For coagulation of arsenious sulphide sol, which of the following salt solutions will be most effective?

We know from colloid chemistry that arsenious sulphide sol, written as $$As_2S_3$$, is obtained by passing hydrogen sulphide through an arsenious oxide solution. In the sol particles $$S^{2-}$$ ions dominate the surface, so every colloidal particle carries a net negative charge.

For bringing about coagulation, ions of opposite charge (called counter-ions) are required; therefore positively charged ions will neutralise the negative charge on the $$As_2S_3$$ sol.

The Hardy-Schulze rule states the principle to compare the effectiveness of different counter-ions. We state the rule first:

$$\text{Coagulating\ power} \propto z^6$$

where $$z$$ is the valency of the ion that carries a charge opposite to that of the sol particles. In words, the greater the valency of the counter-ion, the greater is its power to cause coagulation.

Now we list the positive ions supplied by each electrolyte in the options and write down their valencies:

$$\begin{aligned} Na_3PO_4 &\longrightarrow 3\,Na^+ \quad \text{valency } z = 1 \\ NaCl &\longrightarrow Na^+ \quad \text{valency } z = 1 \\ AlCl_3 &\longrightarrow Al^{3+} \quad \text{valency } z = 3 \\ BaCl_2 &\longrightarrow Ba^{2+} \quad \text{valency } z = 2 \end{aligned}$$

Because $$P \propto z^6$$, we substitute the different valency values one by one to see the relative coagulating powers:

$$\begin{aligned} P_{Na^+} &\propto (1)^6 = 1 \\ P_{Ba^{2+}} &\propto (2)^6 = 64 \\ P_{Al^{3+}} &\propto (3)^6 = 729 \end{aligned}$$

Clearly, $$Al^{3+}$$ gives the largest numerical value, meaning it has the greatest tendency to neutralise the negative charge on the $$As_2S_3$$ sol particles. The other cations, $$Ba^{2+}$$ and $$Na^{+}$$, have much lower coagulating powers.

Therefore, among the given salt solutions, $$AlCl_3$$ will be the most effective in bringing about coagulation of the arsenious sulphide sol.

Hence, the correct answer is Option C.