Explain the brown ring test for nitrate ions.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
The brown ring test for nitrates depends on the ability of $Fe^{2+}$ to reduce nitrates to nitric oxide $(NO)$,which then reacts with $Fe^{2+}$ to form a brown ring complex.
To a sample containing $NO_{3}^{-}$ ions,a dilute ferrous sulphate solution is added,and then concentrated sulphuric acid is added carefully along the sides of the test tube. $A$ brown ring at the interface between the solution and sulphuric acid layers indicates the presence of nitrate ions in the solution.
$NO_{3}^{-} + 3 Fe^{2+} + 4 H^{+} \rightarrow NO + 3 Fe^{3+} + 2 H_{2}O$
$[Fe(H_{2}O)_{6}]^{2+} + NO \rightarrow [Fe(H_{2}O)_{5}(NO)]^{2+} + H_{2}O$ (Brown complex)

Explore More

Similar Questions

Which anion's aqueous solution gives a white precipitate with $AgNO_3$,which is soluble in excess $NH_4OH$?

The incorrect statement in respect of the chromyl chloride test is:

The salt which responds to dilute and concentrated $H_2SO_4$ is

Which of the following radicals does not liberate gas with ($Zn$ + dil. $HCl$) on warming?

Difficult
View Solution

Which ion's salt solution produces a violet layer when chlorine water and chloroform are added?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo