Question

Which mixture of the solutions will lead to the formation of negatively charged colloidal [Agl]I- Sol.? (

• A. \(50 mL\ of\  1 M AgNO3+50 mL \ of 1.5 \ M KI\)
• B. \(50 \ mL\ of \ 1 M\ AgNO_3+50 \ mL \ of \ 2MKI\)
• C. \(50mL \ of 2\ M AgNO_3+50 \ mL \  of 1.5 \ M KI \)
• D. \(50 \ mL\ of\ 0.1 M AgNO_3+50\ mL\ of 0.1 \ M KI\)

Answer 1

The Correct Option is A, B

To determine which mixture of solutions will lead to the formation of a negatively charged colloidal \([ \text{AgI} ]^- \) sol, we need to understand the basic principle of colloidal sol formation in this context. 

The formation of a negatively charged colloidal sol occurs when there is an excess of iodide ions \((\text{I}^-)\) because these surplus ions adsorb onto the surface of the silver iodide \((\text{AgI})\) precipitate, imparting it with a negative charge.

Let's examine each option:

1. Option 1: \(50 \, \text{mL}\) of \(1 \, \text{M} \, \text{AgNO}_3 + 50 \, \text{mL} \, \text{of} \, 1.5 \, \text{M} \, \text{KI}\)
   • Moles of \(\text{AgNO}_3 = \frac{50 \times 1}{1000} = 0.05\) moles
   • Moles of \(\text{KI} = \frac{50 \times 1.5}{1000} = 0.075\) moles
   • Since the iodide ions are in excess (\(0.075\) moles compared to \(0.05\) moles of \(\text{Ag}^+\)), a negatively charged colloidal \([ \text{AgI} ]^- \) sol will form.
2. Option 2: \(50 \, \text{mL}\) of \(1 \, \text{M} \, \text{AgNO}_3 + 50 \, \text{mL} \, \text{of} \, 2 \, \text{M} \, \text{KI}\)
   • Moles of \(\text{AgNO}_3 = \frac{50 \times 1}{1000} = 0.05\) moles
   • Moles of \(\text{KI} = \frac{50 \times 2}{1000} = 0.1\) moles
   • The iodide ions are in significant excess (\(0.1\) moles compared to \(0.05\) moles of \(\text{Ag}^+\)), ensuring the formation of a negatively charged colloidal \([ \text{AgI} ]^- \) sol.
3. Option 3: \(50 \, \text{mL} \, \text{of} \, 2 \, \text{M} \, \text{AgNO}_3 + 50 \, \text{mL} \, \text{of} \, 1.5 \, \text{M} \, \text{KI}\)
   • Moles of \(\text{AgNO}_3 = \frac{50 \times 2}{1000} = 0.1\) moles
   • Moles of \(\text{KI} = \frac{50 \times 1.5}{1000} = 0.075\) moles
   • In this case, the silver ions are in excess, leading to the formation of a positively charged colloidal \([ \text{AgI} ]^+ \) sol, not a negatively charged one.
4. Option 4: \(50 \, \text{mL} \, \text{of} \, 0.1 \, \text{M} \, \text{AgNO}_3 + 50 \, \text{mL} \, \text{of} \, 0.1 \, \text{M} \, \text{KI}\)
   • Moles of \(\text{AgNO}_3 = \frac{50 \times 0.1}{1000} = 0.005\) moles
   • Moles of \(\text{KI} = \frac{50 \times 0.1}{1000} = 0.005\) moles
   • Both reactants are in stoichiometric proportion, which will not favor the formation of a negatively charged sol since neither ion is in excess.

Conclusion: The mixtures that lead to the formation of a negatively charged colloidal \([ \text{AgI} ]^- \) sol are:

• \(50 \, \text{mL} \, \text{of} \, 1 \, \text{M} \, \text{AgNO}_3 + 50 \, \text{mL} \, \text{of} \, 1.5 \, \text{M} \, \text{KI}\)
• \(50 \, \text{mL} \, \text{of} \, 1 \, \text{M} \, \text{AgNO}_3 + 50 \, \text{mL} \, \text{of} \, 2 \, \text{M} \, \text{KI}\)