Problem 21 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Solubility product constant \(\left(K_{s p}\right)\) of salts of types \(M X, M X_{2}\) and \(M_{3} X\) at temperature \(T\) are \(4.0 \times 10^{-8}, 3.2 \times 10^{-14}\) and \(2.7 \times 10^{-15}\), respectively. Solubilities (mol \(\mathrm{dm}^{-3}\) ) of the salts at temperature ' \(T\) are in the order (a) \(M X>M X_{2}>M_{3} X\) (b) \(\mathrm{M}_{3} \mathrm{X}>\mathrm{MX}_{2}>\mathrm{MX}\) (c) \(M X_{2}>M_{3} X>M X\) (d) \(M X>M_{3} X>M X_{2}\)

Step-by-Step Solution

Answer

(a) MX > MX2 > M3X

Step 1: Write the Reactions

First, write the dissociation equations for the salts:1. For salt MX: \( \text{MX} \rightleftharpoons \text{M}^+ + \text{X}^- \) 2. For salt MX\(_2\): \( \text{MX}_2 \rightleftharpoons \text{M}^{2+} + 2\text{X}^- \)3. For salt M\(_3\)X: \( \text{M}_3\text{X} \rightleftharpoons 3\text{M}^{3+} + \text{X}^{3-} \)

Step 2: Define Solubility in Terms of Ksp

For each salt, we compute solubility \(s\) using their respective equations:1. For MX: \( K_{sp} = s^2 \)2. For MX\(_2\): \( K_{sp} = 4s^3 \)3. For M\(_3\)X: \( K_{sp} = 27s^4 \)

Step 3: Solve for Solubility "s"

Calculate solubility \(s\) for each salt:1. For MX: \( s = \sqrt{4.0 \times 10^{-8}} \) \( s = 2.0 \times 10^{-4} \)2. For MX\(_2\): \( s = \sqrt[3]{\frac{3.2 \times 10^{-14}}{4}} \) \( s \approx 2.0 \times 10^{-5} \)3. For M\(_3\)X: \( s = \sqrt[4]{\frac{2.7 \times 10^{-15}}{27}} \) \( s \approx 1.0 \times 10^{-5} \)

Step 4: Compare and Order Solubilities

Order the calculated solubilities in decreasing order:1. MX (\(2.0 \times 10^{-4}\))2. MX\(_2\) (\(2.0 \times 10^{-5}\))3. M\(_3\)X (\(1.0 \times 10^{-5}\))Hence, the order of solubility is: \( MX > MX_2 > M_3X \).

Key Concepts

Ionic Equilibrium

Ionic equilibrium is a fundamental aspect of chemistry, especially when we discuss solutions and their ability to dissolve salts. It refers to a state where the rates of forward and reverse reactions are equal, leading to constant concentrations of ions in solution.
It is when salts like those mentioned in the exercise—MX, MX

• At equilibrium, the rate of the salt dissolving in solution equals the rate at which ions precipitate back into the solid form.
• This balance is crucial for determining solubility and involves interactions between the solute and solvent molecules.

For any ionic compound in solution, we can express an equilibriumm equation based on the dissociation of the ionic solid. This dissociation reaches an equilibrium state, impacting the concentration of ions (ionic equilibrium), thus directly affecting the solubility of the compound in question.

Dissociation Equations

Dissociation equations help to understand how an ionic solid separates into its constituent ions in a solution. For instance, when a salt is added to a solution, it splits into positive and negative ions.
Taking the example from the exercise, MX, MX

• The process of dissociation is simply splitting into ions: \( \text{MX} \rightleftharpoons \text{M}^+ + \text{X}^- \) and similarly for other types of salts.
• For each dissociation equation, an equilibrium is established that depends on the intrinsic properties of the salt and the conditions of the solution.

This process can be represented mathematically to explore the solubility product \(K_{sp}\) which directly helps in determining the solubility (\(s\)). Each type of salt dissociates differently and influences the equilibrium state, impacting the ability of the solution to hold more ions.

Chemical Equilibria

Chemical equilibria involve the balance between ions and their undissolved forms in a saturated solution. When salts dissolve, they reach a chemical equilibrium that defines how many ions can exist in the solution at any time. This balance reflects a dynamic process based on reaction rates.

• Equilibria explain how solubility depends on the solubility product constant \(K_{sp}\), which is unique to each compound and influenced by temperature.
• During equilibrium, there is a continuous interconversion between ionic pairs and undissolved salt, maintaining a stable yet dynamic state.

Understanding chemical equilibria highlights why different salts have varying solubilities, as demonstrated by distinct \(K_{sp}\) values. \(K_{sp}\) represents the product of the molar concentrations of the ions, each raised to the power of its stoichiometric coefficient in the balanced dissociation equation. This makes it a key player in answering how and why solubilities differ across different salts.