Which mixture of the solutions will lead to the formation of negatively charged colloidal [AgI]I
−
sol?
- A50mL of 1M AgNO
3
+50mL of 1.5M KI - B50mL of 1M AgNO
3
+50mL of 2M KI - C50mL of 2M AgNO
3
+50mL of 1.5M KI - D50mL of 0.1M AgNO
3
+50mL of 0.1M KI
Solution & Step-by-step Explanation
To form a negatively charged sol of Silver Iodide ([AgI]I
−
), excess KI must be present so that I
−
ions are adsorbed on the surface of AgI particles.
A: Moles of KI> Moles of AgNO
3
(0.075>0.05). Negative Sol.
B: Moles of KI> Moles of AgNO
3
(0.1>0.05). Negative Sol.
Wait, typically the first significant excess is the intended answer. Both A and B provide negative sols. In NEET 2019, both (1) and (2) were considered technically correct depending on the final key.
−
), excess KI must be present so that I
−
ions are adsorbed on the surface of AgI particles.
A: Moles of KI> Moles of AgNO
3
(0.075>0.05). Negative Sol.
B: Moles of KI> Moles of AgNO
3
(0.1>0.05). Negative Sol.
Wait, typically the first significant excess is the intended answer. Both A and B provide negative sols. In NEET 2019, both (1) and (2) were considered technically correct depending on the final key.