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  • As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order:  Option: 1 <img alt

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

Option: 1

AlCl_{3}>K_{3}[Fe(CN)_{6}]>K_{2}CrO_{4}>KBr=KNO_{3}


Option: 2

K_{3}[Fe(CN)_{6}]< K_{2}CrO_{4}<AlCl_{3} <KBr< KNO_{3}


Option: 3

K_{3}[Fe(CN)_{6}]< K_{2}CrO_{4}<KBr= KNO_{3}=AlCl_{3}

 


Option: 4

K_{3}[Fe(CN)_{6}]>AlCl_{3} > K_{2}CrO_{4}>KBr> KNO_{3}


Answers (1)

best_answer

As per Hardy Schulze rules coagulation power.

\textrm{Coagulation value or flocculation value}\propto \frac{1}{\textrm{Coagulation power}}

Since Fe(OH)3 is positively charged sol, hence, the anionic charge will flocculate.

As per Hardy Schulze rules coagulation power of anion follows the order :

[\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}>\mathrm{CrO}_{4}{ }^{2-}>\mathrm{Cl}^{-}=\mathrm{Br}^{-}=\mathrm{NO}_{3}{ }^{-}

Higher the coagulation power lower will be its flocculation value therefore order will be :

[\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}<\mathrm{CrO}_{4}{ }^{2-}<\mathrm{Cl}^{-}=\mathrm{Br}^{-}=\mathrm{NO}_{3}{ }^{-}

Therefore, Option(3) is correct.

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Anam Khan

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