Question

Which salt responds to dilute and concentrated ${{H}_{2}}S{{O}_{4}}$:
[A] $Cs{{F}_{2}}$
[B] $Ba{{(N{{O}_{3}})}_{2}}$
[C] $N{{a}_{2}}S{{O}_{4}}$
[D] $N{{a}_{3}}P{{O}_{4}}$

Answer 1

Hint: When we add an acid to a salt, a new acid and a new salt is formed. The reaction differs depending upon the concentration of the acid added. When sulphuric acid is added to one of the given salts, it forms a sulphate precipitate and that will be the correct answer.

Complete step by step answer:
Sulphuric acid is a strong acid and depending on its concentration, it forms different salts.
We generally use sulphuric acid as an oxidizing agent, but it is also used as a dehydrating agent.
Concentrated sulphuric acid is generally 98% sulphuric acid and 2% water whereas dilute sulphuric acid is 80% to 90% water.
 Let us go through each option to see which salt responds to both dilute and concentrated sulphuric acid.

In option [D] we have $N{{a}_{3}}P{{O}_{4}}$-
When aqueous sulphuric acid is added to trisodium phosphate, we get sodium sulphate and phosphoric acid as the product. We can write the reaction as-$3{{H}_{2}}S{{O}_{4}}(aq)+2N{{a}_{2}}P{{O}_{4}}(aq)\to 3N{{a}_{2}}S{{O}_{4}}(aq)+2{{H}_{3}}P{{O}_{4}}(aq)$

In option [C] we have $N{{a}_{2}}S{{O}_{4}}$-
When we add aqueous sulphuric acid to sodium sulphate, we get sodium hydrogen sulphate. We can write the reaction as-
$N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}S{{O}_{4}}(aq)\to 2NaHS{{O}_{4}}(aq)$
In option [B] we have $Ba{{(N{{O}_{3}})}_{2}}$-

When we add sulphuric acid to barium nitrate, we get a precipitate of barium sulphate and it reacts to both aqueous and dilute sulphuric acid. We can write the reactions as-
$Ba{{(N{{O}_{3}})}_{2}}+{{H}_{2}}S{{O}_{4}}(aq)\to BaS{{O}_{4}}\downarrow +2HN{{O}_{3}}(aq)$

Barium sulphate is insoluble in water. Therefore, we get a white precipitate of barium sulphate in aqueous sulphuric acid. However, if we add concentrated sulphuric acid in barium sulphate, the same reaction takes place but we cannot see the precipitate because in the product we obtain nitric acid which is miscible and the barium sulphate also dissolves in the solution. We can write the reaction as-
     $Ba{{(N{{O}_{3}})}_{2}}+{{H}_{2}}S{{O}_{4}}(conc.)\to BaS{{O}_{4}}\downarrow +2HN{{O}_{3}}$

As barium nitrate reacts to both concentrated and dilute sulphuric acid it is the correct answer.
Therefore, the correct answer is $Ba{{(N{{O}_{3}})}_{2}}$
So, the correct answer is “Option B”.

Note: Even though barium nitrate forms no precipitate in the concentrated sulphuric acid solution but still it is the correct answer because it responds in addition to concentrated as well as dilute sulphuric acid. As we know concentrated sulphuric acid also has some amount of water therefore, we may be able to see slight precipitation in it even though it is generally not visible.