Question

The coagulating power of electrolytes having ions Na$^+$, Al$^{3+$ and Ba$^{2+}$ for As$_2$S$_3$ sol increases in the order}

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

Hint:

Hardy–Schulze rule: \begin{itemize} \item Coagulation power $\propto$ charge of counter ion \item For negative sol $\Rightarrow$ higher valent cations more effective \end{itemize}

Answer 1

The Correct Option is B

Concept: According to Hardy–Schulze rule:

• Coagulating power depends on valency of oppositely charged ion
• Higher charge \(\Rightarrow\) greater coagulation

Step 1: Nature of sol As$_2$S$_3$ sol is negatively charged. Thus, cations act as coagulating ions. Step 2: Compare valencies Na$^+$ (1$^+$) Ba$^{2+}$ (2$^+$) Al$^{3+}$ (3$^+$) Higher charge \(\Rightarrow\) higher coagulating power. Conclusion: \[ \text{Na}^+ < \text{Ba}^{2+} < \text{Al}^{3+} \]