Some physical properties of solutions differ in important ways from those of the pure solvent. For example, pure water freezes at 0°C, but aqueous solutions freeze at lower temperatures. Ethylene glycol is added to the water in radiators of cars as an antifreeze, to lower the freezing point of the solution. It also raises the boiling point of the solution above that of pure water, permitting operation of the engine at a higher temperature.
The lowering of the freezing point and the raising of the boiling point are examples of physical properties of solutions that depend on the quantity (concentration) but not the kind of solute particles. Such properties are called colligative properties. (Colligative means "depending on the collection"; colligative properties depend on the collective effect of the number of solute particles.) In addition to freezing-point lowering and boiling-point elevation, there are two other colligative properties: vapor-pressure reduction and osmotic pressure.
The effect of a nonvolatile solute on the vapor pressure of a volatile solvent is illustrated by the simple experiment shown in Figure 13.16. Two beakers are placed side by side in a sealed enclosure. One beaker contains pure water, the other an equal volume of an aqueous solution of sugar. Gradually the volume of the sugar solution increases, while the volume of the pure water decreases. Eventually all the water transfers to the sugar solution, as shown in Figure 13.16(b). How do we explain this result?
FIGURE 13.16 Experiment showing that a solution possesses a lower vapor pressure than does the pure solvent under the same conditions of temperature and pressure.
Let's refer back to Section 11.5, particularly Figure 11.20. Recall that the vapor pressure over a liquid is the result of a dynamic equilibrium: The rate at which molecules leave the liquid surface for the gas phase equals the rate at which gas-phase molecules return to the surface of the liquid. A nonvolatile solute added to the liquid reduces the capacity of the solvent molecules to move from the liquid phase to the vapor phase, as shown in Figure 13.17. At the same time, however, there is no change in the rate at which solvent molecules in the gas phase return to the liquid. The shift in equilibrium due to the solute reduces the vapor pressure over the solution. The vapor pressure over the pure solvent is thus higher than that over the solution.
FIGURE 13.17 A nonvolatile solute reduces the rate of vaporization of the solvent.
Return to the experiment shown in Figure 13.16. You should now see why the solvent transfers from the beaker holding pure water to the one holding the sugar solution. The vapor pressure necessary to achieve equilibrium with the pure solvent is higher than that required with the solution. Consequently, as the pure solvent seeks to reach equilibrium by forming vapor, the solution seeks to reach equilibrium by removing molecules from the vapor phase. A net movement of solvent molecules from the pure solvent to the solution results. The process continues until no free solvent remains.
The extent to which a nonvolatile solute lowers the vapor pressure is proportional to its concentration. Doubling the concentration of solute doubles its effect. In fact, the reduction in vapor pressure is roughly proportional to the total concentration of solute particles, whether they are neutral or charged. For example, 1.0 mol of a nonelectrolyte, such as glucose, produces essentially the same reduction in vapor pressure in a given quantity of water as does 0.5 mol of NaCl, a strong electrolyte. Both solutions have 1.0 mol of particles because 0.5 mol of NaCl dissociates to give 0.5 mol of Na+(aq) and 0.5 mol of Cl-(aq).
Quantitatively, the vapor pressure of solutions containing nonvolatile solutes is given by Raoult's law, Equation 13.9, where PA is the vapor pressure of the solution, XA is the mole fraction of the solvent, and is the vapor pressure of the pure solvent:
For example, the vapor pressure of water is 17.5 torr at 20°C. Imagine holding the temperature constant while adding glucose, C6H12O6, to the water so that the resulting solution has = 0.80 and = 0.20. According to Equation 13.9, the vapor pressure of water over the solution will be 80 percent of that of pure water:
Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr (Appendix B).
SOLUTION To apply Raoult's law, Equation 13.9, we must calculate the mole fraction of water in the solution:
We now use Raoult's law to calculate the vapor pressure of water for the solution:
The vapor pressure of the solution has been lowered by 0.6 torr relative to that of pure water.
The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult's law is obeyed, what is the mole fraction of ethylene glycol in the solution? Answer: 0.290
Recall that an ideal gas is described by the ideal-gas equation. Similarly, an ideal solution is a solution that obeys Raoult's law. Real solutions best approximate ideal behavior when the solute concentration is low and when the solute and solvent have similar molecular sizes and similar types of intermolecular attractions.
