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 of \(\mathrm{CaF}_2\) is \(4.14 \times 10^{-11}\) (option a).

Step 1: Calculate Molar Solubility

First, convert the given solubility of \( \text{CaF}_2 \), which is \( 1.7 \times 10^{-3} \text{ g/100 cm}^3 \), into moles per liter (molar solubility). The molar mass of \( \text{CaF}_2 \) is \( 40.08 + 2\times19.00 = 78.08 \text{ g/mol} \). Convert \( 1.7 \times 10^{-3} \text{ g/100 cm}^3 \) to grams per liter by multiplying by 10. Therefore, \(1.7 \times 10^{-2} \text{ g/L} \). Then convert grams to moles: \( \frac{1.7 \times 10^{-2} \text{ g/L}}{78.08 \text{ g/mol}} \approx 2.18 \times 10^{-4} \text{ mol/L} \).

Step 2: Write the Dissociation Reaction

The dissociation of \( \text{CaF}_2 \) in water can be represented as: \[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2 \text{F}^-(aq) \].Let \( s \) be the molar solubility calculated previously: \[ \text{[Ca}^{2+}] = s \] and \[ \text{[F}^-] = 2s \].

Step 3: Setup the Ksp Expression

Using the dissociation equilibrium, write the expression for the solubility product \( K_{sp} \) as: \[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \] Substitute the expressions for the ion concentrations: \[ K_{sp} = (s)(2s)^2 = 4s^3 \].

Step 4: Calculate the Solubility Product

Substitute \( s = 2.18 \times 10^{-4} \text{ mol/L} \) into the \( K_{sp} \) expression: \[ K_{sp} = 4(2.18 \times 10^{-4})^3 = 4.14 \times 10^{-11} \].

Step 5: Choose the Correct Answer

From the calculated \( K_{sp} = 4.14 \times 10^{-11} \), compare it with the choices provided. The correct answer is option (a) \( 4.14 \times 10^{-11} \).

Key Concepts

Molar Solubility

Molar solubility is a measure of how many moles of a substance can dissolve completely in a given volume of solvent, typically water, to form a saturated solution. It’s a crucial step in solving solubility product problems.
For example, with the compound \( \text{CaF}_2 \), the given solubility was \(1.7 \times 10^{-3} \, \text{g/100 cm}^3\). To find the molar solubility, which is expressed in \( \text{mol/L} \), we start by converting grams per 100 cm³ to grams per liter. Simply multiply the grams value by 10 due to unit conversion factors.
Then, use the compound's molar mass, which is the sum of the atomic masses of calcium and twice that of fluorine, to convert the solubility from grams to moles.

• \(\text{Molar mass of } \text{CaF}_2 = 40.08 + 2 \times 19.00 = 78.08 \, \text{g/mol}\)
• \(1.7 \times 10^{-2} \, \text{g/L} \div 78.08 \, \text{g/mol} \approx 2.18 \times 10^{-4} \, \text{mol/L}\)

Thus, the molar solubility \(s\) of \(\text{CaF}_2\) is calculated to be \(2.18 \times 10^{-4} \, \text{mol/L}\).

Dissociation Reaction

In chemistry, a dissociation reaction refers to the process in which a compound breaks down into its ions when dissolved in water. For \(\text{CaF}_2\), the dissociation occurs as shown:\[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2 \text{F}^-(aq) \]This means that when \(\text{CaF}_2\) dissolves, one calcium ion and two fluoride ions separate and go into solution. This balanced equation is essential for setting up the proper solubility product, \(K_{sp}\), expression.
The molar solubility \(s\) calculated earlier helps us understand how many moles of \(\text{Ca}^{2+}\) and \(\text{F}^-\) ions would be present in equilibrium:

• \([\text{Ca}^{2+}] = s = 2.18 \times 10^{-4} \, \text{mol/L}\)
• \([\text{F}^-] = 2s = 4.36 \times 10^{-4} \, \text{mol/L}\)

Knowing the stoichiometry is crucial because it defines the proportion of ions produced. It informs us how to set the concentration terms appropriately in the \(K_{sp}\) expression.

Ksp Expression

The solubility product constant, or \(K_{sp}\), is an equilibrium constant for the dissociation reaction in a saturated solution. It helps us predict the solubility of substances.
For \(\text{CaF}_2\), its \(K_{sp}\) expression is derived by considering the concentrations of the ions at equilibrium. Write it as follows:\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]Given that the molar solubility \(s\) represents \([\text{Ca}^{2+}]\) and \([\text{F}^-] = 2s\), plug these into the expression:\[ K_{sp} = (s)(2s)^2 = 4s^3 \]This equation helps understand how the concentration of \(\text{F}^-\) ions is squared due to the stoichiometry from the dissociation reaction. Simplifying the product, we get a model to calculate \(K_{sp}\) using the known \(s\) value.

Ion Concentrations

Ion concentrations are vital for understanding the outcome of a compound dissolving in a solvent. They greatly inform the calculation of the solubility product constant.
From the dissociation reaction \(\text{CaF}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^-\), we identify:

• Calcium ions: \([\text{Ca}^{2+}] = s = 2.18 \times 10^{-4} \, \text{mol/L}\)
• Fluoride ions double in concentration as they are produced in a 2:1 ratio: \([\text{F}^-] = 2s = 4.36 \times 10^{-4} \, \text{mol/L}\)

Ion concentrations derived from the molar solubility are inserted into the \(K_{sp}\) expression, providing a means to determine \(K_{sp}\).
We use \[ K_{sp} = 4s^3 \] and substitute in the value we’ve calculated for \(s\):
\[ K_{sp} \approx 4 \times (2.18 \times 10^{-4})^3 = 4.14 \times 10^{-11} \]This displays how ion concentrations link directly to the calculated \(K_{sp}\), showing the relationship between how much compound can dissolve and its tendency to disassociate into ions.