Problem 100 - Chemistry The Central Science

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 \(K_{\mathrm{sp}}=6.8 \times 10^{-27}\) . As discussed in the Chemistry and Life box on page \(746,\) 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 expression for hydroxyapatite is given by: \[K_{sp_{HA}} = [\mathrm{Ca^{2+}}]^5[\mathrm{PO_4^{3-}}]^3[\mathrm{OH^-}]\] The solubility constant expression for fluoroapatite is given by: \[K_{sp_{FA}} = [\mathrm{Ca^{2+}}]^5[\mathrm{PO_4^{3-}}]^3[\mathrm{F^-}]\] (b) The molar solubilities of hydroxyapatite and fluoroapatite are approximately \(2.76 \times 10^{-7}\) mol/L and \(2.05 \times 10^{-20}\) mol/L, respectively.

Step 1: Hydroxyapatite solubility constant expression

The formula for hydroxyapatite is given as \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\). Dissolving in water, the reaction can be written as: \[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(s) \rightleftharpoons 5\mathrm{Ca^{2+}}(aq) + 3\mathrm{PO_4^{3-}}(aq) + \mathrm{OH^-}(aq)\] The solubility constant expression for hydroxyapatite is given by: \[K_{sp_{HA}} = [\mathrm{Ca^{2+}}]^5[\mathrm{PO_4^{3-}}]^3[\mathrm{OH^-}]\]

Step 2: Fluoroapatite solubility constant expression

The formula for fluoroapatite is given as \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\). Dissolving in water, the reaction can be written as: \[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}(s) \rightleftharpoons 5\mathrm{Ca^{2+}}(aq) + 3\mathrm{PO_4^{3-}}(aq) + \mathrm{F^-}(aq)\] The solubility constant expression for fluoroapatite is given by: \[K_{sp_{FA}} = [\mathrm{Ca^{2+}}]^5[\mathrm{PO_4^{3-}}]^3[\mathrm{F^-}]\] (b) Calculating the molar solubility of hydroxyapatite and fluoroapatite

Step 3: Molar solubility of hydroxyapatite

Let the molar solubility (concentration of dissolved ions) of hydroxyapatite be \(x\). Then, the concentrations of the ions in the saturated solution are: \[[\mathrm{Ca^{2+}}] = 5x\] \[[\mathrm{PO_4^{3-}}] = 3x\] \[[\mathrm{OH^-}] = x\] We can now substitute these values into the solubility constant expression: \[K_{sp_{HA}} = (5x)^5(3x)^3x\] We are given the value of \(K_{sp_{HA}} = 6.8 \times 10^{-27}\). Solving for \(x\): \[6.8 \times 10^{-27} = (5x)^5(3x)^3x\] We can solve this equation to find the molar solubility of hydroxyapatite, \(x\), which is approximately \(2.76 \times 10^{-7}\) mol/L.

Step 4: Molar solubility of fluoroapatite

Similarly, let the molar solubility of fluoroapatite be \(y\). The concentrations of the ions in the saturated solution are: \[[\mathrm{Ca^{2+}}] = 5y\] \[[\mathrm{PO_4^{3-}}] = 3y\] \[[\mathrm{F^-}] = y\] We can substitute these values into the solubility constant expression: \[K_{sp_{FA}} = (5y)^5(3y)^3y\] We are given the value of \(K_{sp_{FA}} = 1.0 \times 10^{-60}\). Solving for \(y\): \[1.0 \times 10^{-60} = (5y)^5(3y)^3y\] We can solve this equation to find the molar solubility of fluoroapatite, \(y\), which is approximately \(2.05 \times 10^{-20}\) mol/L. Thus, the molar solubilities of hydroxyapatite and fluoroapatite are approximately \(2.76 \times 10^{-7}\) mol/L and \(2.05 \times 10^{-20}\) mol/L, respectively.

Key Concepts

Hydroxyapatite

Hydroxyapatite is a naturally occurring mineral form of calcium apatite, which is highly significant in the biological realm, especially for the human body. Its chemical formula is represented as \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\). This compound forms the primary component of bone mineral and the tooth enamel, providing them with strength and rigidity due to its crystalline structure.
When hydroxyapatite dissolves in water, it breaks down into its respective ions:

• 5 calcium ions \(\mathrm{Ca^{2+}}\)
• 3 phosphate ions \(\mathrm{PO_4^{3-}}\)
• 1 hydroxide ion \(\mathrm{OH^-}\)

The presence of these ions in a saturated solution is then used to define its solubility constant, called the solubility product \(K_{sp}\). The expression for the solubility constant of hydroxyapatite is: \[K_{sp_{HA}} = [\mathrm{Ca^{2+}}]^5[\mathrm{PO_4^{3-}}]^3[\mathrm{OH^-}]\]These ions collectively determine how easily hydroxyapatite can dissolve, which is crucial for various natural processes, including remineralization of tooth enamel.

Fluoroapatite

Fluoroapatite is another mineral that is closely related to hydroxyapatite, differing primarily by the presence of fluoride instead of a hydroxide ion. Its chemical formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\). This difference renders fluoroapatite even less soluble than hydroxyapatite, as indicated by its much smaller solubility product \(K_{sp}\) of \(1.0 \times 10^{-60}\).
The dissolution of fluoroapatite in water results in:

• 5 calcium ions \(\mathrm{Ca^{2+}}\)
• 3 phosphate ions \(\mathrm{PO_4^{3-}}\)
• 1 fluoride ion \(\mathrm{F^-}\)

The solubility constant expression for this mineral is:\[K_{sp_{FA}} = [\mathrm{Ca^{2+}}]^5[\mathrm{PO_4^{3-}}]^3[\mathrm{F^-}]\]The addition of fluoride ions leads to a structure that is more resistant to dissolution, hence fluoroapatite's important role in dental health, particularly through the use of fluoridated water and toothpaste.

Molar Solubility

Molar solubility is an important concept in chemistry that refers to the number of moles of a solute that can dissolve per liter of solution until the solution reaches saturation. This is directly linked to the solubility product \(K_{sp}\), which provides insights into how soluble a particular compound is.
For hydroxyapatite, the molar solubility \(x\) is estimated from the equation:\[6.8 \times 10^{-27} = (5x)^5(3x)^3x\]Solving this equation gives a molar solubility of approximately \(2.76 \times 10^{-7}\) mol/L, indicating that just a small amount can dissolve in water.
For fluoroapatite, the molar solubility \(y\) follows a similar pattern:\[1.0 \times 10^{-60} = (5y)^5(3y)^3y\]Resulting in a very small molar solubility of roughly \(2.05 \times 10^{-20}\) mol/L. This demonstrates why fluoroapatite is significantly more resistant to dissolution than hydroxyapatite, making it valuable for dental applications.