What is `[NH_(4)^(+)]` in a solution that contain `0.02` M `NH_(3)(K_(b)=1.8xx10^(-5))` and `0.01` M KOH?

Options

A. `9xx10^(-6)`
B. `1.8xx10^(-5)`
C. `3.6xx10^(-5)`
D. None of these

Answer

To find the concentration of the ammonium ion \([NH_4^+]\) in a solution containing \(0.02\) M ammonia \((NH_3)\) and \(0.01\) M KOH, we can follow these steps:

### Step 1: Understand the dissociation of KOH
KOH is a strong base and will completely dissociate in solution:
\[ \text{KOH} \rightarrow \text{K}^+ + \text{OH}^- \]
At \(t = 0\), the concentration of \(\text{OH}^-\) is \(0.01\) M.

### Step 2: Write the equilibrium expression for ammonia
Ammonia (\(NH_3\)) is a weak base and can react with water to form ammonium ions \((NH_4^+)\) and hydroxide ions \((OH^-)\):
\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \]
The equilibrium constant \(K_b\) for this reaction is given by:
\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} \]
Given \(K_b = 1.8 \times 10^{-5}\).

### Step 3: Set up the initial concentrations
- Initial concentration of \(NH_3 = 0.02\) M
- Initial concentration of \(OH^- = 0.01\) M (from KOH)

At equilibrium, let \(x\) be the amount of \(NH_3\) that dissociates:
- Concentration of \(NH_3\) at equilibrium: \(0.02 - x\)
- Concentration of \(NH_4^+\) at equilibrium: \(x\)
- Concentration of \(OH^-\) at equilibrium: \(0.01 + x\)

### Step 4: Substitute into the \(K_b\) expression
Substituting the equilibrium concentrations into the \(K_b\) expression:
\[ 1.8 \times 10^{-5} = \frac{x(0.01 + x)}{0.02 - x} \]

### Step 5: Make assumptions for simplification
Since \(K_b\) is small, we can assume that \(x\) is very small compared to \(0.02\) and \(0.01\). Therefore:
- \(0.02 - x \approx 0.02\)
- \(0.01 + x \approx 0.01\)

This simplifies our equation to:
\[ 1.8 \times 10^{-5} = \frac{x(0.01)}{0.02} \]

### Step 6: Solve for \(x\)
Rearranging gives:
\[ x = \frac{1.8 \times 10^{-5} \times 0.02}{0.01} \]
\[ x = 3.6 \times 10^{-5} \]

### Step 7: Conclusion
Thus, the concentration of the ammonium ion \([NH_4^+]\) in the solution is:
\[ [NH_4^+] = 3.6 \times 10^{-5} \, \text{M} \]