Problem 155 - A Complete Resource Book in Chemistry for JEE Main

Question

\(100 \mathrm{ml}\) of \(0.3 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}\) is mixed with \(100 \mathrm{ml}\) of \(0.2\) M \(\mathrm{NaOH} . \mathrm{K}_{b}\) of \(\mathrm{NH}_{4} \mathrm{OH}\) is \(1.8 \times 10^{-5} .\) The degree of dissociation of \(\mathrm{NH}_{4} \mathrm{OH}\) is (a) \(1.02 \times 10^{-2}\) (b) \(1.8 \times 10^{-5}\) (c) \(1.8 \times 10^{-4}\) (d) \(1.02 \times 10^{-4}\)

Step-by-Step Solution

Answer

The degree of dissociation of NH₄OH is (c) 1.8 × 10⁻⁴.

Step 1: Calculate the Total Volume

The total volume when 100 ml of 0.3 M NH\(_4\)OH is mixed with 100 ml of 0.2 M NaOH is \(200 \text{ ml}\) or \(0.2 \text{ L}\).

Step 2: Calculate the Final Concentration of NaOH

The concentration of NaOH after mixing is calculated using the formula: \( C_1V_1 = C_2V_2 \). The initial concentration is \(0.2 \text{ M}\) with initial volume \(100 \text{ ml}\):\[ (0.2 \text{ M}) \times (100 \text{ ml}) = C_2 \times (200 \text{ ml}) \]Solving for \(C_2\):\[ C_2 = 0.1 \text{ M}\]Thus, the final concentration of NaOH is \(0.1 \text{ M}\).

Step 3: Adjust NH₄OH Concentration for Dilution

The concentration of NH\(_4\)OH after dilution is calculated similarly. The initial concentration is \(0.3 \text{ M}\):\[ (0.3 \text{ M}) \times (100 \text{ ml}) = C_f \times (200 \text{ ml}) \]Solving for \(C_f\):\[ C_f = 0.15 \text{ M}\]Thus, the final concentration of NH\(_4\)OH is \(0.15 \text{ M}\).

Step 4: Calculate Degree of Dissociation

For NH\(_4\)OH in the presence of a common ion (OH\(^-\)), the expression becomes:\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]} \]Assume that the concentration of OH\(^-\) is mainly from NaOH (\(0.1 \text{ M}\)). The equilibrium equation can be simplified to:\[ K_b = \alpha \times [OH^-]\]Where \(\alpha\) is the degree of dissociation. Thus, rearrange:\[ \alpha = \frac{K_b}{[OH^-]} \]Given, \(K_b = 1.8 \times 10^{-5}\), the degree of dissociation:\[ \alpha = \frac{1.8 \times 10^{-5}}{0.1} = 1.8 \times 10^{-4} \]

Step 5: Match with Given Options

Comparing the calculated degree of dissociation \(\alpha = 1.8 \times 10^{-4}\) with the provided options, we find the correct answer is (c) \(1.8 \times 10^{-4}\).

Key Concepts

Dilution

Dilution is the process of reducing the concentration of a solute in a solution, usually by mixing it with more solvent. In this exercise, we are mixing two solutions: 100 ml of 0.3 M NH\(_4\)OH and 100 ml of 0.2 M NaOH. When these solutions are combined, their total volume becomes 200 ml, effectively halving the concentration of the individual solutes. This new, combined solution has a lower concentration due to the increased total volume compared to the original solutions.

To calculate the new concentration of each solution after dilution, you use the formula:

• \[ C_f = \frac{C_i \cdot V_i}{V_f} \]

Where:

• \( C_f \) is the final concentration,
• \( C_i \) is the initial concentration,
• \( V_i \) is the initial volume, and
• \( V_f \) is the final volume.

For instance, the final concentration of NH\(_4\)OH after dilution is calculated by substituting these values into the equation: \( C_f = \frac{(0.3 \, \text{M}) \times (100 \, \text{ml})}{200 \, \text{ml}} = 0.15 \, \text{M}\). This process of dilution allows us to adjust the concentrations of our reactants without changing their total number of moles.

Concentration Calculation

Concentration calculation is crucial in chemistry, especially when dealing with reaction mixtures. Concentration tells us how much solute is present in a given quantity of solvent, most often expressed in molarity (M), which is moles of solute per liter of solution.

In this exercise, after determining the total volume, we calculated the final concentration for each solution using the dilution concept. The initial concentration of NH\(_4\)OH is 0.3 M, and of NaOH, it is 0.2 M. After mixing, their volumes were summed to obtain the final concentrations.

The formula used was:

• \[ C_2 = \frac{C_1 \cdot V_1}{V_2} \]

For NaOH after dilution, this becomes: \( C_2 = \frac{(0.2 \, \text{M}) \times (100 \, \text{ml})}{200 \, \text{ml}} \), resulting in a final concentration of 0.1 M. Similarly, the concentration of NH\(_4\)OH is recalculated to be 0.15 M.
These calculations allow us to quantify the amount of each reactant present in the mixture, which is essential for understanding how they will behave in subsequent chemical reactions.

Equilibrium Constant

The equilibrium constant (\(K_b\)) is a fundamental concept in chemistry that describes the ratio of products to reactants at equilibrium for a given reaction. It indicates how far a reaction goes towards completion. For weak bases like NH\(_4\)OH in this exercise, the equilibrium constant is crucial in determining the degree of dissociation.

The degree of dissociation, \( \alpha \), describes how much a reactant (NH\(_4\)OH) dissociates into its ions in solution. In the presence of a common ion (such as OH\(^-\) from NaOH), the dissociation of NH\(_4\)OH is affected. Using the expression for \(K_b\):

• \[ K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]} \]

We can simplify it to \( K_b = \alpha \times [OH^-] \) if we assume that the concentration of OH\(^-\) predominates from NaOH.

Rearranging gives us the formula for the degree of dissociation:

• \[ \alpha = \frac{K_b}{[OH^-]} \]

By substituting \(K_b = 1.8 \times 10^{-5} \) and \([OH^-] = 0.1 \text{ M}\), we find \( \alpha = 1.8 \times 10^{-4} \).
This calculation shows how weak bases dissociate in the presence of common ions, illustrating an essential technique in chemistry to predict the extent of reactions at equilibrium.