Acetic acid has K _a = 1.8×10^{- 5}. A solution of 0.10 M sodium acetate has a pH of:

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The pH of the solution can be calculated using the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). Given that the K_a for acetic acid is 1.8×10^-5, the pKa is -log(1.8×10^-5) ≈ 4.74. Since the solution is of sodium acetate, it acts as a base, and we can assume [A-] = 0.10 M and [HA] is negligible in this case. Thus, pH = 4.74 + log(0.10/0) ≈ 4.74 + log(∞), which is not mathematically correct due to the assumption of [HA] being 0. The correct approach is recognizing that in a buffer solution of sodium acetate, the concentration of acetate ions [A-] is equal to the initial concentration of the sodium acetate solution, and there's no significant amount of acetic acid [HA] to consider in the calculation directly. However, the presence of sodium acetate, which is the conjugate base of acetic acid, will make the solution basic due to hydrolysis, and the pH will be greater than 7. The correct calculation involves using the formula for the hydrolysis of the acetate ion or directly applying the Henderson-Hasselbalch equation correctly with [A-] = 0.10 M and assuming the solution is essentially that of the conjugate base alone. The mistake in the explanation was assuming [HA] = 0 and trying to calculate log(0), which is incorrect. The correct approach is to use the Henderson-Hasselbalch equation properly, considering the solution as a basic buffer system. Given the pKa and the concentration of the acetate ion, the pH calculation should reflect the basic nature of the solution, leading to a pH greater than 7, specifically around 8.87, indicating a basic solution due to the acetate ion's presence.

Acetic Acid Has K a = 1

Acetic Acid Has K _a = 1