Question

From the given electrolytes, indicate which one will bring about the coagulation of a gold sol quickest and in the least of molar concentration?

• A. $NaCl$
• B. $MgS{O}_4$
• C. $A{l}_2{\left(S{O}_4\right)}_3$
• D. $K_4\left[Fe{\left(CN\right)}_6\right]$

Answer 1

The correct option is C $A{l}_2{\left(S{O}_4\right)}_3$

COAGULATION OR FLOCCULATION :

The colloidal sols are stable by the presence of electric charges on the colloidal particles. Because of the electric repulsion the particles do not come close to one another and coalesce. The removal of charge by any means will lead to the aggregation of particles and hence precipitation immediately. The process by means of which the particles of the dispersed phase in a sol are precipitated is known as coagulation or flocculation.

As we know,

i) the ions carrying charge opposite to that of sol particles are effective in causing coagulation of sol. these arecalled active ions<br>ii) Coagulating power of an electrolyte is directly proportional to the fourth power of the valency of the active ions.

Among all given compounds,

NaCl = 2 ion

$MgS{O}_4$ = 2 ion

$A{l}_2(S{O}_4{)}_3$ = 5 ion

$A{l}_2(S{O}_4{)}_3$ provides maximum number of ions (5) in solution. So, it will bring the coagulation of a solution quickest and in the least of molar concentration.