Question

State True or False.
$N{H_4}Cl$ is added in $III$ group basic radicals to suppress the ionization of $N{H_4}OH$.
A.True
B.False

Answer 1

Hint: A radical is defined in terms of an atom, molecule, or ion that has an unpaired valence electron. Due to some exceptions, the unpaired electrons make radicals highly chemically reactive. The common ion effect is the decrease in solubility of ionic precipitate on adding to a solution of the soluble compound with an ion in common as the precipitate.

Complete step by step answer:
Now we will discuss the given statement in detail.
Group III cations are $F{e^{3 + }},C{a^{2 + }},A{l^{3 + }}$. These cations precipitate as hydroxides when we add $N{H_4}OH$ to the aqueous solution of the mixture.
But if we add $N{H_4}Cl$ before $N{H_4}OH$ due to the common ion effect the concentration of $NH_4^ + $ ion gets suppressed or decreased in the solution. So the only hydroxides of group III which have less ${K_{sp}}$ value get precipitated and other hydroxides whose ${K_{sp}}$ value is more do not get precipitated.
Thus, we can say that $N{H_4}Cl$ is added in $III$ group basic radicals to suppress the ionization of $N{H_4}OH$.
After discussing we can conclude that the given statement is true.
Hence, the correct option is (A).

Note:
The common-ion-effect is a consequence of Le Chatelier's principle for the equilibrium reaction of the ionic association or dissociation. Here $N{H_4}Cl$ is a strong electrolyte and $N{H_4}OH$ is a weak base. The increase in the precipitation of the salt is due to the addition of more amount of one or more ions present in the salt and it also reduces the concentration of both ions of the salt until the solubility equilibrium is reached. To keep the concentration of ${{O}}{{{H}}^{{ - }}}$ less, we add $N{H_4}Cl$. The effect is based on the fact that both the original salt and the other added chemical have one ion in common with each other.