Question:medium

A colloidal solution is subjected to an electric field. The colloidal particles move toward the anode. In the coagulation of this solution using NaCl, BaCl$_2$ and AlCl$_3$ separately, which one is the correct order of coagulation power?

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Remember the Hardy-Schulze rule: for coagulation of colloids, the effective ions are those with charge opposite to the colloidal particles. Higher the valency of these effective ions, greater their coagulating power.
Updated On: Jul 14, 2026
  • NaCl \textgreater BaCl$_2$ \textgreater AlCl$_3$
  • BaCl$_2$ \textgreater AlCl$_3$ \textgreater NaCl
  • NaCl \textgreater AlCl$_3$ \textgreater BaCl$_2$
  • AlCl$_3$ \textgreater BaCl$_2$ \textgreater NaCl
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The Correct Option is D

Solution and Explanation

Another way to rank these electrolytes is through the idea of flocculation value - the minimum amount of electrolyte needed to coagulate a fixed amount of sol. A stronger coagulating ion needs a smaller amount to do the job, so coagulating power and flocculation value move in opposite directions.

  1. \(Al^{3+}\): Carrying three positive charges, this ion neutralises the negative surface charge on the colloidal particles most efficiently, so only a tiny concentration of \(AlCl_3\) is needed to bring about coagulation. This gives it the lowest flocculation value and, therefore, the highest coagulating power.
  2. \(Ba^{2+}\): With two positive charges, this ion needs a somewhat larger concentration than \(Al^{3+}\) to achieve the same neutralising effect, placing \(BaCl_2\) in the middle both in flocculation value and coagulating power.
  3. \(Na^+\): Carrying just one positive charge, this ion is the least efficient at neutralising the negative surface, so a comparatively large concentration of \(NaCl\) is required. It therefore has the highest flocculation value and the weakest coagulating power of the three.

Ranking the three cations from lowest to highest flocculation value places \(Al^{3+}\) first, \(Ba^{2+}\) second, and \(Na^+\) last, which is the same as ranking their electrolytes from strongest to weakest coagulating power.

The correct answer is \(AlCl_3\) > \(BaCl_2\) > \(NaCl\).

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