# What is the overall effect of adding a solute to a solution?

Feb 10, 2017

Several colligative properties conspire to give the following three most common effects:

• Lower vapor pressure above the solution of the solvent in solution relative to that of the solvent by itself
• Higher boiling point of the solution relative to that of the pure solvent
• Lower freezing point of the solution relative to that of the pure solvent

Osmotic pressure $\Pi = c R T$ is affected as well, but is not all that interesting.

Vapor pressure for the solvent above the solution is given by Raoult's law for ideal solutions:

$\boldsymbol{{P}_{i} = {\chi}_{i} {P}_{i}^{\text{*}}}$

where ${\chi}_{i} = {n}_{i} / {n}_{\text{tot}}$ is the mol fraction, ${P}_{i}$ is the partial pressure of the solvent above the solution, and ${P}_{i}^{\text{*}}$ is the pure vapor pressure of the solvent above itself.

When we consider component $j \ne i$ in a two-component solution, we have that ${\chi}_{i} = 1 - {\chi}_{j}$. Therefore:

${P}_{i} = \left(1 - {\chi}_{j}\right) {P}_{i}^{\text{*}}$

Since ${\chi}_{i} \le 1$, it follows that $1 - {\chi}_{j} < 1$. Therefore, ${P}_{i} < {P}_{i}^{\text{*}}$, and the vapor pressure of the solvent decreases due to adding any solute to the solution.

Boiling point elevation occurs as follows:

$\boldsymbol{\Delta {T}_{b} = i {K}_{b} m}$,

where:

• $\Delta {T}_{b} = {T}_{b} - {T}_{b}^{\text{*}} > 0$ is the change in boiling point due to adding solute. Of course, ${T}_{b}$ is the boiling point of the solution, and ${T}_{b}^{\text{*}}$ is the boiling point of the solvent by itself.
• $i$ is the van't Hoff factor that is approximately how many particles go into solution for every solute formula unit.
• ${K}_{b} = \text{0.512"^@ "C"cdot"kg/mol}$ is the boiling point elevation constant.
• $m$ is the $\text{mol solute"/"kg solvent}$ (the molality).

From the equation we can see that as concentration of solute increases, ${T}_{b}$ increases.

Freezing point depression is analogous. The only difference is the numbers used:

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

Whatever $\Delta {T}_{f}$ you get, it should always be negative. If it numerically does not turn out that way, check your work or set your sign to $\left(-\right)$.