Question:
When a dilute \(AgNO_3\) solution is mixed with excess of dilute \(KI\), the colloidal solution formed is
When a dilute \(AgNO_3\) solution is mixed with excess of dilute \(KI\), the colloidal solution formed is
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In precipitate sols, the ion present in excess is preferentially adsorbed on the colloidal particles and determines the charge on the sol.
Updated On: Jun 18, 2026
- \(AgNO_3/I^-\)
- \(AgI\)
- \(AgI/Ag^+\)
- \(AgI/I^-\)
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The Correct Option is D
Solution and Explanation
Step 1: Write the precipitation reaction.
When silver nitrate reacts with potassium iodide, \[ AgNO_3 + KI \rightarrow AgI \downarrow + KNO_3 \] A precipitate of silver iodide is formed.
Step 2: Consider the reagent present in excess.
The question states that \(KI\) is present in excess.
Therefore, after precipitation, iodide ions remain in solution.
Step 3: Determine the charge on colloidal particles.
According to the preferential adsorption theory, the ion present in excess gets adsorbed on the surface of the precipitate.
Since \(I^-\) ions are present in excess, they are adsorbed on the surface of \(AgI\) particles.
Thus, the colloidal particles become negatively charged.
The colloidal sol is represented as \[ AgI/I^- \]
Step 4: Final conclusion.
Hence, the colloidal solution formed is \[ \boxed{AgI/I^-} \] Therefore, the correct option is (4).
When silver nitrate reacts with potassium iodide, \[ AgNO_3 + KI \rightarrow AgI \downarrow + KNO_3 \] A precipitate of silver iodide is formed.
Step 2: Consider the reagent present in excess.
The question states that \(KI\) is present in excess.
Therefore, after precipitation, iodide ions remain in solution.
Step 3: Determine the charge on colloidal particles.
According to the preferential adsorption theory, the ion present in excess gets adsorbed on the surface of the precipitate.
Since \(I^-\) ions are present in excess, they are adsorbed on the surface of \(AgI\) particles.
Thus, the colloidal particles become negatively charged.
The colloidal sol is represented as \[ AgI/I^- \]
Step 4: Final conclusion.
Hence, the colloidal solution formed is \[ \boxed{AgI/I^-} \] Therefore, the correct option is (4).
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