Problem 62 - Chemistry: An Atoms-Focused Approach

Question

Effects of Excess Fluoridation on Teeth Too much fluoride might lead to the formation of calcium fluoride according to the reaction $$\begin{aligned}\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s)+& 10 \mathrm{F}^{-}(a q) \rightleftharpoons \\\& 5 \mathrm{CaF}_{2}(s)+3 \mathrm{PO}_{4}^{3-}(a q)+\mathrm{OH}^{-}(a q)\end{aligned}$$ Write the equilibrium constant expression for the reaction. Given the \(K_{\mathrm{sp}}\) values for the following two reactions, calculate \(K\) for the reaction between \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})\) and fluoride ion that forms \(\mathrm{CaF}_{2}\) \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s) \rightleftharpoons 5 \mathrm{Ca}^{2+}(a q)+3 \mathrm{PO}_{4}^{3-}(a q)+\mathrm{OH}^{-}(a q)\) \(K_{\mathrm{sp}}=2.3 \times 10^{-59}\) \(\mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \quad K_{\mathrm{sp}}=3.9 \times 10^{-11}\)

Step-by-Step Solution

Answer

Answer: The equilibrium constant (K) for the reaction is approximately \(4.15 \times 10^9\).

Step 1: Write the equilibrium constant expression for the main reaction

Given the main reaction: $$\begin{aligned}\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s)+& 10 \mathrm{F}^{-}(a q) \rightleftharpoons \\\& 5 \mathrm{CaF}_{2}(s)+3 \mathrm{PO}_{4}^{3-}(a q)+\mathrm{OH}^{-}(a q)\end{aligned}$$ The equilibrium constant expression for this reaction can be written as: $$K=\frac{[\mathrm{CaF}_{2}]^{5} \cdot [\mathrm{PO}_{4}^{3-}]^{3} \cdot [\mathrm{OH}^{-}]}{[\mathrm{F}^{-}]^{10}}$$

Step 2: Use given Ksp values to find the equilibrium concentrations

We have the following Ksp values given: $$\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s) \rightleftharpoons 5 \mathrm{Ca}^{2+}(a q)+3 \mathrm{PO}_{4}^{3-}(a q)+\mathrm{OH}^{-}(a q)$$ $$K_{\mathrm{sp1}}=2.3 \times 10^{-59}$$ (1) $$\mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q)$$ $$K_{\mathrm{sp2}}=3.9 \times 10^{-11}$$ (2) From reaction (1): $$[\mathrm{Ca}^{2+}]^5 \cdot [\mathrm{PO}_{4}^{3-}]^3 \cdot [\mathrm{OH}^{-}] = 2.3 \times 10^{-59}$$ From reaction (2): $$[\mathrm{Ca}^{2+}] \cdot [\mathrm{F}^{-}]^2 = 3.9 \times 10^{-11}$$

Step 3: Solve for K of the main reaction using equilibrium concentrations from Ksp values

From the main reaction and its equilibrium constant expression, we have: $$K=\frac{[\mathrm{CaF}_{2}]^{5} \cdot [\mathrm{PO}_{4}^{3-}]^{3} \cdot [\mathrm{OH}^{-}]}{[\mathrm{F}^{-}]^{10}}$$ Based on the Ksp values, we have: $$[\mathrm{CaF}_{2}]^{5} \cdot [\mathrm{PO}_{4}^{3-}]^{3} \cdot [\mathrm{OH}^{-}] = (5[\mathrm{Ca}^{2+}])^{5} \cdot (3[\mathrm{PO}_{4}^{3-}])^{3} \cdot [\mathrm{OH}^{-}= 2.3 \times 10^{-59}$$ (1) $$[\mathrm{Ca}^{2+}] \cdot [\mathrm{F}^{-}]^2 =3.9 \times 10^{-11}$$ (2) Now, using expressions (1) and (2) to solve for K, we get $$K=\frac{2.3 \times 10^{-59}}{(3.9 \times 10^{-11})^{10}}$$

Step 4: Calculate the equilibrium constant K for the main reaction

Finally, we calculate the value of K: $$K=\frac{2.3 \times 10^{-59}}{(3.9 \times 10^{-11})^{10}} \approx 4.15 \times 10^{9}$$ So the equilibrium constant for the reaction involving the formation of calcium fluoride and the dissolution of \(\mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}(\mathrm{OH})\) in the presence of excess fluoride is approximately \(4.15 \times 10^9\).

Key Concepts

Equilibrium Constant

In every chemical reaction, there's a tendency to reach a state where the rate of the forward reaction equals the rate of the backward reaction. At this point, the system is said to be in equilibrium. An equilibrium constant, represented as \( K \), gives us valuable insights into the position of equilibrium for a particular reaction.

For the reaction involving calcium fluoride and excess fluoride, the equilibrium constant is defined by the concentration of the products over the concentration of the reactants, each raised to the power of their coefficients in the balanced equation. In this reaction, the formula is:

• \( K = \frac{[\mathrm{CaF}_{2}]^{5} \cdot [\mathrm{PO}_{4}^{3-}]^{3} \cdot [\mathrm{OH}^{-}]}{[\mathrm{F}^{-}]^{10}} \)

It's crucial to note that pure solids and liquids are not included in the equilibrium expression, only aqueous ions and gases are.

Solubility Product Constant

The solubility product constant, known as \( K_{\mathrm{sp}} \), is a special type of equilibrium constant that applies to the dissolving of sparingly soluble salts. It tells us about the extent to which a compound will dissolve in water.

The reactions provided in the exercise help us understand this concept:

• Calcium hydroxylphosphate reaction: \( K_{\mathrm{sp1}} = 2.3 \times 10^{-59} \)
• Calcium fluoride reaction: \( K_{\mathrm{sp2}} = 3.9 \times 10^{-11} \)

The very low \( K_{\mathrm{sp}} \) values indicate that both compounds are not very soluble in water, meaning they precipitate out instead of remaining in solution. In these systems, the \( K_{\mathrm{sp}} \) helps determine at which point the concentrations of ions combine to form solid precipitates, maintaining equilibrium in solutions.

Calcium Fluoride

Calcium fluoride (\( \mathrm{CaF}_{2} \)) is a crystalline solid compound that occurs naturally as the mineral fluorite. It plays a significant role in both natural and industrial processes. In chemistry, particularly in solutions, calcium fluoride is studied as it illustrates how ionic compounds dissolve and reach equilibrium.

When \( \mathrm{CaF}_{2} \) dissolves in water, it separates into calcium ions (\( \mathrm{Ca}^{2+} \)) and fluoride ions (\( \mathrm{F}^{-} \)). The reaction of its formation:

• \( \mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2 \mathrm{F}^{-}(aq) \)

Understanding this dissolution process helps explain the role of calcium fluoride in various biological and environmental contexts, like its effects in dental health. As seen in the exercise, an excess of fluoride ions can lead to the formation of calcium fluoride, showcasing the balance between saturation and precipitation in aqueous solutions.