Question

Consider the following reaction involving two acids shown below : formic acid and HF.

Which of the following statements about this reaction are true?

• A. Formic acid is the strongest Bronsted acid in the reaction
• B. HF is the strongest Bronsted acid in the reaction
• C. KF is the strongest Bronsted base in the reaction
• D. $KO_2CH$ is the strongest Bronsted base in the reaction
• E. The equilibrium favours the reactants

Answer 1

Answer: (B), (D), (E)

B, D, E

Answer 2

The reaction is: $KF + HCOOH \rightleftharpoons K^+HCO_2^- + HF$ This can be written in terms of ions and molecules as: $F^- + HCOOH \rightleftharpoons HCO_2^- + HF$

We are given the $pK_a$ values: $pK_a(HCOOH) = 3.8$ $pK_a(HF) = 3.2$

1. Acid Strength Comparison: A lower $pK_a$ value indicates a stronger acid. Since $pK_a(HF) = 3.2$ and $pK_a(HCOOH) = 3.8$, we have $pK_a(HF) < pK_a(HCOOH)$. Therefore, $HF$ is a stronger acid than $HCOOH$.

2. Base Strength Comparison: The conjugate base of $HCOOH$ is $HCO_2^-$ (formate ion). The conjugate base of $HF$ is $F^-$ (fluoride ion). The strength of a conjugate base is inversely related to the strength of its conjugate acid. Since $HF$ is a stronger acid than $HCOOH$, its conjugate base $F^-$ is a weaker base than the conjugate base of $HCOOH$, which is $HCO_2^-$. Thus, $HCO_2^-$ is a stronger base than $F^-$.

3. Equilibrium Position: In an acid-base reaction, the equilibrium favors the formation of the weaker acid and the weaker base. In this reaction, $HCOOH$ is the weaker acid and $F^-$ is the weaker base. Therefore, the equilibrium favors the reactants ($HCOOH + F^-$).

Evaluating the statements:

• (A) Formic acid is the strongest Bronsted acid in the reaction: False. $HF$ is the stronger acid ($pK_a = 3.2$) compared to formic acid ($pK_a = 3.8$).
• (B) HF is the strongest Bronsted acid in the reaction: True. As established, $HF$ has a lower $pK_a$ than $HCOOH$, making it the stronger acid.
• (C) KF is the strongest Bronsted base in the reaction: False. $KF$ dissociates to give $K^+$ and $F^-$. $F^-$ is a Bronsted base. The other base in the reaction is $HCO_2^-$. Since $HCO_2^-$ is a stronger base than $F^-$, $KF$ (which provides $F^-$) is not the strongest Bronsted base.
• (D) $KO_2CH$ is the strongest Bronsted base in the reaction: True. $KO_2CH$ (potassium formate) dissociates to give $K^+$ and $HCO_2^-$. $HCO_2^-$ is the conjugate base of $HCOOH$. As determined, $HCO_2^-$ is the stronger base in this reaction.
• (E) The equilibrium favours the reactants: True. The equilibrium shifts towards the side of the weaker acid ($HCOOH$) and weaker base ($F^-$), which are the reactants.