Question

For coagulating $A{s}_2{S}_3$ colloidal sol, which of the following will have the highest coagulation value :

• A. $NaCl$
• B. $KCl$
• C. $BaC{l}_2$
• D. $AlC{l}_3$

Answer 1

The correct option is D $AlC{l}_3$

Hardy-Schulze Rule:
Greater the valence of flocculating ion added, greater is its power to cause coagulation.
Coagulating power of bivalent ion is 20-80 times higher than monovalent ion. Coagulating power of trivalent ion is many more times higher than divalent ion.

$A{s}_2{S}_3$ is negative sol.

So, cation with higher positive charge will have greater power to cause coagulation.

Among the given electrolytes, $AlC{l}_3$ have the higher positve charge $\left(A{l}^{+3}\right)$

Therefore, $A{l}^{3+}$ has highest coagulation power for $A{s}_2{S}_3$.

Hence, Option (d) is correct