500 CHAPTER 12 Intermolecular F or ces and the Physical Properties of Liquids and Solids Additional Problems 12.98 Name the kinds of attractive forces that must be overcome in order to (a) boil liquid ammonia, (b) melt solid phosphorus (P 4) , (c) dissolve Cs I in liquid HF , (d) melt potassium metal. 12.99 Which of the following properties indicates very strong intermolecular forces in a liquid: (a) very low surface tension, (b) very low cr itical temperature, (c) very low boiling point, (d) very low vapor pressure? 12.100 At - 35°C, liquid HI has a higher vapor pre ssure than liquid HF. Explain. 12.101 Ba sed on the following properties of el ementa l boron, classify it as one of the crystalline solids discussed in Section 12.4: high melting point (2300°C), po or conductor of heat and electricity, in soluble in water, very hard substance. 12.102 R efen ing to Figure 12.34, determine the stable phase of CO 2 at (a) 4 atm and -60 °C and (b) 0.5 atm and -20 °C. 12.103 Which of the following substances has the highest polarizability: CH 4 , H z, CCI 4 , SF 6 , HzS? 12.104 Given two complementary stra nd s of DNA containing 100 12.105 12.106 12.107 12.108 12.109 12.110 12.111 ba se pairs each, calculate the ratio of two separate strands to hydrogen-bonded double helix in solution at 300 K. (Hint: The formula for calculating this ratio is e - t!. ElRT, where 6.E is the energy difference between hydrogen-bonded double-strand DNAs and single-strand DNAs and R is the gas con stant.) Assume the energy of hydrogen bonds per base pair to be 10 kJ/mo\. The average distance between base pairs mea sured parallel to o the axis of a DNA molecule is 3.4 A. The average molar mass of a pair of nucleotides is 650 g/mo \. Estima te the len gt h in centimeters of a DNA molecule of molar ma ss 5.0 X 10 9 g/mo\. Roughly how many base pairs are contained in this molecule? A CO z fire extinguisher is located on the outside of a building in Massachu se tts. During the winter months, one can hear a sloshing sound when the extinguisher is gently shaken. In the summertime there is often no s ound when it is shaken. Ex plain. Assume that the extinguisher has no leaks and that it ha s not been used. What is the va por pre ssure of mercury at its normal boiling point (357° C)? A flask of water is connected to a powerful vacuum pump. When the pump is turned on, the water begins to boi\. After a few minutes, the sa me water begins to freeze. Eve ntuall y, the ice disappears. Explain what happens at each step. The liquid-vapor boundary line in the phase diagram of any substance always stops abruptly at a certain point. Why ? The interionic distance of several alkali halide crystals are as follows: Crystal Interionic dista nce (pm) NaCI 282 NaBr 299 NaI KCI KEr 324 315 330 KI 353 Plot lattice energy versus the reciprocal interi on ic distance. How would you explain the plot in terms of the dependence of lattice energy on the distance of separation between ions? What law governs this interacti on? ( For lattice energies, see Table 8.1.) Which ha s a greater density, crystalline Si0 2 or amorphous Si0 2 ? Why ? 12. 11 2 A student is given four solid sa mpl es labeled W, X, Y, and Z. All have a metallic luster. She is told that the solids could be gold, lead sulfide, mica (which is quartz, or SiO z ), and iodine. The results of her investigations are: (a) W is a good el ec trical conductor; X, Y, and Z are poor electrical conductors. (b) When the solids are hit with a hammer , W flattens out, X shatters into many pieces, Y is smashed into a powder, and Z is not affected. (c) When the solids are heated with a Bun sen burner, Y melts with so me sublimation, but X, W, and Z do not melt. (d) In treatment with 6 M HN0 3 , X dissolves; there is no effect on W, Y, or Z. On the basis of these test results, identify the solids. 