A buffer solution was made by adding `15.0 g` of `CH_(3) COOH` and `20.5gCH_(3)COONa`. The buffer is diluted to `1.0L`.
Calculate the `pH` of solution.
Given: `pK_(a) of CH_(3)COOH = 4.74, log ((13)/(12)) = 0.035`

Explanation

To calculate the pH of the buffer solution made by adding 15.0 g of acetic acid (CH₃COOH) and 20.5 g of sodium acetate (CH₃COONa), we will follow these steps:

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

1. **Molecular weight of CH₃COOH (acetic acid)**:
- C: 12.01 g/mol × 2 = 24.02 g/mol
- H: 1.01 g/mol × 4 = 4.04 g/mol
- O: 16.00 g/mol × 2 = 32.00 g/mol
- Total = 24.02 + 4.04 + 32.00 = **60.06 g/mol**

2. **Moles of CH₃COOH**:
\[ \text{Moles of CH₃COOH} = \frac{15.0 \text{ g}}{60.06 \text{ g/mol}} \approx 0.2497 \text{ moles} \]

3. **Molecular weight of CH₃COONa (sodium acetate)**:
- C: 12.01 g/mol × 2 = 24.02 g/mol
- H: 1.01 g/mol × 3 = 3.03 g/mol
- O: 16.00 g/mol × 2 = 32.00 g/mol
- Na: 22.99 g/mol
- Total = 24.02 + 3.03 + 32.00 + 22.99 = **82.04 g/mol**

4. **Moles of CH₃COONa**:
\[ \text{Moles of CH₃COONa} = \frac{20.5 \text{ g}}{82.04 \text{ g/mol}} \approx 0.2495 \text{ moles} \]

### Step 2: Calculate the concentrations of CH₃COOH and CH₃COONa in 1 L solution

Since the buffer solution is diluted to a total volume of 1.0 L:

1. **Concentration of CH₃COOH**:
\[ [\text{CH₃COOH}] = \frac{0.2497 \text{ moles}}{1.0 \text{ L}} = 0.2497 \text{ M} \]

2. **Concentration of CH₃COONa**:
\[ [\text{CH₃COONa}] = \frac{0.2495 \text{ moles}}{1.0 \text{ L}} = 0.2495 \text{ M} \]

### Step 3: Use the Henderson-Hasselbalch equation to calculate pH

The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

Substituting the values:
- Given \( \text{pK}_a = 4.74 \)
- \([\text{Salt}] = [\text{CH₃COONa}] = 0.2495 \text{ M}\)
- \([\text{Acid}] = [\text{CH₃COOH}] = 0.2497 \text{ M}\)

Calculating the ratio:
\[ \frac{[\text{Salt}]}{[\text{Acid}]} = \frac{0.2495}{0.2497} \approx 0.9992 \]

Taking the logarithm:
\[ \log(0.9992) \approx -0.0008 \quad (\text{Using } \log(1 + x) \approx x \text{ for small } x) \]

Now substituting back into the equation:
\[ \text{pH} = 4.74 + (-0.0008) \approx 4.7392 \]

### Final Answer:
The pH of the buffer solution is approximately **4.74**.

---