When silver nitrate solution is added to potassium iodide solution then the sol produced is :

A g N O_{3} / N O_{3}^{-}

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A g I / I^{-}

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K I / N O_{3}^{-}

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A g I / A g^{+}

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Solution

The correct option is B $A g I / I^{-}$
When silver nitrate is added to excess $K I$ , $A g I$ gets precipitated.
$A g N O_{3} (a q) + K I (a q) \to A g I (s) ↓ + A g N O_{3} (a q)$

The precipitated silver iodide adsorbs iodide ions $(I^{-})$ from the dispersion medium and negatively charged colloidal sol is formed $A g I / I^{-}$ .

Hence, option (d) is correct.

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Similar questions

what happens when potassium iodide is added to silver nitrate solution?

Question 1

(i)What is observed when a solution of potassium iodide solution is added to a solution of lead nitrate?

(ii)Name the type of reaction.

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