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given the equation representing a reversible reaction:

CH3COOH(aq) + H2O <—> CH3COO-(aq) + H3O+(aq)

according to one acid-base theory, the two H+ donors in the equation are

Respuesta :

sebassandin

Answer:

- Acetic acid (CH₃COOH) and hydronium ion (H₃O⁺)

Explanation:

Hello,

In this case, based on the acid-base theory which states that acids are known as H⁺ donors, if we consider the direct reaction:

[tex]CH_3COOH(aq) + H_2O \rightarrow CH_3COO^-(aq) + H_3O^+(aq)[/tex]

It is clear that the acetic acid is the first H⁺ donor as it losses one H⁺ to turn into the acetate ion. Moreover, if we consider the inverse reaction:

[tex]CH_3COO^-(aq) + H_3O^+(aq)\rightarrow CH_3COOH(aq) + H_2O[/tex]

It is also clear that the hydronium ion is the second H⁺ donor as it losses one H⁺ to turn into water.

Best regards.

mickymike92

Based on the reversible reaction, the two H+ donors in the equation are acetic acid, CH3COOH and hydronium ion H3O+.

What are protons?

Protons are hydrogen atoms which have lost an electron to form hydrogen ions.

Acids according to the Bronsted-Lowry definition is a s substance which donates protons.

In the reversible react ion given, the H+ donor is acetic acid, CH3COOH.

The H+ donor in the backward reaction is the hydronium ion H3O+.

Therefore, the two H+ donors in the equation are acetic acid, CH3COOH and hydronium ion H3O+.

Learn more about protons at: https://brainly.com/question/15309966

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