Question

Which of the following electrolytes is most effective in the coagulation of a negative sol like \(As_2S_3\)?

• A. \(NaCl\)
• B. \(MgCl_2\)
• C. \(AlCl_3\)
• D. \(KCl\)

Hint:

Since $As_2S_3$ is a negative sol, only the cation of each salt matters for coagulation; the common anion (chloride) can be ignored right away. Compare the cations by their charge, and remember coagulating power grows very steeply with valency, not in simple proportion to it.

Answer 1

The Correct Option is C

Concept: According to the Hardy–Schulze rule, the coagulating power of an electrolyte depends on the valency of the ion having charge opposite to that of the colloidal particles.
• For a negative sol, the cation is responsible for coagulation.
• Higher the valency of the cation, greater will be its coagulating power.

Step 1: Identify the charge on the sol. The sol \(As_2S_3\) is a negatively charged sol. Therefore, the coagulating ion must be a cation.

Step 2: Compare the valency of cations in the given electrolytes. \[ NaCl \rightarrow Na^+ \quad (\text{valency } = 1) \] \[ MgCl_2 \rightarrow Mg^{2+} \quad (\text{valency } = 2) \] \[ AlCl_3 \rightarrow Al^{3+} \quad (\text{valency } = 3) \]

Step 3: Apply Hardy–Schulze rule. \[ Al^{3+} > Mg^{2+} > Na^+ \] Thus, \(Al^{3+}\) has the highest coagulating power. \[ \boxed{AlCl_3} \]

Answer 2

Concept:

• In a negative sol, only the cation of the added electrolyte causes coagulation; the anion carries the same sign as the sol and plays no role, so a common anion across all options can be ignored.
• The quantitative form of the Hardy-Schulze rule states that coagulating power rises very sharply with the valency of the effective ion, roughly in proportion to the sixth power of its charge, not in simple proportion to the charge itself.

Step 1: Identify which ion of each electrolyte is responsible for coagulation.
$As_2S_3$ forms a negatively charged sol, so coagulation is caused by the cation of each added salt.

Step 2: Remove the common ion from consideration.
All four salts share $Cl^-$ as their anion, so $Cl^-$ cannot explain any difference in coagulating power between the options. Only the cation and its charge matter.

Step 3: List the cation and its valency supplied by each salt.
$NaCl \to Na^+$, valency $z = 1$
$KCl \to K^+$, valency $z = 1$
$MgCl_2 \to Mg^{2+}$, valency $z = 2$
$AlCl_3 \to Al^{3+}$, valency $z = 3$

Step 4: Rank the coagulating power using the sixth-power relation.
Coagulating power $\propto z^6$
For $z = 1$: power $\propto 1$
For $z = 2$: power $\propto 64$
For $z = 3$: power $\propto 729$
$Al^{3+}$ has by far the largest coagulating power among the given cations.

Final Answer: $AlCl_3$