Sol of silver iodide AgI produced in excess of a solution of potassium iodide. Determine the charge of the particles in obtained sol, write the formula of its micelles and specify which of the electrolyte: potassium chloride, barium nitrate or aluminum sulfate - will have the greatest capacity for coagulating the resulting sol.

Since the sol of silver iodide \(AgI\) is prepared in excess of potassium iodide \(KI\), iodide ions \(I^-\) are adsorbed on the surface of \(AgI\) particles.

So the sol particles have a negative charge.

The micelle can be written as:

\[\left[(AgI)_m \cdot nI^- \cdot (n-x)K^+\right]^{x-} \cdot xK^+\]

For a negatively charged sol, the coagulating ions are positive ions. Their coagulating power increases with charge:

\[Al^{3+} > Ba^{2+} > K^+\]

Therefore, among \(KCl\), \(Ba(NO_3)_2\), and \(Al_2(SO_4)_3\), the greatest coagulating capacity will be shown by aluminum sulfate \(Al_2(SO_4)_3\).

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