Step 1: Understanding the Concept:
A buffer solution resists changes in pH upon dilution or the addition of small amounts of acid or base.
For an acidic buffer (weak acid + its salt with a strong base), the pH is calculated using the Henderson-Hasselbalch equation.
The question also tests the fundamental property that moderate dilution does not significantly alter the pH of a buffer because the salt-to-acid ratio remains constant.
Step 2: Key Formula or Approach:
The Henderson-Hasselbalch equation:
\[ pH = pK_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]
Since both salt and acid are in the same total volume, we can use the ratio of their millimoles (\(n\)):
\[ pH = pK_a + \log \left( \frac{n_{\text{salt}}}{n_{\text{acid}}} \right) \]
Step 3: Detailed Explanation:
1. pH Before Dilution:
- Millimoles of weak acid (\(CH_3COOH\)), \(n_{\text{acid}} = 100\text{ ml} \times 0.01\text{ M} = 1.0\text{ mmol}\).
- Millimoles of salt (\(CH_3COONa\)), \(n_{\text{salt}} = 200\text{ ml} \times 0.02\text{ M} = 4.0\text{ mmol}\).
- \(pK_a = 4.74\).
\[ pH = 4.74 + \log \left( \frac{4.0}{1.0} \right) = 4.74 + \log(4) \]
\[ pH = 4.74 + 2 \log(2) = 4.74 + 2(0.301) \]
\[ pH = 4.74 + 0.602 = 5.342 \approx 5.34 \]
2. pH After Dilution:
- Addition of \(700\text{ ml}\) of water changes the total volume to \(1000\text{ ml}\).
- New concentrations: \([Acid] = 1/1000 = 0.001\text{ M}\), \([Salt] = 4/1000 = 0.004\text{ M}\).
- The ratio \([Salt]/[Acid]\) is still \(0.004 / 0.001 = 4\).
- Therefore, the \(\log\) term remains unchanged, and the pH remains 5.34.
Buffer solutions are characterized by their ability to maintain pH as long as the concentrations of the components are not diluted to the point where the dissociation of water becomes significant.
Step 4: Final Answer:
The pH values before and after dilution are 5.34 and 5.34 respectively. The correct option is (D).