Question

On adding $AgN{O}_3$ solution into KI solution, a negatively charged colloidal sol is obtained when they are in?

• A. $50$ mL of $0.1$M $AgN{O}_3+40$mL of $0.1$M KI
• B. $50$ mL of $0.1$M $AgN{O}_3+50$mL of $0.2$M KI
• C. $50$ mL of $0.2$M $AgN{O}_3+50$mL of $0.2$M KI
• D. None of these

Answer 1

The correct option is B $50$ mL of $0.1$M $AgN{O}_3+50$mL of $0.2$M KI

In order to form a negatively charged colloidal sol on adding $AgN{O}_3$ solution into KI solution, KI should be in excess.

Among all options, KI is in excess in option B as shown:

50 mL of 0.1 M $AgN{O}_3$ + 40 mL of 0.1 M KCl

m moles of $AgN{O}_3=50\times 0.1=5$

m moles of $KI=50\times 0.2=10$

KI is in excess

So, $I^{-}$ is adsorbed on AgI sol