Problem 55 - A Complete Resource Book in Chemistry for JEE Main

Question

For sparingly soluble salt \(\mathrm{A}_{\mathrm{p}} \mathrm{B}_{\mathrm{q}}\), the relationship of its solubility product \(\left(\mathrm{L}_{s}\right)\) with its solubility (s) is (a) \(\mathrm{L}_{\mathrm{s}}=\mathrm{s}^{\mathrm{p}^{+} q} \cdot \mathrm{p}^{\mathrm{p}} \cdot \mathrm{q}^{\mathrm{q}}\) (b) \(\mathrm{L}_{\mathrm{s}}=\mathrm{s}^{\mathrm{p}^{+} q}{\underline{\phantom{xx}}}^{q} \mathrm{p}^{\mathrm{q}} \cdot \mathrm{q}^{\mathrm{p}}\) (c) \(\mathrm{L}_{\mathrm{s}}=\mathrm{s}^{\mathrm{pq}} \mathrm{p}^{\mathrm{p}} \cdot \mathrm{q}^{q}\) (d) \(L_{s}=s^{p q} \cdot(p q)^{p+q}\)

Step-by-Step Solution

Answer

The correct answer is (a): \( L_s = s^{p+q} \cdot p^p \cdot q^q \).

Step 1: Understanding the chemical equation

For the sparingly soluble salt \( A_pB_q \), the dissolution reaction in water can be represented as \( A_pB_q \rightleftharpoons pA^{n+} + qB^{m-} \). This equation shows that for each unit of the salt that dissolves, \( p \) moles of \( A^{n+} \) and \( q \) moles of \( B^{m-} \) are produced.

Step 2: Writing the solubility product expression

The solubility product \( L_s \) of the salt \( A_pB_q \) is defined as the product of the equilibrium concentrations of the ions, each raised to the power of its coefficient in the balanced equation: \( L_s = [A^{n+}]^p[B^{m-}]^q \).

Step 3: Relating concentrations to solubility

If \( s \) is the molar solubility of the salt, it means \( s \) moles of \( A_pB_q \) dissolve to produce \( ps \) moles of \( A^{n+} \) and \( qs \) moles of \( B^{m-} \). This gives us the concentrations as \([A^{n+}] = ps\) and \([B^{m-}] = qs\).

Step 4: Substituting into the solubility product expression

Substitute \([A^{n+}] = ps\) and \([B^{m-}] = qs\) into the expression for \( L_s \):\[L_s = (ps)^p(qs)^q = p^p \cdot q^q \cdot s^{p+q}\]

Step 5: Matching the options

Look through the options to see which matches the derived expression. From the analysis, option (a) corresponds to the derived formula since: \( L_s = s^{p+q} \cdot p^p \cdot q^q \).

Key Concepts

Sparingly Soluble Salts

Sparingly soluble salts are substances that only dissolve slightly in water. Unlike salts that dissolve completely, their solubility is much lower, meaning only a small amount of the salt will dissociate into ions in a solution. These salts are important in chemistry because their behavior is closely studied in reactions and solubility equilibrium studies. For example, when a sparingly soluble salt like calcium sulfate is added to water, it only partially dissolves, forming a limited concentration of calcium and sulfate ions.

This limited solubility is characterized by the ionic components of the salt reaching an equilibrium state with the undissolved salt. Sparingly soluble salts are crucial because they help us understand processes such as precipitation reactions and are used in applications like water purification and pharmaceuticals.

Equilibrium Concentrations

Equilibrium in chemistry refers to a state where the concentrations of reactants and products remain constant over time. For sparingly soluble salts, it signifies that the rate of dissolving and precipitating is equal. When such a salt is in water, it dissolves until equilibrium is reached, meaning the solutions' ion concentrations do not change any further.

This state is mathematically described by the solubility product constant, often denoted as \( K_{sp} \). For instance, in our given salt \( A_pB_q \), the equilibrium is represented as the combination of its constituent ions: \( A^{n+} \) and \( B^{m-} \). At equilibrium, the product of their concentrations raised to the power of their stoichiometric coefficients equals the solubility product \( L_s \). Ensuring equilibrium helps predict whether a solute will precipitate or remain dissolved, which is vital for practical applications like predicting mineral formation.

Solubility Expression

The solubility expression is a formula that represents the solubility product in terms of equilibrium ion concentrations. For a sparingly soluble salt \( A_pB_q \), this expression is derived from the balanced chemical equation that describes its dissolution. The solubility expression is given as \( L_s = [A^{n+}]^p[B^{m-}]^q \), where each concentration is raised to the power of its corresponding stoichiometric coefficient.

This expression is key to understanding the relationship between the solubility of a salt and how it dissociates into ions. When dealing with solubility expressions, it is important to remember that they are specific to the compound's ion ratios. By using a solubility expression, chemists can calculate the molar solubility of sparingly soluble salts, predict whether a precipitate will form, and understand how changes in concentration can affect equilibrium conditions.

Ionic Concentration Representation

Representing ionic concentrations is a fundamental aspect of understanding solubility equilibrium. When a sparingly soluble salt is in solution, its ions exist in limited amounts. These ion concentrations are often described using the salt's molar solubility, denoted as \( s \). For the salt \( A_pB_q \), if \( s \) moles of the salt dissolve, then it produces \( ps \) moles of \( A^{n+} \) and \( qs \) moles of \( B^{m-} \).

This means that the ionic concentrations can be written as \([A^{n+}] = ps\) and \([B^{m-}] = qs\) for the solution at equilibrium. This representation allows chemists to understand and predict the outcome of mixing solutions, as it provides a clear picture of how the sparingly soluble salts will behave. Knowing the concentrations also helps in calculating the widespread Ksp values seen in tables, offering insights into different salt solubility levels. Such concentration representations are indispensable in the study of solution chemistry.