Question

Which of the following is the most effective in causing coagulation of ferric hydroxide solution?
(A) $KCl$
(B) $KN{O_3}$
(C) ${K_2}S{O_4}$
(D) ${K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$

Answer 1

Hint: Ferric hydroxide is a positive charged colloidal solution. It coagulates negatively charged particles. Higher the negative charge more effective is the coagulation.

Complete answer:
Coagulation is a process of aggregation together the colloidal particles so as to change them into large sized particles which ultimately settle as a precipitate. The quantity of the electrolyte which is required to coagulate a definite amount of a colloidal solution depends upon the valency of the coagulating ion. (Ion having a charge opposite to that of the colloidal particles). This observation of hardy and schulze is known as Hardy Shulze law. The main points of which may be stated as follows:
(i) The effective ions of the electrolyte in bringing about coagulation are those which carry charge opposite to that of the colloidal particles. These ions are called coagulating ions or flocculating ions.
(ii) Greater is the valency of coagulating or the flocculating greater is its power to bring about coagulations.
The correct option is ${K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$. Because it breaks into $3{K^ + }$ and ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{ - 3}}$ which is among the all given options is highly negatively charged particle.

Hence the correct option is (D).

Note: For a negatively charged solution like that of $A{S_2}{S_3}$ the coagulating powers of the cations are in the order:
$A{l^{ + 3}} > B{a^{ + 2}} > N{a^ + }$
For a positively charged solution like that of $Fe{\left( {OH} \right)_3},$ the coagulating powers of anions are in the order:
${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} > PO_4^{3 - } > SO_4^{2 - } > C{l^ - }$