---
title: "A student prepares a solution of acetic acid, \\(\\text{CH}_3\\text{COOH(aq)}\\), by dissolving enough acid in water to create a \\(0.18 \\text{ M}\\) solution at \\(25^\\circ\\text{C}\\). Given that the acid dissociation constant for acetic acid is \\(K_a = 1.8 \\times 10^{-5}\\) at \\(25^\\circ\\text{C}\\), what is the percent ionization of acetic acid in this solution?"
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url: "https://nerd-notes.com/ubq/119279/"
date_modified: "2026-08-19T12:39:38+00:00"
---

# A student prepares a solution of acetic acid, \(\text{CH}_3\text{COOH(aq)}\), by dissolving enough acid in water to create a \(0.18 \text{ M}\) solution at \(25^\circ\text{C}\). Given that the acid dissociation constant for acetic acid is \(K_a = 1.8 \times 10^{-5}\) at \(25^\circ\text{C}\), what is the percent ionization of acetic acid in this solution?

A student prepares a solution of acetic acid, \(\text{CH}_3\text{COOH(aq)}\), by dissolving enough acid in water to create a \(0.18 \text{ M}\) solution at \(25^\circ\text{C}\). Given that the acid dissociation constant for acetic acid is \(K_a = 1.8 \times 10^{-5}\) at \(25^\circ\text{C}\), what is the percent ionization of acetic acid in this solution?

- **A.** \(0.0018\%\)
- **B.** \(0.10\%\)
- **C.** \(1.0\%\)
- **D.** \(10\%\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119279/*