Question

Flocculation value of \(\mathrm{BaCl_2}\) is much less than that of \(\mathrm{KCl}\) for sol A and flocculation value of \(\mathrm{Na_2SO_4}\) is much less than that of \(\mathrm{NaBr}\) for sol B. The correct statement among the following is :

• A. Both the sols A and B are negatively charged
• B. Sol A is positively charged and Sol B is negatively charged
• C. Both the sols A and B are positively charged
• D. Sol A is negatively charged and sol B is positively charged

Hint:

By the Hardy-Schulze rule, the ion with charge opposite to the sol and of higher valence has the lowest flocculation value.

Answer 1

The Correct Option is D

Flocculation value is the minimum concentration of electrolyte needed to coagulate a sol; it is lower for the ion of higher charge having a sign opposite to that of the sol (Hardy-Schulze rule).
Sol A: \(\mathrm{BaCl_2}\) has a much lower flocculation value than \(\mathrm{KCl}\). The cations are \(\mathrm{Ba^{2+}}\) and \(\mathrm{K^+}\), and the divalent \(\mathrm{Ba^{2+}}\) is the more effective, so the coagulating ion is a cation and sol A is negatively charged.
Sol B: \(\mathrm{Na_2SO_4}\) has a much lower flocculation value than \(\mathrm{NaBr}\). The cation \(\mathrm{Na^+}\) is common, so the difference comes from the anions: \(\mathrm{SO_4^{2-}}\) (divalent) is more effective than \(\mathrm{Br^-}\), so the coagulating ion is an anion and sol B is positively charged.
Answer: (4) Sol A is negatively charged and sol B is positively charged