Topic of the Question:
The topic of this question is ionic equilibrium, specifically focusing on the common ion effect on the dissociation of a weak base in the presence of a strong base.
Step 1 : Understanding the Question:
We are asked to calculate the equilibrium concentration of ammonium ions ($[\text{NH}_4^+]$) in an aqueous solution that contains both a weak base ($0.02\text{ M }\text{NH}_3$) and a strong base ($0.01\text{ M KOH}$). The base dissociation constant ($K_b$) for ammonia is given as $1.8 \times 10^{-5}$.
Step 2 : Key Formulas and Approach:
The strong base KOH dissociates completely in solution: $\text{KOH}(aq) \rightarrow \text{K}^+(aq) + \text{OH}^-(aq)$.
The weak base ammonia undergoes partial dissociation in water: $\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)$.
The equilibrium expression for ammonia dissociation is: $K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$.
Due to the common ion effect of $\text{OH}^-$ ions from KOH, the dissociation of ammonia is highly suppressed, allowing us to use simplifying approximations.
Step 3 : Detailed Explanation:
Since KOH is a strong electrolyte, it dissociates completely, contributing $0.01\text{ M}$ of $\text{OH}^-$ ions to the solution.
We set up the equilibrium concentration table for ammonia. Let the concentration of ammonia that dissociates be $x \text{ M}$.
At equilibrium, the concentrations of the species are: $[\text{NH}_3] = 0.02 - x \text{ M}$, $[\text{NH}_4^+] = x \text{ M}$, and $[\text{OH}^-] = 0.01 + x \text{ M}$.
Because $K_b$ is small ($1.8 \times 10^{-5}$) and the presence of the common ion $\text{OH}^-$ further suppresses the dissociation, $x$ is extremely small.
We can apply the following approximations: $0.02 - x \approx 0.02 \text{ M}$ and $0.01 + x \approx 0.01 \text{ M}$.
Substituting these approximations into the equilibrium constant expression: $K_b = \frac{(x)(0.01)}{0.02}$.
We set this equal to the given value of $K_b$: $1.8 \times 10^{-5} = \frac{x \cdot 0.01}{0.02} = 0.5 \cdot x$.
Solving for $x$: $x = 1.8 \times 10^{-5} \times 2 = 3.6 \times 10^{-5}\text{ M}$.
Since $x$ represents the equilibrium concentration of ammonium ions, we have: $[\text{NH}_4^+] = 3.6 \times 10^{-5}\text{ M}$.
Step 4 : Final Answer:
The equilibrium concentration of ammonium ions is $3.6 \times 10^{-5}\text{ M}$, which corresponds to Option (A).