A reddish brown solution (containing $F{e^{3 + }}$) is obtained by:
(A) The addition of a small amount of $FeC{l_3}$ solution to freshly prepared $Fe{(OH)_3}$ precipitate.
(B) The addition of $Fe{(OH)_3}$ to the freshly prepared $FeC{l_3}$ solution.
(C) The addition of $N{H_4}OH$ to the $FeC{l_3}$ solution dropwise.
(D) The addition of $NaOH$ to the $FeC{l_3}$ solution dropwise.
Answer
299.7k+ views
Hint: Formation of a reddish brown solution is formed due to the adsorption of the common ion of $FeC{l_3}$ in $Fe{(OH)_3}$ sol. The process takes place by adding a suitable electrolyte to ionise the ions which causes the formation of a reddish brown solution.
Complete Step by Step Solution:
Deflocculation or peptization is a process in which the precipitate is converted into colloid by shaking it with a suitable electrolyte. The electrolyte added in the process is known as a peptizing agent and the formation of a reddish brown solution from precipitate is an example of peptization.
In the peptization process, the charged ions of the electrolyte are adsorbed by the precipitate forming either positive or negative charge which results in breaking the precipitate into smaller particles which are as small as the size of colloid particles.
Thus, when a small quantity of ferric chloride solution ($FeC{l_3}$) is added to the freshly precipitated ferric hydroxide ($Fe{(OH)_3}$).
The process can be represented with the help of a chemical equation as follows:
$FeC{{l}_{3}}\rightleftharpoons F{{e}^{3+}}+3C{{l}^{-}}$
$Fe{(OH)_3} + F{e^{3 + }} \to \left[ {Fe{{(OH)}_3}} \right]F{e^{3 + }}$
Where, $\left[ {Fe{{(OH)}_3}} \right]F{e^{3 + }}$ is the reddish brown solution.
So for the given question, the right answer is option (A).
Note: It is important to note that due to the above process, an electrical double layer is formed in the solution which consist of positively charged ion ($F{e^{3 + }}$ ion) is a fixed part of the electrical double layer whereas the ion to balance the positive charge i.e., $C{l^ - }$ ions are the mobile part of the electrical double layer formed in the solution.
Complete Step by Step Solution:
Deflocculation or peptization is a process in which the precipitate is converted into colloid by shaking it with a suitable electrolyte. The electrolyte added in the process is known as a peptizing agent and the formation of a reddish brown solution from precipitate is an example of peptization.
In the peptization process, the charged ions of the electrolyte are adsorbed by the precipitate forming either positive or negative charge which results in breaking the precipitate into smaller particles which are as small as the size of colloid particles.
Thus, when a small quantity of ferric chloride solution ($FeC{l_3}$) is added to the freshly precipitated ferric hydroxide ($Fe{(OH)_3}$).
The process can be represented with the help of a chemical equation as follows:
$FeC{{l}_{3}}\rightleftharpoons F{{e}^{3+}}+3C{{l}^{-}}$
$Fe{(OH)_3} + F{e^{3 + }} \to \left[ {Fe{{(OH)}_3}} \right]F{e^{3 + }}$
Where, $\left[ {Fe{{(OH)}_3}} \right]F{e^{3 + }}$ is the reddish brown solution.
So for the given question, the right answer is option (A).
Note: It is important to note that due to the above process, an electrical double layer is formed in the solution which consist of positively charged ion ($F{e^{3 + }}$ ion) is a fixed part of the electrical double layer whereas the ion to balance the positive charge i.e., $C{l^ - }$ ions are the mobile part of the electrical double layer formed in the solution.
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