Q19.123CP - Chemistry: Molecular Nature Of Matter And Change

Question

Tooth enamel consists of hydroxyapatite, ${\mathbf{Ca}}_{\mathbf{5}}{\left({\mathrm{PO}}_4\right)}_{\mathbf{3}}\mathbf{OH}$ ${\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{37}}$. Fluoride ion added to drinking water reacts with   ${\mathbf{\text{Ca}}}_{\mathbf{\text{5}}}{\left({\text{PO}}_{\text{4}}\right)}_{\mathbf{\text{3}}}\mathbf{\text{OH}}$ to form the more tooth decay–resistant fluoroapatite Fluoridated water has dramatically decreased cavities among children. Calculate the solubility of ${\mathbf{\text{Ca}}}_{\mathbf{\text{5}}}{\left({\text{PO}}_{\text{4}}\right)}_{\mathbf{\text{3}}}{\mathbf{\text{OH and of Ca}}}_{\mathbf{\text{5}}}{\left({\text{PO}}_{\text{4}}\right)}_{\mathbf{\text{3}}}\mathbf{\text{F}}$  in water.

Step-by-Step Solution

Answer

The solubility of hydroxyapatite and fluorapatite is given below.

 $$\begin{gathered} {\mathrm{S}}_{\mathrm{hydroxyapatite}}=2.7\times {10}^{-5}\mathrm{M} \\ {\mathrm{S}}_{\mathrm{fluoroapatite}}= 6.1\times {10}^{-8} \mathrm{M} \end{gathered}$$

Explanation of the Concept

Solubility product: A compound's solubility product equals the product of the concentrations of the ions in equilibrium, each raised to the power of its coefficient in the equilibrium equation. The product of solubility is denoted as.

Dissolution equation of the hydroxyapatite

First, we can write the hydroxyapatite dissolution equation

${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\text{OH}\left(s\right)\rightleftharpoons {\text{5Ca}}^{\text{2 + }}\left(aq\right){\text{ + 3PO}}_{\text{4}}^{\text{3 - }}\left(aq\right){\text{ + OH}}^{\text{ - }}\left(aq\right)$ 

Let the solubility of each ion be S.

Now we must calculate the solubility of each component of the equation:

 $$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}={\left(5\mathrm{S}\right)}^5\times {\left(3\mathrm{S}\right)}^3\times \mathrm{S} \\ 6.8\times {10}^{-37}=84375{\mathrm{S}}^9 \\ \mathrm{S}=2.7\times {10}^{-5}\mathrm{M} \end{gathered}$$

Dissolution equation of the fluorapatite

We can write the fluorapatite dissolution equation here

 ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\text{F}\rightleftharpoons {\text{5Ca}}^{\text{2 + }}{\text{ + 3PO}}_{\text{4}}^{\text{3 - }}{\text{ + F}}^{\text{ - }}$

Let the solubility of each ion be S.

The principle is the same as in the preceding case:

 $$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}={\left(5\mathrm{S}\right)}^5\times {\left(3\mathrm{S}\right)}^3\times \left(\mathrm{S}\right) \\ 1\times {10}^{-60}=84375{\mathrm{S}}^9 \\ \mathrm{S}=6.1\times {10}^{-8}\mathrm{M} \end{gathered}$$