In the preparation of AgI sol, the excess fo `AgNO_(3)` is added to potassium iodide solution. The particles of the sol will acquire

Options

A. Nagative charge
B. Positive charge
C. No charge
D. Unpredictable

Answer

To solve the question regarding the preparation of AgI sol when excess AgNO₃ is added to potassium iodide (KI) solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants:**
- We have two reactants: Silver Nitrate (AgNO₃) and Potassium Iodide (KI).

2. **Dissociation of Reactants:**
- When KI is dissolved in water, it dissociates into potassium ions (K⁺) and iodide ions (I⁻).
- Similarly, AgNO₃ dissociates into silver ions (Ag⁺) and nitrate ions (NO₃⁻).

3. **Reaction Between Ions:**
- When excess AgNO₃ is added to the KI solution, the Ag⁺ ions will react with the I⁻ ions to form silver iodide (AgI) precipitate.
- The reaction can be represented as:
\[ \text{Ag}^+ (aq) + \text{I}^- (aq) \rightarrow \text{AgI} (s) \]

4. **Formation of AgI Sol:**
- The AgI formed is a colloidal sol. Since we have excess AgNO₃, there will be more Ag⁺ ions present in the solution.

5. **Charge on the Sol:**
- The AgI particles will have I⁻ ions on their surface due to the reaction with excess Ag⁺ ions. This leads to the overall charge of the sol.
- Since the sol particles acquire I⁻ ions, the AgI sol will be negatively charged.

6. **Conclusion:**
- Therefore, the particles of the AgI sol will acquire a negative charge due to the presence of excess iodide ions on their surface.

### Final Answer:
The particles of the AgI sol will acquire a negative charge.

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