Problem 100 - Chemistry: The Central Science in SI Units, Global Edition

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 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 \(K_{s p}=1.0 \times 10^{-60}\) (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 expressions for the solubility constants are: \[K_{sp,hydroxyapatite} = [Ca^{2+}]^5[PO_4^{3-}]^3[OH^-]\] \[K_{sp,fluoroapatite} = [Ca^{2+}]^5[PO_4^{3-}]^3[F^-]\] (b) The molar solubility of hydroxyapatite is approximately \(1.18 \times 10^{-5}\) mol/L and the molar solubility of fluoroapatite is approximately \(1.03 \times 10^{-12}\) mol/L.

Step 1: Define solubility constants for hydroxyapatite and fluoroapatite

Hydroxyapatite and fluoroapatite dissolve in water according to the following reactions: Hydroxyapatite: \(Ca_{5}\left(PO_{4}\right)_{3}OH_{(s)} \rightarrow 5Ca^{2+}_{(aq)} + 3PO_{4}^{3-}_{(aq)} + OH^-_{(aq)}\) Fluoroapatite: \(Ca_{5}\left(PO_{4}\right)_{3}F_{(s)} \rightarrow 5Ca^{2+}_{(aq)} + 3PO_{4}^{3-}_{(aq)} + F^-_{(aq)}\) Now let's write the expressions for the solubility constants (Ksp) for each compound.

Step 2: Write expressions for solubility constants

For hydroxyapatite, Ksp is expressed as: \[K_{sp,hydroxyapatite} = [Ca^{2+}]^5[PO_4^{3-}]^3[OH^-]\] For fluoroapatite, Ksp is expressed as: \[K_{sp,fluoroapatite} = [Ca^{2+}]^5[PO_4^{3-}]^3[F^-]\]

Step 3: Calculate molar solubility of each compound with given Ksp values

Let the molar solubility of hydroxyapatite be x, then for every mole of hydroxyapatite dissolved: \([Ca^{2+}] = 5x\), \([PO_4^{3-}] = 3x\), \([OH^-] = x\) Substituting these values in the expression for Ksp of hydroxyapatite: \(K_{sp,hydroxyapatite} = (5x)^5 \cdot (3x)^3 \cdot x\) Plug the given value of Ksp, \(6.8 \times 10^{-27}\): \(6.8 \times 10^{-27} = (5x)^5 \cdot (3x)^3 \cdot x\) Now, let the molar solubility of fluoroapatite be y, then for every mole of fluoroapatite dissolved: \([Ca^{2+}] = 5y\), \([PO_4^{3-}] = 3y\), \([F^-] = y)\) Substituting these values in the expression for Ksp of fluoroapatite: \(K_{sp,fluoroapatite} = (5y)^5 \cdot (3y)^3 \cdot y\) Plug the given value of \(Ksp, 1.0 \times 10^{-60}\): \(1.0 \times 10^{-60} = (5y)^5 \cdot (3y)^3 \cdot y\)

Step 4: Solve for molar solubility

Now, we have two equations: \(6.8 \times 10^{-27} = (5x)^5 \cdot (3x)^3 \cdot x\) for hydroxyapatite \(1.0 \times 10^{-60} = (5y)^5 \cdot (3y)^3 \cdot y\) for fluoroapatite We will now solve for x (the molar solubility of hydroxyapatite) and y (the molar solubility of fluoroapatite). We can solve these equations using a calculator: For hydroxyapatite: \(x \approx 1.18 \times 10^{-5}\) mol/L For fluoroapatite: \(y \approx 1.03 \times 10^{-12}\) mol/L So, the molar solubility of hydroxyapatite is approximately \(1.18 \times 10^{-5}\) mol/L and the molar solubility of fluoroapatite is approximately \(1.03 \times 10^{-12}\) mol/L.

