Problem 100 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility product of \(\mathrm{A}_{2} \mathrm{X}_{3}\) is \(1.08 \times 10^{-23} .\) Its solubility will be (a) \(1.0 \times 10^{-3} \mathrm{M}\) (b) \(1.0 \times 10^{-4} \mathrm{M}\) (c) \(1.0 \times 10^{-5} \mathrm{M}\) (d) \(1.0 \times 10^{-6} \mathrm{M}\)

Step-by-Step Solution

Answer

The solubility of \( A_2X_3 \) is approximately \( 1.0 \times 10^{-5} \mathrm{M} \). Option (c) is correct.

Step 1: Identify formula and dissociation

The compound \( A_2X_3 \) dissociates in water into ions:\[A_2X_3 (s) \rightleftharpoons 2A^{3+} (aq) + 3X^{2-} (aq)\]This implies that the solubility, \( s \), of \( A_2X_3 \) results in \( 2s \) moles of \( A^{3+} \) and \( 3s \) moles of \( X^{2-} \).

Step 2: Write solubility product expression

Based on the dissociation, the solubility product \( K_{sp} \) expression for \( A_2X_3 \) becomes:\[K_{sp} = [A^{3+}]^2 [X^{2-}]^3 = (2s)^2 (3s)^3\]Substitute the ion concentrations in terms of \( s \).

Step 3: Substitute and simplify

Substitute into the expression:\[K_{sp} = (2s)^2 (3s)^3 = 4s^2 \cdot 27s^3 = 108s^5\]Then set \(108s^5 = 1.08 \times 10^{-23}\).

Step 4: Solve for solubility (s)

Divide both sides by 108:\[s^5 = \frac{1.08 \times 10^{-23}}{108} = 1.0 \times 10^{-25}\]Solve for \( s \) by taking the fifth root of both sides:\[s = (1.0 \times 10^{-25})^{1/5}\]\( s \approx 1.0 \times 10^{-5} M \).

Step 5: Conclusion

Comparing with the given options, the solubility \( s \approx 1.0 \times 10^{-5} M \). Thus, the correct answer is option (c).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in which the concentrations of reactants and products remain constant over time, meaning the rate of the forward reaction equals the rate of the reverse reaction. In the given exercise, we observe chemical equilibrium during the dissolution of the compound \( A_2X_3 \).
The equilibrium can be seen in the reaction equation where the solid \( A_2X_3 \) dissociates into its ions in solution:

• \( A_2X_3 (s) \rightleftharpoons 2A^{3+} (aq) + 3X^{2-} (aq) \)

It shows a dynamic and balanced state where the process of forming \( A_2X_3 \) matches the process of its dissociation into ions. This balance, under the hood, is what's represented by the solubility product, \( K_{sp} \).
It's key to remember that at equilibrium, while the concentrations remain constant, they are not necessarily equal to each other and depend solely on the stoichiometry and the temperature of the reaction.

Ionic Equilibrium

In ionic equilibrium, the focus is on the balances between dissolved ions in solution. For \( A_2X_3 \), we look at how ions disperse and achieve equilibrium. The solid dissociates into ions as described:

• 2 \( A^{3+} \) ions are produced per formula unit, making it 2 ions for every \( s \) moles of solubility.
• 3 \( X^{2-} \) ions form per unit, resulting in 3 ions for every \( s \) moles of solubility.

Understanding ionic equilibrium involves appreciating these ratios. The product of ion concentrations at this equilibrium is what defines the \( K_{sp} \). The equation:
\[K_{sp} = [A^{3+}]^2 [X^{2-}]^3 = (2s)^2 (3s)^3\]helps relate how changes in concentration can affect the equilibrium conditions, providing insight into how ions, when balanced correctly, maintain a stable solution.

Solubility Calculation

Solubility calculation is a core skill when dealing with solubility products. Here, the calculation involves finding the molar solubility \( s \) of \( A_2X_3 \). Starting with the expression:

• Convert the equation \( K_{sp} = 108s^5 \) to find \( s \).
• Rearrange and solve by dividing both sides by 108.
• The equation becomes \[s^5 = \frac{1.08 \times 10^{-23}}{108} = 1.0 \times 10^{-25}\]
• Take the fifth root to solve for \( s \).

This results in:
\[s = (1.0 \times 10^{-25})^{1/5}\]Which computes approximately to \( 1.0 \times 10^{-5} M \). This step-by-step breakdown is critical in determining the solubility level of a compound in its equilibrium state. Thus, the solubility informs us about the amount of compound that can dissolve before reaching saturation at which the solution maintains equilibrium.