Question

Identify the correct decreasing order of precipitation power of flocculating ion added, from following.

• A. $\text{Al}^{3+} \gt \text{Na}^+ \gt \text{Ba}^{2+}$
• B. $\text{Ba}^{2+} \gt \text{Al}^{3+} \gt \text{Na}^+$
• C. $\text{Al}^{3+} \gt \text{Ba}^{2+} \gt \text{Na}^+$
• D. $\text{Na}^+ \gt \text{Ba}^{2+} \gt \text{Al}^{3+}$

Hint:

The precipitation power increases dramatically with charge, not just linearly! For instance, a trivalent ion like $\text{Al}^{3+}$ can be up to 500 to 1000 times more effective at coagulating a negative sol than a monovalent ion like $\text{Na}^+$. Always sort by the highest absolute charge first.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the correct decreasing sequence of coagulation or precipitation efficiency among three different cationic species.

Step 2: Key Formula or Approach:
According to the

Hardy-Schulze Rule, the flocculating or precipitating capacity of an ion added to a lyophobic colloidal solution is directly proportional to the magnitude of its valency (ionic charge): $$\text{Precipitation Power} \propto (\text{Ionic Charge})^n$$

Step 3: Detailed Explanation:
Let's look at the absolute charge magnitudes carried by each of the given cations: Aluminum ion: $\text{Al}^{3+}$ (Charge magnitude = 3) Barium ion: $\text{Ba}^{2+}$ (Charge magnitude = 2) Sodium ion: $\text{Na}^+$ (Charge magnitude = 1) Since the Hardy-Schulze rule states that a higher charge magnitude provides vastly superior power to neutralize the opposing colloidal surface charges and cause rapid aggregation, the relative precipitation order must follow the valency trend: $$\text{Al}^{3+} \gt \text{Ba}^{2+} \gt \text{Na}^+$$

Step 4: Final Answer:
The correct decreasing order of precipitation power is $\text{Al}^{3+} \gt \text{Ba}^{2+} \gt \text{Na}^+$, which matches option (C).