A buffer solution contains $100$ ml of $0.01$ (M) $CH_3COOH$ and $200$ ml of $0.02$ (M) $CH_3COONa$. $700$ ml of water is added subsequently to the buffer solution. The pH before and after dilution are [given, $pK_a = 4.74$; $\log 2 = 0.301$]

Initial millimoles of $CH_3COOH = 100 \times 0.01 = 1$ mmol

Initial millimoles of $CH_3COONa = 200 \times 0.02 = 4$ mmol

Using the Henderson-Hasselbalch equation for the buffer solution:

$pH = pK_a + \log \left( \dfrac{[\text{Salt}]}{[\text{Acid}]} \right)$

$pH = 4.74 + \log \left( \dfrac{4}{1} \right)$

$pH = 4.74 + 2 \log 2 = 4.74 + 2(0.301) = 4.74 + 0.602 = 5.342 \approx 5.34$

When $700$ ml of water is added, the volume of the solution increases, but the ratio of the number of moles of salt to acid remains unchanged. Therefore, the pH of the buffer solution remains constant upon dilution.

pH after dilution $= 5.34$

Answer: $5.34, 5.34$