Have you ever wondered what happens when we add salt to water when cooking pasta? In addition to giving it flavor, salt increases the boiling point of the solution, allowing you to cook your spaghetti faster in salt water!
Boiling point elevation is a colligative property. So, without further ado, let's dive into the colligative properties of solutions!
Colligative Properties: Definition
To start, let's look at the definition of colligative properties.
The colligative properties of solutions are those properties that vary according to the number of solute particles present in the solution.
For example, if we added radiator antifreeze (solute) to the water (solvent) in a car, the freezing point of the car's cooling system would become lower, keeping the car's system from freezing and becoming damaged when the temperature falls.
Colligative properties are considered to be physical properties since they are dependent on the quantity but not on the chemical identity of the solute. So, it doesn't matter what kind of solute you are adding to a solvent, just how much of it you are adding!
When more solute particles are added to the solvent, the intermolecular forces present in the solvent are disrupted, changing the solution's colligative properties.
Types of Colligative: Properties
There are different types of colligative properties you should know about. These are boiling point elevation, freezing point depression, vapor pressure depression and osmotic pressure. The magnitude of the changes in most of these properties depends on the molality of the solute particles.
Molality is the number of moles of solute in each kilogram (kg) of solvent.
The formula for determining molality is as follows:
$$ \text{m = }\frac{\text{mol of solute}}{\text{kg of solvent}} $$
What is the molality (concentration) of a NaCl solution where 2.0 moles of NaCl has been dissolved in 0.5 kg of water (H2O)?
To answer this question, all we need to do is plug in the given moles of salt and kilograms (kg) of water into the formula for molality.
$$ \text{m = }\frac{\text{mol of solute}}{\text{kg of solvent}} = \frac{\text{2.0 mol of NaCl}}{\text{0.5 kg of water}} = 4.0\text{ m } $$
Now, let's look at the effects of a solute in each of these colligative properties. The effect that the addition of solutes to a particular solution has on boiling point (Tb) is that as more solute gets added, the higher the boiling point of a solution becomes. In other words, the addition of a nonvolatile solute will result in a solution with a higher boiling point than the pure solvent.
The freezing point (Tf) of a solution decreases with the addition of more and more solute to a solution. Think of it this way: when more solute is added to a solution, the boiling point increases and the freezing point decreases, basically expanding the limits at which that solution is in a liquid state!
The third colligative property is vapor pressure (VA). When solute is added, the vapor pressure of the solvent decreases because the solute molecules get attached to the solvent molecules and prevent them from escaping into the gas phase.
Lastly, we have osmotic pressure (π), and yes, it has something to do with osmosis.
The process of osmosis occurs when water molecules (H2O) diffuse across a partially permeable membrane from an area of high H2O molecule concentration to an area of low H2O molecule concentration.
Now, in terms of solute concentration, we say that there will be a high concentration of water where the solute concentration is lower. Basically, water will travel from where there is lower solute concentration to where there is higher solute concentration.
For example, let's say that you have a container separated in the middle by a semi-permeable membrane. On one side of the container you have a solution, and on the other side of the container you have a pure solvent. Since this semi-permeable membrane allows the passage of some particles, you will start seeing that the level of the pure solvent side will start decreasing, whereas the solution side level would increase. This difference in levels that is seen when the system reaches equilibrium is the osmotic pressure.
Colligative Properties: Formula
When dealing with solutions and colligative properties, the van't Hoff factor becomes important. This factor takes into account that electrolytes tend to dissociate into ions when dissolved in water.
The van't Hoff factor is referred to as the number of ions a compound dissociates into per formula unit.
- For electrolytes ⇒ \( i \) > 1
- For non-electrolytes ⇒ \( i \) = 1
Let's look at some examples. Sodium chloride (NaCl) is a strong electrolyte that completely dissociates into one sodium ion (Na+) and one chlorine ion (Cl-). Therefore, the van't Hoff factor in this case would be equal to 2. Methanol, CH3OH, on the other hand, is a non-electrolyte. For non-electrolytes, the van hoff factor is always equal to 1.
What is the van't Hoff factor for the strong base Ba(OH)2?
