Question

Explain with the help of a balanced equation, the brown ring test for nitric acid.

Answer 1

1. Brown ring test is used to detect the Nitrate ions (${\mathbf{NO}}_{\mathbf{3}}^{\mathbf{-}}$) present in a solution.
2. In Brown ring test, a solution of nitrate is taken to which saturated solution of Iron(II) Sulphate or Ferrous sulphate is added.
3. To the mixture, few drops of conc. Sulphuric acid is added which results in the formation of brown ring between the junction.
4. The balanced chemical reaction followed in this test is shown below:$\underset{\left(\mathrm{Iron}\right(\mathrm{II}\left)\mathrm{sulphate}\right)}{6{\mathrm{FeSO}}_4\left(\mathrm{s}\right)}+\underset{(\mathrm{Sulphuric}\mathrm{acid})}{3{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{2{\mathrm{HNO}}_3\left(\mathrm{aq}\right)}\to \underset{\left(\mathrm{Iron}\right(\mathrm{III})\mathrm{sulphate})}{3{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{s}\right)}+\underset{\left(\mathrm{Water}\right)}{4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{(\mathrm{Nitric}\mathrm{oxide})}{2\mathrm{NO}\left(\mathrm{g}\right)}$
5. In the above reaction Ferrous sulphate reacts with Sulphuric acid and Nitric acid to form Iron (III) sulphate, water and Nitric oxide.
6. Further, the remaining Ferrous sulphate reacts with Nitrogen oxide to form Nitroso Ferrous sulphate (brown colored compound).
7. The chemical reaction is as follows:
   $\underset{(\mathrm{Iron}(\mathrm{II})\mathrm{sulphate})}{{\mathrm{FeSO}}_4\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{Oxide})}{\mathrm{NO}\left(\mathrm{g}\right)}\to \underset{(\mathrm{Nitroso}\mathrm{ferrous}\mathrm{sulphate})}{{\mathrm{FeSO}}_4.\mathrm{NO}\left(s\right)}$