Question

Which of the following will form a negatively charged colloidal solution?
(A) 100 ml 0.1 M $AgN{{O}_{3}}$ + 100 ml 0.1 M KI
(B) 100 ml 0.2 M $AgN{{O}_{3}}$ + 100 ml 0.1 M KI
(C) 100 ml 0.1 M $AgN{{O}_{3}}$ + 100 ml 0.2 M KI
(D) 100 ml 0.2 M $AgN{{O}_{3}}$ + 200 ml 0.1 M KI

Answer 1

Hint: To solve this question more precisely, we need to know the reaction taking place in between $AgN{{O}_{3}}$ and KI. Knowing this, we will get an idea about the negatively charged colloidal solution quickly.
Colloidal solution is the mixture where the substances are regularly suspended in a fluid.

Complete answer:
Let us study the given reaction to answer the illustration;
$\begin{align}
  & AgN{{O}_{3}}\rightleftarrows A{{g}^{+}}+N{{O}_{3}}- \\
 & KI\rightleftarrows {{K}^{+}}+{{I}^{-}} \\
\end{align}$
When $AgN{{O}_{3}}$will precipitate as AgI;
$A{{g}^{+}}+{{I}^{-}}\to AgI$
The remaining KI will be absorbed by AgI and will form the negatively charged colloidal solution (as AgI will be hence surrounded by the ${{I}^{-}}$ ions).
Analysing all the options, we will get the correct option. For example, option (C);
Millimoles of $AgN{{O}_{3}}$ = $100\times 0.1=10$
Millimoles of KI = $100\times 0.2=20$
which shows the excess of KI.

Therefore, option (C) is correct.

Note:
We can easily predict the ionic equation of two given elements i.e. whether they are anion or cation to answer the question.
When the KI is present in excess of that of $AgN{{O}_{3}}$, it will give us the negatively charged colloidal solution. Hence, option (A); giving equal concentration, option (B); giving $AgN{{O}_{3}}$in excess and option (D); giving equal concentration, can never be the answers.