Qualitative Analysis Self Test

Iron(III) ions would form a reddish-brown precipitate when aqueous sodium hydroxide solution is added dropwise to it. This is because iron(III) ions react with sodium hydroxide to form iron(III) hydroxide, which is a reddish-brown precipitate. Copper(II) ions do not form a reddish-brown precipitate, but rather a blue precipitate. Iron(II) ions and zinc ions do not form precipitates with sodium hydroxide.

When acidified barium nitrate solution is added to sulfate ions in solution, a white precipitate of barium sulfate is formed. This is because barium sulfate is insoluble in water and forms a solid precipitate when barium ions react with sulfate ions.

When aqueous ammonia is added to a solution of zinc nitrate until in excess and heated, a white precipitate is formed. This white precipitate is zinc hydroxide, which is insoluble in water. However, when excess aqueous ammonia is added, it dissolves the white precipitate to form a colorless solution. This is because zinc hydroxide reacts with excess aqueous ammonia to form a complex ion called tetraamminezinc(II) ion, which is soluble in water. On heating, a colorless and pungent gas, which is ammonia gas, is evolved due to the decomposition of the tetraamminezinc(II) ion.

Aluminium ions, Lead(II) ions, and Zinc ions would form a white precipitate that is soluble in excess sodium hydroxide solution. This is because these ions can react with sodium hydroxide to form their respective hydroxide compounds, which are white precipitates. However, these precipitates are soluble in excess sodium hydroxide solution due to the formation of soluble complex ions.

Chlorine and sulfur dioxide are the correct answers because both gases are colorless and have a pungent odor. Additionally, they both have the ability to turn moist blue litmus paper pink and bleach it.