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Question

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


A

AlCl3>K3[Fe(CN)6]>K2CrO4>KBr=KNO3

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B

K3[Fe(CN)6]<K2CrO4<AlCl3<KBr<KNO3

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C

K3[Fe(CN)6]<K2CrO4<KBr=KNO3=AlCl3

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D

K3[Fe(CN)6]>AlCl3>K2CrO4>KBr>KNO3

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Solution

The correct option is C

K3[Fe(CN)6]<K2CrO4<KBr=KNO3=AlCl3


The explanation for the correct options:

C) K3[Fe(CN)6]<K2CrO4<KBr=KNO3=AlCl3

  • 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.
  • Flocculationvalue=1/Coagulationpower
  • The compounds AlCl3,KNO3,KBr which contain one negative ion.

AlCl3Al3++3Cl-KNO3K++NO3-KBrK++Br-

  • The compound K2CrO4 contains two negative ions K2CrO42K++CrO42-
  • The compound K2[Fe(CN)6] contains three negative ions K2[Fe(CN)6]3K++[Fe(CN)6]3-.

The explanation for the incorrect options:

A. AlCl3>K3[Fe(CN)6]>K2CrO4>KBr=KNO3

The compounds have K2[Fe(CN)6],K2CrO4 a high flocculation value than AlCl3.

B. K3[Fe(CN)6]<K2CrO4<AlCl3<KBr<KNO3

These three compounds AlCl3,KNO3,KBr contains the same number of negative ions. Thus, the flocculation value is the same.

D. K3[Fe(CN)6]>AlCl3>K2CrO4>KBr>KNO3

The flocculation value of K2CrO4 is higher than AlCl3. As K2CrO4 carries two negative ions while AlCl3 contains one negative ion.
Thus, option C is correct.


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