Problem 144 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility of \(\mathrm{CaF}_{2}\) in water at \(298 \mathrm{~K}\) is \(1.7 \times 10^{-3}\) gm per \(100 \mathrm{~cm}^{3} .\) The solubility product of \(\mathrm{CaF}_{2}\) at 298 \(\mathrm{K}\) is (a) \(4.14 \times 10^{-11}\) (b) \(4.14 \times 10^{11}\) (c) \(4.14 \times 10^{-6}\) (d) \(4.14 \times 10^{6}\)

Step-by-Step Solution

Answer

The solubility product \( K_{sp} \) of \( \text{CaF}_2 \) at 298 K is \( 4.14 \times 10^{-11} \). (Option a)

Step 1: Convert Solubility to Molarity

First, we need to convert the solubility of \( \text{CaF}_2 \) from grams per 100 cm³ to moles per liter. The molar mass of \( \text{CaF}_2 \) is calculated as \( 40.08 + 2 \times 19.00 = 78.08 \text{ g/mol} \). So, the solubility in mol/L is: \[ \frac{1.7 \times 10^{-3} \text{ g/100 cm}^3}{78.08 \text{ g/mol}} \times 10 = 2.1777 \times 10^{-4} \text{ M} \]

Step 2: Expression for Solubility Product

The solubility product (\( K_{sp} \)) expression for \( \text{CaF}_2 \) dissociating into \( \text{Ca}^{2+} \) and \( 2 \text{F}^- \) is given by \[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \].

Step 3: Assign Concentrations from Solubility

Given the dissociation \( \text{CaF}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^- \), the concentration of \( \text{Ca}^{2+} \) ions is \( s \) and that of \( \text{F}^- \) ions is \( 2s \), where \( s = 2.1777 \times 10^{-4} \text{ M} \).

Step 4: Calculate the Solubility Product

Plug the concentrations into the solubility product expression: \[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = (2.1777 \times 10^{-4})(2 \times 2.1777 \times 10^{-4})^2 \] Simplifying this gives: \[ K_{sp} = (2.1777 \times 10^{-4})(4.3554 \times 10^{-4})^2 \] \[ K_{sp} = (2.1777 \times 10^{-4}) \times (1.898 \times 10^{-7}) \approx 4.14 \times 10^{-11} \]

Step 5: Select the Correct Answer

Based on the calculated solubility product, the correct choice from the options is (a) \( 4.14 \times 10^{-11} \).

Key Concepts

Solubility

Solubility refers to the ability of a substance to dissolve in a solvent, such as water. It indicates how much of a substance can be dissolved within a specific amount of solvent to form a solution. Break this down further, and solubility is generally expressed in terms of grams of solute per 100 milliliters of solvent at a specified temperature. In the case of calcium fluoride ( C\( \text{CaF}_2 \)), its limited solubility in water is essential in understanding its dissociation and the eventual use of its solubility product.

• Solubility provides vital information about the extent to which the ions will dissociate in a solution.
• The solubility product, \( K_{sp} \), is derived from the concentrations of the ions in the solution.

When you know the solubility of a compound, you can determine the concentration of ions that form when that compound dissolves.

Calcium Fluoride (CaF2)

Calcium fluoride is an inorganic compound with the chemical formula \( \text{CaF}_2 \). It's commonly found as the mineral fluorite and is known for its uses in various industrial processes minus being significant in the production of optical components.

• Because of its relatively low solubility in water, \( \text{CaF}_2 \) dissociates into calcium ions \( \text{Ca}^{2+} \) and fluoride ions \( \text{F}^- \) to a limited extent.
• This characteristic is crucial in applications where minimal fluoride release is desired, like in drinking water treatment.

Understanding its solubility aids in comprehending how \( \text{CaF}_2 \) behaves chemically and physically in water.

Molarity

Molarity is a way to express the concentration of a solute in a solution. It's the number of moles of solute per liter of solution, commonly represented as M. For example, a 1 M solution has one mole of a solute in a liter of solution.

• To find the solubility of \( \text{CaF}_2 \) in terms of molarity, you need to know the compound's molar mass and use it to convert grams into moles.
• For \( \text{CaF}_2 \), the molar mass is \( 78.08 \text{ g/mol} \).

This conversion is essential in writing the solubility product expression because it deals with molar concentrations.

Dissociation Equation

A dissociation equation represents how a compound splits into ions in solution. For calcium fluoride, the equation is:\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \]

• This equation indicates that one formula unit of \( \text{CaF}_2 \) produces one \( \text{Ca}^{2+} \) ion and two \( \text{F}^- \) ions.
• Understanding this equation helps in forming the solubility product expression \( K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \).

The equation balances the number of ions on both sides, ensuring that the charge and matter are conserved during the dissociation process.