Many solutions do not obey Raoult's law exactly: They are not ideal solutions. If the intermolecular forces between solvent and solute are weaker than those between solvent and solvent and between solute and solute, then the solvent vapor pressure tends to be greater than predicted by Raoult's law. Conversely, when the interactions between solute and solvent are exceptionally strong, as might be the case when hydrogen bonding exists, the solvent vapor pressure is lower than Raoult's law predicts. Although you should be aware that these departures from ideal solution occur, we will ignore them for the remainder of this chapter.
In Sections 11.5 and 11.6 we examined the vapor pressures of pure substances and how they can be used to construct phase diagrams. How will the phase diagram of a solution, and hence its boiling and freezing points, differ from those of the pure solvent? The addition of a nonvolatile solute lowers the vapor pressure of the solution. Thus, as shown in Figure 13.19, the vapor-pressure curve of the solution (blue line) will be shifted downward relative to the vapor-pressure curve of the pure liquid (black line); at any given temperature, the vapor pressure of the solution is lower than that of the pure liquid. Recall that the normal boiling point of a liquid is the temperature at which its vapor pressure equals 1 atm. (For more information, see Section 11.5) At the normal boiling point of the pure liquid, the vapor pressure of the solution will be less than 1 atm (Figure 13.19). Therefore, a higher temperature is required to attain a vapor pressure of 1 atm. Thus, the boiling point of the solution is higher than that of the pure liquid.
The increase in boiling point relative to that of the pure solvent, Tb, is directly proportional to the number of solute particles per mole of solvent molecules. We know that molality expresses the number of moles of solute per 1000 g of solvent, which represents a fixed number of moles of solvent. Thus, Tb is proportional to molality:
The magnitude of Kb, which is called the molal boiling-point-elevation constant, depends only on the solvent. Some typical values for several common solvents are given in Table 13.4.
For water, Kb is 0.52°C/m; therefore, a 1 m aqueous solution of sucrose or any other aqueous solution that is 1 m in nonvolatile solute particles will boil at a temperature of 0.52°C higher than that of pure water. It is important to realize that the boiling-point elevation is proportional to the number of solute particles present in a given quantity of solution. When NaCl dissolves in water, 2 mol of solute particles (1 mol of Na+ and 1 mol of Cl-) are formed for each mole of NaCl that dissolves. Therefore, a 1 m solution of NaCl in water causes a boiling-point elevation twice as large as a 1 m solution of a nonelectrolyte such as sucrose.
The lower vapor pressure of a solution relative to the pure liquid also affects the freezing point of the solution. The explanation is a bit more subtle than that for boiling-point elevation, but let's proceed as follows.
When a solution freezes, crystals of pure solvent usually separate out; the solute molecules are not normally soluble in the solid phase of the solvent. For example, when aqueous solutions are partially frozen, the solid that separates out is almost always pure ice. As a result, the part of the phase diagram in Figure 13.19 that represents the vapor pressure of the solid is the same as that for the pure liquid. The vapor pressure curves for the liquid and solid phases meet at the triple point. (For more information, see Section 11.6) In Figure 13.19, we see that the triple point of the solution must be at a lower temperature than that in the pure liquid because of lower vapor pressure of the solution compared to that of the pure liquid.
The freezing point of a solution is the temperature at which the first crystals of pure solvent begin to form in equilibrium with the solution. Recall from Section 11.6 that the line representing the solid-liquid equilibrium rises nearly vertically from the triple point. Because the triple-point temperature of the solution is lower than that of the pure liquid, the freezing point of the solution will also be lower than that of the pure liquid.
Like the boiling-point elevation, the decrease in freezing point, Tf, is directly proportional to the molality of the solute:
The values of Kf, the molal freezing-point-depression constant, for several common solvents are given in Table 13.4. For water, Kf is 1.86°C/m; therefore, a 0.5 m aqueous solution of NaCl, or any aqueous solution that is 1 m in nonvolatile solute particles, will freeze 1.86°C lower than pure water. The freezing-point lowering caused by solutes explains the use of antifreeze in cars (Sample Exercise 13.7) and the use of calcium chloride, CaCl2, to melt ice on roads during winter.
Automotive antifreeze consists of ethylene glycol, C2H6O2, a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass percent solution of ethylene glycol in water.
