[given, $pK_a = 4.74$ ; $\log 2 = 0.301$]
Buffer pH Calculation: Before Dilution
To calculate the initial pH of the buffer solution, we use the Henderson-Hasselbalch equation:
$pH = pK_a + \log\frac{[\text{Conjugate Base}]}{[\text{Acid}]}$
First, calculate the moles of the acid ($CH_3COOH$) and the conjugate base ($CH_3COO^-$ from $CH_3COONa$):
- Moles of $CH_3COOH$ = Volume $\times$ Molarity = $100\text{ ml} \times \frac{1\text{ L}}{1000\text{ ml}} \times 0.01\text{ mol/L} = 0.001\text{ mol}$
- Moles of $CH_3COONa$ (source of $CH_3COO^-$) = Volume $\times$ Molarity = $200\text{ ml} \times \frac{1\text{ L}}{1000\text{ ml}} \times 0.02\text{ mol/L} = 0.004\text{ mol}$
Since the volume term cancels out in the ratio within the Henderson-Hasselbalch equation, we can use the moles directly:
$pH_{initial} = pK_a + \log\frac{\text{Moles of } CH_3COO^-}{\text{Moles of } CH_3COOH}$
$pH_{initial} = 4.74 + \log\frac{0.004\text{ mol}}{0.001\text{ mol}}$
$pH_{initial} = 4.74 + \log(4)$
Using $\log(4) = \log(2^2) = 2 \log(2) = 2 \times 0.301 = 0.602$:
$pH_{initial} = 4.74 + 0.602 = 5.342$
Buffer pH Calculation: After Dilution
When water is added to a buffer solution, both the acid and its conjugate base are diluted by the same factor. Therefore, the ratio of their concentrations (or moles) remains unchanged.
Initial total volume = $100\text{ ml} + 200\text{ ml} = 300\text{ ml}$.
Water added = $700\text{ ml}$.
Final volume = $300\text{ ml} + 700\text{ ml} = 1000\text{ ml}$.
The moles of $CH_3COOH$ ($0.001\text{ mol}$) and $CH_3COO^-$ ($0.004\text{ mol}$) do not change.
New concentration of $CH_3COOH = \frac{0.001\text{ mol}}{1\text{ L}} = 0.001\text{ M}$
New concentration of $CH_3COO^- = \frac{0.004\text{ mol}}{1\text{ L}} = 0.004\text{ M}$
Applying the Henderson-Hasselbalch equation again:
$pH_{final} = pK_a + \log\frac{[\text{New Conjugate Base}]}{[\text{New Acid}]}$
$pH_{final} = 4.74 + \log\frac{0.004\text{ M}}{0.001\text{ M}}$
$pH_{final} = 4.74 + \log(4)$
$pH_{final} = 4.74 + 0.602 = 5.342$
Thus, the pH before dilution is approximately $5.34$ and the pH after dilution is also approximately $5.34$.
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