Acetic acid ionizes in water as follows:
CH3COOH + H2O --> CH3COO– + H3O+
<--
Fewer than 1% of ethanoic acid molecules are ionized at any instant. The acetate ion (CH3COO– ) is therefore ____.
A.a poor hydrogen-ion acceptor
B.a good hydrogen-ion acceptor
C.a poor hydrogen-ion donor
D.a good hydrogen-ion donor
I believe this is D?
A good hydrogen ion donor, I believe, would show more than 1% ionization. Therefore, CH3OOH is a weak acid, and a poor hydrogen-ion donor.
ok...thank you
You are correct! The acetate ion (CH3COO–) is a good hydrogen-ion donor. In the ionization of acetic acid, it donates a hydrogen ion (H+) to form the hydronium ion (H3O+). This means that the acetate ion acts as an acid, releasing hydrogen ions into the solution.
To determine the answer, we need to understand the behavior of the acetate ion (CH3COO–) in water.
In the given equation, acetic acid (CH3COOH) donates a proton (H+) to water, forming the hydronium ion (H3O+) and the acetate ion (CH3COO–). We are told that fewer than 1% of ethanoic acid molecules are ionized at any instant. This means that the majority of acetic acid molecules remain unionized (CH3COOH).
Since the acetate ion is formed as a result of the ionization of acetic acid, and only a small percentage of acetic acid molecules ionize, we can conclude that the acetate ion (CH3COO–) is a poor hydrogen-ion donor.
Therefore, the answer is C) a poor hydrogen-ion donor.