Which mixture of the solutions will lead to the formation of negatively charged colloidal (AgI)I- sol. ?

Question

Which mixture of the solutions will lead to the formation of negatively charged colloidal (AgI)I- sol. ? (A) 50 mL of 2 M AgNO3 + 50 mLof 1.6 M KI (B) 50 mL of 0.1 M AgNO3 + 50 mL of 0.1 MKI (C) 50 mL of 1M AgNO3 + 50 mL of 1.5 M Kl (D) 50 mL of 1M AgNO3 + 50 mL of 2M KI

Tardigrade Admin · Accepted Answer

Generally charge present on the colloid is due to adsorption of common ion from dispersion medium. Millimole of KI is maximum in option (2) (50

\times

2 = 100) so act as solvent and anion

I^{-}

is adsorbed by the colloid AgI formed

D.P AgNO X_{3} + (excess) KI Negatively charged colloid Agl + KNO X_{3}

Q. Which mixture of the solutions will lead to the formation of negatively charged colloidal $(A g I) I^{-}$ sol. ?

50 mL of 2 M $AgNO X_{3}$ + 50 mLof 1.6 M KI

50 mL of 0.1 M $AgNO X_{3}$ + 50 mL of 0.1 MKI

50 mL of 1M $AgNO X_{3}$ + 50 mL of 1.5 M Kl

50 mL of 1M $AgNO X_{3}$ + 50 mL of 2M KI

Q. Which mixture of the solutions will lead to the formation of negatively charged colloidal (AgI)I− sol. ?

50 mL of 2 M AgNOX3​ + 50 mLof 1.6 M KI

50 mL of 0.1 M AgNOX3​ + 50 mL of 0.1 MKI

50 mL of 1M AgNOX3​ + 50 mL of 1.5 M Kl

50 mL of 1M AgNOX3​ + 50 mL of 2M KI

Q. Which mixture of the solutions will lead to the formation of negatively charged colloidal $(A g I) I^{-}$ sol. ?

50 mL of 2 M $AgNO X_{3}$ + 50 mLof 1.6 M KI

50 mL of 0.1 M $AgNO X_{3}$ + 50 mL of 0.1 MKI

50 mL of 1M $AgNO X_{3}$ + 50 mL of 1.5 M Kl

50 mL of 1M $AgNO X_{3}$ + 50 mL of 2M KI