# 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)

• 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
• @IvanNeretin $\ce{HgI2}$ solute aqueous$\ce{KI}$ solvent – Nilay Ghosh Feb 28 '16 at 3:25