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The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is
[Given pK_a of CH₃COOH = 4.57]

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2.57

3.57

5.57

4.57

answer is D.

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Detailed Solution

$\begin{array}{l} P^{H} = P^{Ka} + \log_{10} \frac{[acid]}{[salt]} \\ P^{H} = 4 .57 + \log_{10} (\frac{50 \times 0 .1}{0 .01 \times 50}) \\ P^{H} = 4 .57 + 1 = 5 .57 \end{array}$

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