Problem 112 - A Complete Resource Book in Chemistry for JEE Main

Question

If the solubility of sodium hexafluoroaluminate is 'a' mol/litre, its solubility product is (a) \(a^{8}\) (b) \(27 \mathrm{a}^{4}\) (c) \(180 \mathrm{a}^{3}\) (d) \(2916 \mathrm{a}^{8}\)

Step-by-Step Solution

Answer

The correct option is (b) \(27a^4\).

Step 1: Write the Dissociation Equation

First, we need to write the dissociation equation for sodium hexafluoroaluminate. It dissociates as \( Na_3[AlF_6] \rightarrow 3Na^+ + [AlF_6]^{3-} \).

Step 2: Express Molar Solubility

Let the molar solubility of \( Na_3[AlF_6] \) be \( a \) mol/L. Therefore, at equilibrium, the concentration of \( Na^+ \) will be \( 3a \) mol/L and the concentration of \( [AlF_6]^{3-} \) will be \( a \) mol/L.

Step 3: Write the Expression for Solubility Product

The solubility product \( K_{sp} \) for the dissociation of \( Na_3[AlF_6] \) is given by the product of the ion concentrations raised to the power of their stoichiometric coefficients: \( K_{sp} = [Na^+]^3 imes [[AlF_6]^{3-}] \).

Step 4: Substitute the Ion Concentrations

Substitute the ion concentrations from Step 2 into the expression for \( K_{sp} \). Thus, \( K_{sp} = (3a)^3 imes a = 27a^3 imes a = 27a^4 \).

Step 5: Identify Correct Option

Now, compare the calculated solubility product with the given options. The correct solubility product matches option (b), which is \( 27a^4 \).

Key Concepts

Dissociation Equation

Understanding the dissociation equation is key in calculating the solubility product of a compound like sodium hexafluoroaluminate. When sodium hexafluoroaluminate, denoted as \( Na_3[AlF_6] \), dissolves in water, it separates into its constituent ions. This process is known as dissociation. The equation for this reaction is:

• \( Na_3[AlF_6] \rightarrow 3Na^+ + [AlF_6]^{3-} \)

This equation tells us that one formula unit of sodium hexafluoroaluminate dissociates into three sodium ions \( (Na^+) \) and one hexafluoroaluminate ion \( ([AlF_6]^{3-}) \). This knowledge allows us to understand how many moles of each ion form per mole of compound that dissolves, which is crucial for further calculations. Emphasizing the stoichiometry of the reaction, three moles of sodium ions are formed for every mole of the compound, highlighting the importance of the dissociation process in solubility calculations.

Molar Solubility

Molar solubility refers to the number of moles of solute that can dissolve in one liter of solution before reaching saturation. In the case of sodium hexafluoroaluminate \( Na_3[AlF_6] \), if its molar solubility is denoted by \( a \), this means \( a \) moles of the compound can dissolve per liter of solution. Once dissolved, it contributes to the concentration of the ions in the solution. Since the dissociation produces 3 \( Na^+ \) ions for every formula unit of \( Na_3[AlF_6] \) dissolved, the sodium ion concentration at equilibrium will be \( 3a \) mol/L. Additionally, the concentration of the \( [AlF_6]^{3-} \) ions will be simply \( a \) mol/L. The molar solubility is pivotal as it provides a direct measure of how much of the compound dissolves before the solution becomes saturated.

Ion Concentrations

Ion concentrations in a dissociation process determine how the ions contribute to the solubility product expression. For sodium hexafluoroaluminate, the ion concentrations are derived from its molar solubility. When \( Na_3[AlF_6] \) dissolves, the dissociation equation shows that each mole breaks into 3 moles of \( Na^+ \) and 1 mole of \( [AlF_6]^{3-} \). If the molar solubility is \( a \), then:

• The concentration of sodium ions \( [Na^+] \) is \( 3a \) mol/L.
• The concentration of hexafluoroaluminate ions \( [[AlF_6]^{3-}] \) is \( a \) mol/L.

Understanding these concentrations is crucial to writing the solubility product expression, as these values are plugged into the formula raised to the power of their respective stoichiometric coefficients. This computation helps to determine the product of the concentrations, which characterizes the solubility product.

Stoichiometric Coefficients

Stoichiometric coefficients are numerical values in chemical equations that provide the exact proportions of reactants and products involved in a chemical reaction. In the dissociation of sodium hexafluoroaluminate, represented by:

• \( Na_3[AlF_6] \rightarrow 3Na^+ + [AlF_6]^{3-} \)

the stoichiometric coefficients are 1 for \( Na_3[AlF_6] \), 3 for \( Na^+ \), and 1 for \( [AlF_6]^{3-} \). These coefficients are vital in calculating the solubility product \( K_{sp} \) as they determine the power to which the concentration of each ion is raised. In the solubility product expression\( K_{sp} = [Na^+]^3 \times [[AlF_6]^{3-}] \), the sodium ion concentration is cubed due to its coefficient, reflecting how the concentration of ions affects the solubility product calculation. Understanding these coefficients helps in accurately describing how the dissolving process affects overall ion balance in the equilibrium state.