# What is the effect of the van't Hoff factor on the boiling point elevation or freezing point depression for a given electrolytic solution?

Jul 12, 2017

Well, usually $i \ge 1$, so in many cases, it will accentuate the boiling point elevation or freezing point depression from the pure solvent.

For non-electrolytes, we would normally write

$\Delta {T}_{f} = {T}_{f} - {T}_{f}^{\text{*}} = - {K}_{f} m$ $\text{ } \boldsymbol{\left(1\right)}$

$\Delta {T}_{b} = {T}_{b} - {T}_{b}^{\text{*}} = {K}_{b} m$ $\text{ "" } \boldsymbol{\left(2\right)}$

where:

• ${T}_{f}$ and ${T}_{b}$ are the freezing and boiling points, respectively, of the solution. $\text{*}$ indicates pure solvent.
• ${K}_{f}$ and ${K}_{b}$ are the freezing point depression and boiling point elevation constants, respectively, of the solvent.
• $m$ is the molality of the solution in $\text{mol solute/kg solvent}$.

When we account for electrolytes, i.e. those substances that dissociate in solution, we write in the van't Hoff factor $i$ to account for the effective number of solute particles per formula unit:

$\Delta {T}_{f} = {T}_{f} - {T}_{f}^{\text{*}} = - \textcolor{red}{\underline{i}} {K}_{f} m$ " "bb((1"'"))

$\Delta {T}_{b} = {T}_{b} - {T}_{b}^{\text{*}} = \textcolor{red}{\underline{i}} {K}_{b} m$ " "" "bb((2"'"))

And thus, $i \ge 1$ for typical electrolytes, increasing the magnitude to which the boiling points increase and freezing points decrease.