Determine the acid dissociation constant for a 0.10 M acetic acid solution that has a pH of 2.87.Acetic acid is a weak monoprotic acid and the equilibrium equation of interest is CH₃COOH(aq)+ H₂O(l)⇌ H₃O⁺(aq)+ CH₃CO₂^-(aq). A)1.3 × 10^-2 B)1.3 × 10^-3 C)1.8 × 10^-5 D)1.8 × 10^-6

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The pH of the solution is 2.87, which means the concentration of H+ ions is 10^-2.87. For a 0.10 M acetic acid solution, the dissociation is small, so we can assume the concentration of acetic acid remains approximately 0.10 M. The acid dissociation constant (Ka) is calculated using the formula Ka = [H+][A-]/[HA], where [H+] = 10^-2.87, [A-] = [H+], and [HA] ≈ 0.10 M. Plugging in the values, Ka ≈ (10^-2.87)^2 / 0.10 = 1.35 × 10^-5, which is closest to 1.8 × 10^-5, option C.

Determine the Acid Dissociation Constant for a 0