Question

# (a) Explain with the help of a balanced equation, the brown ring test for nitric acid. (b) Why is freshly prepared ferrous sulphate solution used in the ring test? (c) Brown ring disappears if the test tube is disturbed.

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Solution

## Take freshly prepared FeSO4 and add it to a test tube containing dilute nitric acid. Concentrated sulphuric acid is then added down the sides of the test tube till a dark brown layer is formed at the junction of the two layers. The brown ring is formed due to the formation of complex compounds by ferrous sulphate, nitric acid and water molecules. Being heavier, concentrated sulphuric acid settles down and iron (II) sulphate layer remains above it, resulting in the formation of the brown ring at the junction. $6{\mathrm{FeSO}}_{4}+3{\mathrm{H}}_{2}{\mathrm{SO}}_{4}+2{\mathrm{HNO}}_{3}\to 3{\mathrm{Fe}}_{2}{\left({\mathrm{SO}}_{4}\right)}_{3}+2\mathrm{NO}+4{\mathrm{H}}_{2}\mathrm{O}\phantom{\rule{0ex}{0ex}}{\mathrm{FeSO}}_{4}+\mathrm{NO}+5{\mathrm{H}}_{2}\mathrm{O}\to \left[\mathrm{Fe}\left(\mathrm{NO}\right){\left({\mathrm{H}}_{2}\mathrm{O}\right)}_{5}\right]{\mathrm{SO}}_{4}$ Freshly prepared ferrous sulphate solution is used in the ring test because the solution gets oxidised quickly to give ferric sulphate by reaction with the air and hence, no brown ring is formed. When the test tube is shaken the concentrated H2SO4 may further mix with water and the heat evolved decomposes the ring. Hence, if the test tube is disturbed , the brown ring will disappear.

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