Question

An arsenious sulphide sol carries a negative charge. The maximum precipitating power of this sol is possessed by:
(A) ${{K}_{2}}S{{O}_{4}}$
(B) $CaC{{l}_{2}}$
(C) $N{{a}_{3}}P{{O}_{4}}$
(D) $AlC{{l}_{3}}$

Answer 1

To answer this question we should know about the coagulating power of a sol and Hardy-Schulze rule. When a charged sol comes in the vicinity of the charge opposite to the charge it is carrying then it gets coagulated or precipitated.

Complete step by step solution:

Let’s look at the answer:
- The arsenious sulphide sol given in the question carries a negative charge on it. It will get precipitated when a positive charge will come into its contact.
- The process of precipitation of a colloidal solution by the addition of an electrolyte in excess is termed as its coagulation or flocculation.
- Negatively charged sols are precipitated by cations while positively charged sols are precipitated by anions. This is governed by Hardy-Schulze rule as:
“According to Hardy-Schulze rule, the greater the valency of the flocculating ion (ion carrying opposite charge to the sol) the greater will be its coagulating power.”
Now, in ${{K}_{2}}S{{O}_{4}}$, K is the cation and it carries +1 charge on it. In $CaC{{l}_{2}}$, Ca is the cation and it carries a +2 charge on it. In $N{{a}_{3}}P{{O}_{4}}$, Na is the cation and it carries +1 charge on it. In $AlC{{l}_{3}}$, Al is the cation and it carries a +3 charge on it.
Therefore, $AlC{{l}_{3}}$will cause maximum precipitation of the given sol.
Hence, the answer to the given question is option (D)

Note: While answering the above question, the charges on the various ions should be remembered carefully. If a positively charged sol is given in the question then its coagulation will depend on the amount of charge on the negative ion. Greater the negative charge, maximum will be the coagulation.