Problem 133 - A Complete Resource Book in Chemistry for JEE Main

Question

There sparingly soluble salts \(\mathrm{A}_{2} \mathrm{X}, \mathrm{AX}\) and \(\mathrm{AX}_{3}\) have the same solubility product. Their solubilities will be in the order (a) \(\mathrm{AX}_{3}>\mathrm{AX}>\mathrm{A}_{2} \mathrm{X}\) (b) \(\mathrm{AX}_{3}>\mathrm{A}_{2} \mathrm{X}>\mathrm{AX}\) (c) \(\mathrm{AX}>\mathrm{AX}_{3}>\mathrm{A}_{2} \mathrm{X}\) (d) \(\mathrm{AX}>\mathrm{A}_{2} \mathrm{X}>\mathrm{AX}_{3}\)

Step-by-Step Solution

Answer

The solubilities are in the order (a) \(AX_{3} > AX > A_{2} X\).

Step 1: Understanding Solubility Product (Ksp)

The solubility product (\(K_{sp}\)) is the equilibrium constant for the dissolution of a sparingly soluble salt. For a salt \(A_xB_y\), it dissociates as \(xA^{n+} + yB^{m-}\) and \(K_{sp} = [A^{n+}]^x[B^{m-}]^y\). The higher the concentration of ions, the higher the solubility of the salt.

Step 2: Expressing Ksp for each Salt

For \(A_2X\), which dissociates into \(2A^+ + X^{2-}\), the solubility \(s\) leads to concentrations \([A^+] = 2s\) and \([X^{2-}] = s\), thus \(K_{sp} = (2s)^2(s) = 4s^3\). For \(AX\), it dissociates into \(A^+ + X^-\), giving \(K_{sp} = s^2\). For \(AX_3\), it dissociates into \(A^{3+} + 3X^-\), giving \(K_{sp} = s(3s)^3 = 27s^4\).

Step 3: Solving for Solubility (s)

Since \(K_{sp}\) is the same for each salt, let's assume it equal to \(k\). We have three equations: \(4s_{A_2X}^3 = k\), \(s_{AX}^2 = k\), and \(27s_{AX_3}^4 = k\). Solving for each solubility: \(s_{AX} = \sqrt{k}\), \(s_{A_2X} = \sqrt[3]{k/4}\), \(s_{AX_3} = \sqrt[4]{k/27}\).

Step 4: Comparing Solubilities

To determine the order of solubilities, compare \(s_{AX}, s_{A_2X},\) and \(s_{AX_3}\). Through simplification, since \(s_{AX} = \sqrt{k}\) and \(s_{AX_3} = \sqrt[4]{k/27}\), it follows that \(s_{AX} > s_{AX_3}\). Also, \(s_{AX_3} \gt s_{A_2X}\) because \(s_{AX_3} = \sqrt[4]{k/27}\) is greater than \(s_{A_2X} = \sqrt[3]{k/4}\) after addressing the constants in the context of equal \(k\).

Key Concepts

Sparingly Soluble Salts

Sparingly soluble salts are compounds that do not dissolve completely in water. Instead, they dissolve just enough to reach an equilibrium between the dissolved ions and the undissolved salt. This equilibrium is represented by a special type of constant known as the solubility product constant, or simply, the solubility product. In practical terms, they may seem almost insoluble, but they do release a small quantity of ions into the solution. This is key in understanding reactions where such salts are involved.

• If you have a salt that doesn’t dissolve much, only a few ions are available in the solution, making it sparingly soluble.
• Common examples include salts like calcium sulfate and silver chloride.
• Even though the solubility is low, these salts are important in predicting how they will behave in mixtures and reactions.

Equilibrium Constant

The equilibrium constant is a fundamental concept in chemistry that applies to reactions in equilibrium. For sparingly soluble salts, the equilibrium constant directly relates to their solubility product, symbolized as \( K_{sp} \). It refers to the point where the rate of dissolution equals the rate of precipitation, meaning no net change in the amount of dissolved ions occurs in the solution.
It is calculated by the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients in the balanced dissolution equation. In simpler terms:

• For a salt \( AX \) dissociating into ions \( A^+ \) and \( X^- \), it's represented as \( K_{sp} = [A^+][X^-] \).
• This value shows how much of the salt can actually dissolve – its solubility.
• Understanding \( K_{sp} \) helps in predicting whether a precipitate will form when solutions are mixed.

Dissolution Process

The dissolution process reflects how a salt dissolves in a solvent, forming a solution. For sparingly soluble salts, this process comes to a stop when the concentration of ions in the solution matches the value determined by the solubility product. The dissolution process can be described by chemical equations showing the dissociation of ions, which are essential in properly understanding the behavior of the compound in various conditions.

• Imagine placing a powdery salt into water – what dissolves breaks into its ions, and this is the dissolution process in action.
• The extent of this process is limited by the solubility product; when this point is reached, adding more salt won’t result in more dissolution because a dynamic equilibrium is established.
• This concept is crucial in fields such as pharmaceuticals, where the solubility of drugs in bodily fluids affects their absorption rate.