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0.5 L of $0.5 M {CH}_{3} COOH$ and 0.5 L of $0.25 M {CH}_{3} COONa$ are available. What maximum volume of a buffer with a $pH = 4.58$ can be prepared using only these two resources?

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0.85 L

0.65 L

0.45 L

0.95 L

answer is A.

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Detailed Solution

$\begin{array}{l} 4.58 = 4.74 + \log \frac{[{AcO}^{-}]}{[AcOH]} \\ \Rightarrow [{AcO}^{-}] = 0.7 \times [AcOH] \\ \Rightarrow 0.5 \times 0.25 = 0.7 \times V \times 0.5 \\ V = 0.35 L \end{array}$

Total volume $= 0.5 + 0.35 = 0.85 L$

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