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Chapter 16: Problem 83

Tooth enamel is composed of the mineral hydroxyapatite. The \(K_{\mathrm{sp}}\) of hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\( is \)6.8 \times 10^{-37}$ . Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\) , forms. The \(K_{\mathrm{sp}}\) of this substance is \(1 \times 10^{-60}\) . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

Short Answer

Expert verified
The solubility of hydroxyapatite in pure water is \(2.02 \times 10^{-7}\) mol/L. Adding acid increases its solubility by reacting with OH- ions and shifting the equilibrium. On the other hand, fluorapatite has a solubility of \(1.89 \times 10^{-8}\) mol/L, making it more resistant to dissolution than hydroxyapatite. When fluoride is added to drinking water, it forms a less soluble layer of fluorapatite on tooth enamel, making teeth more resistant to acid dissolution and reducing tooth decay. This provides the rationale for fluoridating drinking water to protect teeth.

Step by step solution

01

Write the solubility equilibrium for hydroxyapatite

Write the solubility equilibrium for hydroxyapatite (\(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}\mathrm{OH}\)) in water: \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}\mathrm{OH} \rightleftharpoons 5\mathrm{Ca^{2+}} + 3\mathrm{PO_{4}^{3-}} + \mathrm{OH^{-}}\)
02

Set up the \(K_{sp}\) equation for hydroxyapatite

Based on the coefficients in the equilibrium equation, set up a \(K_{sp}\) equation for hydroxyapatite: \(K_{sp} = [\mathrm{Ca^{2+}}]^5 [\mathrm{PO_{4}^{3-}}]^3 [\mathrm{OH^{-}}]\)
03

Calculate the solubility of hydroxyapatite in water

Let the solubility of hydroxyapatite in water be represented by 's'. Now we can rewrite the \(K_{sp}\) equation in terms of 's': \(K_{sp} = (5s)^5 (3s)^3 (s) = 15^5 \times 3^3 s^9\) Now we can solve for 's' using the given value of \(K_{sp} = 6.8 \times 10^{-37}\): \(6.8 \times 10^{-37} = 15^5 \times 3^3 s^9\) \(\Rightarrow s = \sqrt[9]{\frac{6.8 \times 10^{-37}}{15^5 \times 3^3}} = 2.02 \times 10^{-7} \ \mathrm{mol/L}\) So, the solubility of hydroxyapatite in pure water is \(2.02 \times 10^{-7}\) moles per liter.
04

Discuss the effect of adding acid on the solubility of hydroxyapatite

Adding acid to a solution containing hydroxyapatite increases the concentration of H+ ions. The presence of acid will cause the dissolution of hydroxyapatite to increase because the protons will react with the OH- ions in the hydroxyapatite, shifting the equilibrium to produce more dissolved hydroxyapatite ion species. As a result, the solubility of hydroxyapatite will increase in the presence of acid.
05

Write the solubility equilibrium for fluorapatite

Write the solubility equilibrium for fluorapatite (\(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}\mathrm{F}\)) in water: \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}\mathrm{F} \rightleftharpoons 5\mathrm{Ca^{2+}} + 3\mathrm{PO_{4}^{3-}} + \mathrm{F^{-}}\)
06

Set up the \(K_{sp}\) equation for fluorapatite

Based on the coefficients in the equilibrium equation, set up a \(K_{sp}\) equation for fluorapatite: \(K_{sp} = [\mathrm{Ca^{2+}}]^5 [\mathrm{PO_{4}^{3-}}]^3 [\mathrm{F^{-}}]\)
07

Calculate the solubility of fluorapatite in water

Let the solubility of fluorapatite in water be represented by 's'. Now we can rewrite the \(K_{sp}\) equation in terms of 's': \(K_{sp} = (5s)^5 (3s)^3 (s) = 15^5 \times 3^3 s^9\) Now we can solve for 's' using the given value of \(K_{sp} = 1 \times 10^{-60}\): \(1 \times 10^{-60} = 15^5 \times 3^3 s^9\) \(\Rightarrow s = \sqrt[9]{\frac{1 \times 10^{-60}}{15^5 \times 3^3}} = 1.89 \times 10^{-8} \ \mathrm{mol/L}\) So, the solubility of fluorapatite in pure water is \(1.89 \times 10^{-8}\) moles per liter.
08

Discuss the rationale for fluoridation of drinking water

The solubility of hydroxyapatite (\(2.02 \times 10^{-7}\) mol/L) is much higher than that of fluorapatite (\(1.89 \times 10^{-8}\) mol/L), meaning that fluorapatite is less soluble and more resistant to dissolution. When fluoride is added to drinking water, it reacts with the hydroxyapatite in tooth enamel to form the less soluble fluorapatite, making teeth more resistant to acid dissolution and, therefore, reducing the risk of tooth decay. The significant decrease in solubility between hydroxyapatite and fluorapatite is the rationale for fluoridating drinking water to protect teeth.

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Most popular questions from this chapter

Approximately 0.14 g nickel(II) hydroxide, Ni(OH) \(_{2}(s),\) dissolves per liter of water at \(20^{\circ} \mathrm{C}\) . Calculate \(K_{\text { sp }}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

The solubility of the ionic compound \(\mathrm{M}_{2} \mathrm{X}_{3},\) having a molar mass of \(288 \mathrm{g} / \mathrm{mol},\) is $3.60 \times 10^{-7} \mathrm{g} / \mathrm{L}\( . Calculate the \)K_{\mathrm{sp}}$ of the compound.

Calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) is relatively insoluble in water \(\left(K_{\mathrm{sp}}=2 \times 10^{-9}\right) .\) However, calcium oxalate is more soluble in acidic solution. How much more soluble is calcium oxalate in 0.10\(M \mathrm{H}^{+}\) than in pure water? In pure water, ignore the basic properties of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}.\)

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid–base properties. a. \(P b I_{2}, K_{s p}=1.4 \times 10^{-8}\) b. \(\operatorname{CdCO}_{3}, K_{s p}=5.2 \times 10^{-12}\) c. $\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{s p}=1 \times 10^{-31}$

For which salt in each of the following groups will the solubility depend on \(\mathrm{pH}\) ? $\begin{array}{ll}{\text { a. AgF, AgCl, AgBr }} & {\text { c. } \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}, \operatorname{Sr}\left(\mathrm{NO}_{2}\right)_{2}} \\ {\text { b. } \mathrm{Pb}(\mathrm{OH})_{2}, \mathrm{PbCl}_{2}} & {\text { d. } \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Ni}(\mathrm{CN})_{2}}\end{array}$
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