5.12 Colligative Properties
Properties of pure liquids change when a solution is formed. The greater the concentration of the solute(s) in the solution, the larger the change in these properties. These properties are known as colligative properties meaning the change in property of the solution does not depend upon the identity of the solute, it only depends upon the number of solute particles.
5.12.1 Vapor Pressure Lowering
The vapor pressure of a solvent is lower than that of the pure liquid. The higher the concentration of the solute (i.e. the lower the mole fraction of the solvent), the more the vapor pressure of the solvent is lowered.
Raoult’s Law
Raoult’s Law relates the vapor pressure, P, of a solution, the solvent concentration (mole fraction: χ) and the vapor pressure of the pure solvent P°.
Raoult’s Law can only really be applied to solutions that
- closely resemble ideal solutions (i.e. dilute solutions)
- contain solutes that do not change when they dissolve (i.e. non-electrolytes).
- solutes that are non-volatile
The lowered vapor pressure of the solution is directly proportional to the concentration of the solution. The more solute, the lower the vapor pressure of the solution.
Raoult’s Law works well for solutions that behave ideally (i.e. dilute solutions). For solutions containing non-volatile solutes, the vapor pressure of the solution (Psolution) is directly proportional to the mole fraction of the solvent times the vapor pressure of the pure solvent (P°solvent). The more concentrated the solution, the smaller the mole fraction of the solvent resulting in a lower vapor pressure.
](https://upload.wikimedia.org/wikipedia/commons/c/c7/Raoult.jpg)
Figure 5.26: François-Marie Raoult
For solutions containing nonvolatile solutes (ones that exert a trivial, i.e., approximately zero, vapor pressure), the vapor pressure of the solution is only dependent on the mole fraction of the solvent and the vapor pressure of the solvent give as
\[P_{\mathrm{solution}} = \chi_{\mathrm{solvent}} P_{\mathrm{solvent}}^{\circ}\]
What about electrolytes?
I have inserted this narrative after a good question was asked in a previous semester about Raoult’s Law and Van’t Hoff factor.
Colligative properties deal with the total number of solute particles in solution. However, Raoult’s Law only works well for solutions that are nearly ideal and solutions that contain solutes that do not change their nature when they dissolve In practice, very dilute solutions could be considered as ideal and solutes should be non-electrolytes (or treated as non-electrolytes).
We could, in theory, imagine a variation of Raoult’s law that considers electrolytes (shown below). However, we will not be using this equation for any practical purpose other than class discussion.
The mole fraction of the solvent should account for the number of solute particles including those that dissociate (van’t Hoff factor, i) such that
\[\chi_{\mathrm{solvent}} = \dfrac{n_{\mathrm{solvent}}}{\left (i \times n_{\mathrm{solute}} \right ) + n_{\mathrm{solvent}}}\]
Practice
122 g of sugar, sucrose (C12H22O11; m.w. = 342.3 g mol–1), a non-volatile solute, is added to 350 g of water at 25 °C. What is the vapor pressure (in atm) of the solution? P°water = 0.0313 atm
Solution
Note that sugar is a non-electrolyte and given that it is non-volatile, its vapor pressure is treated to be zero (Pˆsucrose = 0).
Moles of solute
\[n_{\mathrm{solute}} = 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.30~\mathrm{g}} \right ) = 0.356~\mathrm{mol}\]
Moles of solvent
\[n_{\mathrm{solvent}} = 350~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 19.42~\mathrm{mol}\] Mole fraction of solvent
\[\begin{align*} \chi_{\mathrm{water}} &= \dfrac{n_{\mathrm{solvent}}}{n_{\mathrm{solution}}} \\ &= \dfrac{19.42~\mathrm{mol~H_2O}}{0.356~\mathrm{mol~C_{12}H_{22}O_{11}}+19.42~\mathrm{mol~H_2O}} \\ &= 0.982 \end{align*}\]
Vapor pressure of solution
\[\begin{align*} P_{\mathrm{solution}} &= \chi_i P_{\mathrm{solvent}}^{\circ} \\ &= 0.982 \times 0.0313~\mathrm{atm} \\ &= 0.0307~\mathrm{atm} \end{align*}\]
5.12.2 Freezing Point Depression
The freezing point of a solution is lower than that of the pure liquid. The higher the concentration of the solution, the more the freezing point is lowered. This relationship can be given as
\[\Delta T_{\mathrm{f}} = i m K_{\mathrm{f}}\]
where i is the van’t Hoff factor of the solute, m is the concentration of the solution, Kf is the molal freezing point depression constant for the solvent (or cryoscopic constant), and ΔTf is the change in the freezing point from the pure solvent.
