1. 50 mL of 1 M AgNO ₃ + 50 mL of 1.5 M KI 2. 50 mL of 1 M AgNO ₃ + 50 mL of 2 M KI 3. 50 mL of 2 M AgNO ₃ + 50 mL of 1.5 M KI 4. 50 mL of 0.1 M AgNO ₃ + 50 mL of 0.1 M KI

Answer: 1. 50 mL of 1 M AgNO ₃ + 50 mL of 1.5 M KI or 2. 50 mL of 1 M AgNO ₃ + 50 mL of 2 M KI In the sol [AgI]I^- , AgI adsorbs the common ion I^- , so KI must be in excess. (1) AgNO3 = 50 mmol, KI = 75 mmol: excess I^- , negative sol. (2) AgNO3 = 50 mmol, KI = 100 mmol: excess I^- , negative sol. (3) AgNO3 = 100 mmol, KI = 75 mmol: excess Ag^+ , positive sol. (4) Equal amounts: AgI precipitates, no sol. Options (1) and (2) both fit.

Topic: Lyophilic and lyophobic sols· Principles Related to Practical Chemistry

Which mixture of the solutions will lead to the formation of negatively charged colloidal [AgI]I^- sol. ?

1. 50 mL of 1 M AgNO\(_3\) + 50 mL of 1.5 M KI

2. 50 mL of 1 M AgNO\(_3\) + 50 mL of 2 M KI

3. 50 mL of 2 M AgNO\(_3\) + 50 mL of 1.5 M KI

4. 50 mL of 0.1 M AgNO\(_3\) + 50 mL of 0.1 M KI

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Answer: 1. 50 mL of 1 M AgNO\(_3\) + 50 mL of 1.5 M KI or 2. 50 mL of 1 M AgNO\(_3\) + 50 mL of 2 M KI

In the sol \([\mathrm{AgI}]\mathrm{I^-}\), AgI adsorbs the common ion \(\mathrm{I^-}\), so KI must be in excess.

(1) AgNO3 = 50 mmol, KI = 75 mmol: excess \(\mathrm{I^-}\), negative sol.
(2) AgNO3 = 50 mmol, KI = 100 mmol: excess \(\mathrm{I^-}\), negative sol.
(3) AgNO3 = 100 mmol, KI = 75 mmol: excess \(\mathrm{Ag^+}\), positive sol.
(4) Equal amounts: AgI precipitates, no sol.

Options (1) and (2) both fit.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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Which mixture of the solutions will lead to the formation of negatively charged colloidal [AgI]I^- sol. ?

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