Question

As per Hardy-Schulze formulation, the flocculation value of the following for ferric hydroxide sols are in the order:
(a)- $AlC{{l}_{3}}$ > ${{K}_{3}}[Fe{{(CN)}_{6}}]$ > ${{K}_{2}}Cr{{O}_{4}}$ > KBr = $KN{{O}_{3}}$
(b)- ${{K}_{3}}[Fe{{(CN)}_{6}}]$ <${{K}_{2}}Cr{{O}_{4}}$ <$AlC{{l}_{3}}$ = KBr = $KN{{O}_{3}}$
(c)- ${{K}_{3}}[Fe{{(CN)}_{6}}]$ <${{K}_{2}}Cr{{O}_{4}}$ <$AlC{{l}_{3}}$ < KBr <$KN{{O}_{3}}$
(d)- ${{K}_{3}}[Fe{{(CN)}_{6}}]$> $AlC{{l}_{3}}$> ${{K}_{2}}Cr{{O}_{4}}$> KBr > $KN{{O}_{3}}$

Answer 1

Hint: The Hardy-Schulze rule is the relation between the flocculation value and coagulation power in which it states that the Flocculation value is inversely proportional to coagulation power. By calculating the charge on the ion we can easily find the coagulation power.

Complete answer:
When the colloidal particles combine or aggregate to form particles of larger size so that they can settle down at the bottom is known as coagulation or precipitation. This process rarely occurs without any external factors, so we have to use a little amount of electrolyte for coagulation.
Coagulation or flocculation or precipitation value is the minimum amount of electrolyte added in the colloidal solution for the coagulation process.
The Hardy-Schulze rule is the relation between the flocculation value and coagulation power in which it states that the Flocculation value is inversely proportional to coagulation power.
$Flocculation\text{ }value=\dfrac{1}{Coagulation\text{ }power}$
The given sol is Ferric hydroxide, and it has positive charge, therefore, it will show maximum flocculation value to negatively charged ions.
By calculating the charge on the ion in the given compounds we can easily find the coagulation power.
The dissociation of the ions will be as follows:
$AlC{{l}_{3}}\to A{{l}^{3+}}+3C{{l}^{-}}$
$KN{{O}_{3}}\to {{K}^{+}}+NO_{3}^{-}$
$KBr\to {{K}^{+}}+B{{r}^{-}}$
${{K}_{2}}Cr{{O}_{4}}\to 2{{K}^{+}}+CrO_{4}^{2-}$
${{K}_{2}}[Fe{{(CN)}_{6}}]\to 3{{K}^{+}}+{{[Fe{{(CN)}_{6}}]}^{3-}}$
So, the lesser the negative charge the greater will be flocculation value. $AlC{{l}_{3}}$, KBr, and $KN{{O}_{3}}$ have single negative charged ions, therefore they will have high flocculation value. ${{K}_{2}}Cr{{O}_{4}}$ has two negative charge ion while ${{K}_{3}}[Fe{{(CN)}_{6}}]$ has three negative charge ion.
The order will be:
${{K}_{3}}[Fe{{(CN)}_{6}}]$<${{K}_{2}}Cr{{O}_{4}}$<$AlC{{l}_{3}}$= KBr =$KN{{O}_{3}}$

Therefore, the correct answer is an option (b).

Note:
If the sol is given is negatively charged like $A{{s}_{2}}{{S}_{3}}$, then we will see the charge of the positive ions of the electrolytes added into the sol, in this also as the positive charge increases the flocculation value decreases.