Question

Identify the Anion present in the compound L on reacting with Barium chloride solution gives a white precipitate insoluble in dilute hydrochloric acid or dilute nitric acid?

Answer 1

The Anion present is Sulphate ion, SO₄^2-

The compound L is Sodium sulphate.

When Sodium sulphate reacts with barium chloride solution giving a white precipitate of barium sulphate insoluble in dilute hydrochloric acid or dilutes nitric acid.

Na₂SO₄ (aq) + BaCl₂(aq) ⇢ BaSO₄ (s) + 2NaCl(aq)

Na₂SO₄ on ionization gives Na⁺ and SO₄ ^2-.(In order for these two polyatomic ions to bond the charges must be equal and opposite, it will take two sodium ions to balance the one sulphate ion.)

BaCl₂ on ionization gives Ba⁺ and Cl^-.

On reaction with these two components(Sodium sulphate and barium chloride) The cation (Na+) combines with (Cl^-) anion .

Similarly the anion SO₄^2- combines with the cation Ba⁺.

Therefore the anion present in the given compound L is SO₄^2-.