Which of the following electrolyte will have maximum flocculation value for Fe(OH)3 sol

A
$NaCl$
B
${Na}_{2} S$
C
${({NH}_{4})}_{3} {PO}_{4}$
D
$K_{2} {SO}_{4}$

Solution:

$Flocculation value \propto \frac{1}{Coagulating power}$

Greater the coagulating power of effective ion, lower will be the flocculation value of the electrolyte.

$Fe (OH)_{3}$ is a positively charged sol. According to Hardy Schulze rule, greater the charge on the oppositely charged ion of the electrolyte added, more effective it is in bringing coagulation.

Among the four electrolytes, $NaCl$ has the lowest coagulating power, so its flocculation value will be maximum.

Which of the following electrolyte will have maximum flocculation value for $Fe (OH)_{3}$ sol

Solution:

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