Problem 8 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

What is the molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in \(0.2 \mathrm{M} \mathrm{NaOH}\) solution? Given that, solubility product of \(\mathrm{Al}(\mathrm{OH})_{3}=2.4 \times 10^{-24}\) : (a) \(3 \times 10^{-19}\) (b) \(12 \times 10^{-21}\) (c) \(3 \times 10^{-22}\) (d) \(12 \times 10^{-23}\)

Step-by-Step Solution

Answer

The molar solubility of \(\mathrm{Al}( \mathrm{OH})_3\) in \(0.2 \mathrm{M} \mathrm{NaOH}\) is \(3 \times 10^{-22}\) (option (c)).

Step 1: Write the Dissolution Equation

Write the dissolution reaction for aluminum hydroxide, \(\mathrm{Al}(\mathrm{OH})_3\), in water: \[ \mathrm{Al}(\mathrm{OH})_3(s) \rightleftharpoons \mathrm{Al}^{3+}(aq) + 3\mathrm{OH}^-(aq) \]

Step 2: Expression for Solubility Product, Ksp

The solubility product (\(K_{sp}\)) for the reaction is the product of the concentrations of the ions, each raised to the power of its stoichiometric coefficient:\[ K_{sp} = [\mathrm{Al}^{3+}][\mathrm{OH}^-]^3 \]Given \( K_{sp} = 2.4 \times 10^{-24} \).

Step 3: Set Up the Equilibrium Concentration

Let \( s \) be the molar solubility of \( \mathrm{Al}(\mathrm{OH})_3 \) in the \(0.2 \, \mathrm{M}\) \( \mathrm{NaOH} \) solution. This means \([\mathrm{Al}^{3+}] = s\) and \([\mathrm{OH}^-] = 0.2 + 3s\). Since \( s \) is very small compared to \(0.2\), we can approximate \([\mathrm{OH}^-] \approx 0.2\).

Step 4: Substitute Into Ksp Expression

Substitute the concentrations into the \(K_{sp}\) expression:\[ 2.4 \times 10^{-24} = s \times (0.2)^3 \]This simplifies to:\[ 2.4 \times 10^{-24} = s \times 0.008 \]

Step 5: Solve for Molar Solubility

Solve the equation for \( s \):\[ s = \frac{2.4 \times 10^{-24}}{0.008} \]\[ s = 3.0 \times 10^{-22} \]

Step 6: Answer Verification

According to the provided options, the molar solubility value \( 3.0 \times 10^{-22} \) corresponds to option \((c)\).

Key Concepts

Solubility Product

The solubility product, known as \(K_{sp}\), plays a critical role in understanding the solubility of compounds in solution. It is the equilibrium constant for a solid substance dissolving in an aqueous solution. The solubility product gives us a snapshot of the concentration of ions in solution when the solid is at equilibrium. This is particularly important for sparingly soluble compounds, like \(\mathrm{Al} (\mathrm{OH})_{3}\).

To calculate \(K_{sp}\), you take the product of the concentrations of the resultant ions, each raised to the power of their stoichiometric coefficients from the balanced dissolution equation. For example, when \(\mathrm{Al} (\mathrm{OH})_{3}\) dissolves, it results in \(\mathrm{Al}^{3+}\) and \(\mathrm{OH}^{-}\) ions. Thus, the solubility product is represented as \([\mathrm{Al}^{3+}][\mathrm{OH}^{-}]^3\).

Understanding \(K_{sp}\) helps us predict whether a precipitate will form in a given solution. If the ion product is less than \(K_{sp}\), the solution is unsaturated, meaning more solute can dissolve. Conversely, if it exceeds \(K_{sp}\), the solution is supersaturated, and the excess ions will precipitate to form solid.

Equilibrium Concentration

Equilibrium concentration is a concept referring to the concentration of reactants and products in a chemical reaction that are in balance during equilibrium. In the context of solubility, this implies the concentration of ions present in a solution when the dissolution process has reached a stable state, with no net change over time.

When dealing with the molar solubility of a compound, such as \(\mathrm{Al} (\mathrm{OH})_{3}\), it's crucial to calculate the equilibrium concentrations of the resulting ions. In simple terms, we set the concentration of \(\mathrm{Al}^{3+}\) equal to \(s\) and determine \([\mathrm{OH}^-]\) as \(0.2 + 3s\) due to the added \(\mathrm{NaOH}\) in the solution. However, because \(s\) is often very small, particularly with highly insoluble substances, we approximate \([\mathrm{OH}^-] \approx 0.2\).

Such approximations simplify calculations and are valid in scenarios where the additional hydroxide ions from the dissolution reaction do not significantly alter the concentration provided by a strong base like \(\mathrm{NaOH}\). This approach makes it easier to use the \(K_{sp}\) expression to find the solubility at equilibrium.

Dissolution Reaction

A dissolution reaction is an essential starting point for determining the solubility of a compound. It represents how a compound breaks down into its constituent ions in solution. For aluminum hydroxide, the dissolution reaction is \(\mathrm{Al} (\mathrm{OH})_3(s) \rightleftharpoons \mathrm{Al}^{3+}(aq) + 3\mathrm{OH}^-(aq)\).

Each component of the reaction is important:

• \(\mathrm{Al} (\mathrm{OH})_3\), a solid in its undissolved form, indicates the beginning of the dissolution.
• \(\mathrm{Al}^{3+}\), the aluminum ion, is what's produced along with hydroxide ions when the compound dissolves.
• The 3:1 ratio of \(\mathrm{OH}^-\) to \(\mathrm{Al}^{3+}\) is crucial for setting up the solubility product equation.

Understanding the dissolution reaction allows us to see how the concentration of ions is affected as a solid reaches its equilibrium state in solution. This comprehension is key in computations that lead to the molar solubility, as we follow changes from the solid phase to the aqueous phase, ultimately describing how much of the compound can dissolve before equilibrium is reestablished.