CH3CH2COOH(aq) + H2O(l) <=> CH3CH2COO-(aq) + H3O+(aq) Propanoic acid, CH3CH2COOH, is a carboxylic ac — Equilibrium Chemistry Question

CH₃CH₂COOH and CH₃CH₂COO^-
acid base
OR
H₃O⁺ and H₂O
acid base

1 point is earned for writing (or naming) either of the Brønsted^-Lowry conjugate acid-base pairs with a clear indication of which is the acid and which is the base.

[H₃O⁺] = 10^(-pH) = 10^(-2.79) = 1.6 x 10^(-3) M

[CH₃CH₂COO^-] = [H₃O⁺]
AND [CH₃CH₂COOH] = 0.20 M - [H₃O⁺], OR [CH₃CH₂COOH] ≈ 0.20 M (state or assume that [H₃O⁺] << 0.20 M)

Ka = ([CH₃CH₂COO^-][H₃O⁺]) / [CH₃CH₂COOH] = (1.6 x 10^(-3))^2 / 0.20 = 1.3 x 10^(-5)

1 point is earned for correctly solving for [H₃O⁺].
1 point is earned for the Ka expression for propanoic acid OR 1 point is earned for substituting values into the Ka expression.
1 point is earned for correctly solving for the value of Ka.

(i) False. The conjugate base of a weak acid undergoes hydrolysis at equivalence to form a solution with a pH > 7.
CH₃CH₂COO^-(aq) + H₂O(l) <=> CH₃CH₂COOH(aq) + OH^-(aq)

1 point is earned for noting that the statement is false AND providing a supporting explanation.

(ii) True. HCl is a strong acid that ionizes completely. Fewer moles of HCl are needed to produce the same [H₃O⁺] as the propanoic acid solution, which only partially ionizes.

1 point is earned for noting that the statement is true and providing a supporting explanation.

moles of NaOH = (0.173 mol NaOH / 1 L NaOH) * 0.02052 L NaOH = 3.55 x 10^(-3) mol NaOH
Since the acid is monoprotic, 1 mol NaOH reacts with 1 mol of acid:
moles of propanoic acid = 3.55 x 10^(-3) mol acid

Molarity = 3.55 x 10^(-3) mol acid / 0.02500 L acid = 0.142 M

OR

Since CH₃CH₂COOH is monoprotic and, at the equivalence point, moles H⁺ = moles OH^-:
Ma * Va = Mb * Vb
Ma * (25.00 mL) = (0.173 M) * (20.52 mL)
Ma = 0.142 M

1 point is earned for correctly calculating the number of moles of acid that reacted at the equivalence point.
1 point is earned for the correct molarity of acid.

Disagree with the student’s claim.
From part (b) above, Ka for propanoic acid is 1.3 x 10^(-5), so pKa = -log(1.3 x 10^(-5)) = 4.89. Because 4.83 is so close to 4.89, the pH at the equivalence point in the titration of butanoic acid should be close enough to the pH in the titration of propanoic acid to make the original indicator appropriate for the titration of butanoic acid.

1 point is earned for disagreeing with the student’s claim and making a valid justification using pKa, Ka, or pH arguments.
1 point is earned for numerically comparing either: the two pKa values, the two Ka values, or the two pH values at the equivalence point.