The `pH` of a solution resulting from the addition of `12.5mL` of `0.1M HCl` to `50mL` of a solution containing 0.15M `CH_(3)COOH` & `0.2M` `CH_(3)COONa` will be (Given : `pK_(a)` of `CH_(3)COOH=4.74`)

Options

A. `4.74`
B. `lt4.74`
C. `gt4.74`
D. `gt 9.26`

Explanation

To solve the problem of determining the pH of a solution resulting from the addition of 12.5 mL of 0.1 M HCl to 50 mL of a solution containing 0.15 M acetic acid (CH₃COOH) and 0.2 M sodium acetate (CH₃COONa), we can follow these steps:

### Step 1: Calculate the initial moles of CH₃COOH and CH₃COONa

1. **Calculate moles of CH₃COOH:**
\[ \text{Moles of CH₃COOH} = \text{Molarity} \times \text{Volume} = 0.15 \, \text{M} \times 50 \, \text{mL} = 0.15 \times 0.050 \, \text{L} = 0.0075 \, \text{mol} = 7.5 \, \text{mmol} \]
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