#### xaktly | Chemistry | equilibrium

Colligative properties

### Dissolved solutes can alter some of the properties of solvents.

Adding non-volatile salts to a solvent can change a few key properties of the solvent:

Vapor pressure can be reduced

The boiling point can be elevated, and

The freezing / melting point can be depressed.

This occurs for both ionic solutes like salts, and nonionic solutes, such as sugars. On this page we'll describe the results of adding solutes to liquid solvents, review some theores that can help us get quantitative about the increases and decreases of these properties, and discuss why these phenomena occur.

The ideas in this section apply to "ideal" solutions. We often have to begin forming our theories of physical phenomena with simplified systems like that, adding complications later as we better understand what's going on.

One definition of an ideal solution is one in which the interactions between solute and solvent particles is no different from the intereactions within pure solvent or pure solute.

There are other definitions, but for our purposes, we'll note that non-ideality is more likely to arise when a solution is more concentrated with solute, so we'll mostly stipulate that dilute solutions are closer to ideal than concentrated ones.

### Vapor pressure reduction

The reason for vapor pressure reduction (often called "lowering," but I'm not a fan of that grammar), is really pretty simple. First, consider a beaker of pure solvent, such as water. Below is a highly schematic view. At the surface, occaisionally a water molecule has enough translational kinetic energy to escape the fluid and fly off as a gaseous molecule. The number of such events at a given temperature in a given time results in a force that can be detected over the liquid, the vapor pressure (pressure is force divided by area — in this case, the area of the liquid surface).

Now in an aqueous solution containing solute molecules (red in the figure below), some of the solute particles will be at the surface. This effectively crowds the solute molecules out, leaving a smaller surface from which they can escape. The presence of the solute particles in no way changes the rate at which water molecules leave the surface, it's just that they have access to less surface. This turns out to be an overly simplistic model, but it gets most of the explanation right.

#### Modeling vapor pressure reduction

The simplest model of vapor pressure reduction is called Raoult's law. Raoult proposed that the vapor pressure of the liquid containing the solute is directly (linearly) proportional to the vapor pressure of the pure solvent, and that the decrease is proportional to the mole fraction of solvent.

It looks like this:

$$P_S = \chi_S \cdot P_S^o$$

where PS is the vapor pressure of the solvent with solute, XS is the mole fraction of solvent, and PSo is the vapor pressure of the pure solvent at the temperature of interest. Recall that vapor pressure increases with temperature, and that when the vapor pressure is equal to the atmospheric pressure, the solvent is boiling.

Here's what Raoult's law looks like in the ideal case.

As the mole fraction of solute increases, the mole fraction of solvent decreases. Less surface is available for escape from the liquid, so the vapor pressure decreases, theoretically to zero when all of the solvent is gone.

#### Deviations from Raoult's law

It turns out that the identity of the solute and how it interacts — whether the intermolecular forces are attractive or repulsive — matters. When the solvent and solute share attractive forces, there is some "stickiness" for solvent particles leaving the surface near solute particles, and that reduces the vapor pressure.

Conversely, when the intermolecular forces are repulsive, solute particles near the surface weaken the attraction of solvent particles to it, and they can escape more easily. Thus the vapor pressure is increased.

This graph shows some of the deviations.

• A positive deviation arises when repulsive solvent-solute interactions dominate.
• A negative deviation arises from attractive forces keeping solvent particles in solution, and

When, as illustrated by the dashed lines, there is a clear maximum or minimum in the vapor pressure vs. mole-fraction curve, the two compounds form an azeotrope, a subject for another section.

One thing that leads to non-ideal behavior is highly-concentrated solutions. The lower the solute concentration, the closer to ideal behavior. High concentrations mean many more chances for solute-solute and solute-solvent interactions.

### Example 1

Calculate the change in vapor pressure of water after adding 117 g of solid NaCl to 1 L of pure water.

Solution: First we need to calculate the mole fraction of NaCl and water in the final solution. 58.45 g of NaCl is one mole, so we're adding two moles of NaCl.

Now for the number of moles of water, we recall that the density of water is about* 1g/ml, so 1L of water has a mass of 1000 g.

$$1000 \; g \; H_2O \left( \frac{1 \text{ mol }H_2O}{18 \; g \; H_2O} \right) = 55.6 \text{ mol } H_2O$$

Now the mole fraction of NaCl is

$$\chi_{NaCl} = \frac{2}{2 + 55.6} = 0.0347$$

* It varies with temperature, and is actually somewhat less (0.9982˚C) at 20˚C.

