Question

When only a little quantity of HgCl2(s) is added to excess KI(aq) to obtain a clear solution, which of the following is true for this solution? (no volume change on mixing). The reaction is 4KI(aq.) + HgCl2(s) → K2 [Hgl4] (aq.) + 2KCl (aq.)

• A. Its boiling and freezing points remain same
• B. Its boiling point is lowered
• C. Its vapour pressure become lower
• D. Its boiling point is raised
• E. Its freezing point is lowered.

Answer 1

Answer: (B)

B

Answer 2

The reaction is $4KI(aq.) + HgCl_2(s) \rightarrow K_2[HgI_4](aq.) + 2KCl(aq.)$.

Initially, the solution contains $KI(aq)$, which dissociates into $K^+$ and $I^-$ ions. For every mole of $KI$, there are 2 moles of ions.

Let's consider the change in the number of ions involved in the reaction for 1 mole of $HgCl_2$:

Reactants: 4 moles of $KI$ yield $4 \times 2 = 8$ moles of ions.

Products: 1 mole of $K_2[HgI_4]$ yields $1 \times 3 = 3$ moles of ions ($2K^+ + [HgI_4]^{2-}$). 2 moles of $KCl$ yield $2 \times 2 = 4$ moles of ions ($2K^+ + 2Cl^-$). Total ions produced from reaction = $3 + 4 = 7$ moles of ions.

Thus, for every 8 moles of ions consumed from $KI$, 7 moles of ions are produced. This indicates a net decrease in the number of ions in the solution due to the reaction ($8 \rightarrow 7$).

Since $HgCl_2$ is added in "little quantity" to "excess KI", the reaction proceeds, and the total number of solute particles (ions) in the solution will decrease.

Colligative properties depend on the number of solute particles:

1. Vapour Pressure: A decrease in solute particles leads to an increase in vapour pressure (less lowering).

2. Boiling Point: A decrease in solute particles leads to a decrease in boiling point elevation ($\Delta T_b$). Since $T_b^{solution} = T_b^{pure} + \Delta T_b$, a decrease in $\Delta T_b$ means the boiling point of the solution is lowered (closer to pure solvent's boiling point).

3. Freezing Point: A decrease in solute particles leads to a decrease in freezing point depression ($\Delta T_f$). Since $T_f^{solution} = T_f^{pure} - \Delta T_f$, a decrease in $\Delta T_f$ means the freezing point of the solution is raised (closer to pure solvent's freezing point).

Based on this analysis:

(A) Its boiling and freezing points remain same - Incorrect. (B) Its boiling point is lowered - Correct. (C) Its vapour pressure become lower - Incorrect (it becomes higher). (D) Its boiling point is raised - Incorrect. (E) Its freezing point is lowered - Incorrect (it is raised).