17.6: Colligative Properties
Lesson Objectives
The student will:
 explain what the term colligative means and list colligative properties.
 indicate what happens to the boiling point, the freezing point, and the vapor pressure of a solvent when a solute is added to it.
 explain why a solution has a lower vapor pressure than the pure solvent of that solution.
 explain why a solution has an elevated boiling point and a depressed freezing point.
 define the van’t Hoff factor.
 calculate boiling point elevations and freezing point depressions for both nonelectrolyte and electrolyte solutions.
 calculate molar masses from freezing point depression data.
Vocabulary
 boiling point elevation
 colligative property
 freezing point depression
 van't Hoff factor
Introduction
People who live in colder climates have seen the trucks put salt on the roads when snow or ice is forecast. Why do they do that? When planes fly in cold weather, the planes need to be deiced before liftoff. Why is that done? It turns out that pure solvents differ from solutions in their boiling points and freezing points. In this lesson, you will understand why these events occur. You will also learn to calculate exactly how much of an effect a specific solute can have on the boiling point or freezing point of a solution.
Vapor Pressure Lowering
An enclosed liquid will reach vapor pressure equilibrium with its vapor in the space above the liquid. Vapor pressure equilibrium is reached when the rate of evaporation and the rate of condensation become equal. The vapor pressure depends on the temperature of the liquid, so raising the temperature increases the rate of evaporation and therefore the vapor pressure of the liquid. When the temperature of the liquid becomes high enough for the vapor pressure to equal the surrounding pressure, the liquid will boil. When the surrounding pressure is \begin{align*}1.00 \ \mathrm{atm}\end{align*}
In a pure solvent, all the molecules at the surface are solvent molecules. Remember that only the molecules on the surface of a liquid are able to evaporate. For the pure solvent, then, the entire surface area is available for evaporation, and the forces that need to be overcome are the attractive forces between the solvent molecules. Adding a solute to a solvent lowers the vapor pressure of the solvent. There are two equally valid explanations for why the addition of a solute lowers the vapor pressure of a solution. One explanation is that since some of the surface molecules are solute molecules, these solute molecules take up some of the surface area, making less surface area available for evaporation (see illustration below). Therefore, the rate of evaporation of the solvent will be lower, resulting in a lower vapor pressure at the same temperature. The other explanation says that the attractive forces between the solvent molecules and the solute molecules are greater than the attractive forces between solvent molecules, so the solvent molecules will not evaporate at as high a rate. Once again, the vapor pressure will be lowered.
The amount that the vapor pressure is lowered is related to the molal (m) concentration of the solute. Suppose a pure solvent has a vapor pressure of \begin{align*}20. \ \mathrm{mm \ of \ Hg}\end{align*}
Experiments with this phenomenon demonstrate that as long as the solute is a nonelectrolyte, the effect is the same regardless of what solute is used. The effect is related only to the number of particles of solute, not the chemical composition of the solute. Vapor pressure lowering is an example of a colligative property. Colligative properties are properties that are due only to the number of particles in solution and not to the chemical properties of the solute.
Boiling Point Elevation
The boiling point of a liquid occurs when the vapor pressure above the surface of the liquid equals the surrounding pressure. At 1 atm of pressure, pure water boils at \begin{align*}100^\circ\mathrm{C}\end{align*}
Freezing Point Depression
The effect of adding a solute to a solvent has the opposite effect on the freezing point of a solution as it does on the boiling point. Recall that the freezing point is the temperature at which the liquid changes to a solid. At a given temperature, if a substance is added to a solvent (such as water), the solutesolvent interactions prevent the solvent from going into the solid phase, requiring the temperature to decrease further before the solution will solidify. A common example is found when salt is used on icy roadways. Here the salt is put on the roads so that the water on the roads will not freeze at the normal \begin{align*}0^\circ\mathrm{C}\end{align*} but at a lower temperature, as low as \begin{align*}9^\circ\mathrm{C}\end{align*}. The deicing of planes is another common example of freezing point depression in action. A number of solutions are used, but commonly a solution such as ethylene glycol or a less toxic monopropylene glycol is used to deice an aircraft. The aircrafts are sprayed with the solution when the temperature is predicted to drop below the freezing point. The freezing point depression, then, is the difference between the freezing points of the solution and the pure solvent.
The Mathematics of Boiling Point and Freezing Point Changes
The amount to which the boiling point increases or the freezing point decreases depends on the amount solute that is added to the solvent. A mathematical equation can be used to calculate the boiling point elevation or the freezing point depression. Remember the solution has a higher boiling point, so to find the boiling point elevation you would subtract the boiling point of the solvent from the boiling point of the solution. For example, the boiling point of pure water at \begin{align*}1.0 \ \mathrm{atm}\end{align*} is \begin{align*}100.^\circ\mathrm{C}\end{align*}, while the boiling point of a 2% saltwater solution is about \begin{align*}102^\circ\mathrm{C}\end{align*}. Therefore, the boiling point elevation would be \begin{align*}2^\circ\mathrm{C}\end{align*}. In comparison, the freezing point depression is found by subtracting the freezing point of the solution from the freezing point of the pure solvent.
Both the boiling point elevation and the freezing point depression are related to the molality of the solutions. Looking at the formulas for the boiling point elevation and freezing point depression, we can see similarities between the two.
Boiling point elevation:


