When dilute aqueous solution of `AgNO_3` (excess) is added to KI solution, positively charged sol of Agl in formed due to adsorption of

Options

A. `K^(+)`
B. `Ag^(+)`
C. `l^(-)`
D. `NO_(3)^(-)`

Answer

To solve the question, we need to analyze the reaction that occurs when a dilute aqueous solution of silver nitrate (AgNO₃) is added to potassium iodide (KI) solution, and how this leads to the formation of a positively charged sol of silver iodide (AgI).

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We start with a dilute aqueous solution of silver nitrate (AgNO₃) and potassium iodide (KI).
- AgNO₃ dissociates into Ag⁺ ions and NO₃⁻ ions in solution, while KI dissociates into K⁺ ions and I⁻ ions.

2. **Write the Dissociation Reactions**:
- AgNO₃ → Ag⁺ + NO₃⁻
- KI → K⁺ + I⁻

3. **Reaction Between Ag⁺ and I⁻**:
- When excess AgNO₃ is added to the KI solution, the Ag⁺ ions react with the I⁻ ions to form silver iodide (AgI):
- Ag⁺ + I⁻ → AgI (s)

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

5. **Charge on the Sol**:
- The AgI particles will adsorb excess Ag⁺ ions from the solution onto their surface. This adsorption of positively charged Ag⁺ ions onto the surface of the AgI particles gives the sol a net positive charge.
- Therefore, the sol of AgI formed is positively charged due to the adsorption of excess Ag⁺ ions.

6. **Conclusion**:
- The positively charged sol of AgI is formed due to the adsorption of excess Ag⁺ ions present in the solution.

### Final Answer:
When a dilute aqueous solution of AgNO₃ is added to KI solution, a positively charged sol of AgI is formed due to the adsorption of excess Ag⁺ ions.

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