The pH of a buffer solution prepared by adding 10 mL of `0.1 ` M `CH_(3) COOH` and 20 mL `0.1` M sodium acetate will be ( given : `pK_(a) ` of `CH_(3)COOH = 4.74` )

Options

A. `4.05`
B. `3.04`
C. `5.04`
D. `3.05`

Explanation

To find the pH of the buffer solution prepared by mixing 10 mL of 0.1 M acetic acid (CH₃COOH) and 20 mL of 0.1 M sodium acetate (CH₃COONa), we can use the Henderson-Hasselbalch equation:

### Step-by-Step Solution:

**Step 1: Calculate the millimoles of acetic acid (CH₃COOH).**
- Volume of acetic acid = 10 mL
- Concentration of acetic acid = 0.1 M
- Millimoles of acetic acid = Volume (mL) × Concentration (M)

\[ \text{Millimoles of CH}_3\text{COOH} = 10 \, \text{mL} \times 0.1 \, \text{M} = 1.0 \, \text{mmol} \]

**Step 2: Calculate the millimoles of sodium acetate (CH₃COONa).**
- Volume of sodium acetate = 20 mL
- Concentration of sodium acetate = 0.1 M
- Millimoles of sodium acetate = Volume (mL) × Concentration (M)

\[ \text{Millimoles of CH}_3\text{COONa} = 20 \, \text{mL} \times 0.1 \, \text{M} = 2.0 \, \text{mmol} \]

**Step 3: Use the Henderson-Hasselbalch equation to find the pH.**
- The Henderson-Hasselbalch equation is given by:

\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]

- Here, pKₐ of acetic acid (CH₃COOH) = 4.74
- Base = sodium acetate (CH₃COONa) = 2.0 mmol
- Acid = acetic acid (CH₃COOH) = 1.0 mmol

Substituting the values into the equation:

\[ \text{pH} = 4.74 + \log\left(\frac{2.0}{1.0}\right) \]

\[ \text{pH} = 4.74 + \log(2) \]

- The value of \(\log(2) \approx 0.301\)

\[ \text{pH} = 4.74 + 0.301 = 5.041 \approx 5.04 \]

### Final Answer:
The pH of the buffer solution is approximately **5.04**.

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