Chemistry Question F.5

Seem that this is a HKCEE MC question.
The answer should be B. acidified potassium permanganate solution.

barium chloride solution, it will react with sodium sulphite solution and sodium sulphate solution to give white insoluble barium sulphite and barium sulphate respectively.

Chemical equations:
BaCl2(aq) + Na2SO3(aq) → BaSO3(s) + 2NaCl(aq)
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)

As the two products have the similiar appearance, so barium chloride solution is not an appropriate chemical to distinguish between sodium sulphite solution and sodium sulphate solution. So, answer 1 is eliminated.

For answer 3 potassium iodine solution, the chemical equations go like these:
2KI(aq) + Na2SO3(aq) → K2SO3(aq) + 2NaI(aq)
2KI(aq) + Na2SO4(aq) → K2SO4(aq) + 2NaI(aq)
The reactants and the products are all colourless, so no observable change can be detected, and hence it cannot distinguish between the two solutions. So 3 is eliminated.

For 2 acidified potassium permanganate solution. It is an oxidizing agent. It will react with sodium sulphite solution by oxidizing it. However, it will not react with sodium sulphate solution as S in sulphate has the maximum oxidizing number +6, it cannot be further oxidized.
Chemical equations:
SO3 2-(aq) + MnO4-(aq) + 6H+(aq) + 3e- → SO4 2-(aq) + Mn2+(aq) + 3H2O(l)
The acidified potassium permanganate solution is decolourized when it is added to sodium sulphite solution while it has no colour change (remains purple) when added to sodium sulphate solution. So, only 2 is the answer.