KCET · Chemistry · Surface Chemistry
A sol of \(\mathrm{AgI}\) is prepared by mixing equal volumes of \(0.1 \mathrm{M} \mathrm{AgNO}_{3}\) and \(0.2 \mathrm{M} \mathrm{KI}\), which of the following statement is correct?
- A Sol obtained is a negative sol with \(\mathrm{NO}_{3}^{-}\) adsorbed on AgI.
- B Sol obtained is a positive sol with \(\mathrm{Ag}^{+}\) adsorbed on AgI.
- C Sol obtained is a positive sol with \(\mathrm{K}^{+}\)adsorbed on AgI.
- D Sol obtained is a negative sol with \(\mathrm{I}^{-}{ }^{-}\)adsorbed on AgI.
Answer & Solution
Correct Answer
(D) Sol obtained is a negative sol with \(\mathrm{I}^{-}{ }^{-}\)adsorbed on AgI.
Step-by-step Solution
Detailed explanation
Whenever a sol of silver iodide is prepared by mixing equal volumes of \(0.1 \mathrm{M} \mathrm{AgNO}_{3}\) and \(0.2\) \(\mathrm{M} \mathrm{KI}\), the sol will be a negative sol with \(\mathrm{I}^{-}\) adsorbed on AgI. The reaction will be.
\(\mathrm{AgNO}_{3}+\mathrm{KI} \longrightarrow \mathrm{AgI}+\mathrm{KNO}_{3}\)
Here, molarity of \(\mathrm{KI}\) is more than that of \(\mathrm{AgNO}_{3}\) and hence, \(\mathrm{I}^{-}\)will be adsorbed on AgI to produce a negative sol.
\(\mathrm{AgNO}_{3}+\mathrm{KI} \longrightarrow \mathrm{AgI}+\mathrm{KNO}_{3}\)
Here, molarity of \(\mathrm{KI}\) is more than that of \(\mathrm{AgNO}_{3}\) and hence, \(\mathrm{I}^{-}\)will be adsorbed on AgI to produce a negative sol.
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