Problem 116 - A Complete Resource Book in Chemistry for JEE Main

Question

For which of the following sparingly soluble salt, the solubility (S) and solubility produce (Ksp) are related by the expressions \(\mathrm{S}=(\mathrm{Ksp} / 4)^{13} ?\) (a) \(\mathrm{BaSO}_{4}\) (b) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) (c) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (d) \(\mathrm{Ag}_{2} \mathrm{PO}_{4}\)

Step-by-Step Solution

Answer

None of the given salts match the unique Ksp expression accurately.

Step 1: Understanding the Solubility Expression

The problem asks which salt has solubility (S) and solubility product (Ksp) related by the formula \( S = (Ksp / 4)^{13} \). This implies a specific relationship between the dissolution products of the salt.

Step 2: Analyzing Ionic Dissociation

Consider the general dissolution of a salt \( A_xB_y \) which dissolves as \( xA^{y+} \) and \( yB^{x-} \), with a Ksp expression \( Ksp = [A^{y+}]^x [B^{x-}]^y \). The stoichiometry affects the relationship between S and Ksp.

Step 3: Checking Solubility Product Relation

In the given expression \( S = (Ksp / 4)^{13} \), equate it with the general formula to match stoichiometry. This unusual form suggests an unusual dissociation where the Ksp involves 4 ions achieving a power of 13.

Step 4: Evaluating Options For Suitable Dissociation

Option (d) \( \mathrm{Ag}_{2} \mathrm{PO}_{4} \) dissociates into \( 2Ag^+ \) and \( PO_4^{3-} \), but the Ksp form \( Ksp = [Ag^+]^2[PO_4^{3-}] \) deviates from the unusual form needed, despite involving three ions.

Step 5: Identification of the Unique Formula

Given \( S \) is derived from \( (Ksp / 4) \), the higher power implies more complexity than basic dissociation. None of the straightforward solubility formulas match this expression, indicating no correct option based on typical structure.

Key Concepts

Sparingly Soluble Salts

Sparingly soluble salts are a fascinating group of compounds that only dissolve to a very small extent in water. This limited solubility results in a dynamic equilibrium between the undissolved solid and its ions in solution.
These salts are crucial in many natural processes and industrial applications. When trying to understand the solubility of these salts, the concept of the solubility product constant, or Ksp, becomes essential. The Ksp is a mathematical representation depicting the maximum amount of the salt that can dissolve in water to form a saturated solution.
Each sparingly soluble salt has a characteristic Ksp value, which is determined by the nature of the salt and the interactions between its ions. As you dive into solubility exercises, pay attention to how profoundly each salt can affect the chemical equilibrium in a solution. Recognizing these characteristics is key to predicting solubility and solving complex solubility problems.

Ionic Dissociation

Ionic dissociation is the process where a compound breaks apart into its constituent ions when dissolved in water. For sparingly soluble salts, understanding how they dissociate is vital to calculating the solubility product.
For example, a typical salt represented as \(A_xB_y\) will dissociate into \(xA^{y+}\) ions and \(yB^{x-}\) ions. This dissociation affects the solubility product expression \(K_{sp} = [A^{y+}]^x [B^{x-}]^y\). This equation indicates that the concentration of ions raised to the power of their stoichiometric coefficient will provide the Ksp value.
In our original exercise context, interpreting unusual dissociation patterns is crucial. The expression \(S = (Ksp / 4)^{13}\) hints at a complex dissociation requirement that isn't typically seen with common salts, indicating a higher degree of ionic fragmentation or an advanced interaction involving 4 principal ions.
Consider how the ions interact and separate because these interactions directly affect the calculation of solubility levels.

Stoichiometry in Solubility Equilibria

In chemistry, stoichiometry plays a pivotal role in understanding solubility equilibria. It involves the balancing of particle numbers and charges, an essential task when dealing with ionic substances.
For sparingly soluble salts, stoichiometry helps to set up the right proportions in dissolution reactions. For instance, if a salt \( ext{Ag}_2 ext{PO}_4 \) is involved, the stoichiometric formula becomes \( K_{sp} = [ ext{Ag}^+]^2[ ext{PO}_4^{3-}]\). It's not just about identifying ratios between compounds; it's primarily about recognizing how these proportions impact solubility products and ion concentration expressions.
The unique expression \(S = (Ksp / 4)^{13}\) from the exercise implies a highly unusual stoichiometric setup, suggesting perhaps an intricate dissolution process with a high level of interaction or complexity beyond simple ions identified in basic stoichiometry. Understanding this notion makes it possible to delve deeper into more advanced solubility discussions, preparing students for typical hurdles in practical and exam scenarios.