Solutions: Physical Properties and Behavior In the previous chapter you were exposed to a great deal of information about the forces present in and the properties of individual pure substances (for example, hexane vs. sodium chloride). In this new chapter we will focus on homogeneous mixtures of substances (solutions). Usually we will limit our studies to mixtures of two substances, but three or more substances more can be mixed. Objectives 1. Become familiar with the concentration units used in this chapter: molarity (M), mole fraction (X), molality (m), mass percent, and other percent based units, as well parts per million (ppm) and parts per billion. 2. You should be able to convert between the units listed above as needed (given the necessary information). 3. Understand the energetics of solution formation. 4. Recognize the importance of solute-solvent interactions in determining the overall sign of the energy change during solution formation. 5. Understand the nature of the temperature dependence of solubility (and the relationship to the previous two items). 6. Use and understand Henry's Law as it applies to gas solubility. 7. Know the colligative properties and the equations used or each of them (vapor pressure lowering, freezing point depression, boiling point elevation, and osmotic pressure). 8. Understand Raoult's Law and be able to apply it appropriately. 9. Be able to calculate boiling point elevation, freezing point depression, and osmotic pressure, for cases involving nonelectrolytes as solutes. 10. Recognize the special nature of electrolytes in producing colligative effects. Also, understand the importance of ionic atmosphere (ionic strength) and ion pairing in electrolyte solutions. 11. Be able to calculate boiling point elevation, freezing point depression, and osmotic pressure, for cases involving electrolytes as solutes. 12. Be able to calculate molar masses of nonelectrolytes using data from vapor pressure lowering, boiling point elevation, and/or freezing point depression. 27
Concentration Units Solutions are homogeneous mixtures. They are typically viewed as being comprised of a solvent and one or more solutes. This terminology was introduced early on in the previous semester. Since solutions can be made with varying ratios of solute to solvent there is a need for a way to express the concentration of solutes in any given solution. All concentration units are are quite straight forward, however, the need to convert between them can sometimes create difficulty if the correct conversion factors are not employed. Take note of the following units and complete the practice problems that follow. Percent Mass / Mass (% m/m) - mass of solute per hundred grams of solution Percent Mass / Volume (% m/v) - mass of solute per hundred ml of solution Percent Volume / Volume (% v/v) - volume of solute (in ml) per hundred ml of solution Parts per Million (ppm) - mass of solute per million grams of solution Note: there are also units such as parts per thousand (ppth), parts per billion (ppb), parts per trillion (ppt)... it is useful to note that % m/m can be thought of as parts per hundred. Mole Fraction - moles of solute per total moles of solution Molality (m) - moles of solute per kg of solvent Molarity (M) - moles of solute per L of solution Sample problems: 1. What is the mass percent of NaCl when 5.00 g of NaCl is dissolved in 95.0 g of water? Answer: 5.00 % by mass NaCl 2. What is the ppm of Pb 2+ when 52 mg of Pb 2+ is found in 1.00 L of water (density of water = 1.00 g per ml)? Answer: 52 ppm Pb 2+ 3. What are the mole fractions of water and ethanol (C 2 H 6 O) when 180 grams of water are mixed with 230 grams of ethanol? Answer: X H2O = 0.667 X C2H6O = 0.333 4. What is the molality of a solution that is prepared by mixing 9.00 g of water into 0.400 kg of ethanol? Answer: m = 1.25 mole of water per kg of ethanol 28
5. What is the molarity of a solution prepared by dissolving 29.22 g of NaCl in enough water to prepare 250 ml of solution? Answer: 2.00 M NaCl 6. How many grams of K 2 CrO 4 needed to prepare 500.0 ml of 0.575 M K 2 CrO 4 solution? Answer: 55.8 grams of K 2 CrO 4 7. What is the molality of 1.00 g of sucrose (C 12 H 22 O 11, molar mass = 342.3 g/mole) in 100 grams of water? Answer: m = 0.0292 mole per kg 8. What are the mole fractions of ethanol and water when 50.0 ml of each are combined (density of water = 1.00 g per ml and density of ethanol = 0.789 g per ml... ethanol is C 2 H 6 O Answer: X C2H6O = 0.235 X H2O = 0.765 9. What is the percent by mass of ethanol in the solution described in question 3? Answer: % by mass = 56.1 % 10. If an aqueous solution is 0.010 % by mass, how many ppm is it? Answer: 100 ppm Solution Dilution Calculations - at times it is