluoroapatite: \(1.0 \times 10^{-60} = (5y)^5 \cdot (3y)^3 \cdot y\) Again, simplify the equation and solve for \(y\), the molar solubility of fluoroapatite: \(1.0 \times 10^{-60} = (15^3)y^{14}\) \(y = 1.32 \times 10^{-9} \, moles/L\) So, the molar solubility of hydroxyapatite is approximately \(1.18 \times 10^{-5} \, moles/L\) and the molar solubility of fluoroapatite is approximately \(1.32 \times 10^{-9} \, moles/L\).

Key concepts

Hydroxyapatite

Hydroxyapatite, with the formula \(\mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3} \mathrm{OH}\), is a naturally occurring mineral that makes up most of the human tooth enamel and bones. It's known for its hardness and stability, which is why it plays a crucial role in dental health and strength. When discussing the solubility of hydroxyapatite, it's essential to understand that despite its relative insolubility, certain conditions such as acidity can lead to its gradual dissolution.

When hydroxyapatite dissolves, it dissociates into calcium ions (\(Ca^{2+}\)), phosphate ions (\(PO_4^{3-}\)), and hydroxide ions (\(OH^-\)). The solubility-constant expression for hydroxyapatite reflects this dissolution process and is used to calculate how much of the compound can dissolve in water at equilibrium. This is important for students to understand as it ties back to the remineralization and demineralization processes that affect tooth enamel over time.

Fluoroapatite

Making Teeth More Resistant to Decay

Fluoroapatite, represented by the formula \(\mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3} \mathrm{F}\), is similar to hydroxyapatite but contains a fluoride ion (\(F^-\)) instead of the hydroxide ion. This small substitution makes fluoroapatite more resistant to acid attack, which is why fluoride is an active ingredient in many toothpaste formulas. This resistance is reflected in its significantly lower solubility-constant value (\(K_{sp}\)), meaning fluoroapatite is less soluble and more stable than hydroxyapatite.

Understanding the difference between hydroxyapatite and fluoroapatite is key for students, especially in fields related to health sciences, as it explains the preventative role of fluoride in tooth decay.

Molar Solubility

The concept of molar solubility is pivotal when discussing solubility equilibria. It refers to the number of moles of a substance that can be dissolved in one liter of solution before the solution becomes saturated. Its value, generally measured in moles per liter (mol/L), indicates how much of a compound will dissolve in water under equilibrium conditions.

Calculating the molar solubility involves using the solubility constant (\(K_{sp}\)) and the dissociation reactions of the substance. By understanding molar solubility, students gain insight into how soluble a compound is, which in turn affects various applications such as medicinal dosages and water treatment processes.

Dissociation Reactions

Breaking Down into Ions

Dissociation reactions play a fundamental role in understanding solubility and the behavior of ionic compounds in water. They describe the process by which a solid ionic compound breaks apart into its individual ions when dissolved. These reactions are represented in chemical equations showing the transformation from the solid state to the aqueous ions.

The study of dissociation reactions is essential for students, particularly in chemistry, as it affects the understanding of reaction mechanisms, solubility, pH levels, and the biological availability of minerals. By mastering dissociation reactions, one can predict the results of mixing various substances in solution, an important skill in both chemistry and environmental science.