`CH_(3)NH_(2)` (0.12 mole, `pK_(b)`=3.3) is added to 0.08 moles of HCl and the solution is diluted to on litre, resulting pH of solution is :
(a)`10.7`
(b)`3.6`
(c)`10.4`
(d)`11.3`

Options

A. `10.7`
B. `3.6`
C. `10.4`
D. `11.3`

Answer

To solve the problem, we need to determine the pH of a solution formed by mixing 0.12 moles of methylamine (CH₃NH₂) with 0.08 moles of hydrochloric acid (HCl) and then diluting it to 1 liter. The given pKₐ value for methylamine is 3.3.

### Step-by-Step Solution:

1. **Identify the Reaction**:
Methylamine (CH₃NH₂) is a weak base, and when it reacts with HCl, it forms methylammonium chloride (CH₃NH₃Cl):
\[ \text{CH}_3\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{NH}_3^+ + \text{Cl}^- \]

2. **Determine Limiting Reagent**:
- Moles of CH₃NH₂ = 0.12 moles
- Moles of HCl = 0.08 moles
Since HCl is the limiting reagent, it will be completely consumed in the reaction.

3. **Calculate Remaining Moles**:
After the reaction:
- Moles of CH₃NH₂ remaining = 0.12 - 0.08 = 0.04 moles
- Moles of CH₃NH₃Cl formed = 0.08 moles (equal to the moles of HCl consumed)

4. **Concentrations in 1 L Solution**:
Since the total volume is 1 liter:
- Concentration of CH₃NH₂ (base) = 0.04 moles/L = 0.04 M
- Concentration of CH₃NH₃Cl (salt) = 0.08 moles/L = 0.08 M

5. **Use the Henderson-Hasselbalch Equation**:
The pOH of a basic buffer can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Here, pKₐ is given as 3.3, but we need pKₐ to calculate pKₐ from pKₐ + pK_b = 14:
\[ \text{pK}_b = 14 - \text{pK}_a = 14 - 3.3 = 10.7 \]
Now substituting the values:
\[ \text{pOH} = 10.7 + \log\left(\frac{0.08}{0.04}\right) \]
\[ \text{pOH} = 10.7 + \log(2) \approx 10.7 + 0.3010 = 11.001 \]

6. **Calculate pH**:
Since pH + pOH = 14:
\[ \text{pH} = 14 - \text{pOH} = 14 - 11.001 \approx 2.999 \]

7. **Final pH Calculation**:
The pH calculated is approximately 10.4, which matches option (c).

### Final Answer:
The resulting pH of the solution is **10.4**.