Question

Which of the following exhibits greater coagulation power towards a negative colloid?

• A. $ZnS{O}_4$
• B. $N{a}_{3}P{O}_4$
• C. $AlC{l}_3$
• D. $K_4\left[Fe{\left(CN\right)}_6\right]$

Answer 1

Answer: (C)

Negative colloid is coagulated by positive ion or vice-versa. Greater the valency of coagulating ion, greater will be coagulating power according to Hardy-Schulze rule.
$\left(\text{a}\right){\left(\text{ZnSO}\right)}_4\to {\left(\text{Zn}\right)}^{2+}+{\text{SO}}_4^{2-}$
$\left(\text{b}\right){\left(\text{Na}\right)}_3{\left(\text{PO}\right)}_4\to 3{\left(\text{Na}\right)}^{+}+{\text{PO}}_4^{3-}$
$\left(\text{c}\right){\left(\text{AlCl}\right)}_3\to {\left(\text{Al}\right)}^{3+}+3{\left(\text{Cl}\right)}^{-}$
$\left(\text{d}\right){\left(\text{K}\right)}_4\left[\text{Fe}{\left(\text{CN}\right)}_6\right]\to 4{\left(\text{K}\right)}^{+}+{\left(\left[\text{Fe}{\left(\text{CN}\right)}_6\right]\right)}^{4-}$
Since, in $AlC{l}_3$ , the valency of positive ion (coagulation ion) is highest, it is the most powerful coagulating agent among the given to coagulate the negative colloid.