SOLUTION In order to use the molal boiling-point-elevation and freezing-point-depression constants, we must express the concentration of the solution as molality. Let's assume for convenience that we have 1000 g of solution. Because the solution is 25.0 mass percent ethylene glycol, the masses of ethylene glycol and water in the solution are 250 and 750 g, respectively. The molality of the solution is obtained as follows:
We now use Equations 13.10 and 13.11 to calculate the changes in the boiling and freezing points:
Notice that the boiling point has been elevated and the freezing point depressed compared to those of the pure solvent, water.
Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol, C10H18O, a fragrant substance found in the leaves of eucalyptus trees. Answer: -65.6°C
List the following aqueous solutions in order of their expected freezing points: 0.050 m CaCl2; 0.15 m NaCl; 0.10 m HCl; 0.050 m HC2H3O2; 0.10 m C12H22O11.
SOLUTION First notice that CaCl2, NaCl, and HCl are strong electrolytes, HC2H3O2 is a weak electrolyte, and C12H22O11 is a nonelectrolyte. The molality of each solution in total particles is as follows:
0.050 m CaCl2 (0.15 m in particles)
0.15 m NaCl (0.30 m in particles)
0.10 m HCl (0.20 m in particles)
0.050 m HC2H3O2 (between 0.050 and 0.10 m in particles)
0.10 m C12H22O11 (0.10 m in particles)
Because the freezing points depend on the total molality of particles in solution, the expected ordering is: 0.15 m NaCl (lowest freezing point), 0.10 m HCl, 0.050 m CaCl2, 0.10 m C12H22O11, 0.050 m HC2H3O2 (highest freezing point).
Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO3)2, 2 mol of KCl, or 3 mol of ethylene glycol, C2H6O2? Answer: 2 mol of KCl
Certain materials, including many membranes in biological systems and synthetic substances such as cellophane, are semipermeable. When in contact with a solution, they permit the passage of some molecules but not others. They often permit the passage of small solvent molecules such as water but block the passage of larger solute molecules or ions. This semipermeable character is due to a network of tiny pores within the membrane.
Consider a situation in which only solvent molecules are able to pass through a membrane. If such a membrane is placed between two solutions of different concentration, solvent molecules move in both directions through the membrane. However, the concentration of solvent is higher in the solution containing less solute than in the more concentrated one. Therefore, the rate of passage of solvent from the less concentrated to the more concentrated solution is greater than the rate in the opposite direction. Thus, there is a net movement of solvent molecules from the less concentrated solution into the more concentrated one. This process is called osmosis. The important point to remember is that the net movement of solvent is always toward the solution with the higher solute concentration.
FIGURE 13.20 Osmosis: (a) Net movement of a solvent from the pure solvent or a solution with low solute concentration to a solution with high solute concentration; (b) osmosis stops when the column of solution on the left becomes high enough to exert sufficient pressure at the membrane to counter the net movement of solvent. At this point the solution on the left has become more dilute, but there still exists a difference in concentrations between the two solutions.
Figure 13.20(a) shows two solutions separated by a semipermeable membrane. Solvent moves through the membrane from right to left, as if the solutions were driven to attain equal concentrations. As a result, the liquid levels in the two arms become uneven. Eventually, the pressure difference resulting from the uneven heights of the liquid in the two arms becomes so large that the net flow of solvent ceases, as shown in Figure 13.20(b). Alternatively, we may apply pressure to the left arm of the apparatus, as shown in Figure 13.21, to halt the net flow of solvent. The pressure required to prevent osmosis is known as the osmotic pressure, , of the solution. The osmotic pressure is found to obey a law similar in form to the ideal-gas law: V = nRT, where V is the volume of the solution, n is the number of moles of solute, R is the ideal-gas constant, and T is the temperature on the Kelvin scale. From this equation, we can write
FIGURE 13.21 Applied pressure on the left arm of the apparatus stops net movement of solvent from the right side of the semipermeable membrane. This applied pressure is known as the osmotic pressure of the solution.
If two solutions of identical osmotic pressure are separated by a semipermeable membrane, no osmosis will occur. The two solutions are said to be isotonic. If one solution is of lower osmotic pressure, it is described as being hypotonic with respect to the more concentrated solution. The more concentrated solution is said to be hypertonic with respect to the dilute solution.
