---
title: "A student investigates the relative acid strengths of three hypohalous acids at \\(298\\text{ K}\\). The identity of each acid and its acid-ionization constant, \\(K_a\\), are recorded in the table below.  | Acid | \\(K_a\\) at \\(298\\text{ K}\\) | | :— | :— | | \\(\\text{HOI(aq)}\\) | \\(2.3 \\times 10^{-11}\\) | | \\(\\text{HOBr(aq)}\\) | \\(2.8 \\times 10^{-9}\\) | | \\(\\text{HOCl(aq)}\\) | \\(3.0 \\times 10^{-8}\\) |  Which of the following correctly ranks the acids in order of increasing acid strength, and provides the correct justification?"
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url: "https://nerd-notes.com/ubq/119789/"
date_modified: "2026-08-21T08:12:21+00:00"
---

# A student investigates the relative acid strengths of three hypohalous acids at \(298\text{ K}\). The identity of each acid and its acid-ionization constant, \(K_a\), are recorded in the table below.

| Acid | \(K_a\) at \(298\text{ K}\) |
| :— | :— |
| \(\text{HOI(aq)}\) | \(2.3 \times 10^{-11}\) |
| \(\text{HOBr(aq)}\) | \(2.8 \times 10^{-9}\) |
| \(\text{HOCl(aq)}\) | \(3.0 \times 10^{-8}\) |

Which of the following correctly ranks the acids in order of increasing acid strength, and provides the correct justification?

A student investigates the relative acid strengths of three hypohalous acids at \(298\text{ K}\). The identity of each acid and its acid-ionization constant, \(K_a\), are recorded in the table below.

| Acid | \(K_a\) at \(298\text{ K}\) |
| :--- | :--- |
| \(\text{HOI(aq)}\) | \(2.3 \times 10^{-11}\) |
| \(\text{HOBr(aq)}\) | \(2.8 \times 10^{-9}\) |
| \(\text{HOCl(aq)}\) | \(3.0 \times 10^{-8}\) |

Which of the following correctly ranks the acids in order of increasing acid strength, and provides the correct justification?

- **A.** \(\text{HOI} < \text{HOBr} < \text{HOCl}\), because \(\text{I}\) has the largest atomic radius, which results in the strongest \(\text{O}-\text{I}\) bond and decreases the extent of ionization.
- **B.** \(\text{HOI} < \text{HOBr} < \text{HOCl}\), because \(\text{Cl}\) is the most electronegative halogen, which polarizes the \(\text{O}-\text{H}\) bond and facilitates the loss of \(\text{H}^+\).
- **C.** \(\text{HOCl} < \text{HOBr} < \text{HOI}\), because \(\text{I}\) has the largest atomic radius, which results in the weakest bond to hydrogen and the greatest extent of ionization.
- **D.** \(\text{HOCl} < \text{HOBr} < \text{HOI}\), because \(\text{HOI}\) has the smallest \(K_a\) value, which corresponds to the greatest extent of ionization at equilibrium.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119789/*