The trend in the coagulating power for the sa | vedclass

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(D) According to the $Hardy-Schulze$ law,the coagulating power of an electrolyte depends on the valency of the active ion (the ion with a charge oppos

Vedclass · Accepted Answer

$MgCl_2 = BaCl_2 > NaCl$

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(D) According to the $Hardy-Schulze$ law,the coagulating power of an electrolyte depends on the valency of the active ion (the ion with a charge opposite to that of the colloidal particles). For negatively charged colloids,the coagulating power of cations follows the order of their valency: $Al^{3+} > Ba^{2+} = Mg^{2+} > Na^+$. Since $BaCl_2$ and $MgCl_2$ both provide divalent cations ($Ba^{2+}$ and $Mg^{2+}$) and $NaCl$ provides a monovalent cation $(Na^+)$,the coagulating power is $MgCl_2 = BaCl_2 > NaCl$.

The trend in the coagulating power for the same concentration of $BaCl_2$,$MgCl_2$,and $NaCl$,respectively,is

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