Qualitative Analysis - Result Question 67
####70. The gas liberated, on heating a mixture of two salts with $NaOH$, gives a reddish brown precipitate with an alkaline solution of $K _2 HgI _4$. The aqueous solution of the mixture on treatment with $BaCl _2$ gives a white precipitate which is sparingly soluble in conc. $HCl$.
On heating the mixture with $K _2 Cr _2 O _7$ and conc. $H _2 SO _4$, red vapours $A$ are produced. The aqueous solution of the mixture gives a deep blue colouration $B$ with potassium ferricyanide solution. Identify the radicals in the given mixture and write the balanced equations for the formation of $A$ and $B$.
$(1991,4 M)$
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Solution:
- Formation of a reddish-brown precipitate with alkaline $K _2 HgI _4$ solution indicates the presence of $NH _4^{+}$ion and the gas liberated is ammonia :
On treatment with $BaCl _2$, a white precipitate is formed which indicates the presence of $SO _4^{2-}$ anion :
$$ SO _4^{2-}+BaCl _2 \longrightarrow \underset{\text { White }}{BaSO _4 \downarrow+2 Cl^{-}} $$
With $K _2 Cr _2 O _7$ and conc. $H _2 SO _4$, red vapour of $CrO _2 Cl _2$ is evolved. This indicates presence of $Cl^{-}$ion.
On treatment with potassium ferricyanide, formation of deep blue solution indicates presence of $Fe$ (II) ion :
$$ Fe^{2+}+Fe(CN) _6^{3-} \longrightarrow \underset{\text { Blue }}{Fe _4\left[Fe(CN) _6\right] _3} $$
