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AP Chemistry 5.11 Catalyst -Exam Style Questions- FRQs- New Syllabus

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Question

Answer the following questions about the glycolate ion, \( \mathrm{C_2H_3O_3^-} \), which acts as a base in aqueous solution. A Lewis diagram for the ion is provided.
A. On the Lewis diagram in part A, circle the atom that accepts the proton when the glycolate ion reacts with water.
When the glycolate ion reacts with water, it forms glycolic acid, \( \mathrm{HC_2H_3O_3} \), according to the following equation. The \( K_b \) expression for the reaction is provided.
\( \mathrm{C_2H_3O_3^-(aq) + H_2O(l) \rightleftharpoons HC_2H_3O_3(aq) + OH^-(aq)} \qquad K_b = \dfrac{[\mathrm{HC_2H_3O_3}][\mathrm{OH^-}]}{[\mathrm{C_2H_3O_3^-}]} \)
B. At \( 25^\circ \mathrm{C} \), a \( 2.5\ \mathrm{M} \) solution of glycolate is found to have \( [\mathrm{OH^-}] = 1.3 \times 10^{-5}\ \mathrm{M} \).
i. Calculate the value of \( K_b \) for the glycolate ion.
ii. Using your answer to part B(i), calculate the value of \( K_a \) for glycolic acid at \( 25^\circ \mathrm{C} \).
Glycolic acid can be produced from the hydrolysis of methyl glycolate, \( \mathrm{C_3H_6O_3} \). A proposed mechanism for the reaction is given.
Step \( 1 \): \( \mathrm{C_3H_6O_3 + H_3O^+ \rightleftharpoons C_2H_5O_3^+ + CH_3OH} \)
Step \( 2 \): \( \mathrm{C_2H_5O_3^+ + H_2O \rightleftharpoons HC_2H_3O_3 + H_3O^+} \)
Overall: \( \mathrm{C_3H_6O_3 + H_2O \rightleftharpoons HC_2H_3O_3 + CH_3OH} \)
C. A student claims that \( \mathrm{H_3O^+} \) is a catalyst for the reaction. Do you agree or disagree? Justify your answer based on the mechanism given.

Most-appropriate topic codes (AP Chemistry):

• Topic \( 2.6 \) — Resonance and Formal Charge — Part A
• Topic \( 8.3 \) — Weak Acid and Base Equilibria — Part B(i)
• Topic \( 8.3 \) — Weak Acid and Base Equilibria — Part B(ii)
• Topic \( 5.11 \) — Catalysis — Part C
▶️ Answer/Explanation

A
For a correct circled atom:
Accept one of the following:

Because of resonance, the two \( \mathrm{C-O} \) bonds on the right are equivalent.
So either of those two \( \mathrm{O} \) atoms can be accepted as the proton-accepting atom.

B(i)
For the correct calculated value:
\( K_b = \dfrac{[\mathrm{HC_2H_3O_3}][\mathrm{OH^-}]}{[\mathrm{C_2H_3O_3^-}]} \)

At equilibrium, \( [\mathrm{HC_2H_3O_3}] = [\mathrm{OH^-}] = 1.3 \times 10^{-5}\ \mathrm{M} \), and
\( [\mathrm{C_2H_3O_3^-}] = 2.5 – 1.3 \times 10^{-5}\ \mathrm{M} \).

\( K_b = \dfrac{(1.3 \times 10^{-5})(1.3 \times 10^{-5})}{2.5 – 1.3 \times 10^{-5}} \approx \dfrac{(1.3 \times 10^{-5})^2}{2.5} = 6.8 \times 10^{-11} \)

So, \( \boxed{K_b = 6.8 \times 10^{-11}} \).

B(ii)
For the correct calculated value, consistent with part B(i):
\( K_a = \dfrac{K_w}{K_b} \)

\( K_a = \dfrac{1.0 \times 10^{-14}}{6.8 \times 10^{-11}} = 1.5 \times 10^{-4} \)

Using \( K_w = K_aK_b \), divide \( K_w \) by the \( K_b \) value from part B(i).
Thus, \( \boxed{K_a = 1.5 \times 10^{-4}} \).

C
For the correct answer and a valid justification:
Agree. \( \mathrm{H_3O^+} \) is consumed in step \( 1 \) and regenerated in step \( 2 \), which is consistent with the behavior of a catalyst.

A catalyst participates in the mechanism but does not appear in the overall reaction.
Since \( \mathrm{H_3O^+} \) is used and then produced again, its net amount does not change.

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