# Question #52f16

Mar 16, 2015

Hola,

What you are describing is known as "the brown ring test" which is a test for the nitrate(V) ion $N {O}_{3}^{-}$.

Iron(II) sulfate is added to a solution which you think may contain nitrate(V) ions. Concentrated sulfuric acid is then slowly added. This is denser than the aqueous layer and sinks to the bottom of the test tube.

If a "brown ring" forms between the two layers then nitrate(V) ions are present.

This is due to a redox reaction between iron(II) and nitrate(V).

Iron(II) gets oxidised to iron(III) and nitrate(V) gets reduced to NO:

$3 F {e}^{2 +} + 4 {H}^{+} + N {O}_{3}^{-} \rightarrow 3 F {e}^{3 +} + N O + 2 {H}_{2} O$

The so - called "brown ring" is due to the formation of a complex ion where NO displaces a water ligand:

${\left[F e {\left({H}_{2} O\right)}_{6}\right]}^{2 +} + N O \rightarrow {\left[F e {\left({H}_{2} O\right)}_{5} \left(N O\right)\right]}^{2 +} + {H}_{2} O$