Question

State True or False.

$N{O}_3^{-}$ solution produce brown ring with $FeS{O}_4$ & dil. $H_{2}S{O}_4$.

• A. True
• B. False

Answer 1

The correct option is A True

When a freshly prepared saturated solution of iorn (II) Sulphate is added to nitrate solution and

than concentrated $H_{2}S{O}_4$ is added slowly from the side of the test tube, a brown ring is obtained

at the junction of two layers.

$2N{O}_3^{-}+4H_{2}S{O}_4+6F{e}^{2+}\to 6F{e}^{3+}+2NO\uparrow +4S{O}_4^{2-}+4H_{2}O$

$FeS{O}_4+6H_{2}O\to \left[Fe\right(H_{2}O{)}_6]S{O}_4$

$\left[Fe\right(H_{2}O{)}_6]S{O}_4+NO\uparrow \to [Fe\left(H_{2}O{)}_{5}NO\right]S{O}_4+H_{2}O$

Brown ring

Bromides and iodides interfere because of the liberated halogen. All of these anions can be removed

by adding an excess of nitrate-free $A{g}_{2}S{O}_4$ and filtering the insoluble the insoluble silver salts.