Question

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

• A. 50 mL of 2 M ${AgNO_3}$ + 50 mLof 1.6 M KI
• B. 50 mL of 0.1 M ${AgNO_3}$ + 50 mL of 0.1 MKI
• C. 50 mL of 1M ${AgNO_3}$ + 50 mL of 1.5 M Kl
• D. 50 mL of 1M ${AgNO_3}$ + 50 mL of 2M KI

Answer 1

Answer: (D)

50 mL of 1M ${AgNO_3}$ + 50 mL of 2M KI

Answer 2

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
{ \underset{\text{ D.P}}{{AgNO_3 }}$+$\underset{\text{ (excess)}}{{ KI}}$->$\underset{\text{ Negatively charged colloid}}{{Agl }} + KNO_3}