Question

When excess of $AgN{O_3}$ is treated with KI solution, AgI forms:
A. Positively charged sol
B. Negatively charged sol.
C. Neutral solution
D. True solution

Answer 1

Hint:The colloidal sols are divided into positively charged sol and negatively charged sol. The positively charged sol are metal hydroxides and the negative charged sols are of metal sulphide.

Complete answer:
When a dilute solution of potassium iodide(KI) is added to dilute solution of silver nitrate, a positive charge sol to silver iodide (AgI) is generated. This formation of positive charge sol is because of absorption of silver ions($A{g^ + }$).from the dispersion medium on the precipitate formed by silver iodide (AgI).
The reaction of silver nitrate with potassium iodide is shown below.
$AgN{O_3}(excess) + KI \to AgI\xrightarrow{{AgN{O_3}}}AgI/A{g^ + }$
When a dilute solution of silver nitrate (AgI) is added to excess dilute solution of potassium iodide, silver iodide with negative charged sol is generated. This formation of negative charged sol is due to the absorption of iodide ion (${I^ - }$) from the dispersion medium on the precipitate formed by the silver iodide.
$AgN{O_3} + KI(excess) \to AgI\xrightarrow{{AgN{O_3}}}AgI/{I^ - }$
When silver nitrate is reacted with excess potassium iodide, the colloidal particles formed get released towards the anode.
Thus, when excess of $AgN{O_3}$ is treated with KI solution, AgI forms a positively charged sol.
Therefore, the correct option is (A).

Note:
Make sure about the amount of silver nitrate used. The slight difference in the amount will affect the formation of sol. The classification of colloidal sol is dependent on the nature of the charge which is present in the particles of the dispersed phase.