Problem 270 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility product of a salt having general formula \(\mathrm{MX}_{2}\) in water is, \(4 \times 10^{-12}\) \([2005]\) The concentration of \(\mathrm{M}^{2+}\) ions in the aqueous solution of the salt is (a) \(1.6 \times 10^{-4} \mathrm{M}\) (b) \(2.0 \times 10^{-6} \mathrm{M}\) (c) \(1.0 \times 10^{-4} \mathrm{M}\) (c) \(4.0 \times 10^{-10} \mathrm{M}\)

Step-by-Step Solution

Answer

The concentration of \( \mathrm{M}^{2+} \) ions is \( 1.0 \times 10^{-4} \mathrm{M} \) (option c).

Step 1: Understanding the Dissociation

The salt with the formula \( \mathrm{MX}_2 \) dissociates in water as follows: \( \mathrm{MX}_2 (s) \rightarrow \mathrm{M}^{2+} (aq) + 2\mathrm{X}^{-} (aq) \). This tells us that for every 1 mole of salt that dissolves, 1 mole of \( \mathrm{M}^{2+} \) ions and 2 moles of \( \mathrm{X}^{-} \) ions are produced.

Step 2: Defining the Solubility Product Expression

The solubility product expression (\( K_{sp} \)) for \( \mathrm{MX}_2 \) is: \( K_{sp} = [\mathrm{M}^{2+}][\mathrm{X}^-]^2 \). Given that \( K_{sp} = 4 \times 10^{-12} \), we can use this expression to find the concentration of \( \mathrm{M}^{2+} \) ions.

Step 3: Setting up Variables for Molar Concentrations

Let the solubility of \( \mathrm{MX}_2 \) be \( s \) mol/L. At equilibrium, \( [\mathrm{M}^{2+}] = s \) and \( [\mathrm{X}^-] = 2s \), because 1 mole of \( \mathrm{MX}_2 \) gives 1 mole of \( \mathrm{M}^{2+} \) and 2 moles of \( \mathrm{X}^- \).

Step 4: Substitute into the Solubility Product

Substitute \( [\mathrm{M}^{2+}] = s \) and \( [\mathrm{X}^-] = 2s \) into the solubility product expression: \[ K_{sp} = s \cdot (2s)^2 = 4s^3 \].

Step 5: Solve for \( s \)

We know \( K_{sp} = 4 \times 10^{-12} \). Therefore, \[ 4s^3 = 4 \times 10^{-12} \].Dividing both sides by 4, we get \[ s^3 = 10^{-12} \].Taking the cube root of both sides, \[ s = (10^{-12})^{1/3} = 10^{-4} \].

Step 6: Identify \( [\mathrm{M}^{2+}] \) Concentration

The concentration of \( \mathrm{M}^{2+} \) is equal to \( s \), which is \( 1.0 \times 10^{-4} \mathrm{M} \).

Key Concepts

Solubility Equilibrium

Solubility equilibrium describes the balance between a solid compound and its dissolved ions in a solution. For a salt like \( \mathrm{MX}_2 \), it dissolves in water to establish an equilibrium between the solid \( \mathrm{MX}_2 \) and its dissociated ions: \( \mathrm{M}^{2+} \) and \( \mathrm{X}^{-} \). At equilibrium, the rate of dissolution of \( \mathrm{MX}_2 \) is equal to the rate at which its ions recombine to form the solid. This dynamic balance ensures that the concentrations of the ions remain stable over time, unless external conditions change.
Understanding solubility equilibrium is essential for predicting how much of a salt will dissolve under particular conditions and how the presence of common ions or changes in temperature might affect solubility.

Ion Concentration

In the context of solubility product calculations, ion concentration is pivotal to finding how much a substance dissociates in solution. For the salt \( \mathrm{MX}_2 \), it dissociates to produce \( \mathrm{M}^{2+} \) ions and \( \mathrm{X}^{-} \) ions. The relationship is as follows: for every mole of \( \mathrm{MX}_2 \) that dissolves, 1 mole of \( \mathrm{M}^{2+} \) and 2 moles of \( \mathrm{X}^{-} \) are generated. Thus, if the solubility (\( s \)) is known, the concentrations of these ions can be determined.
The concentration of \( \mathrm{M}^{2+} \) ions at equilibrium is equivalent to the solubility \( s \), while \([\mathrm{X}^-] = 2s \). Knowledge of these concentrations helps in calculating the solubility product and understanding the solubility behavior of different salts.

Chemical Dissociation

When a salt like \( \mathrm{MX}_2 \) dissolves in water, it undergoes chemical dissociation into its constituent ions: \( \mathrm{M}^{2+} \) and \( \mathrm{X}^{-} \). This dissociation is an essential process where the strong ionic bonds in the salt crystal are overcome by the interaction with water molecules, allowing ions to move freely in solution.
The extent of dissociation determines the concentrations of ions, which are critical for calculating properties such as the solubility product (\( K_{sp} \)).

• Complete dissociation is assumed for finding \( K_{sp} \), meaning all dissolved \( \mathrm{MX}_2 \) contributes to the ions in solution.
• Dissociation directly affects the equilibrium expressions used and aids in determining how much of a salt can dissolve before reaching equilibrium.

Understanding chemical dissociation provides insights into the behavior of ionic compounds in aqueous solutions and helps predict reactions in more complex chemical systems.