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The question asks for the equilibrium dissociation equation for the dissociation of the weak acid CH₃COOH (acetic acid).

Acetic acid dissociates in water as follows:

\[
\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+
\]

The equilibrium constant expression for this dissociation (which is the acid dissociation constant, \(K_a\)) is given by:

\[
K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]}
\]

This matches the first option:

\[
\boxed{\frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]}}
\]

Would you like further details on how equilibrium constants are calculated, or any additional explanations?

Here are 5 related questions for further learning:
1. How is the \(K_a\) value of a weak acid like CH₃COOH determined experimentally?
2. What is the significance of the \(K_a\) value in understanding the strength of a weak acid?
3. How does the presence of water influence the dissociation of acetic acid?
4. Can you explain how Le Chatelier’s principle applies to the dissociation of CH₃COOH?
5. How would the dissociation of a strong acid differ in terms of its equilibrium expression?

**Tip**: In equilibrium expressions, the concentrations of pure solids and liquids (like water) are not included since their concentrations do not change.

What is the equilibrium dissociation equation for the dissociation of the weak acid CH₃COOH?

Understand the equilibrium dissociation equation for the weak acid CH₃COOH (acetic acid), including the key concepts of acid dissociation, equilibrium constants (Kₐ), and how to express them using chemical formulas.

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