# Acetic acid has a boiling point of 118.5^@ "C" and a K_b of 3.08^@ "C/m". What is the boiling point of a "3.20 m" solution of Ca(NO_3)_2 in acetic acid?

Apr 1, 2017

Consider that the given boiling point is for pure acetic acid. Thus, the boiling point to solve for is for acetic acid after some solute was dumped into it. That is, we have the boiling point elevation of acetic acid:

$\boldsymbol{\Delta {T}_{b} = {T}_{b} - {T}_{b}^{\text{*}} = i {K}_{b} m}$,

where:

• ${T}_{b}$ is the boiling point, and $\text{*}$ indicates the pure substance.
• $i$ is the van't Hoff factor, indicating how many ions are effectively in solution.
• ${K}_{b} = {3.08}^{\circ} \text{C/m}$ is the boiling point elevation constant, specific to the solvent.
• $m$ is the MOLALITY, not to be confused with the MOLARITY. Molality is in $\text{mol solute/kg solvent}$.

Hence, we can find the new boiling point immediately:

${T}_{b} = {T}_{b}^{\text{*}} + i {K}_{b} m$

The van't Hoff factor for "Ca"("NO"_3)_2 is approximately $3$ (write out the chemical equation and convince yourself that there are three ions). Thus (assuming the $\text{3.20 m}$ solution is even possible...):

color(blue)(T_b) = 118.5^@ "C" + (3)(3.08^@ "C/m")("3.20 m")

$\approx \textcolor{b l u e}{{148.07}^{\circ} \text{C}}$

Keep note that the van't Hoff factor is reduced due to ion-pairing in solution, generating some ${\text{CaNO}}_{3}^{+}$ species. As a result, the true new boiling point is a bit lower than was calculated above.