Question

On adding $\text{AgN}{{\text{O}}_{\text{3}}}$ solution into KI solution, a negatively charged colloidal sol is obtained when they are in?
(A) $\text{50 mL of 0}\text{.1M AgN}{{\text{O}}_{\text{3}}}\text{+ 40mL of 0}\text{.1M KI}$
(B) $\text{50 mL of 0}\text{.1M AgN}{{\text{O}}_{\text{3}}}\text{+ 50mL of 0}\text{.2M KI}$
(C) $\text{50 mL of 0}\text{.2M AgN}{{\text{O}}_{\text{3}}}\text{+ 50mL of 0}\text{.2M KI}$
(D) None of these

Answer 1

In order to solve this question, we need to have an idea about the reaction, which is what the question is mentioning about. Knowing the formation of the solution, in reaction form, will give us an idea about the negatively charged colloidal sol.

Complete step by step answer:
When $\text{AgN}{{\text{O}}_{\text{3}}}$ is fully precipitated as AgI, the remaining KI is absorbed by AgI and it forms a negatively charged colloidal sol. This can be derived from the given chemical equation:
$\text{A}{{\text{g}}^{+}}\text{ + }{{\text{I}}^{-}}\text{ }\xrightarrow{{}}\text{ AgI}$
$\text{AgI + }{{\text{I}}^{-}}\text{ }\to \text{ }{{\left[ \text{Ag}{{\text{I}}_{2}} \right]}^{-}}$
$\therefore \left[ \text{A}{{\text{g}}^{\text{+}}} \right]\text{ }\left[ {{\text{I}}^{\text{-}}} \right]$

In case of Option A, $\text{50 mL of 0}\text{.1M AgN}{{\text{O}}_{\text{3}}}\text{+ 40mL of 0}\text{.1M KI}$, $\left( 50\times 0.1 \right)\text{ }=\text{ }5\text{ }milli\text{ }moles$ of $\text{AgN}{{\text{O}}_{\text{3}}}$are formed. Also, $\left( 40\times 0.1 \right)\text{ }=\text{ }4\text{ }milli\text{ }moles$ of KI is formed. Hence, there is an excess of $\text{AgN}{{\text{O}}_{\text{3}}}$ formed. This leads to the formation of positively charged colloidal sol and hence does not satisfy the requirements of the question.

In case of Option B, $\text{50 mL of 0}\text{.1M AgN}{{\text{O}}_{\text{3}}}\text{+ 50mL of 0}\text{.2M KI}$, $\left( 50\times 0.1 \right)\text{ }=\text{ }5\text{ }milli\text{ }moles$ of $\text{AgN}{{\text{O}}_{\text{3}}}$are formed. Also, $\left( 50\times 0.2 \right)\text{ }=\text{ }10\text{ }milli\text{ }moles$ of KI is formed. Hence, there is an excess of KI formed. This leads to the formation of negatively charged colloidal sol and hence it does satisfy the requirements of the question.

In case of Option C, $\text{50 mL of 0}\text{.2M AgN}{{\text{O}}_{\text{3}}}\text{+ 50mL of 0}\text{.2M KI}$, $\left( 50\times 0.2 \right)\text{ }=\text{ }10\text{ }milli\text{ }moles$ of $\text{AgN}{{\text{O}}_{\text{3}}}$are formed. Also, $\left( 50\times 0.2 \right)\text{ }=\text{ }10\text{ }milli\text{ }moles$ of KI is formed. Hence, there is an equilibrium maintained for both $\text{AgN}{{\text{O}}_{\text{3}}}$ and KI. This leads to the formation of neither positively charged nor negatively charged colloidal sol and hence does not satisfy the requirements of the question.

Hence we can clearly see that on adding $\text{AgN}{{\text{O}}_{\text{3}}}$ solution into KI solution, a negatively charged colloidal sol is obtained when they are in
$\text{50 mL of 0}\text{.1M AgN}{{\text{O}}_{\text{3}}}\text{+ 50mL of 0}\text{.2M KI}$.
So, the correct answer is “Option B”.

Note: We can predict the ionic equation of two elements, knowing the kind of ions they are going to form. This means whether they will be cation or anion. In a colloidal mixture, the substances are regularly suspended in a fluid.