Brown Ring Test – Experimental Procedure, Result, Practice Problems and FAQ

By adding barium chloride (BaCl₂) solution, we may quickly determine whether a sample contains soluble sulphate (say sodium sulphate, Na₂SO₄) ions. The existence of sulphate ions in the solution is confirmed by the formation of a white precipitate of barium sulphate (BaSO₄) when barium chloride reacts with the soluble sulphate ions.

Can we identify the presence of nitrate ions in a sample using a similar method?

The response is "no." To verify the presence of nitrate ion, we need to perform a challenging process known as the "Brown ring test."

Nitrate ions cannot be detected using confirmation tests involving precipitation since, in general, all metal nitrates are soluble in water. Therefore, the precipitation method is only applied to anions that can solidify as precipitates with a certain metal. Because nitrates are so soluble, it is rare to find metal ores in the form of nitrates.

This concept page will explain how the Brown ring test can be used to detect the presence of nitrate ions in a sample.

TABLE OF CONTENTS

• Brown Ring Test
• Experimental Procedure
• Practice Problems
• Frequently Asked Questions – FAQ

Brown Ring Test

Brown ring test is a chemical test carried out to detect the presence of nitrate ions in a given sample. It is also known as the nitrate test.

• It is a common qualitative test performed on any solution which results in the formation of the brown-coloured ring in order to confirm the presence of nitrate ions.

• A brown ring is formed due to the addition of ferrous sulphate in the given solution that contains nitrate ions and is further acidified by adding concentrated sulphuric acid to the entire mixture.

• A brown-coloured ring is generally formed at the junction of ferrous sulphate and sulphuric acid. This ring indicates the presence of nitrates in the given solution.

• The brown ring test or nitrate test is a reduction reaction. The nitrate is reduced to nitric oxide (NO) by ferrous (II) ions from ferrous sulphate, which is oxidised to ferric (III) ions. The nitric oxide thus formed reacts with the ferrous (II) left behind to form a nitrosyl complex. Here, nitric oxide is reduced to NO-.

• This test is effective for up to 2.5 micrograms and a concentration of 1 in every 25,000 parts.

Experimental Procedure

Aim: To detect the presence of nitrate ions in the given sample.

Apparatus Required: Test tubes, freshly prepared ferrous sulphate solution, concentrated sulphuric acid solution.

Theory: The brown ring test is based on the reaction of ferrous sulphate with concentrated sulphuric acid. The reaction which takes place in the experiment is overall a reduction reaction. The nitrate, if present in the given sample, reacts with ferrous sulphate solution and gets reduced to nitric oxide. Ferrous (Fe²⁺) ions are oxidised to ferric (Fe³⁺) ions.

The nitric oxide that got formed as a product of the reduction, now reacts with the remaining ferrous (Fe²⁺) ions to give a nitrosyl complex.

Where [Fe(H₂O)5NO]SO₄ is the nitrosyl complex.

The nitric oxide is reduced to give NO^-. This leads to the formation of a brown-coloured ring at the junction of the two layers. The brown ring thus formed is the confirmation of the presence of nitrate ion in the given solution.

Procedure: The brown ring test proceeds step by step as given below.

1. Take the sample containing the nitrate ions in a test tube.

2. Ferrous sulphate is prepared fresh and taken in another test tube.

3. The freshly prepared ferrous sulphate solution is mixed with the test tube containing the given sample.

4. Concentrated sulphuric acid is added along the sides of the same test tube.

5. The test tube must be tilted a little while concentrated sulphuric acid is added. This is done so that the acid goes into the test tube’s bottom, resulting in the formation of a brown-coloured ring exactly at the junction of the two layers.

6. Once the brown ring is formed in the test tube, do not shake the test tube. This might make the brown ring disappear into the solution.

Observation: A brown-coloured ring forms at the junction of the two layers formed by the mixture of the given compound, ferrous sulphate solution and sulphuric acid. The brown ring confirms the presence of nitrate ions in the given sample of the compound.

