Question

In an electrical field, the particles of a colloidal system move towards cathode. The coagulation of the same sol is studied using $K_2{SO}_4\left(I\right),{Na}_3{PO}_4\left(II\right),K_{4}Fe{\left(CN\right)}_6\left(III\right)$ and $NaCl\left(IV\right)$. Their coagulating power should be:

• A. $\left(I\right)>\left(II\right)>\left(III\right)>\left(IV\right)$
• B. $\left(III\right)>\left(II\right)>\left(I\right)>\left(IV\right)$
• C. $\left(III\right)>\left(I\right)>\left(II\right)>\left(IV\right)$
• D. $\left(IV\right)>\left(III\right)>\left(I\right)>\left(II\right)$
• E. $\left(IV\right)>\left(I\right)>\left(II\right)>\left(III\right)$

Answer 1

Answer: (B)

$\left(III\right)>\left(II\right)>\left(I\right)>\left(IV\right)$

Valency of different given ions are:
I) $K_{2}S{O}_4$ = $2\times 39+32+16\times 4=174$
II) $N{a}_{3}P{O}_4$ = $22\times 3+30+16\times 4=160$
III) $K_{4}Fe(CN{)}_6=39\times 4+55+285\times 6=1921$
IV) $NaCl=22+35=57$
According to Hardy Schulze rule, greater the valency of the active ion or flocculating ion, greater will be its coagulating power”.
Thus coagulating power in given solutions in descending order should be highest for $K_{4}Fe(CN{)}_6$, then $K_{2}S{O}_4$ then $N{a}_{3}P{O}_4$ and lowest for $NaCl$.
Thus Option B is correct.