To begin, we need to understand the qualitative analysis of nitrate ions is performed through Brown Ring Experiment. The brown ring test is conducted by adding iron(II) sulphate to a nitrate solution, then slowly adding strong sulfuric acid until the acid forms a layer under the aqueous solution. The existence of the nitrate ion will be shown by the formation of a brown ring at the intersection of the two layers. This test will be hampered by the presence of nitrite ions.
Water can dissolve all common nitrates, nitrites, thiosulphates and acetates. Confirmatory test for these anions can be performed using respective salt’s water extract. Water extract can be made by boiling 1-2 g of the above mixture in a boiling tube with 10-15 $\mathrm{cm^3}$ of distilled water for a minute or two. Any remaining residue is filtered. Water extract is the name given to the filtrate.
In a test tube, place 3 $\mathrm{cm^3}$ of water extract. 6 $\mathrm{cm^3}$ of concentrated sulphuric acid was added to completely combine the two liquids. The mixture was chilled under cold running water. Slowly pour 4 $\mathrm{cm^3}$ of saturated $\mathrm{FeSO_4}4 solution down the edge of the test tube, forming a distinct layer on top of the solution in the test tube. At the point where the two liquids come into contact, a brown ring will form:
$$\mathrm{NO_3 ^−(aq) + 4H^+(aq) + 3Fe^{2+}(aq)\:\rightarrow\:3Fe^{3+}(aq) + NO(g) + 2H_2O(l)}$$
$$\mathrm{[Fe(H_2O)_6]^{2+}(aq) + NO(g)\:\rightarrow\:[Fe(H_2O)_5NO]^{2+}(aq) + H_2O]}$$
The capacity of the nitrate ion to oxidise $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$ in acidic solution while producing NO gas is the basis for this test. Because NO is more soluble in water at low temperatures, it combines with $\mathrm{Fe^{2+}}$ present in excess in a well-cooled solution to generate the brown nitrosyliron (I1) complex ion, $\mathrm{[Fe(H_2O)_5NO]^{2+}}$. In this test, nitrite, bromide, and iodide ions interfere.
One cannot simply detect nitrate in presence of nitrite, either by brown ring test or by heating with conc. $\mathrm{H_2SO_4}$, since both these test liberate $\mathrm{NO_2}$. As a result, nitrite must be totally eliminated before testing for nitrate.
Any of the following procedures can be used to eliminate nitrite ions:
Sulphamic acid, $\mathrm{H_2NSO_3H}$ is added to the water extract containing $\mathrm{NO_{3^-}}$ and $\mathrm{NO_{2^-}}$ ions. Dilute $\mathrm{H_2SO_4}$ to acidify the solution. Nitrite will decompose, releasing nitrogen gas into the atmosphere.
$$\mathrm{H_2NSO_3H(aq) + NO_{2^-}(aq)\:\rightarrow\:HNO_2(aq) + H_2NSO_{3^-}(aq)}$$
$$\mathrm{H_2NSO_{3^-}(aq) + HNO_2(aq)\:\rightarrow\: N_2(g) + H^+(aq) + SO_4^{2^-}(aq) + H_2O(l)}$$
Boil 2-3 $\mathrm{cm^3}$ water extract with 1 g solid $\mathrm{NH_4Cl}$ until effervescence stops.
$$\mathrm{NO_{2^-}(aq) + NH_4Cl(aq) \:\rightarrow\: N_2(g) + 2H_2O(l) + Cl^-(aq)}$$
In 2-3 $\mathrm{cm^3}$ water extract, urea, and dilute $\mathrm{H_2SO_4}$ is added, the mixture is boiled until the gaseous evolution stops.
$$\mathrm{NH_2CONH_2(aq) + 2NO_{2^-}(aq) + 2H^+(aq) \:\rightarrow\: 2N_2(g) + CO_2(g) + 3H_2O(l)}$$
(a) One half is used for confirming the presence of nitrate ion.
(b) Using dil. $\mathrm{H_2SO_4}$ the other half is acidified. 1 $\mathrm{cm^3}$ starch solution and a little KI is added. The absence of any blue colour shows that all nitrite ions have been removed. A granulated zinc piece is added to the solution now.
The existence of nitrate ions is confirmed by the appearance of a blue colour:
$$\mathrm{Zn(s) + 2H^+(aq)\:\rightarrow\:Zn^{2^+}(aq) + 2H(g)}$$
$$\mathrm{NO_{3^-}(aq) + 2H(g)\:\rightarrow\:NO_{2^-}(aq) + H_2O(l)}$$
$$\mathrm{2I^-(aq) + 2NO_{2^-}(aq) + 4H^+(aq) \:\rightarrow\:2NO(g) + I_2(s) + 2H_2O(l)}$$
$$\mathrm{I_2(s) + Starch\:\rightarrow\:Blue\:Coloured\:Complex}$$
Because of the colour of liberated bromine and iodine, bromide and iodide interfere with the nitrate ring test. The interfering halide should be ejected before performing the ring test to detect nitrate in the presence of iodide and/or bromide. This may be accomplished by heating 2-3 $\mathrm{cm^3}$ of water extract or sodium carbonate extract in excess chlorine water in a china dish until no more $\mathrm{Br_2}$ or $\mathrm{I_2}$ vapours are produced:
$$\mathrm{2Br^-(aq) + Cl_2(aq)\:\rightarrow\:2Cl^-(aq) + Br_2(g)}$$
$$\mathrm{2I^-(aq) + Cl_2(aq)\:\rightarrow\:2Cl^-(aq) + I_2(g)}$$
To detect the nitrate ion in the combination, the ring test was conducted on the halide free solution.
