Which of the salt-solution is most effective for coagulation of arsenious sulphide?
(a) BaCl2
(b) Al

Q: Why is arsenious sulphide specifically a negatively charged sol?

During the preparation of arsenious sulphide sol (usually by passing $\text{H}_2\text{S}$ gas through a solution of arsenious oxide), the colloidal particles preferentially adsorb sulphide ions ($\text{S}^{2-}$) from the dispersion medium onto their surface, acquiring a net negative charge.

Q: Do the anions in the salt solutions play any role in this coagulation?

No, not in this specific case. Since the sol is negatively charged, it is the oppositely charged ions (the cations) that are responsible for neutralizing the charge and causing coagulation. The anions are essentially spectator ions here.

Q: What exactly does 'coagulation' mean in this context?

Coagulation is the process where the dispersed colloidal particles lose their charge, stop repelling each other, come together to form larger aggregates, and eventually settle down as a precipitate under the influence of gravity.

Q: Why does a higher charge make an ion more effective at coagulating?

A higher charge means a stronger electrostatic force of attraction. An $\text{Al}^{3+}$ ion can neutralize the negative charge of the colloidal particles much more efficiently and rapidly than a $\text{Na}^{+}$ ion, requiring a much lower concentration to achieve the same effect.

Q: What if we had to coagulate a positively charged sol like Ferric hydroxide?

If the sol were positively charged, you would look at the anions provided by the salts. According to the Hardy-Schulze rule, the anion with the highest negative charge (like $\text{PO}_4^{3-}$ from $\text{Na}_3\text{PO}_4$) would be the most effective coagulating agent.

Question

Which of the salt-solution is most effective for coagulation of arsenious sulphide?

Options:

(A) $	ext{BaCl}_2$

(B) $	ext{AlCl}_3$

(C) $	ext{Na}_3	ext{PO}_4$

(D) $	ext{NaCl}$

Sigma Percentile · Accepted Answer

### The Mystery of the Arsenious Sulphide Sol

Imagine you are looking at a beaker filled with a slightly cloudy, yellowish liquid. This is the arsenious sulphide sol, chemically written as $	ext{As}_2	ext{S}_3$. But there is a hidden secret on the surface of these tiny colloidal particles. During their formation, they preferentially adsorb sulphide ions ($	ext{S}^{2-}$). This gives the entire colloidal particle a **net negative charge**.

Because all the particles carry the same negative charge, they repel each other. This electrostatic repulsion is what keeps the colloid stable and prevents the particles from clumping together and settling down.

### The Art of Coagulation

Now, suppose we want to destroy this stability. We want the particles to clump together and precipitate out. This process is called **coagulation** or flocculation. To do this, we need to neutralize that negative charge. How do we neutralize a negative charge? By introducing a positive charge, of course! We need to add an electrolyte that provides cations (positively charged ions).

But not all cations are created equal. This is where the brilliant **Hardy-Schulze Rule** comes into play. The rule states a very intuitive principle: *The greater the valency (charge) of the flocculating ion added, the greater is its power to cause precipitation.*

Think of it like paying off a debt. If the colloidal particle has a large negative debt, a cation with a $+3$ charge pays it off much faster than a cation with a $+1$ charge.

### Analyzing the Contenders

Let's look at the salt solutions provided in the options and see what kind of "currency" (cations) they bring to the table:

*   $	ext{BaCl}_2$ dissociates to give $	ext{Ba}^{2+}$ ions.
*   $	ext{AlCl}_3$ dissociates to give $	ext{Al}^{3+}$ ions.
*   $	ext{Na}_3	ext{PO}_4$ dissociates to give $	ext{Na}^{+}$ ions.
*   $	ext{NaCl}$ dissociates to give $	ext{Na}^{+}$ ions.

### The Final Verdict

We are looking for the most effective coagulating agent for our negatively charged $	ext{As}_2	ext{S}_3$ sol. According to the Hardy-Schulze rule, we need the cation with the highest positive charge.

Comparing the charges:

$$	ext{Al}^{3+} > 	ext{Ba}^{2+} > 	ext{Na}^{+}$$

The Aluminum ion ($	ext{Al}^{3+}$) carries the highest charge. Therefore, it has the maximum coagulating power. It will neutralize the negative sol most rapidly and effectively.

Thus, the salt solution containing $	ext{Al}^{3+}$, which is **$	ext{AlCl}_3$**, is the correct answer.

Animated Solution for Chemistry - Surface Chemistry: Which of the salt-solution is most effective for coagulation of arsenious sulphide?

Select Answer:

Visualized Solution

\text{Nature of Arsenious Sulphide Sol}

\text{Hardy-Schulze Rule}

\text{Identifying the Cations}

\text{Comparing Coagulating Power}

\text{Conclusion}

\text{Food for Thought}

The Sigma Insight: Colloids, Micelles and and Emulsions

The Mystery of the Arsenious Sulphide Sol

The Art of Coagulation

Analyzing the Contenders

The Final Verdict

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