To solve this problem, all we need to do is count the number of ions this compound dissociates into. When Ba(OH)2 dissolves in water, it dissociates into one barium ion (Ba2+) and two hydroxide ions (OH-), for a total of three ions. So, the answer would be i = 3.
Knowing how to figure out the van't Hoff factor of electrolytes and non-electrolyte will be very important when talking about the different formulas involved in the calculation of different colligative properties!
Need a refresher on how to recognize whether a compound is a strong, weak or a non-electrolyte? Check out "Electrolytes"!
Freezing Point Depression
Let's start with freezing point depression. Since the lowering of the freezing point is proportional to the solute's molality, we can determine the change in freezing point by using the formula below:
$$ \Delta \text{T}_{f} = i\times \text{K}_{f}\times \text{m} $$
Where:
- \( \Delta \text{T}_{f} \) is the number of degrees that the freezing point has been lowered to (i.e. the difference in the pure solven't freezing point and the freezing point of the solution).
- \( i \) is the van't Hoff factor
- \( \text{K}_{f} \) is the freezing-point depression constant related to the solvent.
- m is the molality of the solute
Let's put this formula into use and solve a problem.
Calculate the freezing point of a solution prepared by dissolving 4.75 grams of sodium chloride (NaCl) into 100 grams of H2O (The value of \( \text{K}_{f} \) for H2O = 1.86 °C/m).
First, we need to find the van't Hoff factor for NaCl. Since NaCl dissociates into one sodium ion and one chlorine ion, then \( i \) = 2.
We also need to calculate the molality for NaCl.
$$ \text{m = }\frac{\text{mol of solute}}{\text{kg of solvent}} = \frac{\text{4.75 g NaCl } \times \frac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} }{\text{0.1 kg of water}} = 0.8\text{ m NaCl} $$
With all the information needed, we can use the formula above to calculate the aqueous solution's change in freezing point.
$$\Delta{T}_{f}= 2\cdot 1.86\frac{^\circ C}{m} \cdot 0.8\,m= - 2.98°\text{C} $$
Finally, we can use the, \( \Delta \text{T}_{f} \), and the freezing point of pure water (0°C) to calculate freezing point depression, \( \text{T}_{f} \).
$$ \text{T}_{f} = 0^\circ C- 2.98^\circ C= -2.98 °\text{C} $$
Boiling Point Elevation
The formula for boiling point elevation looks similar to the formula used to calculate freezing point depression.
- Keep in mind that, by adding solute to a solution, the boiling point of the solution increases and becomes higher than that of the pure solvent.
$$ \Delta \text{T}_{b} = i\times \text{K}_{b}\times \text{m} $$
Where:
- \( \Delta \text{T}_{b} \) is the number of degrees that the boiling point has been increased to (i.e. the difference in the boiling pt. of the pure solvent and the boiling point of the solution).
- \( i \) is the van't Hoff factor
- \( \text{K}_{b} \) is the boiling-point elevation constant related to the solvent.
- m is the molality of the solute
Let's look at an example.
Calculate the boiling point of 1.5 m NaCl in water (The value of \( \text{K}_{b} \) for H2O = 0.51 °C/m).
In this case, notice that we already have the molality of NaCl and the value for \( \text{K}_{b} \). We also know that, for NaCl, the van't Hoff factor is equal to 2.
So, we can plug in these number to the equation above to find the change in boiling point, and then use the boiling point of pure water (100 °C) to find the boiling point for the aqueous solution of NaCl.
$$ \Delta \text{T}_{b} = 2\cdot 0.51\frac{^\circ C}{m} \cdot{1.5\,m} = 1.53 °\text{C} $$
$$ \text{T}_{b} = 100°\text{C} + 1.53 °\text{C}= 101.53°\text{C} $$
Vapor Pressure Depression
Vapor pressure is related to Raoult's Law, which states that the higher the solute concentration, the lower the vapor pressure will be.
The formula for vapor pressure is given below:
$$ P_{A} = \chi_{A}\text{ }\times \text{ }P_{A}^{*} $$
Where:
- \( P_{A} \) is the vapor pressure contributed by liquid, A, in the solution.
