Among the electrolytes $\mathrm{Na}_2\mathrm{SO}_4, \mathrm{CaCl}_2, \mathrm{Al}_2(\mathrm{SO}_4)_3$ and $\mathrm{NH}_4\mathrm{Cl}$, the most effective coagulating agent for $\mathrm{Sb}_2\mathrm{~S}_3$ sol is

(C) $\mathrm{Al}_2(\mathrm{SO}_4)_3$

(A) $\mathrm{Na}_2\mathrm{SO}_4$

(D) $\mathrm{NH}_4\mathrm{Cl}$

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Surface Chemistry

Among the electrolytes

Na_{2} SO_{4}, CaCl_{2}, Al_{2} (SO_{4})_{3}

and

NH_{4} Cl

, the most effective coagulating agent for

Sb_{2} S_{3}

sol is

(A) $Na_{2} SO_{4}$
(B) $CaCl_{2}$
(C) $Al_{2} (SO_{4})_{3}$ ✓ Correct
(D) $NH_{4} Cl$

Answer: (C)

Solution

Sb_{2} S_{3}

forms a negatively charged colloidal sol. According to the Hardy-Schulze rule, the coagulating power of an ion increases with its charge.

For coagulation of a negative sol, we need cations, and their effectiveness follows the order:

Al^{3 +} > Ca^{2 +} > Na^{+} > NH_{4}^{+}

Since

Al_{2} (SO_{4})_{3}

provides

Al^{3 +}

ions with the highest charge (+3), it is the most effective coagulating agent.

Therefore, the answer is **(C)

Al_{2} (SO_{4})_{3}

**.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option

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