Tooth enamel consists of hydroxyapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) \(\left(K_{\mathrm{sp}}=6.8 \times 10^{-37}\right) .\) Fluoride ion added to drinking water reacts with \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) to form the more tooth decay-resistant fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\left(K_{\mathrm{sp}}=1.0 \times 10^{-60}\right) .\) Fluoridated water has dramatically decreased cavities among children. Calculate the solubility of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) and of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\) in water.

The solubility product constant, abbreviated as \( K_{sp} \), is a crucial concept in understanding the solubility of compounds. It measures the extent to which a compound can dissolve in water, leading to ions in solution. Essentially, \( K_{sp} \) gives us an equilibrium constant for the dissolution of a sparingly soluble salt.
For a general salt \( AB \rightarrow A^+ + B^- \), the \( K_{sp} \) expression is:
\[ K_{sp} = [A^+][B^-] \]
Here, \( [A^+] \) and \( [B^-] \) represent the molar concentrations of the ions at equilibrium. The higher the \( K_{sp} \) value, the more soluble the compound is. Conversely, a lower \( K_{sp} \) indicates lower solubility.
Understanding \( K_{sp} \) is vital, especially in contexts like dental health, where the solubility of compounds like hydroxyapatite and fluorapatite is a key factor.

Dissociation reactions describe how a compound separates into ions when it dissolves in water. For example, hydroxyapatite dissociates as:
\[ \text{Ca}_5(\text{PO}_4)_3\text{OH} \rightarrow 5\text{Ca}^{2+} + 3\text{PO}_4^{3-} + \text{OH}^- \]
This reaction shows that one unit of hydroxyapatite produces 5 calcium ions, 3 phosphate ions, and 1 hydroxide ion.
Similarly, fluorapatite dissociates as:
\[ \text{Ca}_5(\text{PO}_4)_3\text{F} \rightarrow 5\text{Ca}^{2+} + 3\text{PO}_4^{3-} + \text{F}^- \]
Dissociation reactions are important because they provide the basis for setting up the solubility product expression and calculating the solubility of different compounds. The number of each ion produced directly influences the solubility calculations later on. Understanding these reactions helps to comprehend the overall process of how substances dissolve and establish equilibrium in solutions.

Chemical equilibrium refers to the state in which both the reactants and products of a chemical reaction are present at constant concentrations. This occurs when the forward and reverse reactions proceed at equal rates.
For the dissociation of hydroxyapatite:
\[ \text{Ca}_5(\text{PO}_4)_3\text{OH} \rightleftharpoons 5\text{Ca}^{2+} + 3\text{PO}_4^{3-} + \text{OH}^- \]
At equilibrium, the concentrations of all ions remain constant over time. The solubility product constant \( K_{sp} \) represents the equilibrium condition of this dissociation process. The expression:
\[ K_{sp} = [Ca^{2+}]^5[PO_4^{3-}]^3[OH^-] \]
Quantifies this equilibrium state.
Achieving chemical equilibrium in solubility contexts ensures that we can accurately predict how much of a substance will dissolve in water. It's a crucial concept for solving solubility problems and understanding the behavior of dissolved ions in solutions.

Solubility calculations enable us to determine how much of a substance can dissolve in water. These calculations use the solubility product constant \( K_{sp} \) and the dissociation reactions.
For hydroxyapatite:
\[ 6.8 \times 10^{-37} = (5S)^5(3S)^3(S) \]
Simplify and solve for \( S \):
\[ S^9 = \frac{6.8 \times 10^{-37}}{84375} \]
\[ S = (8.06 \times 10^{-42})^{1/9} \]
\[ S \text{ for hydroxyapatite} \text{ is approximately } 1.32 \times 10^{-5} \text{ mol/L} \]
For fluorapatite:
\[ 1.0 \times 10^{-60} = (5S)^5(3S)^3(S) \]
Solve similarly:
\[ S^9 = \frac{1.0 \times 10^{-60}}{84375} \]
\[ S = (1.18 \times 10^{-65})^{1/9} \]
\[ S \text{ for fluorapatite} \text{ is approximately } 1.05 \times 10^{-7} \text{ mol/L} \]
These calculations illustrate how values of \( K_{sp} \) are used in conjunction with dissociation reactions and algebraic manipulation to find the solubility of compounds in water, helping us understand behaviors like the reduced solubility leading to increased resistance against tooth decay.