Question

Which mixture of the solutions will lead to the formation of negatively charged colloidal $(AgI)I^-$ sol. ?

• A. 50 mL of 2 M ${AgNO_3}$ + 50 mLof 1.6 M KI
• B. 50 mL of 0.1 M ${AgNO_3}$ + 50 mL of 0.1 MKI
• C. 50 mL of 1M ${AgNO_3}$ + 50 mL of 1.5 M Kl
• D. 50 mL of 1M ${AgNO_3}$ + 50 mL of 2M KI

Answer 1

The Correct Option is D

To determine which mixture of solutions will form a negatively charged colloidal (AgI)I^− sol, we need to consider the principle of excess reactant leading to charged colloids:

1. When silver iodide (AgI) is formed, it can exist as a colloidal particle either with a positive charge or a negative charge, depending on the excess of Ag^+ or I^− ions.

2. If there is an excess of iodide ions (I^−), the colloidal particle will get a negative charge, forming (AgI)I^−.

3. To form negatively charged colloidal sol (AgI)I^−, iodide concentration must be greater than the silver ion concentration.

Let's analyze the options:

• Option 1: 50 mL of 2 M {AgNO_3} + 50 mL of 1.6 M KI: Here, moles of {Ag^+} = 0.1, moles of I^− = 0.08. {Ag^+} is in excess, resulting in a positively charged sol.

• Option 2: 50 mL of 0.1 M {AgNO_3} + 50 mL of 0.1 M KI: Moles are equal here, so it is neutral or can depend on minor ion presence.

• Option 3: 50 mL of 1 M {AgNO_3} + 50 mL of 1.5 M KI: Moles of {Ag^+} = 0.05 and moles of I^− = 0.075. I^− is in excess, could form negatively charged sol.

• Option 4: 50 mL of 1 M {AgNO_3} + 50 mL of 2 M KI: Moles of {Ag^+} = 0.05 and moles of I^− = 0.10. Clearly, I^− is in excess, leading to the formation of a negatively charged sol, (AgI)I^−.

Upon comparison, Option 4 provides a greater excess of iodide ions compared to silver ions, favoring the formation of a negatively charged colloidal sol of (AgI)I^−.

Therefore, the correct answer is:

• 50 mL of 1 M {AgNO_3} + 50 mL of 2 M KI