---
title: "A student investigates the relative acid strengths of three halogen-substituted acetic acids in aqueous solution at \\(25\\ ^\\circ\\text{C}\\): fluoroacetic acid (\\(\\text{CH}_2\\text{FCOOH}\\)), chloroacetic acid (\\(\\text{CH}_2\\text{ClCOOH}\\)), and bromoacetic acid (\\(\\text{CH}_2\\text{BrCOOH}\\)). Which of the following lists the acids in order of decreasing acid strength and provides the correct chemical justification?"
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url: "https://nerd-notes.com/ubq/121391/"
date_modified: "2026-08-23T05:02:03+00:00"
---

# A student investigates the relative acid strengths of three halogen-substituted acetic acids in aqueous solution at \(25\ ^\circ\text{C}\): fluoroacetic acid (\(\text{CH}_2\text{FCOOH}\)), chloroacetic acid (\(\text{CH}_2\text{ClCOOH}\)), and bromoacetic acid (\(\text{CH}_2\text{BrCOOH}\)). Which of the following lists the acids in order of decreasing acid strength and provides the correct chemical justification?

A student investigates the relative acid strengths of three halogen-substituted acetic acids in aqueous solution at \(25\ ^\circ\text{C}\): fluoroacetic acid (\(\text{CH}_2\text{FCOOH}\)), chloroacetic acid (\(\text{CH}_2\text{ClCOOH}\)), and bromoacetic acid (\(\text{CH}_2\text{BrCOOH}\)). Which of the following lists the acids in order of decreasing acid strength and provides the correct chemical justification?

- **A.** \(\text{CH}_2\text{BrCOOH} > \text{CH}_2\text{ClCOOH} > \text{CH}_2\text{FCOOH}\), because the larger atomic radius and greater polarizability of \(\text{Br}\) stabilize the negative charge on the conjugate base more effectively than the smaller halogens.
- **B.** \(\text{CH}_2\text{BrCOOH} > \text{CH}_2\text{ClCOOH} > \text{CH}_2\text{FCOOH}\), because the weaker \(\text{C}-\text{Br}\) bond cleaves more readily during ionization to facilitate proton release.
- **C.** \(\text{CH}_2\text{FCOOH} > \text{CH}_2\text{ClCOOH} > \text{CH}_2\text{BrCOOH}\), because the greater electronegativity of \(\text{F}\) withdraws electron density through inductive effects, dispersing the negative charge and stabilizing the conjugate base.
- **D.** \(\text{CH}_2\text{FCOOH} > \text{CH}_2\text{ClCOOH} > \text{CH}_2\text{BrCOOH}\), because the greater electronegativity of \(\text{F}\) concentrates additional electron density onto the carboxyl oxygen, repelling the proton from the \(\text{O}-\text{H}\) bond.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121391/*