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$HCl{{O}_{4}}$, $HN{{O}_{3}}$ and $HCl$ are all very strong acids in aqueous solutions. In a glacial acetic acid medium, their acid strength is such that:
(A) $HCl{{O}_{4}}>HN{{O}_{3}}>HCl$
(B) $HN{{O}_{3}}>HCl{{O}_{4}}>HCl$
(C) $HCl>HCl{{O}_{4}}>HN{{O}_{3}}$
(D) $HCl>HCl{{O}_{4}}\sim HN{{O}_{3}}$

Answer
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Hint: Acid strength means the tendency of acid to release ${{H}^{+}}$ ions very easily. Glacial acetic is a weak acid. Acetic acid dissociates partially and it will be in equilibrium with the dissociated products in the solution.

bComplete step by step solution:
-In the question, it is given that there are three strong acids.
-The given strong acids are $HCl{{O}_{4}}$, $HN{{O}_{3}}$ and $HCl$.
-The names of the given acids are
$HCl{{O}_{4}}$ - Perchloric acid
$HN{{O}_{3}}$ - Nitric acid
$HCl$ - Hydrochloric acid
-Generally Hydrochloric acid is stronger than nitric acid.
-Glacial acetic acid exists in following equilibrium that is why it is called weak acid .$C{{H}_{3}}COOHC{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$
-But in glacial acetic acid medium, there is a presence of ${{H}^{+}}$ ions released by acetic acid.
-The presence of ${{H}^{+}}$ ions disturbs the acidity of the hydrochloric acid. Means ${{H}^{+}}$ from hydrochloric acid release very quickly but the ${{H}^{+}}$ ions in the solution due to the weak acid (Acetic acid) combines with the counterion of the ${{H}^{+}}$ ion released by hydrochloric acid.
-Therefore hydrochloric acid acts as a weak acid in acetic acid medium.
-It also applies to nitric acid.
-Generally perchloric acid is a weak acid but in the acetic acid medium, it is so strong.
-Therefore in glacial acetic acid medium, the acidic strengths of the given acids are as follows.
$HCl{{O}_{4}}>HN{{O}_{3}}>HCl$

So, the correct option is (A).

Note: Generally the acidic strength of the given strong acids is as follows.
$HCl>HN{{O}_{3}}>HCl{{O}_{4}}$
But due to the liberation of ${{H}^{+}}$ ion by the acetic acid, the acidic strength of the given acids changed in the acetic acid medium as follows.
$HCl{{O}_{4}}>HN{{O}_{3}}>HCl$