Question:
The coagulating power of an electrolyte for arsenious sulphide decrease in order
The coagulating power of an electrolyte for arsenious sulphide decrease in order
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Hardy-Schulze rule: The ion opposite in charge to the colloid with higher valence has greater coagulating power.
Updated On: Apr 20, 2026
- \(\mathrm{Na}^+>\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}\)
- \(\mathrm{PO}_4^{3-}>\mathrm{SO}_4^{2-}>\mathrm{Cl}^-\)
- \(\mathrm{Cl}^->\mathrm{SO}_4^{2-}>\mathrm{PO}_4^{3-}\)
- \(\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^+\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Arsenious sulphide (As\(_2\)S\(_3\)) is a negatively charged colloid. Coagulating power follows Hardy-Schulze rule: higher valence of oppositely charged ion has greater coagulating power.
Step 2: Detailed Explanation: As\(_2\)S\(_3\) is negative, so cations cause coagulation. Coagulating power is proportional to charge on the cation. Hence, the order is: \[ \mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^+ \]
Step 3: Final Answer: \[ \boxed{\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^+} \]
Step 2: Detailed Explanation: As\(_2\)S\(_3\) is negative, so cations cause coagulation. Coagulating power is proportional to charge on the cation. Hence, the order is: \[ \mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^+ \]
Step 3: Final Answer: \[ \boxed{\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^+} \]
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