Practice Problems

1. Pick out the oxidising agent in the given reaction.
    

$2{HNO}_3+3H_2{SO}_4+6{FeSO}_4\to 3{Fe}_2({SO}_4{)}_3+2NO+{4H}_{2}O$

1. HNO₃
2. FeSO₄
3. H₂SO₄
4. None of these

Answer: A
Let the oxidation state of Fe in FeSO₄ be ‘x’

x+6-8=0

x= +2

Let the oxidation state of Fe in Fe₂(SO₄)₃ be ‘y’

2y+3(6-8)=0

y= +3

Let the oxidation state of N in HNO₃ be ‘z’

1+z-6=0

z=+5

The oxidation state of N in NO = +2

There is no change in the oxidation state of the elements present in H₂SO₄.

Fe (+2) in FeSO₄Fe(+3) in Fe₂(SO₄)₃. I.e. oxidation state increases, reducing agent.

N(+5) in HNO₃N(+2) in NO. I.e. oxidation state decreases, oxidising agent.

Hence, among the given options, HNO₃ is the oxidising agent.

So, option A is the correct answer.

2. Brown ring formation takes place due to
    

1. Fe(NO₃)₂
2. Fe(NO₃)₃
3. [Fe(H₂O)₅NO]²⁺
4. All of these

Answer: C
The presence of nitrate ions in a solution is detected by the formation of a brown ring between the interface of the layers formed by the mixture of the given compound, ferrous sulphate solution and sulphuric acid. The brown-ring formation takes place due to [Fe(H₂O)₅NO]²⁺.

So, option C is the correct answer.

3. The IUPAC name of the nitrosyl complex [Fe(H₂O)5NO]SO₄is

1. Pentaaquanitrosoniumiron (I) sulphate
2. Pentaaquanitrosoniumiron (II) sulphate
3. Pentaaquanitrosoniumferrate (I) sulphate
4. Pentaaquanitrosoniumferrate (II) sulphate

Answer: A

Solution: The IUPAC name of the nitrosyl complex [Fe(H₂O)5NO]SO₄ is pentaaquanitrosoniumiron(I)sulphate

So, option A is the correct answer.

4. The oxidation state of Fe in the complex [Fe(H₂O)5NO]SO₄ is

1. +1
2. +2
3. +3
4. 0

Answer: A

Solution: Let the oxidation state of Fe be ‘x’

x+{50}+(+1)+(-2)=0

x= +1

So, option A is the correct answer.

Frequently Asked Questions – FAQ

1. Why is sulphuric acid used in the brown ring test?
Answer: If nitrate is present, a brown ring of [Fe(H₂O)5NO]SO₄ forms at the interface of the two layers. When concentrated sulphuric acid is added to a solution of Iron(II)sulphate and a possible nitrate, the acid sinks to the bottom. This is due to the fact that sulphuric acid is denser than the solution. Hence, sulphuric acid helps to form the ring at the junction of two layers.

2. What applications does nitrate have?
Answer: The body also produces nitrate and nitrite. Nitrate is mostly utilised in industry as fertiliser for lawns and crops. Additionally, nitrate and nitrite are employed in the production of several pharmaceutical medications, explosives, munitions, and missiles. They are also utilised to preserve food.

3. Why NO₃^- called nitrate ion?
Answer: If the name of acid ends with -ic acid then the name of its anion ends with -ate ion.

${H{NO}_3 \to NO}_3^{-}$

Nitric acid                                      Nitrate ion

${H_2{SO}_4 \to SO}_4^{2-}$

Sulphuric acid                                Sulphate ion

If the name of acid ends with -us acid then the name of its anion ends with -ite ion.

${H{NO}_2 \to NO}_2^{-}$

Nitrous acid                                       Nitrite ion

${H_2{SO}_3 \to SO}_3^{2-}$

Sulphurous acid                                 Sulphite ion

4. Why is freshly prepared FeSO4 used in brown-ring test?
Answer: Freshly produced FeSO4 is used to test nitrogen because FeSO4 immediately oxidises to ferric sulphate when exposed to the environment and does not support the development of the brown ring.

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