5.12.2.1 Determination of constants for solvents
The molal freezing (Kf) and boiling (Kb) point constants for a solvent can be determined experimentally and are reported as positive values.
We will analyze how Kf is determined for benzene, a common solvent. The normal freezing point of benzene is 5.5 °C.
Below is table (click to expand) a of solutes dissolved in benzene at 1g/100g solvent concentration as well as the resulting change in freezing point13. All listed solutes are neutral and do not dissociate in solution (i.e. they are non-electrolytes; i = 1).
Kf Table
Note: ΔT is a change in freezing point in the negative direction.
| Solute | Formula | m.w. (g mol–1) | ΔT (°C) | m | Kf |
|---|---|---|---|---|---|
| methyl iodide | CH3I | 142 | 0.335 | 0.07042 | 5.04 |
| chloroform | CHCl3 | 119.5 | 0.428 | 0.08368 | 5.11 |
| carbon tetrachloride | CCl4 | 154 | 0.333 | 0.06494 | 5.12 |
| carbon disulfide | CS2 | 76 | 0.654 | 0.13158 | 4.97 |
| ethyl iodide | C2H5I | 156 | 0.331 | 0.06410 | 5.16 |
| ethyl bromide | C2H5Br | 109 | 0.461 | 0.09174 | 5.02 |
| hexane | C6H14 | 86 | 0.597 | 0.11628 | 5.13 |
| ethylene chloride | C2H5Cl | 99 | 0.491 | 0.10101 | 4.86 |
| turpentine (a-pinene) | C10H16 | 136 | 0.366 | 0.07353 | 4.98 |
| nitrobenzene | C6H5NO2 | 123 | 0.390 | 0.08130 | 4.80 |
| naphthalene | C10H8 | 128 | 0.391 | 0.07813 | 5.00 |
| anthracene | C14H10 | 178 | 0.287 | 0.05618 | 5.12 |
| methyl nitrate | CH3NO3 | 77 | 0.640 | 0.12987 | 4.93 |
| dimethyl oxalate | C4H6O4 | 118 | 0.417 | 0.08475 | 4.92 |
| methyl salicylate | C8H8O3 | 152 | 0.339 | 0.06579 | 5.15 |
| diethyl ether | (C2H5)2O | 74 | 0.671 | 0.13514 | 4.97 |
| diethyl sulfide | C4H10S | 90 | 0.576 | 0.11111 | 5.18 |
| ethyl cyanide | C3H5N | 55 | 0.938 | 0.18182 | 5.16 |
| ethyl formate | C3H6O2 | 74 | 0.666 | 0.13514 | 4.93 |
| ethyl valerate | C7H14O2 | 130 | 0.384 | 0.07692 | 5.00 |
| allyl thiocyanate | C4H5NS | 99 | 0.519 | 0.10101 | 5.14 |
| nitroglycerine | C3H5N3O9 | 227 | 0.220 | 0.04405 | 4.99 |
| tributyrin | C15H26O6 | 302 | 0.161 | 0.03311 | 4.87 |
| triolein | C57H104O6 | 884 | 0.056 | 0.01131 | 4.98 |
| acetaldehyde | C2H4O | 44 | 1.107 | 0.22727 | 4.87 |
| chloral | C2HCl3O | 147.5 | 0.342 | 0.06780 | 5.03 |
| benzaldehyde | C7H6O | 106 | 0.473 | 0.09434 | 5.01 |
| camphor | C10H16O | 152 | 0.338 | 0.06579 | 5.14 |
| acetone | C3H6O | 58 | 0.850 | 0.17241 | 4.93 |
| 5-Nonanone | C9H18O | 142 | 0.359 | 0.07042 | 5.10 |
| Mean | 5.02 | ||||
| Min | 4.80 | ||||
| Max | 5.18 | ||||
| Std. Dev | 0.11 |
Freezing points of each solution were measured. It was found that the lowered freezing point of each solution was directly proportional to the concentration of the solution (here given as molality) regardless of the identity of the solute. Here we plot the freezing point of benzene vs concentration.
