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 posted this question ealier and have reread some information and I am getting 2 answers between B and C?
Can anyone clarify? Thank you in advance...
Hello,
A is NOT the correct answer, i just took this test and it is...
Fewer than 1% of ethanoic acid molecules are ionized at any instant. The acetate ion (CH3COO– ) is therefore ____. (1 point)
(0 pts) a poor hydrogen-ion acceptor
(1 pt) a good hydrogen-ion acceptor
(0 pts) a poor hydrogen-ion donor
(0 pts) a good hydrogen-ion donor
A Poor hydrogen ion donor
To determine whether the acetate ion (CH3COO–) is a poor hydrogen-ion acceptor or donor, we need to consider its behavior in water and its ability to accept or donate hydrogen ions.
The equation given shows that acetic acid (CH3COOH) donates a proton (H+) to water, forming the acetate ion (CH3COO–) and the hydronium ion (H3O+). Since acetic acid only ionizes to a small extent (less than 1%), the concentration of the hydronium ion is relatively low compared to the concentration of acetic acid molecules.
Considering this, we can conclude that the acetate ion is a poor hydrogen-ion acceptor (option A) because it does not readily acquire hydrogen ions from the solution. The majority of the acetate ions will remain as acetate ions and not accept hydrogen ions.
Therefore, the correct answer is A. Acetate ion (CH3COO–) is a poor hydrogen-ion acceptor.
Correction, it would be a poor hydrogen-ion acceptor.
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Take a look at wisconsin's website to understand what the correct answer will be.