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(A) The formation of a negatively charged $[AgI]I^{-}$ sol occurs when iodide ions $(I^-)$ are preferentially adsorbed on the surface of $AgI$ particl

Vedclass · Accepted Answer

$50 \ mL$ of $1 \ M AgNO_3 + 50 \ mL$ of $1.5 \ M KI$

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(A) The formation of a negatively charged $[AgI]I^{-}$ sol occurs when iodide ions $(I^-)$ are preferentially adsorbed on the surface of $AgI$ particles.This happens when $KI$ is present in excess relative to $AgNO_3$ in the reaction: $AgNO_3 + KI ightarrow AgI + KNO_3$.In option $A$,the moles of $KI$ are $50 \ mL 	imes 1.5 \ M = 75 \ mmol$,while the moles of $AgNO_3$ are $50 \ mL 	imes 1 \ M = 50 \ mmol$.Since $KI$ is in excess,$I^-$ ions are adsorbed on the $AgI$ precipitate,resulting in a negatively charged $[AgI]I^-$ sol.

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

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