# Does complex formation lead to increase in freezing point?

Original question:-

When mercuric iodide is added to an aqueous solution of Potassium iodide, freezing point increases. Why?

Due to formation of Complex $\ce{K2HgI4}$,number of particle decrease and hence, freezing point increases. $$\ce{ HgI2 + 2KI <=> K2HgI4}$$ No. of particles is inversely proportional to increase in freezing point.

Does complex formation reduces the no. of particles? Does decrease in no. of particles increases the freezing point?

My attempt:-

I assumbed $\ce{HgI2}$ to be solute and $\ce{KI}$ to be solvent.

No. of moles of $\ce{HgI2}$ = $\ce{n_{HgI2}}$

No. of particles = $\ce{N_{HgI2}}$ = $\ce{n_{HgI2}.N_{A}}$

Molality = m = $\frac{n_{HgI2}}{W_{KI}}$ = $\frac{N_{HgI2}}{N_{A}.W_{KI}}$

Difference in freezing point = $\Delta{T_{f}}$ = $\ce{K_{f}.m}$ = $\frac{K_{f}.N_{HgI2}}{N_{A}.W_{KI}}$ ($\ce{K_{f}}$ - cryoscopic constant)

$\therefore$ $\Delta{T_{f}}$ $\propto$ $\ce{N_{HgI2}}$ (provided $\ce{W_{KI}}$ to be constant)

Overall, I am not satisfied with the answer given in the book. Is there any alternate answer to the question? (Even if the answer given is right, Is there any alternate answer with some in-depth analysis?)

• The answer given in the book looks right to me, so any alternate answer would be either similar or wrong. As for your solution, here is the mistake: $\ce{KI}$ is solid, so it can't be the solvent for $\ce{HgI2}$. – Ivan Neretin Feb 27 '16 at 15:51
• @IvanNeretin An aqueous solution of $\ce{KI}$ was made and $\ce{HgI2}$ was made to dissolve in it. – Nilay Ghosh Feb 27 '16 at 17:59
• Yes, I've noticed that. So what is the solvent, really? – Ivan Neretin Feb 27 '16 at 19:17
• @IvanNeretin $\ce{HgI2}$ solute aqueous$\ce{KI}$ solvent – Nilay Ghosh Feb 28 '16 at 3:25
• Is it? What is the molar mass of that solvent? And what is its freezing point? – Ivan Neretin Feb 28 '16 at 7:31