Question

Which of the following cations will have minimum flocculation value for an arsenic sulphide solution:
A. $N{{a}^{+}}$
B. $M{{g}^{2+}}$
C. $C{{a}^{2+}}$
D. $A{{l}^{3+}}$

Answer 1

Hint: Think about the Hardy-Schulz law, and how it relates the flocculation value with coagulation power and the charge present on the cations.

Complete step by step solution:
* Flocculation is basically aggregating small, colloidal particles (negatively charged in this example) and gathering large sized particles which can later separate and settle down as a precipitate. The flocculating ion (positively charged in this example) does this job of aggregating the small, colloidal particles that need to be precipitated.
* The flocculation value is the minimum concentration of these flocculating ions that is required to flocculate all the colloidal particles in the solution.
* The greater the charge on the flocculating ion, the more negatively charged colloidal particles it can neutralize and coagulate. Thus, the greater the charge on the flocculating ion, the greater the coagulation power of that ion.
* If the number of colloidal particles the flocculating ion can neutralize increases, this implies that less amount of the flocculating ion will be required to flocculate all the colloidal particles in the solution. Hence, the greater the charge on the ion, the less the flocculation value.
Thus, as the Hardy-Schulz law states and we deduced:
\[\text{Charge }\alpha \text{ }\dfrac{1}{\text{flocculation value}}\]

Hence, among the options given, aluminium has the most amount of charge. The answer is ‘D. $A{{l}^{3+}}$’

Note: Please do not get confused between flocculation and coagulation, they refer to the same phenomenon but different aspects of it. Coagulation is the neutralization of charged colloidal particles, and flocculation is the aggregation of all the coagulated material and its precipitation.