Suppose a 1.0 L solution containing 0.20 moles of propanoic acid is titrated with 0.40 M KOH.The pH at the equivalence point is 8.99.What is the K_a of propanoic acid?

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The correct answer is A because the pH at the equivalence point is 8.99, which means that the solution is basic. This indicates that the propanoic acid has reacted completely with the KOH to form the propanoate ion, which is a weak base. The K_a of propanoic acid can be calculated using the following equation:K_a = [H+][propanoate-] / [propanoic acid]At the equivalence point, the concentration of propanoic acid is 0 M and the concentration of propanoate ion is 0.20 M. The pH of the solution is 8.99, which means that the concentration of H+ ions is 1.0 x 10^-8.99 M. Substituting these values into the equation, we get:K_a = (1.0 x 10^-8.99 M)(0.20 M) / (0 M) = 1.4 x 10^-5