Question

Assertion: We can distinguish between solutions of dil.Sulphuric acid with dil.HCl and dil.$HN{{O}_{3}}$.
Reason: With Barium Chloride solution sulphuric acid will give white precipitate while hydrochloric acid and nitric acid will not give any precipitate.
A) Both assertion and reason are correct and reason is the correct explanation for assertion.
B) Both assertion and reason are correct but reason is not the correct explanation for assertion.
C) Assertion is correct but reason is incorrect.
D) Both assertion and reason are incorrect.

Answer 1

Hint: Barium Chloride is undergoing double displacement reaction with dil.Sulphuric acid. The anions in $HCl$ and $BaC{{l}_{2}}$ are similar.

Complete answer:
In the question two statements, an assertion statement and the reason statement is given that supports the assertion. So now we have to say whether the statements are correct or wrong and select the correct answer from the above equations.

- So first let’s see the assertion statement. It is given that we can distinguish the solutions of dil.Sulphuric acid with dil.HCl and dil.$HN{{O}_{3}}$ and the reason for this is given as the Barium chloride forms white precipitate with sulphuric acid but no reaction with the nitric acid and chloride acid.

- Let’s check the reactions of $BaC{{l}_{2}}$ with dil.${{H}_{2}}S{{O}_{4}}$,$HN{{O}_{3}}$ and $HCl$
$BaC{{l}_{2}}$ Reacts with ${{H}_{2}}S{{O}_{4}}$ yielding insoluble product $BaS{{ O }_{4}}$. It is a white precipitate and retains in the solution. By this method we could distinguish dil.Sulphuric acid from others.
$BaC{{l}_{2\left( aq \right)}}+dil.{{H}_{2}}S{{O}_{4}}_{\left( aq \right)}\to BaS{{O}_{{{4}_{\left( S \right)}}}}+2HCl$
This type of reaction is a double displacement reaction in which the ions in the one compound is replaced by the ions of the other.

- Now let’s see how nitric acid is reacting with $BaC{{l}_{2}}$.
The $BaC{{l}_{2}}$ reacts with $HN{{O}_{3}}$, which yields $Ba{{(N{{O}_{3}})}_{2}}$ as the product, but is soluble in the solution i.e. all the ions are in the aqueous solution, so we cannot see any characteristic change during the reaction.
$BaC{{l}_{{{2}_{(aq)}}}}+2HN{{O}_{{{3}_{(aq)}}}}\to Ba{{(N{{O}_{3}})}_{{{2}_{\left( aq \right)}}}}+2HCl$

Hence we cannot distinguish $HN{{O}_{3}}$ from dil.${{H}_{2}}S{{O}_{4}}$, and $HCl$
Now let’s see the reaction of $BaC{{l}_{2}}$ with $HCl$. As the anions in the two compounds are similar, a reaction is not possible between them.
So $BaC{{l}_{2}}+HCl\to $ no reaction.

- Also we cannot distinguish between $HCl$ and other acids given.
So we can conclude that the assertion statement given in the question is correct and the reason given here is the correct explanation for the statement.

Therefore, the correct answer is “Option (A)”, Both assertion and reason are correct and reason is the correct explanation for assertion.

Note: For a chemical reaction, which is used as a distinguishing test between the samples given, then they should give any of the characteristic change –
-Either a gas should be evolved during the reaction
-A precipitate should be formed
-or a complex or non-dissociating product should be formed.
If we see any of the observations, then we could differentiate between the products with respect to the products formed.