`50.0` mL of `0.10` M ammonia solution is treated with `25.0` mL of `0.10M HCI`. If `K_(b)(NH_(3))=1.77xx10^(-5)`, the pH of the resulting solution will be

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial moles of ammonia (NH₃) and hydrochloric acid (HCl).

- **For NH₃:**
\[ \text{Moles of NH₃} = \text{Molarity} \times \text{Volume} = 0.10 \, \text{M} \times 50.0 \, \text{mL} = 0.10 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{mol} \]
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