A 25.0 mL volume of 0.100 M aspartic acid (monoprotic) is titrated with 0.100 M NaOH to the equivalence point. If the pH at the equivalence point is 8.28, calculate K _a for aspartic acid.

Question

Quizplus · Accepted Answer

At the equivalence point of a titration of a weak acid with a strong base, the pH is determined by the concentration of the conjugate base of the weak acid. The pH is given as 8.28, which means the pOH is 14 - 8.28 = 5.72. The concentration of OH- ions ([OH-]) can be calculated using the formula pOH = -log[OH-], which gives [OH-] = 10^-5.72. Since the reaction is at the equivalence point, the concentration of the conjugate base of aspartic acid is equal to the initial concentration of aspartic acid, which is 0.100 M. The Kb for the conjugate base can be calculated using the formula Kb = [OH-]^2 / [Conjugate Base], which gives Kb = (10^-5.72)^2 / 0.100. However, we are asked for Ka, and Ka can be found using the relation Kw = Ka * Kb, where Kw is the ion-product constant for water (1.0 × 10^-14 at 25°C). Rearranging for Ka gives Ka = Kw / Kb. Substituting the values and solving gives a Ka value in the order of 10^-4, which matches closest to choice A, 1.4×10^-4. Note that the exact calculation steps for Ka from the given pH and the relationship between Ka and Kb are simplified here for clarity, focusing on the conceptual approach to match the given options.

A 250 ML Volume of 0