Question

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?

• A. NaCl \textgreater BaCl$_2$ \textgreater AlCl$_3$
• B. BaCl$_2$ \textgreater AlCl$_3$ \textgreater NaCl
• C. NaCl \textgreater AlCl$_3$ \textgreater BaCl$_2$
• D. AlCl$_3$ \textgreater BaCl$_2$ \textgreater NaCl

Hint:

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.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question describes a negatively charged colloidal solution and asks for the order of coagulation power of three different electrolytes. This involves applying the Hardy-Schulze rule.

Step 3: Detailed Explanation:
1. Charge of Colloidal Particles: The problem states that the colloidal particles move toward the anode (positive electrode) when subjected to an electric field. This indicates that the colloidal particles are negatively charged.
2. Hardy-Schulze Rule: This rule states that the coagulation power of an effective ion is directly proportional to its valency (charge). For a negatively charged sol, positive ions (cations) are the effective ions that cause coagulation.
3. Effective Ions and their Valencies: The electrolytes provided are NaCl, BaCl$_2$, and AlCl$_3$. The cations they provide are:
* From NaCl: $Na^+$ (valency +1)
* From BaCl$_2$: $Ba^{2+}$ (valency +2)
* From AlCl$_3$: $Al^{3+}$ (valency +3)
4. Order of Coagulation Power: According to the Hardy-Schulze rule, the higher the valency of the effective ion, the greater its coagulation power.
Therefore, the order of coagulation power is:
$Al^{3+}$ \textgreater $Ba^{2+}$ \textgreater $Na^+$
Which means, the order of electrolytes is:
AlCl$_3$ \textgreater BaCl$_2$ \textgreater NaCl.

Step 4: Final Answer:
The correct order of coagulation power is AlCl$_3$ \textgreater BaCl$_2$ \textgreater NaCl.

Answer 2

The colloid here moves toward the anode, which means the particles are negatively charged, so coagulating this sol calls for positively charged ions, and the Hardy-Schulze rule says the coagulating power of an ion rises sharply with its charge. Each ordering below is checked against that rule.

1. \(NaCl\) > \(BaCl_2\) > \(AlCl_3\): This places the singly charged \(Na^+\) ion ahead of the doubly and triply charged ions, which is backwards - higher charge should mean stronger, not weaker, coagulating power.
2. \(BaCl_2\) > \(AlCl_3\) > \(NaCl\): This correctly ranks \(NaCl\) last, but places the doubly charged \(Ba^{2+}\) ahead of the triply charged \(Al^{3+}\), reversing the expected order between those two.
3. \(NaCl\) > \(AlCl_3\) > \(BaCl_2\): This scrambles the order in two places at once, ranking the weakest ion first and swapping the two strongest ions as well.
4. \(AlCl_3\) > \(BaCl_2\) > \(NaCl\): This lines up exactly with rising charge - triply charged \(Al^{3+}\) first, doubly charged \(Ba^{2+}\) second, singly charged \(Na^+\) last - matching the Hardy-Schulze rule precisely.

Since the coagulating ion's power increases with its positive charge, and \(Al^{3+}\) carries the highest charge among the three cations, it must have the strongest effect, followed by \(Ba^{2+}\) and then \(Na^+\).

Therefore, the correct answer is \(AlCl_3\) > \(BaCl_2\) > \(NaCl\).