Question

Why is the concentrated sulphuric acid a weaker acid compared to dilute sulphuric acid?
a.) Less presence of $O{{H}^{-}}$
b.) Easily gives ${{H}^{+}}$
c.) Less presence of ${{H}^{+}}$
d.) All of the above

Answer 1

Hint: We know that strength of an acid depends on the concentration of hydronium ions present in the solution. We know that sulphuric acid is a very strong acid.

Complete answer:
The chemical formula of sulphuric acid is ${{H}_{2}}S{{O}_{4}}$. It can give two ${{H}^{+}}$ ions. And it is a strong acid that completely dissociates to ${{H}^{+}}$and ions. Due to the presence of a higher amount of water sulphuric acid completely dissociates in dilute sulphuric acid. So, it has a higher amount of hydronium (${{H}^{+}}$) ion. Whereas in the presence of a higher amount of sulphuric acid, sulphuric acid doesn't dissociate completely into ions in aqueous solution. And here we know that the strength of acidity depends on the strength of ${{H}^{+}}$ions. So, in dilute solution ${{H}^{+}}$ are more and in concentrated solution ${{H}^{+}}$are less. So, concentrated sulphuric acid is less acidic than dilute sulphuric acid. Below given is the dissociation reaction of sulphuric acid:
In concentrated sulphuric acid:
${{H}_{2}}S{{O}_{4}}(l)+2{{H}_{2}}O\overset{{}}\to{{H}_{3}}{{O}^{+}}(aq)+HSO_{4}^{1}(aq)$
In dilute solution:
${{H}_{2}}S{{O}_{4}}(l)+2{{H}_{2}}O\overset{{}}\to3{{H}_{3}}{{O}^{+}}(aq)+SO_{4}^{2-}(aq)$.

So, from the above explanation we can say that option “C”.

Note: The new bond that results in the formation of ${{H}_{3}}{{O}^{+}}$is much stronger than the H-O bond in ${{H}_{2}}S{{O}_{4}}$ as evidenced by the very exothermic reaction that occurs when concentrated ${{H}_{2}}S{{O}_{4}}$ is added to water.