Question

In the group \[III\] radicals, in place of $N{H_4}Cl$ which of the following can be used
(A) $N{H_4}N{O_3}$
(B) \[{(N{H_4})_2}S{O_4}\]
(C) \[{(N{H_4})_2}C{O_3}\]
(D) $NaCl$

Answer 1

Hint: In systemic schematic qualitative analysis of cations, the cations are divided into various groups according to the selective precipitation of sparingly soluble salts. These six groups are chlorides (Group\[I\]), sulphides in acid medium (Group\[II\]), hydroxides (Group\[III\]), sulphides in alkaline medium (Group\[IV\]), carbonates (Group\[V\]) hydrogen phosphate (Group\[VI\]), $N{H_4}^ + $and ${K^ + }$ do not belong to any particular group. They are said to belong to zero group.


Complete Step by Step Solution:
$N{H_4}Cl$ is added before $N{H_4}OH$ to decrease concentration of $O{H^ - }$ ions by common ion effect. Due to increase in concentration of $N{H_4}^ + $ ions, dissociation of $N{H_4}OH$ will shift backward thus decreasing the concentration of $O{H^ - }$ ion sufficient enough to precipitate group \[III\]cations. Thus,$N{H_4}Cl$can be replaced by a compound that furnishes $N{H_4}^ + $ion. Thus,$N{H_4}N{O_3}$ can replace $N{H_4}Cl$.

But \[{(N{H_4})_2}S{O_4}\] can't be used because precipitation of sulphate would take place in group \[V\]in group \[III\]itself. Same is the reason for\[{(N{H_4})_2}C{O_3}\].$NaCl$can't be used as there are no $N{H_4}^ + $ ions hence no common ion effect.

Note: Up to \[II\] group the medium is acidic but after \[II\] group of analysis, as soon as the solution is made ammoniacal for the precipitation of group \[III\]hydroxides. Both \[{(N{H_4})_2}C{O_3}\]and \[{(N{H_4})_2}S{O_4}\]contain $N{H_4}^ + $ ions that can be used for common ion effect. But the anionic part of these compounds interferes in the process.