- \( \chi_{A} \), is the mole fraction of that particular liquid. The mole fraction is calculated by using the following equation: \( \chi_{A}= \frac{\text{moles of A}}{\text{moles of A + moles of B + ....}} \)
- \( P_{A}^{*} \), is the vapor pressure of liquid, A, when pure.
Calculate the vapor pressure contributed by benzene in a solution of benzene and toluene if the mole fraction of benzene is 0.4 and its vapor pressure in pure form is 190 torr.
By plugging all values into the formula for vapor pressure, we can find the vapor pressure of toluene in the solution.
$$ P_{benzene} = \chi_{benzene}\text{ }\times \text{ }P_{benzene}^{*} $$
$$ P_{benzene} = \text{0.4 }\times \text{ 190 torr} = 76 \text{ torr} $$
Notice that, in solution, the vapor pressure of benzene is lower than when benzene is in its pure form.
Osmotic Pressure
The last formula we will be looking at is the formula for osmotic pressure, which is the following:
$$ \Pi = i\times \text{M}\times \text{R}\times \text{T} $$
Where:
- \( \Pi \) is the osmotic pressure.
- \( i \) is the van't Hoff factor
- M is the molarity
- R is the gas constant (0.08206 L·atm/mol·K)
- T is the temperature in K
Calculate the osmotic pressure at 25°C of the fluid in the diagram below?
The diagram shows a container where 1.3 M of NaCl is on one side, and pure water (H2O) is on the other side. By using the formula above, we can calculate the osmotic pressure. Remember that the van't Hoff factor for NaCl is 2.
$$ \Pi = 2\times \text{1.3 M}\times \text{(0.08206 L}\cdot \text {atm/mol}\cdot \text {K )}\times \text{298 K} = \text {31.8 atm} $$
Remember that the molarity is equal to moles of solute per liter of solution! To learn more about molarity, check out "Molarity and Molality"!
Colligative Properties Examples
Now, let's talk about examples of colligative properties in a chemistry lab. Probably the most common experiment you will encounter in a laboratory to learn about colligative properties involves determining a compound's molar mass by freezing point depression.
For example, if we know the mass of the unknown compound that has been added to a known mass of solvent, and if we are able to find the freezing point change of the solution relative to the pure solvent, then we can calculate the molar mass of the unknown compound by using the equation below!
$$ \text{Molar mass of unknown compound} = \frac{\text{(K}_{f} \times \text{ g})}{ \text {(ΔT}_{f} \times \text{kg of solvent}) } $$
Application of colligative properties
To finish off, let's talk about another daily application of colligative properties. In areas where the temperature falls below freezing point, salt gets added to the icy roads and sidewalks in order to lower the freezing point and cause the ice to melt!
Now, I hope that you feel more confident in your understanding of colligative properties!
Colligative Properties - Key takeaways
- The colligative properties of solutions are those properties that vary according to the amount of solute particles present in the solution.
- There are different types of colligative properties are boiling point elevation, freezing point depression, vapor pressure depression and osmotic pressure.
- The van't Hoff factor is referred to as the number of ions a compound dissociates into per formula unit.
References
- Moore, J. T., & Langley, R. (2021b). McGraw Hill : AP chemistry, 2022. Mcgraw-Hill Education.
- Zumdahl, S. S., Zumdahl, S. A., & Decoste, D. J. (2019). Chemistry. Cengage Learning Asia Pte Ltd.
- Theodore Lawrence Brown, Eugene, H., Bursten, B. E., Murphy, C. J., Woodward, P. M., Stoltzfus, M. W., & Lufaso, M. W. (2018). Chemistry : the central science (14th ed.). Pearson.
- Post, R. (2020). Chemistry : Concepts and Problems, a Self-Teaching Guide. Wiley & Sons, Incorporated, John.
- Experiment 1: Colligative Properties Determination of the Molar Mass of a Compound by Freezing Point Depression. (n.d.). https://www.ulm.edu/chemistry/courses/manuals/chem1010/experiment_01.pdf
- Chad’s Videos - Taking the Stress Out of Learning Science. (n.d.-b). Chad’s Prep -- DAT, MCAT, OAT & Science Prep. Retrieved October 19, 2022, from https://courses.chadsprep.com/courses/take/general-chemistry-1-and-2/pdfs/30158183-chapter-13-study-guide
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