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, \(4 \times 10^{-12}\) \([\mathbf{2 0 0 5}]\) 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 \( \mathrm{M}^{2+} \) is \( 1.0 \times 10^{-4} \mathrm{M} \).

Step 1: Understand the Dissolution Equation

A salt with the formula \( \text{MX}_2 \) dissolves in water according to the equation: \[ \text{MX}_2 \rightarrow \text{M}^{2+} + 2\text{X}^- \]

Step 2: Define Solubility Product (Ksp)

The solubility product (\( K_{sp} \)) is the equilibrium constant for this dissolution: \( K_{sp} = [\text{M}^{2+}][\text{X}^-]^2 \). Given that \( K_{sp} = 4 \times 10^{-12} \).

Step 3: Express Ion Concentrations in Terms of Solubility (s)

Let \( s \) be the solubility of salt in moles per liter. Thus, \( [\text{M}^{2+}] = s \) and \( [\text{X}^-] = 2s \).

Step 4: Substitute into the Ksp Expression

Substituting into the \( K_{sp} \) expression, we have: \[ K_{sp} = s(2s)^2 = 4s^3 \] Set \( 4s^3 = 4 \times 10^{-12} \).

Step 5: Solve for Solubility (s)

To find \( s \), divide both sides by 4: \[ s^3 = 10^{-12} \] Take the cube root of both sides: \[ s = (10^{-12})^{1/3} = 10^{-4} \]

Step 6: Conclusion

The concentration \([\text{M}^{2+}]\) in the aqueous solution is equal to \( s = 1.0 \times 10^{-4} \) M.

Key Concepts

Dissolution Equation

When dealing with the solubility of ionic compounds like salts, it's crucial to understand how these compounds dissolve in water. For a salt with the formula \( \text{MX}_2 \), the dissolution process can be represented by a chemical equation. This equation is expressed as:\[ \text{MX}_2 \rightarrow \text{M}^{2+} + 2\text{X}^- \]Here, \( \text{M} \) represents a metal cation with a \( 2^+ \) charge, while \( \text{X}^- \) represents a halide or anions with a \( 1^- \) charge. This equation illustrates that each formula unit of \( \text{MX}_2 \) when dissolved, produces one \( \text{M}^{2+} \) ion and two \( \text{X}^- \) ions. Understanding this reaction is the first step in working with solubility product problems, as it lays the foundation for determining how many of each type of ion will be present in solution. By knowing the dissolution equation, you can correctly account for all ions generated and begin to relate these to their concentrations in the solution.

Ion Concentration

Ion concentration in a solution is vital for calculating the solubility product. When a salt such as \( \text{MX}_2 \) dissolves, it forms ions in specific proportions based on its dissolution equation. Here:

• \([\text{M}^{2+}] = s\)
• \([\text{X}^-] = 2s\)

where \( s \) represents the solubility of the salt, or the concentration of \( \text{M}^{2+} \) ions in moles per liter. Since each unit of \( \text{MX}_2 \) produces two \( \text{X}^- \) ions, the concentration of \( \text{X}^- \) ions will be twice that of \( \text{M}^{2+} \). It is important to identify the ratio in which the ions are produced, because this directly impacts the expressions used to calculate the solubility product, \( K_{sp} \). Correctly setting up these relationships allows for accurate calculations of ion concentrations.

Equilibrium Constant

The equilibrium constant for the dissolution of a salt is known as the solubility product, \( K_{sp} \). It is a specialized type of equilibrium constant that applies to saturated solutions of sparingly soluble salts. For \( \text{MX}_2 \) dissolving in water, the expression for \( K_{sp} \) is written as:\[ K_{sp} = [\text{M}^{2+}][\text{X}^-]^2 \]Given the chemical equilibrium, the solubility product is constant at a given temperature. By substituting the ion concentrations we formulated earlier into this expression, you get:\[ K_{sp} = s(2s)^2 = 4s^3 \]Given \( K_{sp} \) for \( \text{MX}_2 \) as \( 4 \times 10^{-12} \), you can solve for \( s \), the solubility, by setting\[ 4s^3 = 4 \times 10^{-12} \]After dividing both sides by 4, you find:\[ s = (10^{-12})^{1/3} = 10^{-4} \]Thus, \( s \) indicates the concentration of \( \text{M}^{2+} \) ions in the solution, demonstrating how \( K_{sp} \) helps forecast the extent of salt dissolution in aqueous solutions.