Alternatively, a test tube with 2-3 $\mathrm{cm^3}$ of water extract is filled. 1 $\mathrm{cm^3}$ of KI solution, 1 $\mathrm{cm^3}$ of starch solution, and a few granules of zinc are added to acidify with dil. $\mathrm{H_2SO_4}$.
The presence of nitrate ions in the mixture is confirmed by the presence of a blue colour.
$$\mathrm{Zn(s) + 2H^+(aq)\:\rightarrow\: Zn^{2^+}(aq) + 2H(g)}$$
$$\mathrm{NO_{3^-}(aq) + 2H(g)\:\rightarrow\:NO_{2^-}(aq) + H_2O(l)}$$
$$\mathrm{2I^-(aq) + 2NO_{2^-}(aq) + 4H^+(aq)\:\rightarrow\:2NO(g) + I_2(s) + 2H_2O(l)}$$
$$\mathrm{I_2(s) + Starch\:\rightarrow\:Blue\:Coloured\: Complex}$$
Nitrate is converted to Nitric Oxide, due to $\mathrm{Fe^{2^+}}$ in acidic medium. The excess $\mathrm{Fe^{2^+}}$ creates a compound with the nitric oxide. At the interface of the concentrated acid layer and the aqueous medium containing $\mathrm{Fe^{2^+}}$, this appears as a brown ring.
Q1. What is the nature of Brown Ring Test?
Ans: Brown Ring Test is qualitative in nature. It is used to detect the presence of nitrate in a solution by adding iron(II) sulphate to a nitrate solution, then slowly adding strong sulfuric acid until the acid forms a layer under the aqueous solution.
Q2. What are the chemical reactions involved in Brown Ring Test?
$$\mathrm{NO_3^−(aq)+4H^+(aq)+3Fe^{2^+}(aq) \:\rightarrow\:3Fe^{3^+}(aq)+NO(g)+2H_2O(l)}$$
$$\mathrm{[Fe(H_2O)_6]^{2^+}(aq)+NO(g)\:\rightarrow\:[Fe(H_2O)_5NO]^{2^+}(aq)+ H_2O]}$$
Q3. What are the ions that could interfere with Brown Ring Test for nitrate?
Ans: Anions such as $\mathrm{NO_2^-,\: Br^-}$ and $\mathrm{Cl^-}$ interfere with Brown Ring Test for nitrate.
Q4. How would you remove NO2- from a solution containing both $\mathrm{NO_{3^-}}$ and $\mathrm{NO_{2^-}}$?
Ans: The $\mathrm{NO_{2^-}}$ can be removed from the solution containing both $\mathrm{NO_{3^-}}$ and $\mathrm{NO_{2^-}}$ using any of the following methods:
(i) Sulphamic acid, $\mathrm{H_2NSO_3H}$ is added to the water extract containing $\mathrm{NO_{3^-}}$ and $\mathrm{NO_{2^-}}$ ions. Dilute $\mathrm{H_2SO_4}$ to acidify the solution. Nitrite will decompose, releasing nitrogen gas into the atmosphere.
$$\mathrm{H_2NSO_3H(aq) + NO_{2^-}(aq)\:\rightarrow\:HNO_2(aq) + H_2NSO_{3^-}(aq)}$$
$$\mathrm{H_2NSO_{3^-}(aq) + HNO_2(aq)\:\rightarrow\:N_2(g) + H^+(aq) + SO_4^{2^-}(aq) + H_2O(l)}$$
(ii) Boil 2-3 $\mathrm{cm^3}$ water extract with 1 g solid $\mathrm{NH_4Cl}$ until effervescence stops.
$$\mathrm{NO_{2^-}(aq) + NH_4Cl(aq)\:\rightarrow\:N_2(g) + 2H_2O(l) + Cl^-(aq)}$$
(iii) In 2-3 $\mathrm{cm^3}$ water extract, urea, and dilute $\mathrm{H_2SO_4}$ is added, the mixture is boiled until the gaseous evolution stops.
$$\mathrm{NH_2CONH_2(aq) + 2NO_{2^-}(aq) + 2H^+(aq)\:\rightarrow\:2N_2(g) + CO_2(g) + 3H_2O(l)}$$
Q5. What happens as KI solution, starch solution and Zn granules to acidifed solution containing nitrate?
Ans: 1 $\mathrm{cm^3}$ of KI solution, 1 $\mathrm{cm^3}$ of starch solution, and a few granules of zinc are added to acidify with dil. $\mathrm{H_2SO_4}$. The presence of nitrate ions in the mixture is confirmed by the presence of a blue colour.
$$\mathrm{Zn(s) + 2H^+(aq)\:\rightarrow\:Zn^{2^+}(aq) + 2H(g)}$$
$$\mathrm{NO_{3^-}(aq) + 2H(g)\:\rightarrow\: NO_{2^-}(aq) + H_2O(l)}$$
$$\mathrm{2I^-(aq) + 2NO_{2^-}(aq) + 4H^+(aq)\:\rightarrow\:2NO(g) + I_2(s) + 2H_2O(l)}$$
$$\mathrm{I_2(s) + Starch\:\rightarrow\:Blue\:Coloured\: Complex}$$
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