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# State Hardy-Schulze. Between $A{l_2}{(S{O_4})_3}$ and ${(N{H_4})_3}P{O_4}$ which is a better coagulating agent for a negative sol?

Last updated date: 18th Mar 2023
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Hint :The coagulation capacity or flocculate capacity of an electrolyte increases if charge is increased and the law is called Hardy-Schulze law. According to this rule, greater the valency of active ions, greater is its coagulating power. That means the coagulating power of an electrolyte is directly proportional to the valency of the active ions.

Coagulation is the process of aggregating together the colloidal particles in order to change them into large sized particles which finally settle as a precipitate. Flocculating ion is the ion that is responsible for the neutralization of charge on the particles of the colloid. Flocculation value is the minimum concentration of electrolyte that is required to cause flocculation of a sol. The power of an electrolyte to precipitate a colloidal solution is called flocculating power. For negative sols, we require positive sols. $A{l^{ + 3}}$has more positive charge than $NH_4^ +$. So, $A{l_2}{(S{O_4})_3}$ is a better coagulation agent than${(N{H_4})_3}P{O_4}$.
Therefore, $A{l_2}{(S{O_4})_3}$ is a better coagulation agent than${(N{H_4})_3}P{O_4}$.

Note :
Flocculating agents are either inorganic salts or water-soluble organic polymers and they work by shrinking the double layer or by neutralizing the surface charge of the suspended particles. Remember that an ion with the maximum charge will have minimum coagulating value. The electrolyte having minimum coagulating power will have the maximum flocculation value. Note that flocculation power is inversely proportional to the coagulating power of ions.