Question

The coagulating power of an electrolyte for arsenious sulphide decrease in order

• A. \(\mathrm{Na}^+>\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}\)
• B. \(\mathrm{PO}_4^{3-}>\mathrm{SO}_4^{2-}>\mathrm{Cl}^-\)
• C. \(\mathrm{Cl}^->\mathrm{SO}_4^{2-}>\mathrm{PO}_4^{3-}\)
• D. \(\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^+\)

Hint:

Hardy-Schulze rule: The ion opposite in charge to the colloid with higher valence has greater coagulating power.

Answer 1

The Correct Option is D

Step 1: Underlying 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: 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: Conclusion: \[ \boxed{\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^+} \]