Problem 208 - A Complete Resource Book in Chemistry for JEE Main

Question

Three sparingly soluble salts \(\mathrm{M}_{2} \mathrm{~B}, \mathrm{MB}\) and \(\mathrm{MB}_{3}\) have the same solubility product. Their solubilities will be in the order (a) \(\mathrm{MB}_{3}>\mathrm{M}_{2} \mathrm{~B}>\mathrm{MB}\) (b) \(\mathrm{MB}>\mathrm{M}_{2} \mathrm{~B}>\mathrm{MB}_{3}\) (c) \(\mathrm{MB}_{3}>\mathrm{MB}>\mathrm{M}_{2} \mathrm{~B}\) (d) \(\mathrm{MB}>\mathrm{MB}_{3}>\mathrm{M}_{2}^{2} \mathrm{~B}\)

Step-by-Step Solution

Answer

(b) \( \mathrm{MB} > \mathrm{M}_2\mathrm{B} > \mathrm{MB}_3 \).

Step 1: Understand Solubility Product

The solubility product, denoted as \( K_{sp} \), is a constant for a sparingly soluble salt at a given temperature. It represents the maximum product of the ionic concentrations that can be achieved in a solution at equilibrium.

Step 2: Write Dissolution Reactions and Ksp Expressions

- For \( ext{M}_2 ext{B} \), the dissolution reaction is \( ext{M}_2 ext{B}(s) \rightleftharpoons 2 ext{M}^+(aq) + ext{B}^{2-}(aq) \), and \( K_{sp} = [ ext{M}^+]^2[ ext{B}^{2-}] \).- For \( ext{MB} \), the dissolution reaction is \( ext{MB}(s) \rightleftharpoons ext{M}^+(aq) + ext{B}^-(aq) \), and \( K_{sp} = [ ext{M}^+][ ext{B}^-] \).- For \( ext{MB}_3 \), the dissolution reaction is \( ext{MB}_3(s) \rightleftharpoons ext{M}^{3+}(aq) + 3 ext{B}^-(aq) \), and \( K_{sp} = [ ext{M}^{3+}][ ext{B}^-]^3 \).

Step 3: Express Each Solubility in Terms of S

For each compound, let \( s \) be the solubility in mol/L:- \( ext{M}_2 ext{B} \): \( K_{sp} = (2s)^2(s) = 4s^3 \).- \( ext{MB} \): \( K_{sp} = s^2 \).- \( ext{MB}_3 \): \( K_{sp} = s(3s)^3 = 27s^4 \).

Step 4: Solve for Solubility S in Terms of Ksp

- For \( ext{M}_2 ext{B} \), \( s = \left(\frac{K_{sp}}{4}\right)^{1/3} \).- For \( ext{MB} \), \( s = (K_{sp})^{1/2} \).- For \( ext{MB}_3 \), \( s = \left(\frac{K_{sp}}{27}\right)^{1/4} \).

Step 5: Compare Solubilities

We compare the expressions for \( s \) obtained:- \( ext{MB} > ext{M}_2 ext{B} > ext{MB}_3 \) because: - \( (K_{sp})^{1/2} \) is larger than \((K_{sp}/4)^{1/3} \). - \( (K_{sp}/4)^{1/3} \) is larger than \((K_{sp}/27)^{1/4} \).

Key Concepts

Sparingly Soluble Salts

Sparingly soluble salts are those which dissolve minimally in a solvent like water. When it comes to these salts, only a small fraction actually dissociates into ions. Consequently, an equilibrium is established between the undissolved salt and the dissolved ions in solution. The solubility product constant, or \( K_{sp} \), is a crucial factor for determining their behavior. This constant represents the maximum product of the ionic concentrations that can exist in solution at equilibrium. The lower the solubility product, the less soluble the salt. Yet, even sparingly soluble salts play important roles in various chemical processes, from industrial manufacturing to biological systems.

Dissolution Reactions

Dissolution reactions describe how substances, like sparingly soluble salts, dissolve in a solvent such as water. The chemical equations for these reactions demonstrate how the solid separates into its constituent ions. For sparingly soluble salts such as \( \text{M}_2 \text{B} \), \( \text{MB} \), and \( \text{MB}_3 \), we characterize their dissolution as follows:
- \( \text{M}_2 \text{B}(s) \rightleftharpoons 2\text{M}^+(aq) + \text{B}^{2-}(aq) \)
- \( \text{MB}(s) \rightleftharpoons \text{M}^+(aq) + \text{B}^-(aq) \)
- \( \text{MB}_3(s) \rightleftharpoons \text{M}^{3+}(aq) + 3\text{B}^-(aq) \)
The balance of these ions in solution dictates the solubility product. Each equation has its own \( K_{sp} \) which is derived from the concentrations of the dissociated ions raised to the power of their stoichiometric coefficients. Understanding these reactions is essential for predicting how much of the salt will dissolve under given conditions.

Comparing Solubilities

Comparing the solubilities of different sparingly soluble salts requires evaluating their solubility products and the dissolution reactions. Once we determine the expression for solubility \( s \) in terms of \( K_{sp} \), it becomes easier to compare differing compounds. For example, let's consider the expressions:
- For \( \text{M}_2 \text{B} \), \( s = \left(\frac{K_{sp}}{4}\right)^{1/3} \)\- For \( \text{MB} \), \( s = (K_{sp})^{1/2} \)\- For \( \text{MB}_3 \), \( s = \left(\frac{K_{sp}}{27}\right)^{1/4} \)
Comparing these, we see that \( \text{MB} \) has the largest solubility because \( (K_{sp})^{1/2} \) is greater than the expressions for \( \text{M}_2 \text{B} \) and \( \text{MB}_3 \). This is followed by \( \text{M}_2 \text{B} \), and \( \text{MB}_3 \) has the smallest solubility. Therefore, the order of solubilities is \( \text{MB} > \text{M}_2 \text{B} > \text{MB}_3 \). Such comparisons offer insights into which salts may be more prone to dissolution under certain conditions, affecting both practical applications and theoretical studies.