Osmosis plays a very important role in living systems. For example, the membranes of red blood cells are semipermeable. Placement of a red blood cell in a solution that is hypertonic relative to the intracellular solution (the solution within the cells) causes water to move out of the cell, as shown in Figure 13.22. This causes the cell to shrivel, a process known as crenation. Placement of the cell in a solution that is hypotonic relative to the intracellular fluid causes water to move into the cell. This causes rupturing of the cell, a process known as hemolysis. People needing replacement of body fluids or nutrients who cannot be fed orally are given solutions by intravenous (IV) infusion, which feeds nutrients directly into the veins. To prevent crenation or hemolysis of red blood cells, the IV solutions must be isotonic with the intracellular fluids of the cells.
FIGURE 13.22 Osmosis through the semipermeable membrane of a red blood cell: (a) crenation caused by movement of water from the cell; (b) hemolysis caused by movement of water into the cell.
The average osmotic pressure of blood is 7.7 atm at 25°C. What concentration of glucose, C6H12O6, will be isotonic with blood?
In clinical situations, the concentrations of solutions are generally expressed in terms of mass percentages. The mass percentage of a 0.31 M solution of glucose is 5.3 percent. The concentration of NaCl that is isotonic with blood is 0.16 M because NaCl ionizes to form two particles, Na+ and Cl- (a 0.155 M solution of NaCl is 0.310 M in particles). A 0.16 M solution of NaCl is 0.9 mass percent in NaCl. Such a solution is known as a physiological saline solution.
What is the osmotic pressure at 20°C of a 0.0020 M sucrose, C12H22O11, solution? Answer: 0.048 atm, or 37 torr.
There are many interesting examples of osmosis. A cucumber placed in concentrated brine loses water via osmosis and shrivels into a pickle. If a carrot that has become limp because of water loss to the atmosphere is placed in water, the water moves into the carrot through osmosis, making it firm once again. People who eat a lot of salty food experience water retention in tissue cells and intercellular space because of osmosis. The resultant swelling or puffiness is called edema. Movement of water from soil into plant roots and subsequently into the upper portions of the plant is due at least in part to osmosis. The preservation of meat by salting and of fruit by adding sugar protects against bacterial action. Through the process of osmosis, a bacterium on salted meat or candied fruit loses water, shrivels, and dies.
In osmosis, water moves from an area of high water concentration (low-solute concentration) into an area of low water concentration (high-solute concentration). Such movement of a substance from an area where its concentration is high to an area where it is low is spontaneous. Biological cells transport not only water but also other select materials through their membrane walls. This permits entry of nutrients and allows for disposal of waste materials. In some cases substances must be moved from an area of low concentration to one of high concentration. This movement is called active transport. It is not spontaneous and so requires expenditure of energy by the cell.
The colligative properties of solutions provide a useful means of experimentally determining molar mass. Any of the four colligative properties could be used to determine molar mass. The procedures are illustrated in Sample Exercises 13.10 and 13.11.
A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g in 40.0 g of CCl4. The normal boiling point of the resultant solution increased by 0.357°C. Calculate the molar mass of the solute.
SOLUTION Using Equation 13.10, we have
Thus the solution contains 0.0711 mol of solute per kilogram of solvent. The solution was prepared from 0.250 g of solute and 40.0 g of solvent. The number of grams of solute in a kilogram of solvent is therefore
Notice that a kilogram of solvent contains 6.25 g of solute, which, from the Tb measurement, must be 0.0711 mol. Therefore,
Camphor, C10H16O, melts at 179.8°C; it has a particularly large freezing-point depression-constant, Kf = 40.0°C/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7°C. What is the approximate molar mass of the solute? Answer: 110 g/mol
The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine its molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25°C was found to be 1.54 torr. Calculate the molar mass of the protein.
SOLUTION Using Equation 13.12, we can calculate the molarity of the solution:
Because the volume of the solution is 5.00 mL = 5.00 × 10-3 L, the number of moles of protein must be
The molar mass is the number of grams per mole of the substance. The sample has a mass of 3.50 mg = 3.50 × 10-3 g. The molar mass is the number of grams divided by the number of moles:
Because small pressures can be measured easily and accurately, osmotic pressure measurements provide an excellent way to determine the molar masses of large molecules.
A sample of 2.05 g of the plastic polystyrene was dissolved in enough toluene to form 100 mL of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25°C. Calculate the molar mass of the polystyrene. Answer: 4.20 × 104 g/mol