Single Choice

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

AAlCl$$_3 >$$ K$$_3$$[Fe(CN)$$_6$$] $$>$$ K$$_2$$CrO$$_4 >$$ KBr$$=$$KNO$$_3$$
BK$$_3$$[Fe(CN)$$_6$$] $$<$$ K$$_2$$CrO$$_4 <$$ KBr $$ = \ $$KNO$$_3 = \ $$AlCl$$_3$$
Correct Answer
CK$$_3$$[Fe(CN)$$_6$$] $$<$$ K$$_2$$CrO$$_4$$ $$<$$ AlCl$$_3$$ $$<$$ KBr $$<$$ KNO$$_3$$
DK$$_3$$[Fe(CN)$$_6$$] $$>$$ AlCl$$_3$$ $$>$$ K$$_2$$CrO$$_4$$ $$>$$ KBr $$>$$ KNO$$_3$$

Solution

According to Hardy - Schulze rule
"Greater is the valency of ions, greater is its power to bring about coagulation.

Flocculation value $$= \dfrac{1}{\text{Coagulation power}}$$

$$Fe(OH)_3$$ sol is a positive charge sol means it will show maximum flocculation value to negatively charged ions.

$$\left.\begin{matrix}AlCl_3 \to Al^{3+} + 3Cl^-\\KNO_3 \to K^+ + NO_3^- \\KBr \to K^+ + Br^- \end{matrix}\right]$$ All are single $$-ve$$ charged ions.

$$\left.\begin{matrix} K_2CrO_4 \to 2K^+ + CrO_4^{2-}\end{matrix}\right]$$ Two $$-ve$$ charge

$$\left.\begin{matrix} K_2[Fe(CN)_6] \to 3K^+ + [Fe(CN)_6]^{3-}\end{matrix}\right]$$ Three $$-ve$$ charge

Correct order would be
$$K_3[Fe(CN)_6] < K_2CrO_4 < KBr = KNO_3 = AlCl_3$$

Hence, the correct option is $$\text{B}$$

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