Three unlabeled flasks contain solutions of sodium chloride, barium chloride and sodium iodide.

| March 15, 2021 | education

Three unlabeled flasks contain solutions of sodium chloride, barium chloride and sodium iodide. How can these substances be identified?

1. Divide the contents of the tubes into 2 parts. Silver ion is a qualitative reagent for halogen ions.

Let’s add the silver nitrate solution to the first part of the test tubes. With chlorine ions, it gives a white precipitate. This means that in those test tubes where barium chloride and sodium chloride are located, a white precipitate will fall out. With iodine ions, the silver ion gives a yellow precipitate – silver iodide.

To recognize in which test tube is barium chloride and which is sodium chloride, we use sulfuric acid. Barium ions and sulfate anions (Ba2 + and SO 2-4) form barium sulfate BaSO4, which will precipitate as a white precipitate. Sodium chloride will not react with sulfuric acid, since the reaction is reversible.

BaCl2 + 2AgNO3 = 2AgCl ↓ + Ba (NO3) 2.

NaCl + AgNO3 = AgCl ↓ + NaNO3.

NaI + AgNO3 = AgI ↓ + NaNO3.

BaCl2 + H2SO4 = BaSO4 ↓ + 2HCl.



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