For the coagulation of a negative sol,the spe | vedclass

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Question

(B) According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to the magnitude of its charge.For a negative sol,the

Vedclass · Accepted Answer

$Ba^{2+}$

Vedclass · Answer

(B) According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to the magnitude of its charge.For a negative sol,the coagulating species must be a cation.Comparing the given cations: $Na^{+}$ (charge $+1$) and $Ba^{2+}$ (charge $+2$).Since $Ba^{2+}$ has a higher charge than $Na^{+}$,it will have a higher flocculating power.Therefore,$Ba^{2+}$ is the correct species.

For the coagulation of a negative sol,the species below,that has the highest flocculating power is:

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