Question

there are three bottles of unknown acids for analysis. One of the acids is sulfuric acid, H2SO4. one of the acids is nitric acid HNO3. one of the acids is hydrochloric acid, HCl. obtain a sample from each of the bottles and determine the identity of each acid. Describe the procedure you intend to use to perform this analysis

Answer 1

For the analysis of HCl we are performing the Silver nitrate test. The silver chloride is reacts with chloride ion it will forms white insoluble pricipitate of silver chloride molecule.

For analysis of the H2SO4 we are using Barium chloride test. The barium chloride is react with sulfate ion in sulfuric acid to produce the white insoluble pricipitate of the barium sulfate.

For analysis of the HNO3 we are performing the brown ring test. The brown cloured ring is formed at the two layer junction.

TEST FOR CHLORIDE (HCl)

Using silver nitrate solution

Carrying out the test

This test has to be done in solution. If you start from a solid, it must first be dissolved in pure water.

The solution is acidified by adding dilute nitric acid. (Remember: silver nitrate + dilute nitric acid.) The nitric acid reacts with, and removes, other ions that might also give a confusing precipitate with silver nitrate.

Silver nitrate solution is then added to give:

ion present	observation
F-	no precipitate
Cl-	white precipitate
Br-	very pale cream precipitate
I-	very pale yellow precipitate

The chloride, bromide and iodide precipitates are shown in the photograph:

The chloride precipitate is obviously white, but the other two aren't really very different from each other. You couldn't be sure which you had unless you compared them side-by-side.

All of the precipitates change colour if they are exposed to light - taking on grey or purplish tints.

The absence of a precipitate with fluoride ions doesn't prove anything unless you already know that you must have a halogen present and are simply trying to find out which one. All the absence of a precipitate shows is that you haven't got chloride, bromide or iodide ions present.

The chemistry of the test

The precipitates are the insoluble silver halides - silver chloride, silver bromide or silver iodide.

Silver fluoride is soluble, and so you don't get a precipitate.

Sulfate ions in solution, SO42-, are detected using barium chloride solution. The test solution is acidified using a few drops of dilute hydrochloric acid, and then a few drops of barium chloride solution are added. A white precipitate of barium sulfate forms if sulfate ions are present.

Testing for sulfate ions (H2SO4)

Sulfate ions in solution, SO42-, are detected using barium chloride solution. The test solution is acidified using a few drops of dilute hydrochloric acid, and then a few drops of barium chloride solution are added. A white precipitate of barium sulfate forms if sulfate ions are present.

BaCl2(aq) + H2SO4(aq) → 2HCl(aq) + BaSO4(s)

The hydrochloric acid is added first to remove any carbonate ions that might be present - they would also produce a white precipitate, giving a false positive result.

Barium nitrate solution can be used instead of barium chloride solution. However, nitric acid is added first to acidify the test solution. Sulfuric acid cannot be used because it contains sulfate ions - these would interfere with the second part of the test.

Testing for nitrate ions (HN03)

Nitrate ions (NO3-) can be detected by reducing them to ammonia. This is done by:

• adding sodium hydroxide solution, then aluminium powder or foil
• heating strongly

If nitrate ions are present, ammonia gas is given off. This has a characteristic choking smell. It also turns damp red litmus paper or damp universal indicator paper blue.

A common nitrate test, known as the brown ring test^[1] can be performed by adding iron(II) sulfate to a solution of a nitrate, then slowly adding concentrated sulfuric acid such that the acid forms a layer below the aqueous solution. A brown ring will form at the junction of the two layers, indicating the presence of the nitrate ion.^[2] Note that the presence of nitrite ions will interfere with this test.^[3]

The overall reaction is the reduction of the nitrate ion by iron(II) which is oxidised to iron(III) and formation of a nitrosonium complex where nitric oxide is reduced to NO^−[4].

2HNO₃+ 3H₂SO₄ + 6FeSO₄ --->> 3Fe₂(SO₄)₃ + 2NO + 4H₂O

[Fe(H₂O)₆]SO₄ + NO = [Fe(H₂O)₅(NO)]SO₄+ H₂O