Question

What kind of colloid particle would result if ($ \text{AgNO}_3 $) solution is being added to excess KI solution?

• A. Negatively charged \([ \text{AgI} ]^{\text{-}} \)
• B. Positively charged \([ \text{Ag} ]^{+} \text{Ag} \)
• C. Negatively charged \([ \text{AgNO}_3 ]^{\text{-}} \)
• D. Negatively charged \([ \text{AgI} ]^{\text{-}} \text{NO}_3^{\text{-}} \)

Hint:

In the formation of colloidal particles, excess ions can stabilize the particles and impart a charge to them, such as in the case of \( [\text{AgI}]^- \) formed with excess iodide.

Answer 1

The Correct Option is A

Step 1: Precipitation of AgI.
When silver nitrate (\( \text{AgNO}_3 \)) solution is added to excess potassium iodide (\( \text{KI} \)) solution, silver iodide (\( \text{AgI} \)) precipitates out according to the following reaction: \[ \text{AgNO}_3 + \text{KI} \rightarrow \text{AgI} \, (s) + \text{KNO}_3 \]
Step 2: Formation of colloidal particles.
The precipitate formed, \( \text{AgI} \), can form colloidal particles in the presence of excess \( \text{KI} \). Due to the presence of excess iodide ions (\( \text{I}^- \)), these particles will be negatively charged.
Step 3: Conclusion.
Thus, the colloidal particles formed are negatively charged \( [\text{AgI}]^- \).