Figure 5.27: Freezing point of some benzene solutions.
A linear regression of the data gives the line
\[y = -4.952 + 5.4951\]
a very good fit (R2 = 0.9975).
Therefore, the freezing point of benzene solutions is strongly correlated to the concentration of the solution (i.e. the number of dissolved solute particles) and is not affected by the type of solute being dissolved.
The Kf values are simply the ratio of the change in freezing point vs. the number of particles in solution. These values for benzene are remarkably similar (the average value of Raoult’s findings is 5.02 °C m–1). This is very close the currently accepted value of 5.12 °C m–1.
Calculating Kf
Given that
\[\Delta T = imK_{\mathrm{f}}\]
then
\[K_{\mathrm{f}} = \dfrac{\Delta T}{im}\]
For all species listed in the table above, i = 1. The change in a freezing point (ΔT ) can be measured. Therefore, to obtain Kf, simply punch in the concentration (m) of the solution and its change in freezing point.
Example: Determine the Kf for methyl iodide using the experimental data given above
\[\begin{align*} K_{\mathrm{f}} &= \dfrac{\Delta T}{im}\\[1.5ex] &= \dfrac{0.335~\mathrm{^{\circ}C}}{1\times 0.070~m}\\[1.5ex] &= 4.76~\mathrm{^{\circ}C~}m^{-1} \end{align*}\]
Note: In the table above, Kf for methyl iodide is listed as 5.04 as this is what is reported in Raoult’s original paper. However, we surmise that this is likely a typographical error as we see the value should be 4.76 here.
Kf for water
Below is a table of freezing point temperature changes (in °C) for 0.1 m aqueous solutions as reported by Linus Pauling.1
| Solvent | Formula | ΔT (°C) |
|---|---|---|
| Hydrogen peroxide | H2O2 | 0.186 |
| Methanol | CH3OH | 0.181 |
| Ethanol | CH3CH2OH | 0.183 |
| Dextrose | C6H12O6 | 0.186 |
| Sucrose | C12H22O11 | 0.188 |
Similar with Raoult’s findings, the change in the freezing point (ΔT) shows little variation across a range of solutes. The molal freezing point depression constant for water is currently accepted to be 1.86 °C m–1.
Practice
122 g of sugar, sucrose (C12H22O11;
m.w. = 342.3 g mol–1) is
added to 350 g of water at 25 °C.
What is the freezing point (in °C) of the solution?
Kf,water = 1.86 °C m–1
Solution
Moles of solute
\[n_{\mathrm{solute}} = 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.30~\mathrm{g}} \right ) = 0.356~\mathrm{mol~C_{12}H_{22}O_{11}}\]
Molality of solution
\[m = \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} = \dfrac{0.356~\mathrm{mol~C_{12}H_{22}O_{11}}}{0.350~\mathrm{kg~H_2O}} = 1.017~m\]
Change in freezing point
\[\begin{align*} \Delta T_{\mathrm{f}} &= iK_{\mathrm{f}}m = \left ( 1 \right ) \left ( 1.86~^{\circ}\mathrm{C}~m^{-1} \right ) \left ( 1.017~m \right ) = 1.89~^{\circ}\mathrm{C} \\[1.5ex] \end{align*}\]
Final freezing point of solution
\[\begin{align*} T_{\mathrm{f}} &= 0~^{\circ}\mathrm{C} - 1.89~^{\circ}\mathrm{C} = -1.89~^{\circ}\mathrm{C} \\ \end{align*}\]
Constants
Molal freezing point depression (Kf) and boiling point elevation (Kb) constants (in °C m–1) for some solvents. Many can be found in the CRC Handbook.2
| Cryoscopic and ebullioscopic constants for some solvents (in °C m–1) | ||
| Solvent | Kf | Kb |
|---|---|---|
| Water, H2O | 1.86 | 0.513 |
| Ethanol, CH3CH2OH | 1.99 | 1.22 |
| Chloroform, CHCl3 | 4.68 | 3.63 |
| Benzene, C6H6 | 5.07 | 2.64 |
| Carbon tetrachloride, CCl4 | 29.8 | 5.02 |
5.12.3 Boiling Point Elevation
The boiling point of a solution is higher than that of the pure liquid. The higher the concentration of the solution, the more the boiling point is raised This relationship can be given as
\[\Delta T_{\mathrm{b}} = i m K_{\mathrm{b}}\]
where i is the van’t Hoff factor of the solute, m is the concentration of the solution, Kb is the molal boiling point elevation constant (or ebullioscopic constant) for the solvent, and ΔTb is the change in the boiling point from the pure solvent.
