Qualitative Analysis - Result Question 61
64. A scarlet compound $A$ is treated with conc. $HNO _3$ to give a chocolate brown precipitate $B$. The precipitate is filtered and the filtrate is neutralised with $NaOH$. Addition of $KI$ to the resulting solution gives a yellow ppt $C$. The brown ppt $B$ on warming with conc. $HNO _3$ in the presence of $Mn\left(NO _3\right) _2$ produces a pink coloured solution due to the formation of $D$. Identify $A, B, C$ and $D$. Write the reaction sequence.
(1995, 4M)
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Solution:
$ A \underset{\text { Scarlet }}{\left(Pb _3 O _4\right)} \xrightarrow{C \cdot HNO _3} \underset{\text { Chocolate }}{B\left(PbO _2\right)} \xrightarrow{\text { Filtered }} \underset{\text { Filtrate }}{Pb\left(NO _3\right) _2}$
Filtrate is neutralised with $NaOH$ and on reaction with $KI$, gives yellow ppt of $PbI _2$.
$Pb\left(NO _3\right) _2+2 KI \longrightarrow \underset{\text { Yellow }}{PbI _2 \downarrow}+2 KNO _3$
$PbO _2$ on warming with conc. $HNO _3$ in presence of $Mn\left(NO _3\right) _2$ produced pink solution due to formation of $Pb\left(MnO _4\right) _2$ (I).
$5 PbO _2+2 Mn\left(NO _3\right) _2+4 HNO _3 \longrightarrow Pb\left(MnO _4\right) _2 $ $ +4 Pb\left(NO _3\right) _2+2 H _2 O $
$\Rightarrow A= Pb _3 O _4, B=PbO _2, C=PbI _2 \text { and } D=Pb\left(MnO _4\right) _2 .$
BETA
