Salt analysis-5

Chemistry Level 3

A brown black compound A on treatment with NaOH give a green colour ompound Y which on reaction with chlorine give purple coloured compound Z . The compound behave as strong oxidising agent. In acidic medium it converts Fe2+ to Fe3+ and reduces to ion B.

The conversion B-------->Z can be achieved by:-

1) sodium bismuthate 2) lead dioxide 3) red lead

1,3 1 1,2,3 2 1,2 2,3 3

1 solution

Vitthal Yellambalse
Nov 17, 2016

A purple colored substance that oxidises ${Fe}^{2+}$ is $KMnO_{4}$ . This tells us that:

A is $MnO_{2}$
B is $MnSO_{4}$
Y is $K_{2}MnO_{4}$
Z is $KMnO_{4}$

( For reactions : link )

Due to the fact that manganese must change from an oxidation state of +2 to +7, a strong oxidising agent is required,

$PbO_{2}$ and $NaBiO_{3}$ are very strong oxidising agents and since red lead is $PbO_{2}.2PbO_{2}$ , all the reagents can oxidise $MnSO_{4}$ to $KMnO_{4}$

Salt analysis-5

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