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(A) $1$. The formation of a dark brown ring with freshly prepared ferrous sulphate in an acidic medium is the characteristic test for the nitrate ion

Vedclass · Accepted Answer

$CH_3COO^{-}$ and $NO_3^{-}$

Vedclass · Answer

(A) $1$. The formation of a dark brown ring with freshly prepared ferrous sulphate in an acidic medium is the characteristic test for the nitrate ion $(NO_3^{-})$,known as the Brown Ring Test: $2NO_3^{-} + 4H_2SO_4 + 6Fe^{2+} \rightarrow 6Fe^{3+} + 2NO + 4SO_4^{2-} + 4H_2O$ followed by $[Fe(H_2O)_6]^{2+} + NO \rightarrow [Fe(H_2O)_5(NO)]^{2+} + H_2O$ (brown ring).$2$. The formation of a deep red colour with neutral $FeCl_3$ is the characteristic test for the acetate ion $(CH_3COO^{-})$,which forms ferric acetate: $CH_3COO^{-} + FeCl_3 \rightarrow [Fe_3(OH)_2(CH_3COO)_6]^+$. This red colour disappears on boiling,resulting in a brown-red precipitate of basic ferric acetate: $[Fe_3(OH)_2(CH_3COO)_6]^+ + 4H_2O \rightarrow 3Fe(OH)_2(CH_3COO) \downarrow + 3CH_3COOH + H^+$.$3$. Therefore,the mixture contains $CH_3COO^{-}$ and $NO_3^{-}$.

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