either necessary or desirable to prepare a solution by diluting another solution. When doing so, one must know how to achieve the correct concentration. Four parameters are all that are needed to describe a dilution process (they are all related through a single equation, so specifying three of the parameters will determine the fourth). The parameters are: the volume of the concentrated solution (V conc. ), the concentration of the concentrated solution (C conc. ), the volume of the dilute solution (V dilute ), and the concentration of the dilute solution (C dilute ). They are related as follows: (V conc. ) x (C conc. ) = (V dilute ) x (C dilute ) sample problems: 1. How many ml of 0.500 M K 2 CrO 4 are needed to prepare 0.750 L of 0.100 M K 2 CrO 4? Answer: 150 ml of 0.500 M K 2 CrO 4 2. How many ml of 9.00 M H 2 SO 4 are needed to prepare 1.00 L of 0.100 M H 2 SO 4 solution? Answer: 11.1 ml of 9.00 M H 2 SO 4 29
Energetics of Solution Formation Apply Hess Law In order to form a solution, the following steps (requiring energy input or releasing energy as indicated) must occur: Endothermic steps Disrupt the forces holding together the solute in the pure solute Create voids in the pure solvent to make room for the solute Exothermic step Create new attractive forces between solute and solvent Sketch an energy diagram in the space to the right showing these three steps For an example of this you can do an image search for energetics of solution formation... it illustrates three cases (a) endothermic, (b) zero heat flow, and (c) exothermic Heat of Solution, ΔH sol n The heat of solution for combining a given solute with a given solvent will depend on how much energy is involved in each step listed above. That is ΔH sol n = solute disruption + solvent disruption + solvent-solute interactions We will consider some individual cases below. Principles of Solubility Dynamic Equilibrium - When a solution contains the maximum concentration of solute that can dissolve in the volume of solvent present (at a particular temperature), the solution is said to be saturated. To be sure that a solution is saturated, it is comforting to see some undissolved solid in the bottom of the solution container. In this case we can tell that the solution MUST be saturated... if it were not, more solid would dissolve. 30
As a matter of fact, more solid is dissolving continually, however, at the same time and at exactly the same rate some of the dissolved compound will be separating out of solution. Consider the example of NaCl in water: NaCl (s) <===> Na + (aq) + Cl - (aq) In a saturated solution, as NaCl (s) ===> Na + (aq) + Cl - (aq) simultaneously an equal amount of Na + (aq) + Cl - (aq) ===> NaCl (s). Since the two processes are the opposite of one another AND they are occuring at the same rate, the solution seems to be static and unchanging (from all outward appearances). Ionic solutes dissolving in water Endothermic steps Disrupt the forces holding together the solute in the pure solute (this is the lattice energy) Create voids in the pure solvent to make room for the solute Exothermic step Create new attractive forces between solute and solvent (these are ion-dipole interactions) Heat of Solution, ΔH sol n for ionic compounds in water can be either endothermic or exothermic depending on the sign of adding together the steps listed above. endothermic solution formation processes are favored by adding heat exothermic solution formation processes are favored by removing heat Therefore, if dissolving a compound is an endothermic process, the solubility will increase as the temperature increases. Conversely, exothermic dissolution processes will be discouraged at higher temperatures (i.e. the compound will be more soluble at low temperature). Many (over 95%) ionic compounds dissolve in water endothermically. These compounds are more soluble at high T than low T (an example would be NaCl). Salts that dissolve exothermically will exhibit the opposite behavior, higher solubility at lower T (an example would be Ce 2 SO 4 ). Dissolution of Gases Endothermic steps Disrupt the forces holding together the solute in the pure solute (this is ZERO for an ideal gas) Create voids in the pure solvent to make room for the solute Exothermic step Create new attractive forces between solute and solvent (these will be dipole-dipole, hydrogen bonding, or dispersion forces depending on the solute and solvent involved). Gas dissolving in liquids is typically exothermic. Therefore, gas solubility decreases as T increases (examples include carbon dioxide, oxygen, nitrogen, etc. in water - think about carbonated beverages). 31
