Problem 187 - A Complete Resource Book in Chemistry for JEE Main

Question

A solution containing \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{NH}_{4} \mathrm{OH}\) has a hydroxide ion concentration of \(10^{-6}\) mol litre \(^{-1}\), which of the following hydroxides could be precipitated when this solution is added in equal volume to a solution containing \(0.1 \mathrm{M}\) of metal ions? (a) \(\mathrm{AgOH}\left(\mathrm{K}_{\mathrm{sp}}=5 \times 10^{-3}\right)\) (b) \(\mathrm{Cd}(\mathrm{OH})_{2}\left(\mathrm{~K}_{s \mathrm{p}}=8 \times 10^{-6}\right)\) (c) \(\mathrm{Mg}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=3 \times 10^{-11}\right)\) (d) \(\mathrm{Fe}(\mathrm{OH})_{3}\left(\mathrm{~K}_{3 \mathrm{sp}}=8 \times 10^{-16}\right)\)

Step-by-Step Solution

Answer

Only Mg(OH)₂ will precipitate from the solution.

Step 1: Understand the Problem

We are given the hydroxide ion concentration of a solution and need to determine which metal hydroxides will precipitate when this solution is mixed with a solution of metal ions. Each metal hydroxide has a different solubility product constant (\(K_{sp}\)).

Step 2: Combine Solutions and Adjust Concentrations

When we mix two solutions of equal volume, the concentrations of chemicals in the resultant solution are halved. The initial hydroxide ion concentration is \(10^{-6}\) mol/L, so after mixing, it becomes \(5 \times 10^{-7}\) mol/L. The concentration of metal ions will also be halved from \(0.1\ M\) to \(0.05\ M\).

Step 3: Calculate Ionic Product

To find out whether precipitation occurs, calculate the ionic product (IP) for each hydroxide using the formula:\[IP = [M^{n+}] \cdot [OH^-]^m\]where \([M^{n+}]\) is the concentration of metal ions and \([OH^-]^m\) is the concentration of hydroxide ions. Apply this to each hydroxide.

Step 4: Compare IP with \(K_{sp}\) for Each Hydroxide

Compare calculated ionic products with known \(K_{sp}\) values.- **AgOH:** \(IP = 0.05 \times 5 \times 10^{-7} = 2.5 \times 10^{-8}\), \(K_{sp} = 5 \times 10^{-3}\) \(\Rightarrow\) Since \(IP < K_{sp}\), no precipitation.- **Cd(OH)_2:** \(IP = 0.05 \times (5 \times 10^{-7})^2 = 1.25 \times 10^{-14}\), \(K_{sp} = 8 \times 10^{-6}\) \(\Rightarrow\) Since \(IP < K_{sp}\), no precipitation.- **Mg(OH)_2:** \(IP = 0.05 \times (5 \times 10^{-7})^2 = 1.25 \times 10^{-14}\), \(K_{sp} = 3 \times 10^{-11}\) \(\Rightarrow\) Since \(IP > K_{sp}\), precipitation occurs.- **Fe(OH)_3:** \(IP = 0.05 \times (5 \times 10^{-7})^3 = 6.25 \times 10^{-20}\), \(K_{sp} = 8 \times 10^{-16}\) \(\Rightarrow\) Since \(IP < K_{sp}\), no precipitation.

Key Concepts

Ionic Product

The ionic product (IP) is crucial in understanding the solubility of compounds in a solution. It refers to the product of the concentrations of the ions in the solution, each raised to the power of its stoichiometric coefficient. For a general compound, AB in equilibrium with its ions A and B, the ionic product would be given by:

• \[IP = [A]^m[B]^n\]

where \(m\) and \(n\) are the stoichiometric coefficients of the ions A and B, respectively. The ionic product is used extensively in the context of solubility equilibria.
It helps compare whether a solution is supersaturated (indicating possible precipitation) or unsaturated (indicating no precipitation).
When the ionic product exceeds the solubility product constant \(K_{sp}\), the solution is supersaturated, leading to precipitation. On the other hand, if the ionic product is less than \(K_{sp}\), no precipitation occurs. Understanding this concept is critical for predicting when and how substances will precipitate from a solution.

Hydroxide Ion Concentration

Hydroxide ion concentration \([OH^-]\) is central to the behavior of solutions that involve bases and precipitation reactions. It measures how many hydroxide ions are present in a solution, which, in tandem with other ions, can determine the compound's solubility.
In the context of precipitation reactions, controlling hydroxide ion concentration is essential to drive or avoid the formation of precipitates.
When solutions are combined, as in the given exercise, the hydroxide ion concentration is altered. Mixing equal volumes of different solutions halves each solution's original concentrations.

• For example, if the initial hydroxide concentration is \(10^{-6}\) mol/L, after mixing, the concentration reduces to \(5 \times 10^{-7}\) mol/L.

This reduction is key when recalculating the ionic product to determine if precipitation occurs with given metal ions, as different hydroxides will precipitate under different conditions depending on their solubility product (\(K_{sp}\)).

Precipitation Reactions

Precipitation reactions are valuable in determining the solubility and reactivity of different compounds in a solution. These reactions occur when the ions in two mixed solutions form an insoluble compound --- a precipitate.
The fundamental principle is comparing the calculated ionic product of the ions in the solutions to their respective solubility product constant (\(K_{sp}\)).

• If \(IP > K_{sp}\), a precipitate forms since the solution is supersaturated.
• If \(IP < K_{sp}\), no precipitate forms as the solution is unsaturated.

These reactions allow chemists and students to identify conditions under which substances in a mixture become solids, as seen with copper hydroxide \(\text{Mg(OH)}_2\) in the exercise. Mastery of precipitation reactions comes from understanding this interplay between ionic concentration in a solution and the solubility product of compounds.