Problem 65 - Chemistry: An Atoms-Focused Approach

Question

Some sources give the formula of hydroxyapatite as \(\mathrm{Ca}_{10}\left(\mathrm{PO}_{4}\right)_{6}(\mathrm{OH})_{2} .\) If the \(K_{\mathrm{sp}}\) of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})\) is \(2.3 \times 10^{-59},\) what is the \(K_{\mathrm{sp}}\) of \(\mathrm{Ca}_{10}\left(\mathrm{PO}_{4}\right)_{6}(\mathrm{OH})_{2} ?\)

Step-by-Step Solution

Answer

Answer: The Ksp of Ca10(PO4)6(OH)2 is 5.29 x 10^-118.

Step 1: Write the dissociation equations for both compounds

We need to write the dissociation equations for both compounds: For \(\mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}(\mathrm{OH})\): \(\mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}(\mathrm{OH}) \rightleftharpoons 5\mathrm{Ca^{2+}} + 3\mathrm{PO}_{4}^{3-} + \mathrm{OH}^{-}\) For \(\mathrm{Ca}_{10}(\mathrm{PO}_{4})_{6}(\mathrm{OH})_{2}\): \(\mathrm{Ca}_{10}(\mathrm{PO}_{4})_{6}(\mathrm{OH})_{2} \rightleftharpoons 10\mathrm{Ca^{2+}} + 6\mathrm{PO}_{4}^{3-} + 2\mathrm{OH}^{-}\)

Step 2: Express the Ksp in terms of concentrations for both compounds

Given the dissociation equations, we can now express the \(K_{\mathrm{sp}}\) in terms of concentrations for both compounds: For \(\mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}(\mathrm{OH})\): \(K_{\mathrm{sp1}} = [\mathrm{Ca^{2+}}]^5 [\mathrm{PO}_{4}^{3-}]^3 [\mathrm{OH}^{-}]\) For \(\mathrm{Ca}_{10}(\mathrm{PO}_{4})_{6}(\mathrm{OH})_{2}\): \(K_{\mathrm{sp2}} = [\mathrm{Ca^{2+}}]^{10} [\mathrm{PO}_{4}^{3-}]^6 [\mathrm{OH}^{-}]^2\)

Step 3: Observe the relationship between the two \(K_{\mathrm{sp}}\) expressions

Looking at the two \(K_{\mathrm{sp}}\) expressions, we can observe that the \(K_{\mathrm{sp2}}\) is the square of the \(K_{\mathrm{sp1}}\): \(K_{\mathrm{sp2}} = (K_{\mathrm{sp1}})^2\)

Step 4: Calculate the Ksp of \(\mathrm{Ca}_{10}(\mathrm{PO}_{4})_{6}(\mathrm{OH})_{2}\)

Now that we know the relationship between the two \(K_{\mathrm{sp}}\) values, we can calculate the \(K_{\mathrm{sp}}\) of \(\mathrm{Ca}_{10}(\mathrm{PO}_{4})_{6}(\mathrm{OH})_{2}\) using the given \(K_{\mathrm{sp}}\) of \(\mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}(\mathrm{OH})\): \(K_{\mathrm{sp2}} = (K_{\mathrm{sp1}})^2 = (2.3 \times 10^{-59})^2\) \(K_{\mathrm{sp2}} = 5.29 \times 10^{-118}\) So, the \(K_{\mathrm{sp}}\) of \(\mathrm{Ca}_{10}(\mathrm{PO}_{4})_{6}(\mathrm{OH})_{2}\) is \(5.29 \times 10^{-118}\).

Key Concepts

Hydroxyapatite

Hydroxyapatite is a crucial component found predominantly in human bones and teeth. It plays a significant role in maintaining strength and structural integrity. Chemically, it is known as calcium apatite, and its formula can vary slightly between sources. It is often written as \( \mathrm{Ca}_{10}(\mathrm{PO}_{4})_{6}(\mathrm{OH})_{2} \), reflecting its composition of calcium, phosphate, and hydroxide ions.

Hydroxyapatite forms through a natural process where calcium ions bind with phosphate ions, resulting in a crystal-like structure. This compound not only provides rigidity but also enables the regulation of ionic exchanges within the biological systems. The interplay of ions such as calcium and phosphate is crucial for bone strength and dental health.

In chemistry, understanding hydroxyapatite's solubility and reactivity in aqueous environments is essential. Its dissociation into ions plays a vital role in chemical equilibrium, especially in biological fluids, where it can dissolve or reform under varying conditions.

Chemical Equilibrium

Chemical equilibrium refers to a state where the concentrations of reactants and products remain constant over time. It occurs when the rate of the forward reaction equals the rate of the reverse reaction. In the case of hydroxyapatite dissociation, equilibrium helps predict how much of the compound will dissolve in water.

At equilibrium, the system may appear static, but it is quite dynamic at the molecular level. It is essential in contexts like solubility product constants \( (K_{sp}) \), which quantify the solubility of sparingly soluble compounds. The \( K_{sp} \) indicates the level at which a compound will dissolve to form a saturated solution, where any additional amount will not dissolve.

• Equilibrium Constant \(K_{sp}\): The solubility product constant specifically refers to the equilibrium constant for a solid dissolving in water.
• Importance in Nature: Understanding these constants in biological systems, like bones, helps anticipate changes in conditions such as pH or ionic strength.

Dissociation Equations

Dissociation equations depict how a compound breaks down into its constituent ions in a solution. For hydroxyapatite, it dissociates into calcium ions \((\mathrm{Ca^{2+}})\), phosphate ions \((\mathrm{PO}_4^{3-})\), and hydroxide ions \((\mathrm{OH}^-)\).

These equations are critical for calculating the solubility equilibrium in solutions, which in turn influences the \((K_{sp})\). By knowing the stoichiometry of the dissolution process, you can derive \( K_{sp} \) expressions. Given as power products of ionic concentrations, these expressions depict how much a compound dissolves before reaching equilibrium.

• Example: \( \mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}(\mathrm{OH}) \rightleftharpoons 5\mathrm{Ca^{2+}} + 3\mathrm{PO}_{4}^{3-} + \mathrm{OH}^{-} \)
• Mathematical Representation: Each ion concentration is raised to the power of its coefficient in the dissociation equation.