Question

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

• A. ${\mathrm{AlCl}}_3>{\mathrm{K}}_3\left[\mathrm{Fe}\right(\mathrm{CN}\left)6\right]>{\mathrm{K}}_2{\mathrm{CrO}}_4>\mathrm{KBr}={\mathrm{KNO}}_3$
• B. ${\mathrm{K}}_3[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6]<{\mathrm{K}}_2{\mathrm{CrO}}_4<{\mathrm{AlCl}}_3<\mathrm{KBr}<{\mathrm{KNO}}_3$
• C. ${\mathrm{K}}_3[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6]<{\mathrm{K}}_2{\mathrm{CrO}}_4<\mathrm{KBr}={\mathrm{KNO}}_3={\mathrm{AlCl}}_3$
• D. ${\mathrm{K}}_3[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6]>{\mathrm{AlCl}}_3>{\mathrm{K}}_2{\mathrm{CrO}}_4>\mathrm{KBr}>{\mathrm{KNO}}_3$

Answer 1

The correct option is C

${\mathrm{K}}_3[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6]<{\mathrm{K}}_2{\mathrm{CrO}}_4<\mathrm{KBr}={\mathrm{KNO}}_3={\mathrm{AlCl}}_3$

The explanation for the correct options:

C) ${\mathrm{K}}_3[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6]<{\mathrm{K}}_2{\mathrm{CrO}}_4<\mathrm{KBr}={\mathrm{KNO}}_3={\mathrm{AlCl}}_3$

• According to the Hardy-Schulze formulation, the coagulation is higher when the valency is greater. The ferric hydroxide solution is positively charged, thus showing maximum flocculation towards negatively charged ions.
• $\mathrm{Flocculation}\mathrm{value}=1/\mathrm{Coagulation}\mathrm{power}$
• The compounds ${\mathrm{AlCl}}_3,{\mathrm{KNO}}_3,\mathrm{KBr}$ which contain one negative ion.

$$\begin{gathered} {\mathrm{AlCl}}_3\to \mathrm{Al}{}^{3+}+3{\mathrm{Cl}}^{-} \\ {\mathrm{KNO}}_3\to {\mathrm{K}}^{+}+{{\mathrm{NO}}_3}^{-} \\ \mathrm{KBr}\to {\mathrm{K}}^{+}+{\mathrm{Br}}^{-} \end{gathered}$$

• The compound ${\mathrm{K}}_2{\mathrm{CrO}}_4$ contains two negative ions ${\mathrm{K}}_2{\mathrm{CrO}}_4\to 2{\mathrm{K}}^{+}+{{\mathrm{CrO}}_4}^{2-}$
• The compound ${\mathrm{K}}_2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ contains three negative ions ${\mathrm{K}}_2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]\to 3{\mathrm{K}}^{+}+[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{3-}$.

The explanation for the incorrect options:

A. ${\mathrm{AlCl}}_3>{\mathrm{K}}_3\left[\mathrm{Fe}\right(\mathrm{CN}\left)6\right]>{\mathrm{K}}_2{\mathrm{CrO}}_4>\mathrm{KBr}={\mathrm{KNO}}_3$

The compounds have ${\mathrm{K}}_2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$,${\mathrm{K}}_2{\mathrm{CrO}}_4$ a high flocculation value than ${\mathrm{AlCl}}_3$.

B. ${\mathrm{K}}_3[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6]<{\mathrm{K}}_2{\mathrm{CrO}}_4<{\mathrm{AlCl}}_3<\mathrm{KBr}<{\mathrm{KNO}}_3$

These three compounds ${\mathrm{AlCl}}_3,{\mathrm{KNO}}_3,\mathrm{KBr}$ contains the same number of negative ions. Thus, the flocculation value is the same.

D. ${\mathrm{K}}_3[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6]>{\mathrm{AlCl}}_3>{\mathrm{K}}_2{\mathrm{CrO}}_4>\mathrm{KBr}>{\mathrm{KNO}}_3$

The flocculation value of ${\mathrm{K}}_2{\mathrm{CrO}}_4$ is higher than ${\mathrm{AlCl}}_3$. As ${\mathrm{K}}_2{\mathrm{CrO}}_4$ carries two negative ions while ${\mathrm{AlCl}}_3$ contains one negative ion.
Thus, option C is correct.