Conjugate Acid/Base pair (1 Viewer)

BlueGas

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The answer for this question is D, but can't it be A also? Aren't they both the same thing?

 

Recondit

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Maybe the choices had to be listed repsectively as (acid/conjugate acid) and (conjugate base/base) as the the question says? That would make D the answer and not A
 

InteGrand

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The answer for this question is D, but can't it be A also? Aren't they both the same thing?

I think by the question and answers' wording (and judging by your comment that D is the answer), they're listing the acid first in the answers (otherwise, (A) and (D) would be identical).

So for example, option (A) says that OH is the acid (because it was the first one listed in its pair), which is false.
 

BlueGas

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I think by the question and answers' wording (and judging by your comment that D is the answer), they're listing the acid first in the answers (otherwise, (A) and (D) would be identical).

So for example, option (A) says that OH is the acid (because it was the first one listed in its pair), which is false.
Yeah I get it now, because for A, if CO3 is the acid HCO3 wouldn't because it's conjugate base because a conjugate base is the original acid but with a proton removed.
 

someth1ng

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I think by the question and answers' wording (and judging by your comment that D is the answer), they're listing the acid first in the answers (otherwise, (A) and (D) would be identical).

So for example, option (A) says that OH is the acid (because it was the first one listed in its pair), which is false.
I agree with what you're saying but I think it's a pretty stupid way of making a question - it's too ambiguous and they really should be using a table for something like this.
 

MaccaFacta

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The Bronsted-Lowry definitions for this type of equilibrium say that an acid + base <> conjugate acid + conjugate base. So, in the equation as it is written, H2O is the acid, CO3 2- is the base, HCO3 - is the conjugate acid and OH- is the conjugate base. Speaking as a teacher, I won't be copying and pasting this question into the bank of questions I have on this topic.
 

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