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When barium chloride solution is added to salt solution E, a white precipitate insoluble in dilute hydrochloric acid is obtained. Identify the anion present in the compound.

Last updated date: 19th Jun 2024
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Hint: We know that precipitate is an insoluble solid. It comes from a liquid solution. When an insoluble solid comes out from a liquid solution, it is called precipitation. It can be seen as a suspension also. 2 soluble salts, when mixed, give an insoluble salt. This is precipitation.

Complete step by step answer:
We mentioned above that when two soluble salts are mixed, an insoluble solid would be formed.
When we look at the solubility rules, we can understand that most of the sulphates salts are soluble except barium, calcium and lead.
Similarly, most of the chloride salts are soluble except that of lead, silver and mercury.
In this question, barium chloride is added to E. We know that barium chloride is a soluble salt.
It is mentioned that when ${{BaC}}{{{l}}_{{2}}}$ and E reacts, an insoluble white precipitate is formed. We can say that E should also be soluble to give out precipitate which is an insoluble solid.
As mentioned above the product formed will be a salt of barium and in the salts of barium, sulphate salts are insoluble. Thus, the product can be barium sulphate.
${{BaC}}{{{l}}_{{2}}}$ is a neutral salt which in aqueous solution behaves as a simple salt.
Solution of ${{BaC}}{{{l}}_{{2}}}$ reacts with sulphate ion to produce a thick white precipitate of barium sulphate.
${{B}}{{{a}}^{{{2 + }}}}{{ + S}}{{{O}}_{{4}}}^{{{2 - }}}{{ }} \to {{ BaS}}{{{O}}_{{4}}}$ , where barium sulphate is thick white precipitate which is insoluble in dilute HCl
Therefore, E should be any soluble salts of sulphate. It can be both sodium or potassium sulphates.
${{BaC}}{{{l}}_{{2}}}{{(aq) + N}}{{{a}}_{{2}}}{{S}}{{{O}}_{{4}}}{{(aq)}} \to {{BaS}}{{{O}}_{{4}}}{{(s) + NaCl(aq)}}$ is an example of a precipitation reaction. It is a double displacement reaction. Since barium sulphate is insoluble in water, it precipitates.
We know that sulphate ions are insoluble in HCl since they are weak bases.
Thus, the anion present in sulphate ion ( ${{S}}{{{O}}_{{4}}}^{{{2 - }}}$)

Another reason for precipitate formation is due to the lowering of temperature. When the temperature is lowered, the solubility decreases and as a result precipitation occurs. This type of reaction can be used to detect sulphate ions. Precipitation reaction can be used to detect the presence of a particular ion in a solution. Thus, by knowing the solubility of some salts, we can prepare salt from another salt.