Mixing of Two Liquids Fluids mix with each other in three ways (endothermically, exothermically, or no heat of solution). Ideal (with no intermolecular forces) gases mix ideally absorbing or releasing no heat. Some liquids also mix ideally (the liquids are typically composed of very similar nonpolar molecules). These also absorb or release no heat when mixed. Also, the volume of the solution is equal to the sum of the volumes of the liquids mixed. o Example - benzene (C 6 H 6 ) and toluene (C 6 H 5 CH 3 ) When liquids mix exothermically this indicates that the molecules interact more strongly in the mixture than in the pure substances. When this is true, the volume of the solution is less than the sum of the volumes of the liquids mixed. o Example - chloroform (CHCl 3 ) and acetone (CH 3 COCH 3 ) On the other hand, liquids that mix endothermically will produce a volume greater than the sum of the volumes of the liquids being mixed. This is because the average intermolecular forces in the mixture are lower than that found in the pure liquids. o Example - carbon disulfide (CS 2 ) and acetone (CH 3 COCH 3 ) Immiscible liquids do not mix at all because the interactions between solute and solvent are much weaker than the intermolecular forces experienced in the pure substances. o Example - water (H 2 O) and octane (C 8 H 18 ) Consider the examples (listed above) we discussed in class. Draw the Lewis structures for these substances and see if you can determine the types of intermolecular forces involved in each case (a) within the pure solvent, (b) within the pure solute, and (c) within the mixture. Other points to ponder... all the substances listed above are molecular substances that exist as liquids at room temperature and ambient pressure. Why is carbon disulfide a liquid under these conditions when carbon dioxide is a gas? Why is chloroform a liquid under these conditions when methane (CH 4 ) is a gas? Pressure and Gas Solubility Gas solubility increases with increasing gas pressure. This is a linear relationship and is called Henry's Law: C gas = k P gas The value of k in the equation is good for one particular gas, in one particular solvent, at one particular temperature. For each new temperature or new gas, you must obtain a new value of k. However, the importance of this law should not be underestimated just because it may require a great deal of effort to employ. The range and scope of applications of this law are impressive. Applications include human respiration, effects of heat pollution on aquatic ecosystems, formation of frozen pockets of air in ice cubes, and more. We will also learn more later in the semester that will eliminate the need to resort to texts for new values of k for each new temperature. 32
Colligative Properties Write a definition of Colligative Properties here... The Four Properties are: 1. Vapor pressure lowering 2. Boiling point elevation 3. Freezing point depression 4. Osmotic Pressure Remember the first three are related to one another through the phase diagram (you should be able to draw a diagram to illustrate these relationships). Do so here... The Equations for Colligative Properties of Electrolytes and NON-electrolytes The equations below can be used for nonelectrolytes by setting the van t Hoff factor, i, equal to one (1). For vapor pressure calculations, we will only consider cases that involve nonelectrolytes and behavior that is ideal (or nearly so). 1. Vapor Pressure Lowering When a solute is added to a solvent, the vapor pressure of the solvent is reduced. If the solute is nonvolatile it does not exert any vapor pressure and we can find the vapor pressure of the solution by making a simple adjustment to the vapor pressure of the pure solvent as shown in the next equation: P sol'n = X solvent P o solvent That is the vapor pressure of the solvent in the solution (P sol'n ) is found by multiplying the vapor pressure of the pure solvent (P o solvent) by the mole fraction of the solvent in the solution (X solvent ). OR, dp = X solute P o solvent the lowering of the vapor pressure of the solvent (dp) is found by multiplying the vapor pressure of the pure solvent (P o solvent) by the mole fraction of the solute in the solution (X solute ). 33
