Problem 78 - A Complete Resource Book in Chemistry for JEE Main

Question

Solubility of a \(\mathrm{M}_{2} \mathrm{~S}\) salt is \(3.5 \times 10^{-6}\) then find out solubility product. (a) \(1.7 \times 10^{-6}\) (b) \(1.7 \times 10^{-16}\) (c) \(1.7 \times 10^{-18}\) (d) \(1.7 \times 10^{-12}\)

Step-by-Step Solution

Answer

The solubility product is \(1.7 \times 10^{-16}\). Option (b) is correct.

Step 1: Understand the Dissolution

The salt given is \(\text{M}_2\text{S}\), which dissolves in water according to the reaction: \[\text{M}_2\text{S} (s) \rightarrow 2\text{M}^+ (aq) + \text{S}^{2-} (aq)\].This reaction indicates that for every mole of \(\text{M}_2\text{S}\) that dissolves, it produces 2 moles of \(\text{M}^+\) ions and 1 mole of \(\text{S}^{2-}\) ions.

Step 2: Solubility and Ion Concentrations

The solubility of \(\text{M}_2\text{S}\) is given as \(3.5 \times 10^{-6}\) mol/L. This means:- \(\text{S}^{2-}\) concentration = \(3.5 \times 10^{-6}\) M- \(\text{M}^+\) concentration = \(2 \times 3.5 \times 10^{-6} = 7.0 \times 10^{-6}\) M.

Step 3: Calculate the Solubility Product

The solubility product \(K_{sp}\) for \(\text{M}_2\text{S}\) can be expressed as:\[ K_{sp} = [\text{M}^+]^2[\text{S}^{2-}] \]Substitute the concentrations:\[ K_{sp} = (7.0 \times 10^{-6})^2 \times (3.5 \times 10^{-6}) \]\[ K_{sp} = 49 \times 10^{-12} \times 3.5 \times 10^{-6} \]\[ K_{sp} = 171.5 \times 10^{-18} \]\[ K_{sp} = 1.7 \times 10^{-16} \].

Key Concepts

Dissolution Reaction

When a salt like \( \text{M}_2\text{S} \) dissolves in water, it undergoes a process called dissolution. This process is represented by a chemical equation showing how the solid separates into its ions in solution. For the compound \( \text{M}_2\text{S} \), the dissolution reaction can be written as:\[\text{M}_2\text{S} \rightarrow 2\text{M}^+ + \text{S}^{2-}\]This equation indicates that from one mole of \( \text{M}_2\text{S} \), you produce two moles of \( \text{M}^+ \) ions and one mole of \( \text{S}^{2-} \) ions.
Understanding dissolution reactions is important as it forms the basis for calculating the solubility of ions in solution.

Ion Concentrations

Ion concentrations help us understand how many ions are in a solution after dissolution. For the compound \( \text{M}_2\text{S} \), with a molar solubility of \( 3.5 \times 10^{-6} \) mol/L, you can calculate the concentrations of \( \text{M}^+ \) and \( \text{S}^{2-} \) in the solution. Since each formula unit yields 2 \( \text{M}^+ \) ions, the concentration of \( \text{M}^+ \) will be:

• \( \text{M}^+ \) concentration = \( 2 \times 3.5 \times 10^{-6} = 7.0 \times 10^{-6} \) M
• \( \text{S}^{2-} \) concentration = \( 3.5 \times 10^{-6} \) M

Calculating these concentrations is crucial as they are directly used in determining the solubility product of the compound.

Molar Solubility

Molar solubility refers to the number of moles of a compound that can dissolve in a liter of solution until it achieves equilibrium. For \( \text{M}_2\text{S} \), the molar solubility is provided as \( 3.5 \times 10^{-6} \) mol/L.
This value indicates that at equilibrium, this is the amount of the salt that dissolves to form its ions in solution. The goal is always to find this solubility under specific conditions, such as temperature, because it the basis for calculating other related properties.
Knowing molar solubility allows one to determine how much of a particular solute can be present in a solution without precipitating.

Ionic Product

The ionic product quantifies the product of concentrations of ions in a solution raised to the power of their coefficients in the dissolution equation. For \( \text{M}_2\text{S} \), this refers to the expression:\[[\text{M}^+]^2[\text{S}^{2-}]\]This expression helps to determine whether a solution is saturated, unsaturated, or supersaturated when compared against the solubility product.
In the context of this exercise, the calculation of the ionic product using the concentrations \( (7.0 \times 10^{-6})^2 \) and \( (3.5 \times 10^{-6}) \) resulted in \( 1.7 \times 10^{-16} \). This pivotal value dictates whether precipitation occurs, influencing practical applications in chemistry.

Equilibrium Constants

Equilibrium constants, like the solubility product constant \( K_{sp} \), express the ratio of product and reactant concentrations in a saturated solution under equilibrium conditions.
For \( \text{M}_2\text{S} \), \( K_{sp} \) is calculated using the concentrations of its dissociated ions. The expression for \( K_{sp} \) in this scenario is:\[K_{sp} = [\text{M}^+]^2[\text{S}^{2-}]\]To find \( K_{sp} \), the known ion concentrations are plugged into the formula:
\( K_{sp} = (7.0 \times 10^{-6})^2 \times (3.5 \times 10^{-6}) = 1.7 \times 10^{-16} \)
Calculating and understanding equilibrium constants like \( K_{sp} \) are critical in predicting salt solubility and potential precipitation from solutions.