Question

The coagulating power of electrolytes having ions Na$^+$, Al$^{3+}$ and Ba$^{2+}$ for As$_2$S$_3$ sol increases in the order

• A. Al$^{3+}$ $<$ Ba$^{2+}$ $<$ Na$^{+}$
• B. Na$^{+}$ $<$ Ba$^{2+}$ $<$ Al$^{3+}$
• C. Ba$^{2+}$ $<$ Na$^{+}$ $<$ Al$^{3+}$
• D. Al$^{3+}$ $<$ Na$^{+}$ $<$ Ba$^{2+}$

Hint:

Using a simple frame or just bolding for the box
Key Points: 
As$_2$S$_3$ sol is negatively charged. 
Coagulation of negatively charged sols is caused by cations. 
Hardy-Schulze Rule: Coagulating power increases significantly with increasing charge magnitude of the coagulating ion. 
Cation Charges: Na$^{+}$ (+1), Ba$^{2+}$ (+2), Al$^{3+}$ (+3). 
Increasing Coagulating Power: Na$^{+}$ $<$ Ba$^{2+}$ $<$ Al$^{3+}$.

Answer 1

The Correct Option is B

The coagulation (or flocculation) of a colloidal sol involves neutralizing the charge on the colloidal particles, causing them to aggregate and precipitate. (A) Nature of As₂S₃ Sol: Arsenic sulfide (As₂S₃) sol is a typical example of a negatively charged colloid. The negative charge usually arises from the preferential adsorption of sulfide ions (S^2-) onto the surface of the As₂S₃ particles. (B) Coagulation by Electrolytes: To coagulate a negatively charged sol, electrolytes containing positively charged ions (cations) are added. The cations act as counter-ions, neutralizing the negative charge on the sol particles. (C) Hardy-Schulze Rule: This rule states that the effectiveness of an ion in causing coagulation (its coagulating power) is directly related to the magnitude of its charge. For coagulating a negatively charged sol, the greater the positive charge on the cation, the greater its coagulating power. (D) Comparing the Ions: We are given the cations Na⁺, Ba²⁺, and Al³⁺. Their charges are +1, +2, and +3, respectively. (E) Order of Coagulating Power: According to the Hardy-Schulze rule, the coagulating power follows the order of the charge magnitude: Al³⁺>Ba²⁺>Na⁺ (F) Increasing Order: The question asks for the order in which the coagulating power increases. Therefore, we arrange the ions from the lowest coagulating power to the highest: Na⁺<Ba²⁺<Al³⁺ This corresponds to option (B).