A buffer solution is prepared by mixing `10ml` of `1.0 M` acetic acid & `20 ml` of `0.5 M` sodium acetate and then diluted to `100ml` with distilled water. If the `pK_(a)` of `CH_(3)COOH` is `4.76`. What is the pH of the buffer solution prepared?

Explanation

To find the pH of the buffer solution prepared by mixing acetic acid and sodium acetate, we can follow these steps:

### Step 1: Calculate the number of moles of acetic acid and sodium acetate.
1. **For acetic acid (CH₃COOH)**:
- Volume = 10 mL = 10 × 10⁻³ L
- Molarity = 1.0 M
- Number of moles = Molarity × Volume = 1.0 mol/L × 10 × 10⁻³ L = 0.01 moles

2. **For sodium acetate (CH₃COONa)**:
- Volume = 20 mL = 20 × 10⁻³ L
- Molarity = 0.5 M
- Number of moles = Molarity × Volume = 0.5 mol/L × 20 × 10⁻³ L = 0.01 moles

### Step 2: Calculate the final concentrations after dilution.
1. **Total volume after dilution** = 100 mL = 100 × 10⁻³ L

2. **Final concentration of acetic acid**:
- Concentration = Number of moles / Total volume = 0.01 moles / (100 × 10⁻³ L) = 0.1 M

3. **Final concentration of sodium acetate**:
- Concentration = Number of moles / Total volume = 0.01 moles / (100 × 10⁻³ L) = 0.1 M

### Step 3: Use the Henderson-Hasselbalch equation to find the pH.
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

Where:
- \(\text{pK}_a\) of acetic acid = 4.76
- \([\text{Salt}]\) = concentration of sodium acetate = 0.1 M
- \([\text{Acid}]\) = concentration of acetic acid = 0.1 M

Substituting the values into the equation:
\[ \text{pH} = 4.76 + \log\left(\frac{0.1}{0.1}\right) \]
\[ \text{pH} = 4.76 + \log(1) \]
\[ \text{pH} = 4.76 + 0 \]
\[ \text{pH} = 4.76 \]

### Final Answer:
The pH of the buffer solution is **4.76**.