# As per Hardy-Schulze formulation,

Question:

As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order:

1. $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]<\mathrm{K}_{2} \mathrm{CrO}_{4}<\mathrm{KBr}=\mathrm{KNO}_{3}=\mathrm{AlCl}_{3}$

2. $\mathrm{~K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]<\mathrm{K}_{2} \mathrm{CrO}_{4}<\mathrm{AlCl}_{3}<\mathrm{KBr}<\mathrm{KNO}_{3}$

3. $\mathrm{AlCl}_{3}>\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]>\mathrm{K}_{2} \mathrm{CrO}_{4}>\mathrm{KBr}=\mathrm{KNO}_{3}$

4. $\mathrm{~K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]>\mathrm{AlCl}_{3}>\mathrm{K}_{2} \mathrm{CrO}_{4}>\mathrm{KBr}>\mathrm{KNO}_{3}$

Correct Option: 1

Solution:

According to Hardy-Schulte,

Coagulation value or fluocculation value

$\propto \frac{1}{\text { Coagulation power }}$

order of coagulation power:

$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]>\mathrm{K}_{2} \mathrm{CrO}_{4}>\mathrm{KBr}=\mathrm{KNO}_{3}=\mathrm{AlCl}_{3}$

$\therefore \quad$ order of flocculation value :

$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]<\mathrm{K}_{2} \mathrm{CrO}_{4}<\mathrm{KBr}=\mathrm{KNO}_{3}=\mathrm{AlCl}_{3}$