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 CH_3COOH = 4.57]
5.57
3.57
4.57
2.57
The solution is a mixture of a weak acid (CH_3COOH) and its conjugate base (CH_3COONa), forming an acidic buffer. Using the Henderson-Hasselbalch equation: pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right). Given pK_a = 4.57, [Salt] = 0.10 M, and [Acid] = 0.01 M. pH = 4.57 + \log \left( \frac{0.10}{0.01} \right) = 4.57 + \log(10) = 4.57 + 1 = 5.57.
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