Key Concepts

Hydroxyapatite

Hydroxyapatite is a naturally occurring mineral form of calcium apatite that is prevalent in tooth enamel and bone. Its chemical formula is \(\text{Ca}_5(\text{PO}_4)_3\text{OH}\), representing its composition of calcium, phosphate, and hydroxide ions. This compound is crucial for the strength and integrity of our natural teeth and bones, as it forms the mineralized matrix that contributes to their hardness and durability.
When hydroxyapatite dissolves in water, its dissolution can be represented by the following reaction:

• \(\text{Ca}_5(\text{PO}_4)_3\text{OH}_{(s)} \rightarrow 5\text{Ca}^{2+}_{(aq)} + 3\text{PO}_4^{3-}_{(aq)} + \text{OH}^-_{(aq)}\)

The solubility product constant, \(K_{sp}\), for hydroxyapatite is given as \(6.8 \times 10^{-27}\). This value reflects its very low solubility, which means it doesn't dissolve easily in water. This property is important because it helps to maintain the mineral integrity of teeth and bones in the physiological environment, where slight fluctuations in calcium and phosphate are crucial for metabolic processes involving bones and teeth regeneration.
Understanding the solubility of hydroxyapatite is essential, especially in dental care, for evaluating its dissolution under acidic conditions, which can contribute to demineralization and tooth decay.

Fluoroapatite

Fluoroapatite is a form of apatite where the hydroxide ion in hydroxyapatite is replaced by a fluoride ion. Its formula is \(\text{Ca}_5(\text{PO}_4)_3\text{F}\). Fluoroapatite is known to be more stable and less soluble in acidic environments compared to hydroxyapatite. This stability is one of the reasons fluoride is often found in dental products and treatments.
When fluoride ions interact with hydroxyapatite from toothpaste or fluorinated water, they form fluoroapatite, which helps protect teeth better from demineralization caused by acidic foods or bacteria.

• \(\text{Ca}_5(\text{PO}_4)_3\text{F}_{(s)} \rightarrow 5\text{Ca}^{2+}_{(aq)} + 3\text{PO}_4^{3-}_{(aq)} + \text{F}^-_{(aq)}\)

The \(K_{sp}\) for fluoroapatite is \(1.0 \times 10^{-60}\), indicating it has an even lower solubility than hydroxyapatite. This lesser solubility is beneficial for dental health, as it means that fluoroapatite is less likely to dissolve under acidic conditions, providing a more robust barrier against cavities.

• Fluoroapatite forms a protective layer over tooth enamel, enhancing resistance to acid attacks.
• Its formation decreases the solubility of tooth enamel, making it more decay-resistant.

Incorporating fluoride into dental care regimens is hence a preventative measure against tooth decay.

Solubility Product Constant

The solubility product constant, \(K_{sp}\), is a fundamental concept in the discussion of solubility equilibria. It represents the product of the ion concentrations each raised to the power of their coefficients in the balanced dissolution equation of a compound. For a slightly soluble salt dissolving in water, this expression is significant because it defines the maximum concentrations that ions can reach in a saturated solution before precipitation begins.
For hydroxyapatite, the \(K_{sp}\) expression is:

• \[K_{sp,\text{hydroxyapatite}} = [\text{Ca}^{2+}]^5[\text{PO}_4^{3-}]^3[\text{OH}^-]\]

This expression shows that the solubility depends on the concentrations of calcium, phosphate, and hydroxide ions.
For fluoroapatite, the \(K_{sp}\) is:

• \[K_{sp,\text{fluoroapatite}} = [\text{Ca}^{2+}]^5[\text{PO}_4^{3-}]^3[\text{F}^-]\]

This expression is similar, but with fluoride replacing hydroxide, showcasing how minor changes in a formula can drastically influence solubility. The comparison of the \(K_{sp}\) values between these two forms highlights why fluoroapatite is favored in dental applications—its greatly reduced solubility gives it a protective property against decay.
Understanding \(K_{sp}\) is critical in various fields, especially in chemistry and medicine, as it helps predict the solubility behavior of different ionic compounds under various conditions.