The mole fraction of solvent H2O can be calculated in the same way, but the two have to add to 1, so it's a little easier just to do this:

$$\chi_{H_2O} = 1 - \chi_{NaCl} = 0.9652$$

Now Raoult's law gives us

so the vapor pressure of the solvent is reduced by 0.7 torr, or about 3.5% of its value without solute.

$$P_{H_2O} = 19.3 \text{ torr}$$

Note that this is the vapor pressure calculated from a model assuming no interactions between solvent and solute particles. It's a good start, but we should always expect some deviation from the ideal.

### Practice problems

Calculate the amount of vapor pressure reduction of water at 25˚C after addition of the following solutes to 500 ml of pure water. The vapor pressure of pure water at 25˚C is 23.8 torr.

 1 22.4 g of LiOH Solution Find concentrations \begin{align} 22.4 \text{ g LiOH} &\left( \frac{\text{1 mol LiOH}}{\text{23.94 g LiOH}} \right) \\ &= 0.94 \text{ mol LiOH}\\ \\ 500 \; g \; H_2O &\left( \frac{1 \text{ mol } H_2O}{18 \; g \;H_2O} \right) \\ &= \text{27.78 mol } H_2O \end{align} Now the mole fractions, χ \begin{align} \chi_{LiOH} &= \frac{0.94}{27.78 - 0.94} \\ &= 0.0326 \\ \\ \chi_{H_2O} &= 1 - \chi_{LiOH} \\ &= 1 - 0.0326 = 0.9647 \end{align} Now the pressure: \begin{align} P_S &= \chi_S P_s^o \\ &= 0.9674(23.8 \; torr) \\ &= 23.02 \; torr \end{align} $$\frac{23.8 - 23.02}{23.8} = 3.3 \text{% change}$$ 2 18.0 g of Ca(NO3)2 Solution See solution to #1 for method. The results are: \begin{align} \text{moles solute} &= 0.11 \\ \text{moles solvent} &= 27.78 \\ \chi_{Ca(NO_3)_2} &= 0.0039 \\ \chi_{H_2O} &= 0.9961 \\ \text{VP} &= 23.71 \text{ torr} \\ \text{% change} &= -0.4 \, % \end{align}
 3 225 g of NaOH Solution See solution to #1 for method. The results are: \begin{align} \text{moles solute} &= 5.63 \\ \text{moles solvent} &= 27.78 \\ \chi_{Ca(NO_3)_2} &= 0.1684 \\ \chi_{H_2O} &= 0.8316 \\ \text{VP} &= 19.79 \text{ torr} \\ \text{% change} &= -17 \, % \end{align} 4 100.0 g of PbSO4 Solution See solution to #1 for method. The results are: \begin{align} \text{moles solute} &= 0.33 \\ \text{moles solvent} &= 27.78 \\ \chi_{Ca(NO_3)_2} &= 0.0117 \\ \chi_{H_2O} &= 0.9883 \\ \text{VP} &= 23.52 \text{ torr} \\ \text{% change} &= -1.2 \,% \end{align}

### Boiling point elevation

Boiling point elevation really just follows from vapor pressure reduction. Recall that the boiling point of a liquid is the temperature at which the vapor pressure is equal to the atmospheric pressure, or whatever external pressure is pushing on the liquid. That's why, when we place a beaker of water in a vacuum chamber and evacuate the air, the water will begin to boil — and by the definition, it's truly boiling, although the temperature can still be low.

When the vapor pressure of a liquid is reduced by addition of a solute (see above), that much more heat has to be put into it to get it to boil, so the boiling temperature has to go up.

There is a mathematical model for boiling point elevation. It's a little more complicated than Raoult's law, but working through it might give you some insights into how solutions work.

The boiling point of a liquid is the temperature at which the vapor pressure above the liquid surface is equal to the atmospheric pressure surrounding it. A liquid may be made to boil either by raising its temperature or by reducing the pressure above its surface, or both.

### Modeling boiling point elevation

Making a mathematical model that predicts boiling point elevation is a little more complicated than Raoult's law, but it's still pretty straight forward. We begin by making a linear relationship between the boiling point of a liquid and the concentration of the solution: The change in boiling point, ΔTb, is proportional to the concentration of the solute, this time in molality (moles of solute per Kg of solvent). Molality is useful here because we don't have to worry about changes in solution volume on adding or removing solute, as we would with molarity.

The ebullioscopic constant is simply the constant of proportionality that we use to get the units right, a bridge between temperature units (K) and molality. The ebullioscopic constant is

Here is the dimensional analysis — just looking at the units to see if they make sense:

The resulting units Kg·K·mol-1 will cancel the Kg of the molality to give temperature units for ΔTb, just what we needed.