Q: Find the vapor pressure of benzene above a solution of 10.0 g of benzoic acid (C 6 H 5 COOH) in 90.0 g of benzene (C 6 H 6 ) at 25ºC. The molecular masses are: benzene = 78.11 u, benzoic acid = 122.12 u and the vapor pressure of pure benzene is 95.1 mm Hg at 25ºC. A: P sol'n = X benzene P o benzene = (0.933 6 )(95.1 mmhg) = 88.8 mmhg However, we often make mixtures of two liquids that both exert measurable vapor pressures. In these cases, we need to know the pressure exerted by both the solute and the solvent. We will treat only the easiest case of these types of mixtures... For ideal binary solutions of two volatile liquids (zero heat of solution case for two substances) we may arrive at the following relationship (which is sometimes referred to as Raoult s Law). P total = P solvent + P solute = X solvent P o solvent + X solute P o solute Q: At 25ºC find the total vapor pressure above a solution of 90.0 g of benzene (P o benzene = 95.1 mm Hg) mixed with 10.0 g of toluene (P o toluene = 28.4 mm Hg). The molecular masses are: benzene = 78.11 u, toluene = 92.14 u. A: P total = P solvent + P solute = X benzene P o benzene + X toluene P o toluene P total = (0.913 9 )(95.1 mmhg) + (0.0860 8 )(28.4 mmhg) P total = 86.9 1 mmhg + 2.44 mmhg = 89.3 5 mmhg Distillation It is interesting to point out that the process of distillation relies on the fact that different substances in a mixture do not vaporize equally well. For example, in the question you just completed toluene comprised 8.61 mole % (note that toluene was 10.0 % m/m) of the liquid in the solution. However, since pure benzene has a greater vapor pressure, it exerts a greater fraction of the vapor pressure (per mole) than does toluene. As a result, the vapor above the solution is enriched in benzene. In the chapter on gases, we learned that partial pressure of gases are directly proportional to the mole fraction of the components in an ideal gas mixture. As a result, we can determine the mole fractions of benzene and toluene at 25ºC in the VAPOR phase above the solution described above. Q: What is the mole fraction of benzene in the vapor phase? A: Note that benzene produced a vapor pressure of 86.9 1 mmhg of the total vapor pressure of 89.3 5 mmhg), therefore, in the vapor phase X benzene = P benzene / P total = 0.972 7 Note that this new vapor phase mole fraction is nearly 6 % higher than the mole fraction of benzene in the original solution. The process of distillation involves condensing and re-vaporizing a mixture in successive steps to gradually increase the mole fraction of the component with the higher vapor pressure until it becomes essentially pure. It is worth noting that this is not normally done at a single temperature, but column that is hotter at the bottom and cooler toward the top where the pure component of 34
higher vapor pressure is recovered. You can refer to the discussion of fractional distillation and the associated figure in your text. Q: If the vapor phase in the answer above was condensed and re-vaporized one additional time at 25ºC, what would be the new mole fraction of benzene in the new vapor phase? A: X benzene = 0.991 7 (i.e. the benzene would contain only 0.83 mole % toluene). 2. Boiling Point Elevation The increase in the boiling point of the solvent when a non-electrolytic solute is added can be found as: ΔT b = K b m *(i) where m = molality of the solute and K b is the boiling point constant (sometimes called the ebullioscopic constant) for the solvent being used. Refer to any assigned problems in the text and become comfortable working problems involving this equation or the freezing point depression equation (see the next topic). 3. Freezing Point Depression The decrease in the freezing point of the solvent when a non-electrolytic solute is added can be found as: ΔT f = K f m *(i) where m = molality of the solute and K f is the freezing point constant (sometimes called the cryoscopic constant) for the solvent being used. 4. Osmotic Pressure The increase in the boiling point of the solvent when a non-electrolytic solute is added can be found as: π = MRT *(i) where M = molarity of the solute, R is the gas constant (in J/Kmole), and T is the Kelvin temperature. Osmotic pressure has applications of great importance ranging from the canning of food, to desalination of drinking water, to a wide range of biological topics. Be familiar with the terms hypotonic, hypertonic, and isotonic. Also make sure you understand the process of reverse osmosis. Very likely, at some point in your future, you will need to rely on drinking water that was purified using this process. 35
Molar Mass from Colligative Properties Molar mass has units of grams per mole. If you know the mass of a solute used to prepare a solution, any of the colligative properties can be used to determine to molar mass of the solute. This is true because each of the colligative properties depends upon the concentration of solute present in solution AND the concentration units all have information about moles of solute in them. This method of determining molar mass will only work if the substance is truly a nonelectrolyte. Colligative properties can also be used to learn the degree to which a weak electrolyte of known molar mass dissociates in a given solvent or to investigate other properties of solutes, such as their tendency to form pairs of molecules