Practice
122 g of sugar, sucrose (C12H22O11;
m.w. = 342.3 g mol–1) is
added to 350 g of water at 25 °C.
What is the boiling point (in °C) of the solution?
Kb,water = 0.513 °C m–1
Solution
Moles of solute
\[n_{\mathrm{solute}} = 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.30~\mathrm{g}} \right ) = 0.356~\mathrm{mol~C_{12}H_{22}O_{11}}\]
Molality of solution
\[m = \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} = \dfrac{0.356~\mathrm{mol~C_{12}H_{22}O_{11}}}{0.350~\mathrm{kg~H_2O}} = 1.017~m\]
Change in boiling point
\[\begin{align*} \Delta T_{\mathrm{b}} &= iK_{\mathrm{b}}m = \left ( 1 \right ) \left ( 0.513~^{\circ}\mathrm{C}~m^{-1} \right ) \left ( 1.017~m \right ) = 0.522~^{\circ}\mathrm{C} \end{align*}\]
Final boiling point of solution
\[\begin{align*} T_{\mathrm{b}} &= 100~^{\circ}\mathrm{C} + 0.522~^{\circ}\mathrm{C} = 100.522~^{\circ}\mathrm{C} \end{align*}\]
5.12.4 Osmotic Pressure
A solution and pure solvent are initially separated by an osmotic membrane. Solvent molecules move to higher concentration (osmosis). The osmotic pressure is defined as the external pressure required to give zero net movement of solvent across the membrane.
.](files/ch13/osmotic-pressure.png)
Figure 5.28: Solvent molecules spontaneously flow from low to high concentration. Image from OpenStax.
This can be seen with carrots submersed in salt water.
Figure 5.29: A carrot submerged in salt water (left) and pure water (right) overnight
Osmotic pressure is important in the medical field. Consider blood, a vital solution for our bodies. If the solution is too concentrated, blood cells will become hypertonic and shrivel. If the solution is too dilute, blood cells can become hypotonic.
Figure 5.30: Source: Wikipedia
The osmotic pressure of the solution is higher than that of the pure liquid. The higher the concentration of the solution, the higher the osmotic pressure it exhibits. The relationship can be given as
\[\Pi = iMRT \]
where i is the van’t Hoff factor of the solute, M is the concentration of the solution, R is the gas constant, T is the temperature of the solution, and Π is the osmotic pressure of the solution.
Practice
122 g of sugar, sucrose (C12H22O11; m.w. = 342.3 g mol–1) is added to enough of water to make a 1 L solution (at 25 °C). What is the osmotic pressure (in °C) of the solution?
Solution
Moles of solute
\[n_{\mathrm{solute}} = 122~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{342.30~\mathrm{g}} \right ) = 0.356~\mathrm{mol~C_{12}H_{22}O_{11}}\]
Molarity of solution
\[M = \dfrac{n_{\mathrm{solu}}}{\mathrm{L~soln}} = \dfrac{0.356~\mathrm{mol~C_{12}H_{22}O_{11}}}{1~\mathrm{L~solution}} = 0.356~M\]
Osmotic pressure
\[\begin{align*} \Pi &= iMRT \\[1.5ex] &= (1)(0.356~\mathrm{mol~L^{-1}})(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(298.15~\mathrm{K}) \\[1.5ex] &= 8.71~\mathrm{atm} \end{align*}\]