What is the dissociation equation for acetic acid (CH3COOH) when it is placed in water?
The correct answer and explanation is:
Correct Answer:
CH₃COOH (aq) + H₂O (l) ⇌ CH₃COO⁻ (aq) + H₃O⁺ (aq)
Explanation:
Acetic acid (CH₃COOH) is a weak acid that partially dissociates in water. When acetic acid is added to water, it does not completely break apart into ions. Instead, a reversible reaction takes place. The molecule of acetic acid donates a proton (H⁺) to a water molecule. This results in the formation of a hydronium ion (H₃O⁺) and an acetate ion (CH₃COO⁻). The reaction is represented by a double arrow, showing that it is an equilibrium reaction and not one that goes to completion.
This equilibrium means that in solution, some of the acetic acid molecules remain undissociated while some exist as ions. The position of equilibrium depends on the strength of the acid and the concentration of the solution. Since acetic acid is a weak acid, only a small fraction of its molecules donate protons to water.
This process illustrates a fundamental concept in acid-base chemistry called Bronsted-Lowry theory. According to this theory, an acid is a substance that donates a proton, and a base is a substance that accepts a proton. In this reaction, CH₃COOH acts as the proton donor (acid), and H₂O acts as the proton acceptor (base), forming H₃O⁺.
The acetate ion that is formed, CH₃COO⁻, is the conjugate base of acetic acid. Hydronium, H₃O⁺, is the conjugate acid of water. The presence of H₃O⁺ ions in solution is responsible for the acidic nature of the solution.
Understanding this equilibrium is essential in predicting the pH of a solution and how acetic acid behaves in chemical reactions, especially in buffer systems and titration calculations.