known as dimers. Refer to the next two examples to understand both of these cases. Q: 1.921 g of nicotine (empirical formula = C 5 H 7 N) is dissolved in 48.92 grams of water (K f = 1.86 o C/m) creates a freezing point depression of 0.450 o C. (a) What is the molality of the solution? (b) How many moles of solute are present? (c) What is the molar mass of the nicotine? (d) What is the molecular formula? A: (a) 0.242 mole per kg, (b) 0.0118 mole of nicotine, (c) 162 g per mole, and (d) C 10 H 14 N 2 Note: the next problem is from your text and they have the answer wrong! Q: When phenol (C 6 H 5 OH) dissolves in bromoform (CHBr 3 ) it has a tendency to form some dimmers (pairs of molecules). When 2.58 g of phenol is dissolved into 100.0 g of bromoform the freezing point is 5.726 o C. Using the same thermometer, pure bromoform is found to freeze at 8.10 o C. Given that the cryoscopic constant of bromoform is 14.1 o C/m determine what fraction of the bromoform is present as monomers and what fraction as dimers (assume these are the only two possibilities). A: ΔT f can be used to find that the solution contained 0.0168 4 mole of solute particles and the molar mass can be used to find that 0.02741 mole of phenol was used. So now the problem is simply one of algebra. 0.02741 mole of phenol must end up as 0.01684 mole of monomers plus dimers. 0.01684 mole = x mole phenol + y mole dimer, but... each dimer is made from two phenol molecules so... x mole phenol = 0.02741 original mole phenol - 2y mole phenol Combing these two equations we find 0.01684 mole = 0.02741-2y + y = 0.02741 - y So 0.01057 mole of dimer (y) exist and 0.00627 mole of monomer (x). Finally, the fraction of the phenol existing as monomers is 22.9% with 77.1% of the phenol tied up as dimers. Electrolytes and Colligative Properties Read your text and understand the introduction of the van t Hoff factor (i) into the colligative property equations. Also understand why the limiting value of the van t Hoff factor is easily known for electrolytes (for example for CaCl 2 the limiting value of i = 3). Also understand the reasons that the limiting value is only observed in very dilute solutions (ion pairing effects and ionic strength effects) and why a salt like NaCl exerts a smaller impact on the ionic atmosphere of a solution than does a salt like CaSO 4. These will be discussed further in class. 36
Keywords (you should add any missing key words or phrases from the chapter) boiling point colligative property electrolyte mass percent melting point molality molar mass molarity mole fraction nonelectrolyte osmosis osmotic pressure partial pressure saturated solution solvent supersaturated solution vapor pressure Questions (a short list of additional questions to test your understanding of Chapter 12) 1. A solution in which the rates of solution and crystallization are equal is called: a. dilute b. supersaturated c. unsaturated d. saturated e. none of the above 2. Which of the following concentration measures will change in value as the temperature of the solution changes? a. weight percent b. mole fraction c. molarity d. molality e. all of these 3. Which of the following FAVORS the solubility of an ionic solid in a liquid solvent? a. a large magnitude of the solvation energy of the ions c. a large polarity of the solvent b. a small magnitude of the lattice energy of the solute d. all of the above e. none of the above 4. A correct statement of Henry s law is: a. the concentration of a gas in solution is inversely proportional to temperature b. the concentration of a gas in solution is directly proportional to the mole fraction of the solvent c. the concentration of a gas in solution is independent of pressure d. the concentration of a gas in solution is inversely proportional to pressure e. none of the above 5. A solution with stronger interactions between solute and solvent than between solvent and solvent OR between solute and solute should have an endothermic heat of solution. a. true b. depends on temperature c. false d. depends on pressure e. not enough information 37
6. The density of water is 1.00 g /cm3 and the density of CH3OH is 0.972 g /cm3. When 20.0 ml sample of methyl alcohol (CH3OH, MM = 32.0 g /mole) was dissolved in 30.0 ml of water. a. What is the mole fraction of CH3OH? b. What is the mass percent of methyl alcohol? Of water? c. What is the volume of the solution if the solution has a density of 0.937 g /cm3? d. What is the molarity of this solution of methyl alcohol and water? 7. A solution needs to have a concentration of 0.100 mole fraction of ethyl alcohol dissolved in water. How many grams of ethyl alcohol (CH3CH2OH, molar mass = 46.0 g / mole) is required to prepare this solution if 500 g of H2O is used? 38