Final answer:
The ratio of [A^-]/[HA] at pH 2.75, given the pKa of formic acid is 3.75, is 0.1. The structure of HA for formic acid is H - C(=O) - O - H.
Step-by-step explanation:
To calculate the ratio of [A^-]/[HA] at a given pH, we use the Henderson-Hasselbalch equation, defined as:
\(\text{pH} = \text{pKa} + \log\left(\frac{\text{[A^-]}}{\text{[HA]}}\right)\)
Rearranging this to solve for the ratio gives:
\(\text{[A^-]/[HA]} = 10^{(\text{pH} - \text{pKa})}\)
Given the pH is 2.75 and the pKa of formic acid is 3.75, we can calculate the ratio as follows:
\(\text{[A^-]/[HA]} = 10^{(2.75 - 3.75)} = 10^{-1} = 0.1\)
The structure of HA for formic acid (H-COOH) is:
H - C(=O) - O - H
This structure shows the central carbon atom double-bonded to one oxygen atom and single-bonded to a hydroxyl group (OH) and a hydrogen atom.
This result makes sense because the [A^-]/[HA] ratio is between 1 and 10, so the pH of the buffer must be between the pKa and pKa + 1.