Question

When mercuric iodide is added to aqueous solution of KI:
A. Freezing point is raised
B. Freezing point does not change
C. Freezing point is lowered
D. Boiling point does not change

Answer 1

The addition of any non-volatile solute to any solvent causes the change in the boiling point or the freezing point, whether it gets increasing or decreasing. Depression in freezing point is a colligative property which depends on the number of particles or ions present in the solution.

Complete answer:
When mercuric iodide or $\text{Hg}{{\text{I}}_{2}}$ (solute) reacts with aqueous potassium iodide or $\text{KI}$ (solvent), there appears the odourless yellow crystals of the complex ${{\text{K}}_{2}}\left[ \text{Hg}{{\text{I}}_{4}} \right]$ or mercuric potassium iodide. Its IUPAC name is potassium tetraiodomercurate (II).

When mercuric iodide is added to aqueous solution of Potassium iodide, we have seen the reaction. Let us relate this to freezing point of potassium iodide solution:
When Potassium iodide ionizes in water as $\text{KI}\to {{\text{K}}^{+}}+{{\text{I}}^{-}}$, and when it reacted with $\text{Hg}{{\text{I}}_{2}}$, the number of moles potassium iodide used are 2. So, the number of ionizable particles are 4. The reaction is $2\text{KI}+\text{Hg}{{\text{I}}_{2}}\to {{\text{K}}_{2}}\left[ \text{Hg}{{\text{I}}_{4}} \right]$. But after the reaction, the number of ionizable particles are 3. The particles ionized are $\text{2}{{\text{K}}^{+}}+{{\left[ \text{Hg}{{\text{I}}_{4}} \right]}^{-}}$. So, after the reaction, the number of ionizable particles or ions are decreased. Hence, the colligative properties will change and thus the freezing point of the solution will also change. The number of particles has decreased. The freezing point $\propto \dfrac{1}{\text{decrement of particles}}$ . So, the freezing point of the solution will increase.
​When mercuric iodide is added to aqueous solution of KI, the freezing point is raised.
So, the correct answer is “Option A”.

Additional Information:
The alkaline solution of ${{\text{K}}_{2}}\text{Hg}{{\text{I}}_{4}}$ or Potassium tetraiodomercurate (II) is called Nessler’s reagent. This solution is used for detection of ammonia, the change is observed when pale solution becomes deep yellow. And brown precipitates may also form. The reaction of detection is $\text{N}{{\text{H}}_{4}}^{+}+2{{\left[ \text{Hg}{{\text{I}}_{4}} \right]}^{2-}}+4\text{O}{{\text{H}}^{-}}\to \text{HgO}\text{.Hg}\left( \text{N}{{\text{H}}_{2}} \right)\text{I}\downarrow +7{{\text{I}}^{-}}+3{{\text{H}}_{2}}\text{O}$.

Note: Students will find themselves whether the freezing point will increase or decrease, after knowing that number of ions have decreased. So, this should be remembered that the freezing point of the solution is less than the freezing point of pure solvent. $\vartriangle {{\text{T}}_{\text{f}}}={{\text{T}}_{\left( \text{pure solvent} \right)}}-{{\text{T}}_{\left( \text{solution} \right)}}$ or freezing point $\propto \dfrac{1}{\text{decrement of particles}}$.