Problem 25 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For a sparingly soluble salt \(A_{p} B_{q}\), the relationship of its solubility product \(\left(L_{S}\right)\) with its solubility \((S)\) is [2001S] (a) \(L_{S}=S^{p+q} \cdot p^{p} \cdot q^{q}\) (b) \(L_{S}=S^{p+q} \cdot p^{q} \cdot q^{p}\) (c) \(L_{S}=S^{p q} \cdot p^{p} \cdot q^{q}\) (d) \(L_{S}=S^{p q} \cdot(p q)^{p+q}\)

Step-by-Step Solution

Answer

The correct option is (a): \(L_S = S^{p+q} \cdot p^{p} \cdot q^{q}\).

Step 1: Understanding the Dissolution Equation

Consider a salt with the formula \(A_pB_q\). When this salt dissolves in water, it dissociates into its constituent ions. The dissolution can be written as: \(A_pB_q ightarrow pA^{n+} + qB^{m-}\). The key to solving this problem is to express the ions' concentrations in terms of the solubility \(S\).

Step 2: Expressing Ion Concentrations

When \(A_pB_q\) dissolves, \(S\) moles of the salt would form \(pS\) moles of ion \(A^{n+}\) and \(qS\) moles of ion \(B^{m-}\). Thus, their concentrations are \([A^{n+}] = pS\) and \([B^{m-}] = qS\).

Step 3: Deriving the Solubility Product Expression

The solubility product \(L_S\) is defined as the product of the equilibrium ion concentrations, each raised to the power of their stoichiometric coefficients. Thus, \(L_S = [A^{n+}]^p[B^{m-}]^q = (pS)^p(qS)^q\).

Step 4: Simplifying the Expression

Now expand the expression: \((pS)^p(qS)^q = p^pS^pq \, \text{and} \, q^qS^pq\). Combining these, we get \(L_S = S^{p+q} \, p^p \, q^q\). This matches option (a).

Key Concepts

Sparingly Soluble Salts

Sparingly soluble salts are interesting because they barely dissolve in water. Unlike common salts that readily dissolve, sparingly soluble salts only release a tiny amount of ions into the solution. This limited dissolution makes it challenging to consider their solubility without specific calculations.
When these salts dissolve, they do so to an extent that is much less than highly soluble compounds. Hence, they reach equilibrium quickly between the undissolved solid and the ions in solution. Understanding this helps us determine how much of the salt will dissolve. The measure of a sparingly soluble salt's ability to dissolve in water is known as its solubility product, denoted as \(L_s\).
To study this process, we set up equations that describe the behavior of these ions in solution, which leads us to the next concept.

Dissolution Equation

To understand how sparingly soluble salts dissolve, we write a dissolution equation. For a generic salt \(A_pB_q\), the dissolution equation indicates that the solid compound separates into its constituent ions when added to water:

• The salt \(A_pB_q\) dissociates into \(p\) cations \(A^{n+}\) and \(q\) anions \(B^{m-}\).

This equation is critical to determining how much of each ion will be present in the solution. By determining the dissolution pattern, we can calculate the solubility product, which reflects the extent of ionization in equilibrium. The precise representation of the dissolution process allows us to derive equations to predict the concentration of ions in a solution, albeit in small magnitudes, since they do not dissolve significantly.

Ion Concentration

Once a sparingly soluble salt dissolves, the dissolution process leads to specific concentrations of cations and anions in the solution. The concentration of these ions is directly related to the solubility \(S\) of the salt.
The key is to express the concentration of each ion in terms of \(S\). This is done as follows:

• For the ion \(A^{n+}\), the concentration is represented as \([A^{n+}] = pS\).
• For the ion \(B^{m-}\), the concentration is represented as \([B^{m-}] = qS\).

These expressions allow us to understand how many ions are in the solution at equilibrium, forming the basis for calculating the solubility product \(L_s\). By knowing the ion concentrations at equilibrium, we can delve into further analyses that depend on the power each concentration is raised to, according to their reactions.

Stoichiometric Coefficients

Stoichiometric coefficients play a crucial role in understanding chemical equations and reactions, especially in determining solubility products. In the context of sparingly soluble salts like \(A_pB_q\), these coefficients \(p\) and \(q\) tell us how many ions are formed from each formula unit of the salt.
In the solubility product expression \(L_s = [A^{n+}]^p[B^{m-}]^q\), each concentration of the ions is raised to the power of its respective stoichiometric coefficient. This power indicates how many times each ion's concentration should be counted in the calculation of the solubility product.

• For ion \(A^{n+}\), it is raised to the power \(p\).
• For ion \(B^{m-}\), it is raised to the power \(q\).

By employing the stoichiometric coefficients, chemists accurately predict how slight changes in ion concentration can alter the equilibrium state, making it pivotal in calculating and understanding solubility product values.