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 \( K_{sp} \) is \( 1.7 \times 10^{-16} \), so the correct answer is (b).

Step 1: Understanding the Dissociation

A salt such as \( M_2S \) dissociates in water as \( M_2S \rightarrow 2M^+ + S^{2-} \). This gives us one sulfur ion and two metal ions for each formula unit of the compound.

Step 2: Expressing Solubility in Terms of Ions

Given the solubility of \( M_2S \) as \( 3.5 \times 10^{-6} \), the concentration of sulfur ions \([S^{2-}]\) will be \( 3.5 \times 10^{-6} \) mol/L and the concentration of metal ions \([M^+]\) will be \( 2 \times 3.5 \times 10^{-6} = 7.0 \times 10^{-6} \) mol/L.

Step 3: Writing the Expression for Solubility Product (Ksp)

The solubility product \( K_{sp} \) is determined by the equation: \( K_{sp} = [M^+]^2 [S^{2-}] \). Substitute in the ion concentrations calculated in Step 2: \( K_{sp} = (7.0 \times 10^{-6})^2 \times (3.5 \times 10^{-6}) \).

Step 4: Calculating the Solubility Product

Calculate the values: \((7.0 \times 10^{-6})^2 = 49 \times 10^{-12}\) and \(3.5 \times 10^{-6}\). Multiplying these gives: \(49 \times 3.5 \times 10^{-18} = 171.5 \times 10^{-18} = 1.715 \times 10^{-16}\). Round this to one significant figure matching options: \( K_{sp} = 1.7 \times 10^{-16} \).

Step 5: Matching the Solution to Options

Look through the options provided and find that option (b) \( 1.7 \times 10^{-16} \) matches our calculated solubility product value.

Key Concepts

Ionic Equilibrium

Ionic equilibrium is a fundamental concept in chemistry that deals with the state where ionic compounds are dissociated into ions in a solution, resulting in a dynamic balance between the ions in a solution and the undissolved ionic compound. This balance is crucial for understanding many chemical processes, including solubility, acid-base reactions, and precipitation reactions.

In the context of solubility product calculations, ionic equilibrium helps us understand how salts dissociate into their constituent ions.

• For an ionic compound like \( M_2S \), it dissociates into ions: \( 2M^+ \) and \( S^{2-} \).
• The overall ionic equilibrium is established when the rate at which the salt dissolves equals the rate at which the ions recombine to form the solid.

The solubility product, \( K_{sp} \), is a measure of this equilibrium, defining how much of the salt can dissolve under a given set of conditions. By calculating the concentrations of the dissolved ions, we can find \( K_{sp} \), and determine the state of ionic equilibrium for that compound.

Dissociation in Chemistry

Dissociation is the process by which ionic compounds separate into ions when dissolved in a solvent, such as water. This is essential for making solutions conductive and for facilitating chemical reactions. Understanding dissociation in chemistry, particularly for salts like \( M_2S \), is crucial for determining solubility and equilibrium.

• In water, the compound \( M_2S \) splits into two \( M^+ \) ions and one \( S^{2-} \) ion, a process dictated by the properties of the substances involved and the solubility of the compounds.
• This dissociation enhances the mobility of ions, allowing them to interact with other ions or molecules in the solution, participating in various reactions.

The dissociation equation for \( M_2S \) shows that for each unit of \( M_2S \) that dissolves, it forms two \( M^+ \) and one \( S^{2-} \) ion, illustrating how stoichiometry is key to understanding dissolution.

Solubility Calculations

Solubility calculations are essential in chemistry for predicting how much of a substance will dissolve in a solvent under specific conditions. These calculations involve understanding the relationship between the molecular formula of the compound and its dissociation in solution.

• For \( M_2S \), its solubility tells us the concentration of the solution at which the compound is saturated, meaning no more of it can dissolve at the given conditions.
• The method involves determining the concentration of each ion as a function of the compound's molar solubility.

In this exercise, the solubility of \( M_2S \) is given as \( 3.5 \times 10^{-6} \) mol/L. From this, we calculated the ion concentrations and substituted them into the solubility product formula: \[K_{sp} = [M^+]^2 [S^{2-}]\] Resulting in \( K_{sp} = 1.7 \times 10^{-16} \). These steps illustrate the power of solubility calculations to predict behaviors of salts in solutions.