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)
