Equilibria: Question 8

Worked solution

Step 1: Recall the definition of a conjugate acid-base pair

In the Brønsted-Lowry theory, an acid is a proton ($\text{H}^+$) donor and a base is a proton acceptor. A conjugate acid-base pair consists of two species that differ from each other by exactly one proton. The acid, and the base left behind once it has donated that proton (or, equivalently, the base and the acid formed once it accepts a proton).

Step 2: Identify each species’ role in the forward reaction

$\text{HCOOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HCOO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$

• $\text{HCOOH}$ donates a proton to become $\text{HCOO}^-$, so $\text{HCOOH}$ is the acid and $\text{HCOO}^-$ is its conjugate base.
• $\text{H}_2\text{O}$ accepts that proton to become $\text{H}_3\text{O}^+$, so $\text{H}_2\text{O}$ is the base and $\text{H}_3\text{O}^+$ is its conjugate acid.

Step 3: Check which listed pair differs by exactly one proton

$\text{HCOOH}$ and $\text{HCOO}^-$ differ by exactly one $\text{H}^+$: removing a proton from $\text{HCOOH}$ gives $\text{HCOO}^-$ directly. This is the conjugate acid-base pair formed on the “acid side” of the reaction.

Why the other options are wrong

• B: $\text{H}_2\text{O}$ and $\text{HCOO}^-$ are on the same (product) side conceptually unrelated by a single proton transfer between each other, they belong to different conjugate pairs ($\text{H}_2\text{O}/\text{H}_3\text{O}^+$ and $\text{HCOOH}/\text{HCOO}^-$).
• C: $\text{HCOOH}$ and $\text{H}_3\text{O}^+$ are both acids in this equilibrium (one on each side); neither is the conjugate of the other.
• D: $\text{H}_2\text{O}$ and $\text{HCOOH}$ are both proton donors/acceptors in different pairs but are not themselves related by the loss or gain of a single proton.

Final answer

• $\text{HCOOH}$ and $\text{HCOO}^-$, option A.