Question:

A buffer solution is prepared by mixing 0.1 mol of acetic acid (\( \mathrm{p}K_a = 4.74 \)) and 0.2 mol of sodium acetate in 1 L solution. What is the pH of the buffer?

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\textbf{Tip:} For weak acid–salt buffers, use the Henderson-Hasselbalch equation and remember \( \log 2 \approx 0.30 \).
Updated On: May 30, 2025
  • 4.44
  • 5.04
  • 4.74
  • 5.74
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The Correct Option is B

Approach Solution - 1

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.

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Approach Solution -2

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:

  • Amount of acetic acid = 0.1 mol
  • Amount of sodium acetate = 0.2 mol
  • Volume of the solution = 1 L

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:

  • \([\text{base}] = 0.2 \, \text{mol/L}\)
  • \([\text{acid}] = 0.1 \, \text{mol/L}\)

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.

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