 \begin{align*} \Delta T_b = K_bm\end{align*}

where


 \begin{align*} \Delta T_b = T_{solution}  T_{pure~solvent}\end{align*}



 \begin{align*}K_b\end{align*} = boiling point elevation constant



 \begin{align*}m\end{align*} = molality of the solution.

Freezing point depression:


 \begin{align*} \Delta T_f = K_fm\end{align*}

where


 \begin{align*} \Delta T_f = T_{f(solution)}  T_{f(pure~solvent)}\end{align*}



 \begin{align*}K_f\end{align*} = freezing point depression constant



 \begin{align*}m\end{align*} = molality of the solution.

The boiling point and freezing point constants are different for every solvent and are determined experimentally in the lab. You can find these constants for hundreds of solvents listed in data reference publications for chemistry and physics.
Example:
Antifreeze is used in automobile radiators to keep the coolant from freezing. In geographical areas where winter temperatures go below the freezing point of water, using pure water as the coolant could allow the water to freeze. Since water expands when it freezes, freezing coolant could crack engine blocks, radiators, and coolant lines. The main component in antifreeze is ethylene glycol, \begin{align*}\mathrm{C}_2\mathrm{H}_4(\mathrm{OH})_2\end{align*}. If the addition of an unknown amount of ethylene glycol to \begin{align*}150. \ \mathrm{g}\end{align*} of water dropped the freezing point of the solution by \begin{align*}1.86^\circ\mathrm{C}\end{align*}, what mass of ethylene glycol was used? The freezing point constant, \begin{align*}K_f\end{align*}, for water is \begin{align*}1.86^\circ\mathrm{C/m}\end{align*}.
Solution:


 \begin{align*} \Delta T_f = K_f\mathrm{m}\end{align*}



 \begin{align*}\mathrm{m} = \frac { \Delta T_f} {K_f} = \frac {1.86^\circ\mathrm{C}} {1.86^\circ\mathrm{C/m}} = 1.00 \ \mathrm{m}\end{align*}



 \begin{align*}\mathrm{m} = \frac {\mathrm{mol \ solute}} {\mathrm{kg \ solvent}}\end{align*}



 moles solute = (molality)(kg solvent) = \begin{align*}(1.00 \ \mathrm{mol/kg})(0.150 \ \mathrm{kg}) = 0.150 \ \mathrm{mol}\end{align*}



 mass \begin{align*}\mathrm{C}_2\mathrm{H}_4(\mathrm{OH})_2\end{align*} = (mol)(molar mass) = \begin{align*}(0.150 \ \mathrm{mol})(62.1 \ \mathrm{g/mol}) = 9.32 \ \mathrm{g}\end{align*}

