Question

${{I}^{-}}$ also interferes with the ‘Ring Test’ of $NO_{3}^{-}$. Suggest a chemical reagent that can remove ${{I}^{-}}$.
(A)- $PbC{{l}_{2}}$
(B)- $HgC{{l}_{2}}$
(C)- $FeC{{l}_{2}}$
(D)- None of the above

Answer 1

Ring test in qualitative analysis is used to detect and confirm the presence of nitrate $NO_{3}^{-}$ ions. Mercuric chloride ($HgC{{l}_{2}}$) and ferrous chloride ($FeC{{l}_{2}}$) are soluble in water and dissociates to give $H{{g}^{2+}}$ ions which can combine with iodide (${{I}^{-}}$) ions. Lead chloride ($PbC{{l}_{2}}$) is slightly soluble in water.

Complete step by step solution:
If a salt gives brown fumes on heating with concentrated sulphuric acid (${{H}_{2}}S{{O}_{4}}$), then it indicates the presence of nitrate ions. The brown fumes are due to the evolution of nitrogen oxide, i.e. $N{{O}_{2}}$.
$\begin{aligned} & NaN{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to NaHS{{O}_{4}}+HN{{O}_{3}} \\\ & 4HN{{O}_{3}}\to 4N{{O}_{2}}+{{O}_{2}}+2{{H}_{2}}O \\\ \end{aligned}$
Ring test is performed by adding a freshly prepared solution of ferrous sulphate ($FeS{{O}_{2}}$) into the test containing a solution of nitrate ions. Incline the test tube and add slowly the concentrated ${{H}_{2}}S{{O}_{4}}$ down the side of the tube such that a layer is formed at the interface of two layers, i.e. between ${{H}_{2}}S{{O}_{4}}$ and aqueous layer. The brown ring appears due the formation of a nitrosyl complex of iron (II). The chemical reactions involved are given below:
$\begin{aligned} & NaN{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to NaHS{{O}_{4}}+HN{{O}_{3}} \\\ & 6FeS{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}+2HN{{O}_{3}}\to 3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+4{{H}_{2}}O+2NO \\\ & FeS{{O}_{4}}+NO\to \left[ Fe(NO) \right]S{{O}_{4}} \\\ \end{aligned}$
Iodide (${{I}^{-}}$) ions interferes with the ring test for $NO_{3}^{-}$ and are needed to be removed from the solution of $NO_{3}^{-}$ ions. Iodide ions can be removed by precipitating them using mercuric chloride ($HgC{{l}_{2}}$). $HgC{{l}_{2}}$ on reacting with ${{I}^{-}}$ ions forms orange coloured precipitates of mercuric iodide ($Hg{{I}_{2}}$). The chemical reaction is given below:
$HgC{{l}_{2}}+{{I}^{-}}\to Hg{{I}_{2}}\downarrow +2C{{l}^{-}}$
$PbC{{l}_{2}}$ does not dissociate completely to give $P{{b}^{2+}}$ ions that can combine with ${{I}^{-}}$ ions. And hence $PbC{{l}_{2}}$cannot complexly precipitate all the ${{I}^{-}}$ ions.
$FeC{{l}_{2}}$cannot be used to remove ${{I}^{-}}$ ions because iron (II) iodide formed is soluble in water. This means that ${{I}^{-}}$ ions will remain in the solution and will not be precipitated.

Hence, the correct option is (B).

Note: We are most likely to get confused between the options. So remember that in order to remove the iodide ions from a solution, the reagent must be able to convert the iodide ions from the solution into insoluble precipitates.