Question

Identify the anion present in the following compounds:
Compound L on reacting with barium chloride solution gives a white precipitate insoluble in dilute hydrochloric acid or dilute nitric acid.

Answer 1

Hint: When an ionic compound is formed there are two parts in the compound, the one with a positive charge is known as the cation and the one with a negative charge is known as the anion. So we have to find the negative part in the compound.

Complete Solution :
We can say here the compound is ${{X}_{2}}S{{O}_{4}}$ (valency of X = 1). Because, when the $BaC{{l}_{2}}$(Barium chloride) solution is added to${{X}_{2}}S{{O}_{4}}$ , a white precipitates of $BaS{{O}_{4}}$(Barium sulfate) is formed.
- This white precipitate formed is insoluble in dilute HCl. This test is used for the identification of sulfate ($SO_{4}^{2-}$) radical.
 - To perform this test we have to add a small amount of barium chloride solution to a solution of the test salt. If there is a formation of white precipitate, the salt could be either sulfate or carbonate. Then we have to add dilute HCl in the white precipitate if it does not dissolve, it has a sulfate anion.
Here we can see the reaction involved in this process –
$SO_{4}^{2-}(aq)+B{{a}^{2+}}(aq)+C{{l}^{-}}(aq)\to BaS{{O}_{4}}(s)+C{{l}^{-}}(aq)$
 Therefore, we can conclude that the anion is sulfate ($SO_{4}^{2-}$).

Note: Here we should be aware of the fact that carbonate ($CO_{3}^{2-}$) ions also give a white precipitate of barium carbonate ($BaC{{O}_{3}}$) with barium chloride ($BaC{{l}_{2}}$). But this compound is soluble in dilute HCl. This is due to the fact that when the carbonate is reacted with dilute HCl, there is formation of carbonic acid (${{H}_{2}}C{{O}_{3}}$), which a very stable compound.