Problem 114 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility product of calcium fluoride is \(3.2 \times\) \(10^{-11} \mathrm{M}^{3}\). Its solubility in saturated solution is (a) \(8 \times 10^{-12} \mathrm{M}\) (b) \(2 \times 10^{-4} \mathrm{M}\) (c) \(4 \times 10^{-12} \mathrm{M}\) (d) \(1 \times 10^{-4} \mathrm{M}\)

Step-by-Step Solution

Answer

The correct answer is (b) \(2 \times 10^{-4} \mathrm{M}\).

Step 1: Write the Dissolution Equation

Calcium fluoride (\(\text{CaF}_2\)) dissociates in water as follows:\[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^{-}(aq) \]

Step 2: Write the Expression for the Solubility Product Constant (Ksp)

The solubility product constant for the dissolution of calcium fluoride can be expressed as:\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2 \]

Step 3: Set up the Equation for Solubility (s)

Let the solubility of \(\text{CaF}_2\) be \(s\). Thus, at equilibrium, \([\text{Ca}^{2+}] = s\) and \([\text{F}^{-}] = 2s\) because one mole of \(\text{CaF}_2\) gives one mole of \(\text{Ca}^{2+}\) and two moles of \(\text{F}^{-}\).

Step 4: Substitute and Solve for Solubility (s)

Substitute these concentrations into the Ksp expression:\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2 = s(2s)^2 = 4s^3 \]Given \( K_{sp} = 3.2 \times 10^{-11} \), we have:\[ 4s^3 = 3.2 \times 10^{-11} \]\[ s^3 = \frac{3.2 \times 10^{-11}}{4} = 8 \times 10^{-12} \]Now, solve for \(s\) by taking the cube root:\[ s = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \]

Step 5: Conclusion

The solubility of calcium fluoride in the saturated solution is approximately \( 2 \times 10^{-4} \text{ M} \).

Key Concepts

Calcium Fluoride

Calcium fluoride, known chemically as \( \text{CaF}_2 \), is a sparingly soluble compound that plays a crucial role in various chemical processes. This compound is composed of calcium ions (\( \text{Ca}^{2+} \)) and fluoride ions (\( \text{F}^{-} \)).
When \( \text{CaF}_2 \) is added to water, it dissociates slowly due to its low solubility. The dissociation process results in calcium ions and fluoride ions entering the solution.
Some key features of calcium fluoride include:

• It is found naturally in the form of the mineral fluoride.
• It is used industrially from metallurgical to optical applications.
• In water, it establishes a dynamic equilibrium between the dissolved ions and the solid form.

Understanding calcium fluoride's role in chemical reactions, especially how it dissolves in water, helps in grasping the concept of solubility product and related equilibria.

Dissolution Equation

A dissolution equation represents how a compound dissociates into ions in a solvent, often water. For calcium fluoride, the dissolution equation is expressed as:
\[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^{-}(aq) \]
This equation indicates the breaking down of solid calcium fluoride into one calcium ion and two fluoride ions. The double arrow symbolizes an equilibrium, showing that the process is dynamic. Ions continuously form and precipitate in response to changes in concentration.
When writing such equations:

• Begin with the compound's solid state on the left side.
• Arrow indicates the dissociation process.
• Products are shown with their respective states and stoichiometry on the right.

These equations are fundamental for communicating how a compound interacts with a solvent, providing the foundational framework needed for calculating solubility or understanding equilibrium constants.

Solubility Constant

The solubility constant, or \( K_{sp} \), is a measure of a compound's solubility in a solvent at equilibrium. For calcium fluoride, the \( K_{sp} \) is particularly low, indicating its limited solubility in water.
The expression for calcium fluoride's solubility constant is:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2 \]
This formula arises from the dissolution equation where one mole of \( \text{CaF}_2 \) produces one mole of calcium ions and two moles of fluoride ions in solution. Solubility constants like \( K_{sp} \) are critical in predicting how much solute can dissolve before the solution becomes saturated.
Understanding \( K_{sp} \) involves:

• Recognizing that lower values mean reduced solubility.
• Utilizing it to calculate maximum concentration before precipitation.
• Applying it to predict the shift in equilibrium with ion addition.

Mastering this concept helps in various applications, from calculating solubilities to understanding complex reactions in industrial and natural environments.