For the coagulation of a negative arsenic sul | vedclass

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(B) According to the $Hardy-Schulze$ rule,the coagulating power of an ion is directly proportional to its valency for the coagulation of a sol.For a n

Vedclass · Accepted Answer

$Na^{+} < Ba^{2+} < Al^{3+}$

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(B) According to the $Hardy-Schulze$ rule,the coagulating power of an ion is directly proportional to its valency for the coagulation of a sol.For a negatively charged sol like arsenic sulphide $(As_2S_3)$,the coagulating power of cations increases with the increase in their positive charge.The valencies of the given ions are: $Na^{+} (+1)$,$Ba^{2+} (+2)$,and $Al^{3+} (+3)$.Therefore,the increasing order of coagulating power is $Na^{+} < Ba^{2+} < Al^{3+}$.

For the coagulation of a negative arsenic sulphide sol,what is the increasing order of the coagulating power of the ions $Na^{+}$,$Al^{3+}$,and $Ba^{2+}$?

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