Therefore, \begin{align*}9.32 \ \mathrm{g}\end{align*} of ethylene glycol would have been added to the \begin{align*}150. \ \mathrm{g}\end{align*} of water to lower the freezing point by \begin{align*}1.86^\circ\mathrm{C}\end{align*}.
Remember that colligative properties are due to the number of solute particles in the solution. Adding 10 molecules of sugar to a solvent will produce 10 solute particles in the solution. However, when the solute is an electrolyte, such as \begin{align*}\mathrm{NaCl}\end{align*}, adding 10 molecules of solute to the solution will produce 20 ions (solute particles) in the solution. Therefore, adding enough \begin{align*}\mathrm{NaCl}\end{align*} solute to a solvent to produce a \begin{align*}0.20 \ \mathrm{m}\end{align*} solution will have twice the effect of adding enough sugar to a solvent to produce a \begin{align*}0.20 \ \mathrm{m}\end{align*} solution.
The van't Hoff factor (i) is the number of particles that the solute will dissociate into upon mixing with the solvent. For example, sodium chloride (\begin{align*}\mathrm{NaCl}\end{align*}) will dissociate into two ions, so the van't Hoff factor for \begin{align*}\mathrm{NaCl}\end{align*} is i = 2. For lithium nitrate (\begin{align*}\mathrm{LiNO}_3\end{align*}), i = 2, and for calcium chloride (\begin{align*}\mathrm{CaCl}_2\end{align*}), i = 3.
We can now rewrite our colligative properties formulas to include the van’t Hoff factor.
Boiling point elevation:


 \begin{align*}\Delta T_b = iK_bm\end{align*}

where


 i = van’t Hoff factor



 \begin{align*} \Delta T_b = T_{solution}  T_{pure solvent}\end{align*}



 \begin{align*}K_b\end{align*} = boiling point elevation constant



 \begin{align*}m\end{align*} = molality of the solution.

Freezing point depression:


 \begin{align*}\Delta T_f = iK_fm\end{align*}

where


 i = van’t Hoff factor



 \begin{align*} \Delta T_f = T_{f(solution)}  T_{f(pure solvent)}\end{align*}



 \begin{align*}K_f\end{align*} = freezing point depression constant



 \begin{align*}m\end{align*} = molality of the solution.

We can use this formula for both electrolyte and nonelectrolyte solutions since the van’t Hoff factor for nonelectrolytes is always 1 because they do not dissociate.
Example:
A solution of \begin{align*}10.0 \ \mathrm{g}\end{align*} of sodium chloride is added to \begin{align*}100.0 \ \mathrm{g}\end{align*} of water in an attempt to elevate the boiling point. What is the boiling point of the solution?
Solution:


 \begin{align*} \Delta T_b = iK_bm\end{align*}



 \begin{align*}\mathrm{mol \ NaCl} = \frac {\mathrm{grams}} {\mathrm{molar \ mass}} = \frac {10.0 \ \mathrm{g}} {58.5 \ \mathrm{g/mol}} = 0.171 \ \mathrm{mol}\end{align*}



 molality = \begin{align*} \frac {\mathrm{mol \ solute}} {\mathrm{kg \ solvent}} = \frac {0.171 \ \mathrm{mol}} {0.100 \ \mathrm{kg}} = 1.71 \ \mathrm{m}\end{align*}



 For NaCl, i = 2 \begin{align*}(\mathrm{NaCl} \rightarrow \mathrm{Na}^+ + \mathrm{Cl}^)\end{align*}



 \begin{align*}K_{b(water)} = 0.52^\circ\mathrm{C/m}\end{align*}



 \begin{align*} \Delta T_b = iK_bm\end{align*}



 \begin{align*} \Delta T_b = (2)(0.52^\circ\mathrm{C/m})(1.71 \ \mathrm{m}) = 1.78^\circ\mathrm{C}\end{align*}



 \begin{align*}T_{b(solution)} = T_{b(pure solvent)} + \Delta T_b\end{align*}



 \begin{align*}T_{b(solution)} = 100^\circ\mathrm{C} + 1.78^\circ\mathrm{C} = 101.78^\circ\mathrm{C}\end{align*}

