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Which of the following act as the best coagulating agent for ferric hydroxide solution?
A) magnesium chloride
B) hydrochloric acid
C) aluminium chloride
D) potassium oxalate
E) potassium ferricyanide

Last updated date: 20th Jun 2024
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Hint: Ferric Hydroxide $[Fe(OH)_3]$ is a positively charged solution. So, to coagulate we need a negatively charged electrolyte. By Hardy-Schulze rule- Coagulating power of an electrolyte is directly proportional to the valency of the active ion which causes coagulation.

Complete answer:
Now we have to find out from the given options which one acid as the best coagulating agent for heavy hydroxide.
Before answering the question, let us first understand what a coagulating agent is.
The process of setting of colloidal particles is called coagulation. Coagulation is also known as Precipitation. A coagulating agent is the one which brings about coagulation. The coagulation capacity of different electrolytes is different depending upon the valency of the active ion also called as coagulating ion. The coagulating must have charge opposite to the charge on the colloidal particle. According to Hardy Schulze rule - Greater is the valency of the coagulating Ion greater will be its coagulating power. Hydroxide solution is a positive solution and hence the coagulating agent used must have negative charge.
The decrease in coagulating power of the given options are as follows
\[{\left[ {Fe{\text{ }}{{\left( {CN} \right)}_6}} \right]^{4 - \;\;\;\;}} > \;\;{\left[ {C{r_2}{O_4}} \right]^{2 - \;\;\;\;\;}} > \;\;C{l^ - }\]
Thus, Potassium ferricyanide is the best coagulating agent because it has maximum valency.

Thus, option E is the correct option.

Note: However, the student should note that the coagulation power is inversely proportional to the coagulating value of ion. Smaller the coagulating value of the electrolyte larger is its coagulating power.