How to distinguish a solution of sodium sulphite from a solution of sodium sulphate in the lab

We have to distinguish a solution of sodium sulphite from a solution of sodium sulphate in the laboratory.

Solution of sodium sulphite is acidified with dilute HCl and to that when few drops of barium chloride (BaCl₂) solution is added, white precipitate is formed. The white precipitate is soluble in HCl.

To this solution 2 drops of iodine (I₂) solution is added and brown colour of iodine is discharged as I₂ gets reduced to HI.

The reactions involved in case of sodium sulphite is are:

Na₂SO₃ + BaCl₂ = BaSO₃ ↓ + 2NaCl

                         (white precipitate)

BaSO₃ + 2HCl = BaCl₂ + H₂SO₃

H₂SO₃ + I₂ + H₂O = H₂SO₄ + 2HI

On the other hand, solution of sodium sulphate is acidified with dilute HCl and to that when few drops of barium chloride (BaCl₂) solution is added, white precipitate is formed. The white precipitate of BaSO₄ is formed which is insoluble in HCl.

Na₂SO₄+ BaCl₂ = BaSO₄ ↓ + 2NaCl

                        (white precipitate)

a

explanation:

they are made of gases.

they have a spherical shape.

they make energy through fusion.