Therefore, the boiling point of the solution of \begin{align*}10 \ \mathrm{g}\end{align*} of \begin{align*}\mathrm{NaCl}\end{align*} in \begin{align*}100 \ \mathrm{g}\end{align*} of water is \begin{align*}102^\circ\mathrm{C}\end{align*}.
This video explains the reasoning behind and the math involved in boiling point elevation and freezing point depression (6e): http://www.youtube.com/watch?v=z9LxdqYntlU (14:00).
Lesson Summary
 Colligative properties are properties that are due only to the number of particles in solution and not to the chemical properties of the solute.
 Vapor pressure lowering, boiling point elevation, and freezing point depression are colligative properties.
 The boiling point elevation can be calculated using the formula \begin{align*} \Delta T_b = K_bm\end{align*}, where \begin{align*}\Delta T_b\end{align*} is the boiling point elevation, \begin{align*}K_b\end{align*} is the boiling point elevation constant, and \begin{align*}m\end{align*} is molality.
 The freezing point depression can be calculated using the formula \begin{align*}\Delta T_f = K_fm\end{align*}, where \begin{align*} \Delta T_f\end{align*} is the freezing point depression, \begin{align*}K_f\end{align*} is the freezing point depression constant, and \begin{align*}m\end{align*} is molality.
 For electrolyte solutions, the van’t Hoff factor is added to account for the number of ions that the solute will dissociate into in solution. For nonelectrolyte solutions, the van’t Hoff factor = 1. The equations change to account for this factor (\begin{align*} \Delta T_b = K_bm\end{align*} becomes \begin{align*} \Delta T_b = iK_bm\end{align*} and \begin{align*} \Delta T_f = K_fm\end{align*} becomes \begin{align*} \Delta T_f = iK_fm\end{align*}, where i is the number of particles each solute molecule produces in solution.
Further Reading / Supplemental Links
The following link hosts several videos on how to solve various types of chemistry solutions problem.
Review Questions
 What must be measured in order to determine the freezing point depression?
 From a list of solutions with similar molalities, how could you quickly determine which would have the highest boiling point?
 Why would table salt not be a good solution to use when deicing a plane?
 Identify which of the following statements are true. When a solute is added to a solution, i. the boiling point increases, ii. the boiling point decreases, iii. the freezing point increases, iv. the freezing point decreases.
 i and iii are true
 i and iv are true
 ii and iii are true
 ii and iv are true
 If \begin{align*}25.0 \ \mathrm{g}\end{align*} of sucrose \begin{align*}(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11})\end{align*} is added to \begin{align*}500. \ \mathrm{g}\end{align*} of water, the boiling point is increased by what amount? \begin{align*}(K_b \ (\mathrm{water}) = 0.52^\circ\mathrm{C/m})\end{align*}
 \begin{align*}0.076^\circ\end{align*}
 \begin{align*}0.025^\circ\end{align*}
 \begin{align*}26^\circ\end{align*}
 None of these
 The solubility of seawater (an aqueous solution of \begin{align*}\mathrm{NaCl}\end{align*}) is approximately \begin{align*}0.50 \ \mathrm{m}\end{align*}. Calculate the freezing point of seawater. \begin{align*}(K_f \ (\mathrm{water}) = 1.86^\circ\mathrm{C/m})\end{align*}
 \begin{align*}0.93^\circ\end{align*}
 \begin{align*}0.93^\circ\end{align*}
 \begin{align*}1.86^\circ\end{align*}
 \begin{align*}1.86^\circ\end{align*}
 Determine which of the following solutions would have the lowest freezing point.
 \begin{align*}15 \ \mathrm{g}\end{align*} of ammonium nitrate in \begin{align*}100. \ \mathrm{g}\end{align*} of water.
 \begin{align*}50. \ \mathrm{g}\end{align*} of glucose in \begin{align*}100. \ \mathrm{g}\end{align*} of water.
 \begin{align*}35 \ \mathrm{g}\end{align*} of calcium chloride in \begin{align*}150. \ \mathrm{g}\end{align*} of water.
 A \begin{align*}135.0 \ \mathrm{g}\end{align*} sample of an unknown nonelectrolyte compound is dissolved in \begin{align*}725 \ \mathrm{g}\end{align*} of water. The boiling point of the resulting solution was found to be \begin{align*}106.02^\circ\mathrm{C}\end{align*}. What is the molecular weight of the unknown compound?
 What is the van’t Hoff factor for each of the following:
 \begin{align*}\mathrm{MgCl}_2\end{align*}
 Ammonium sulfate
 \begin{align*}\mathrm{CH}_3\mathrm{OH}\end{align*}
 Potassium chloride
 \begin{align*}\mathrm{KCH}_3\mathrm{COO}\end{align*}
 Calcium chloride is known to melt ice faster than sodium chloride but is not used on roads because the salt itself attracts water. If \begin{align*}15 \ \mathrm{g}\end{align*} of \begin{align*}\mathrm{CaCl}_2\end{align*} was added to \begin{align*}250 \ \mathrm{g}\end{align*} of water, what would be the effect on the freezing point of the solution? \begin{align*}(K_f (\mathrm{water}) = 1.86^\circ\mathrm{C/m})\end{align*}
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