Answer
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Hint: According to the Bronsted theory, a species which will give a proton is an acid. Here, the correct answer has no proton to donate and is therefore not Bronsted acid. It is the conjugate base of a weak acid, acetic acid.
Complete step by step answer:
We have known several theories which explain the concept of acid and bases to us like Arrhenius Theory, Solvent System Theory and many more. These theories explain acid and bases in different ways. One such theory is the Bronsted Lowry Theory.
Bronsted Lowry theory is also known as the protonic theory because it explains the concept of acid and bases on the bases of a proton or hydrogen ion or ${{H}^{+}}$.
According to the Bronsted Lowry theory, “any species that tends to give up a proton (hydron) is an acid, and any substance that tends to accept a proton (hydron) is a base”
Now we will go through each option to check whether it is a Bronsted acid or not.
In the last option we have, $HS{{O}_{3}}^{-}$. In aqueous medium, we can write that-
$HS{{O}_{3}}^{-}+{{H}_{2}}O\to S{{O}_{3}}^{2-}+{{H}_{3}}{{O}^{-}}$
As it can donate a proton therefore, it is a Bronsted acid.
In the third option, we have $HC{{O}_{3}}^{-}$. Again, we can write that-
$HC{{O}_{3}}^{-}+{{H}_{2}}O\to C{{O}_{3}}^{2-}+{{H}_{3}}{{O}^{+}}$
$HC{{O}_{3}}^{-}$ also donates a proton therefore it is also a Bronsted acid.
In the second option we have, $C{{H}_{3}}CO{{O}^{-}}$
It has no proton available to donate therefore it is not a Bronsted acid.
And in the first option we have $N{{H}_{4}}^{+}$, we can write the reaction as-
$N{{H}_{4}}^{+}+{{H}_{2}}O\to N{{H}_{3}}+{{H}_{3}}{{O}^{+}}$
Therefore, it is also a Bronsted acid.
Among the given options, only $C{{H}_{3}}CO{{O}^{-}}$ has no proton to donate and hence is not a Bronsted acid.
Therefore, the correct answer is option [B] $C{{H}_{3}}CO{{O}^{-}}$.
Note:
$C{{H}_{3}}CO{{O}^{-}}$ is the correct option here because it is a Bronsted base and not an acid as it will accept a proton but has none to donate. It is important to remember here that $HC{{O}_{3}}^{-}$ is a Bronsted acid as well as a Bronsted base because it can accept proton and become ${{H}_{2}}C{{O}_{3}}$ as well as donate proton and become $C{{O}_{3}}^{2-}$.
Complete step by step answer:
We have known several theories which explain the concept of acid and bases to us like Arrhenius Theory, Solvent System Theory and many more. These theories explain acid and bases in different ways. One such theory is the Bronsted Lowry Theory.
Bronsted Lowry theory is also known as the protonic theory because it explains the concept of acid and bases on the bases of a proton or hydrogen ion or ${{H}^{+}}$.
According to the Bronsted Lowry theory, “any species that tends to give up a proton (hydron) is an acid, and any substance that tends to accept a proton (hydron) is a base”
Now we will go through each option to check whether it is a Bronsted acid or not.
In the last option we have, $HS{{O}_{3}}^{-}$. In aqueous medium, we can write that-
$HS{{O}_{3}}^{-}+{{H}_{2}}O\to S{{O}_{3}}^{2-}+{{H}_{3}}{{O}^{-}}$
As it can donate a proton therefore, it is a Bronsted acid.
In the third option, we have $HC{{O}_{3}}^{-}$. Again, we can write that-
$HC{{O}_{3}}^{-}+{{H}_{2}}O\to C{{O}_{3}}^{2-}+{{H}_{3}}{{O}^{+}}$
$HC{{O}_{3}}^{-}$ also donates a proton therefore it is also a Bronsted acid.
In the second option we have, $C{{H}_{3}}CO{{O}^{-}}$
It has no proton available to donate therefore it is not a Bronsted acid.
And in the first option we have $N{{H}_{4}}^{+}$, we can write the reaction as-
$N{{H}_{4}}^{+}+{{H}_{2}}O\to N{{H}_{3}}+{{H}_{3}}{{O}^{+}}$
Therefore, it is also a Bronsted acid.
Among the given options, only $C{{H}_{3}}CO{{O}^{-}}$ has no proton to donate and hence is not a Bronsted acid.
Therefore, the correct answer is option [B] $C{{H}_{3}}CO{{O}^{-}}$.
Note:
$C{{H}_{3}}CO{{O}^{-}}$ is the correct option here because it is a Bronsted base and not an acid as it will accept a proton but has none to donate. It is important to remember here that $HC{{O}_{3}}^{-}$ is a Bronsted acid as well as a Bronsted base because it can accept proton and become ${{H}_{2}}C{{O}_{3}}$ as well as donate proton and become $C{{O}_{3}}^{2-}$.
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