To determine the pH of the buffer solution, we use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{p}K_a + \log{\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)} \]
Where:
\( \text{p}K_a = 4.74 \) for acetic acid,
\([\text{A}^-]\) = concentration of acetate ion \( = 0.2 \, \text{mol/L} \),
\([\text{HA}]\) = concentration of acetic acid \( = 0.1 \, \text{mol/L} \).
Substitute these values into the equation:
\[ \text{pH} = 4.74 + \log{\left(\frac{0.2}{0.1}\right)} \]
Simplify the expression:
\[ \text{pH} = 4.74 + \log{(2)} \]
Since \( \log{(2)} \approx 0.301 \):
\[ \text{pH} = 4.74 + 0.301 = 5.04 \]
Therefore, the pH of the buffer solution is 5.04.
To find the pH of the buffer solution, we use the Henderson-Hasselbalch equation, which is:
\( \mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) \)
In this problem, the base is sodium acetate, and the acid is acetic acid. The given values are:
Since the volume is the same for both the acid and the base, their concentration ratios can be directly used in the equation. Thus, the concentrations become:
Substitute these values and the given \( \mathrm{p}K_a \) into the Henderson-Hasselbalch equation:
\( \mathrm{pH} = 4.74 + \log\left(\frac{0.2}{0.1}\right) \)
Calculate the logarithmic part:
\( \log\left(\frac{0.2}{0.1}\right) = \log(2) \approx 0.301 \)
Plug this value back into the equation:
\( \mathrm{pH} = 4.74 + 0.301 = 5.041 \)
Therefore, the pH of the buffer solution is approximately \( 5.04 \), which corresponds to the correct answer.