Question

Which mixture of the solution will lead to the formation of negatively charged colloids of $\left[\mathrm{AgI}\right]{\mathrm{I}}^{-}$ sol?

• A. $50\mathrm{mL}\mathrm{of}1\mathrm{M}{\mathrm{AgNO}}_3+50\mathrm{ml}\mathrm{of}2\mathrm{M}\mathrm{KI}$
• B. $50\mathrm{mL}\mathrm{of}2\mathrm{M}{\mathrm{AgNO}}_3+50\mathrm{ml}\mathrm{of}1.5\mathrm{M}\mathrm{KI}$
• C. $50\mathrm{mL}\mathrm{of}1\mathrm{M}{\mathrm{AgNO}}_3+50\mathrm{ml}\mathrm{of}1.5\mathrm{M}\mathrm{KI}$
• D. $50\mathrm{mL}\mathrm{of}0.1\mathrm{M}{\mathrm{AgNO}}_3+50\mathrm{ml}\mathrm{of}0.1\mathrm{M}\mathrm{KI}$

Answer 1

The correct option is C

$50\mathrm{mL}\mathrm{of}1\mathrm{M}{\mathrm{AgNO}}_3+50\mathrm{ml}\mathrm{of}1.5\mathrm{M}\mathrm{KI}$

Explanation for the correct option:

(C) $50\mathrm{mL}\mathrm{of}1\mathrm{M}{\mathrm{AgNO}}_3+50\mathrm{ml}\mathrm{of}1.5\mathrm{M}\mathrm{KI}$

• The colloidal sol $\mathrm{AgI}$ is prepared by adding $\mathrm{KI}$ solution to the $\mathrm{AgN}{\mathrm{O}}_3$ solution till $\mathrm{KI}$ is in slight excess, iodide ion $\left({\mathrm{I}}^{-}\right)$ will be absorbed in the surface of $\mathrm{AgI}$ particles thereby giving them a negative charge:

$\mathrm{AgI}+{\mathrm{I}}^{-}\to \mathrm{AgI}:{\mathrm{I}}^{-}$

• The reaction that takes place when $\mathrm{AgN}{\mathrm{O}}_3$ and $\mathrm{KI}$ react together can be depicted as:
  $\mathrm{AgN}{\mathrm{O}}_3\left(\mathrm{aq}\right)+\mathrm{KI}\left(\mathrm{s}\right)\to \mathrm{AgI}\left(\mathrm{aq}\right)+\mathrm{KN}{\mathrm{O}}_3\left(\mathrm{aq}\right)$
• Thus, for surrounding the silver iodide with iodide ion, the concentration of $\mathrm{KI}$ must be slightly more than $\mathrm{AgN}{\mathrm{O}}_3$. Thus, the concentration will be $50\mathrm{mL}\mathrm{of}1\mathrm{M}{\mathrm{AgNO}}_3+50\mathrm{ml}\mathrm{of}1.5\mathrm{M}\mathrm{KI}$.

Explanation for incorrect options:

The concentration of $\mathrm{KI}$ must be slightly more than $\mathrm{AgN}{\mathrm{O}}_3$. Thus, options A, B, and C stand incorrect.

Hence, the correct option is C i.e. $50\mathrm{mL}\mathrm{of}1\mathrm{M}{\mathrm{AgNO}}_3+50\mathrm{ml}\mathrm{of}1.5\mathrm{M}\mathrm{KI}$.