Calculate the `pH` after `50.0mL` of 0.1M NH_(3) this solution is treated with `25.0mL` of `0.1M HCI`
`K_(b)` for `NH_(3) = 1.77 xx 10^(-5) (pK_(b) ~~ 4.76)`.

Explanation

To calculate the pH after treating 50.0 mL of 0.1 M NH₃ with 25.0 mL of 0.1 M HCl, we will follow these steps:

### Step 1: Calculate the moles of NH₃ and HCl
- **Moles of NH₃**:
\[ \text{Moles of NH₃} = \text{Volume (L)} \times \text{Concentration (M)} = 0.050 \, \text{L} \times 0.1 \, \text{M} = 0.005 \, \text{mol} \, (5 \, \text{mmol}) \]

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