Question

Colloidal solution is prepared by following reaction $AgN{O}_3\left(\text{excess}\right)+KI\to AgI\downarrow +KN{O}_3.$ The most effective electrolyte in the coagulation of this colloidal solution is:

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

Answer 1

The correct option is A $K_4\left[Fe\right(CN{)}_6]$

According to Hardy-Schulze Rule, the greater the valency of the flocculating ion added, the greater is its power to cause coagulation.
When $KI$ solution is added to excess $AgN{O}_3$ solution
$AgN{O}_3\left(aq\right)+KI\left(aq\right)\to AgI\left(s\right)\downarrow +AgN{O}_3\left(aq\right)$
$AgI$ gets precipitated.
The precipitated silver iodide adsorbs $A{g}^{+}$ions from the dispersion medium and positively charged colloidal sol is formed $\left(AgI/A{g}^{+}\right)$.
So anion with greater valency will have greater power to cause coagulation.
Among the given electrolytes, $K_4\left[Fe\right(CN{)}_6]$ have the higher valence anion $\left[Fe\right(CN{)}_6{]}^{4-}$
$\therefore$ Most effective electrolyte in coagulation of above solution is $K_4\left[Fe\right(CN{)}_6]$.