Now we have to be careful here, because molality of the solute doesn't really tell us how many ions have been introduced into the solution. Each of those ions will contribute to vapor pressure reduction and other colligative properties. For example, when NaCl dissociates we get

NaCl   ⇄   Na+ + Cl-

which yields two moles of ions per mole of solute. Similarly, when Ca(OH)2 is dissolved we get

Ca(OH)2   ⇄   Ca2+ + 2 OH-

or three moles of ions per mole of solute.

We need to modify our equation just a bit to account for the number of moles of ions introduced into solution. Here's the result:

The van't Hoff factor is an average measure of the number of ions in solution, taking solubility of the solutes and any other ion-pairs into account. For NaCl, i = 1.9 - 2.0 (usually 2). In particular,

For soluble salts that yeild three ions, like Fe(NO3)2, i = 3, and so on. for the average three-ion salt, i = 2.3 - 3, lower if the salt is less soluble and higher if it's more soluble. For solutes that don't dissociate, like glucose, i = 1.

You're probably getting the impression that solution science is a little soft, and that's because you're right. Solutions are much more complicated than gases, with many more interactions to account for.

Now we're ready to do a couple of examples.

### Example 2

Calculate the boiling point change upon adding 58.45g of NaCl to 1 liter of pure water.

Solution: This is a common kitchen problem. The idea is that adding some salt to a pot of water raises its boiling temperature. Let's see how much.

First we calculate the molality of the resulting solution. 58.45 g of NaCl is 1 mole of NaCl, so the molality is

$$\frac{\text{1 mol NaCl}}{\text{1 Kg } H_2O} = \text{1 m NaCl}$$

Now Kb for pure water looks like this:

\begin{align} K_b &= \frac{R T_b^2 m}{\Delta H_v} \\ \\ &= \frac{8.314 (373)2 (0.018)}{40650} \\ \\ &= 0.512 \text{ Kg·K} \end{align}

where 8.314 J·mol-1·K-1 is the gas constant in SI units, 373 is the boiling temperature of pure water in K, 0.018 is the molar mass of water in Kg, and 40,650 is the molar heat of vaporization in J/mol.

The van't Hoff number for NaCl is 2 because NaCl is soluble and we don't expect an appreciable concentration of solid NaCl in a 1m solution.

Then the boiling point elevation is

\begin{align} \Delta T_b &= K_b \cdot b_{solution} \cdot i \\ \\ &= (0.512 \text{ K·Kg/mol})(1 \text{ mol/Kg})(2) \\ \\ &= 1.02 \; K \\ \\ &= 1.02˚C \text{ or } 1.84˚F \end{align}

So you'd have to add a whole mole of salt (58 g of table salt is a handful) to a liter of water in order to raise the boiling temperature by a little less than 2 degrees Fahrenheit. That's not a lot for that much salt. It's more likely that addition of salt to boiling water is done for taste than to raise the boiling temperature. Science!

### Example 3

How much Mg(NO3)2 needs to be added to 1 L of water in order to raise the boiling temperature by 4˚C ?

Solution: Mg(NO3)2 is very soluble in water, so we won't need to worry about that. In this problem, we know the temperature rise, so we can solve for the molality first, find that, then back out the number of grams of Mg(NO3)2 when we have it.

The molality of the solute is

\begin{align} \Delta T_b &= K_b \cdot b_{solution} \cdot i \\ \\ b_{solution} &= \frac{\Delta T_b}{K_b \cdot i} \end{align}

We can use Kb of water as calculated in example 2 above, Kb = 0.512. And for i, because Mg(NO3)2 is soluble and splits into three separate ions, we'll use i = 3.

Then

\begin{align} b_{solution} &= \frac{4 \; K}{0.512 \; K\cdot mol^{-1} \cdot 3} \\ \\ &= 2.604 \; mol/Kg \; H_2O \end{align}

Converting that number of moles of solute to grams gives

\begin{align} 2.604 \; mol \; Mg(NO_3)_2 &\left( \frac{148.32 \; g}{1 \; mol \; Mg(NO_3)_2} \right) \\ \\ &= 386 \; g \; Mg(NO_3)_2 \end{align}

That's a lot of salt! It can be done, though. You could actually dissolve more than 1 Kg of Mg(NO3)2 in a liter of water.

### Practice problems

Calculate the change in boiling point of water after addition of the following solutes to 500 ml of pure water.

Roll over or tap the problem box for the solution.

X

### Azeotrope

An azeotrope is a mixture of two liquids which, because they are mixed, have a common boiling temperature.

Water and ethanol form an azeotropic pair, which is why all of the water can never be extracted from ethanol just by boiling and distillation.

xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012-2019, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.