12.113 Which of the following statements are false? (a) Dipole-dipole interactions between mol ecules are greatest if the molecules possess only temporary dipole mom ents. (b) All compounds containing hydrogen atoms can partic ipat e in hydrogen-bond formation. (c) Dispersion forces exist betw ee n all atoms, molecules, and ions. 12.114 The diagram shows a kettle of boiling water on a stove. Identify the phases in regions A and B. 12.115 12.116 12.117 12.118 12.119 12.120 B- A- The south pole of Mars is covered with solid carbon dioxide, which partly sublimes during the summer. The CO 2 vapor recondenses in the winter when the temperature drops to 150 K. Given that the h ea t of sublimation of CO 2 is 25.9 kJ/mol, calculate the atmospheric pressure on the surface of Mars. [Hint: Use Figure 12.34 to detennine the normal sublimation temperature of dry ice and Equation 12.4 which also applies to sublimations.] The properties of gases, liquids, and solids differ in a number of respects. How would you use the kinetic molecular theory (see Section 11.6) to explain the following observations? (a) Ea se of compressibility decreases from gas to liquid to solid. (b) Solids retain a definite shape, but gases and liquids do not. (c) For most substances, the volume of a given amount of material increases as it changes from solid to liquid to gas. Select the substance in each pair that should have the higher boiling point. In each case identify the principal intermolecular forces involved and account briefly for your choice: (a) K 2 S or (CH3)3N, (b) Br 2 or CH 3 CH 2 CH 2 CH 3 . A small drop of oil in water assumes a s ph erical shape. Explain. (Hint: Oil is made up of nonpolar molecule s, which tend to avoid contact with water.) Under the s ame conditions of temperature and density, which of the following gases would you expect to behave less ideally: CH 4 or S0 2? Explain. The fluorides of the second-period elements and their melting points are: LiF, 845°C; BeF 2 , 800°C; BF 3 , -126.7 °C; CF 4 , - 184°C; NF 3 , -206.6 °C; OFz, -223.SoC; F 2 , -2 19.6°C. Classify the type(s) of intermolecular forces present in each compound. 12.121 The standard enthalpy of formation of gaseous molecular iodine is 62.4 kllmol. Use this inf ormation to calc ul ate the molar heat of sublimation of molecular iodine at 25°C. 12.122 The distan ce between Li+ and CI- is 257 pm in solid LiCI and 203 pm in an LiCI unit in the gas phase. Explain the differen ce in the bond lengths. 12.123 Heat of hydration, that i s, the heat chan ge that occurs when ions become hydrated in solution, is lar ge ly due to ion-dipole interactions. The heats of hydration for the alkali metal ions are Li +, - 520 kl lmol; Na +, - 405 kl lmol; K+, - 321 kJ/mo!. Account for the trend in these values. 12.124 If water were a linear molecule, (a) would it still be polar, and (b) would the water molecules still be able to form hydro ge n bonds with one another? 12.125 Calculate the D.H o for the following pr ocesses at 25° C: (a) Br 2(Z) • Br 2(g) and (b) Br 2(g) • 2Br(g). Comme nt on the relative magnitudes of these /::,.H 0 values in terms of the forces in vo lved in each case. (Hint: See Table 8.6.) 12.126 Which liquid would you expect to have a greater viscosit y, water or diethyl ether? The structure of diethyl ether is shown in Problem 12.15. 12.127 A beaker of water is placed in a closed container. Predict the effect on the va por pressure of the water when (a) its temperature is lowered, (b) the vo lum e of the container is doubled, (c) more water is added to the beake r. 12.128 Ozone (0 3 ) is a strong oxidizing agent that can oxidize all the common metals except gold and platinum. A convenient test for ozone is based on its action on mercury. When exposed to ozone, mercury becomes dull looking and st icks to glass tubing (inst ead of flowing freely through it). Write a balanced equation for the reaction. What property of mercury is altered by its interaction with ozone? 12.129 A sample of limestone (CaC0 3 ) is heated in a closed vessel until it is partially decompose d. Write an equation for the reaction, and state how many phases are present. 12.130 Carbon and silicon belong to Group 4A of the periodic table and have the same valence electron configuration (ns 2 np\ Why does silicon dioxide ( Si0 2 ) have a much higher melting point than carbon dioxide (C0 2 )? 12.131 Ga seous or highly volatile liquid anesthetics are often pr eferred in surgical procedures because once inhaled, these vapors can quickly enter the bloodstr ea m through the alveoli and then ent er the brain. Several common gaseous anesthetics are shown with their boiling points: Br F · \ I H-C-C-F ci p Halothane 50°C F CI F I I I F-C - C -O -C -F I I I F H H Isoflurane 48.5°C F F F I I I F-C-O-C-C - Cl I I I H F H Enflurane 56.5°C Ba sed on intermolecular force considerations, explain the advantages of using these anesthetics. (Hi nt : The brain barrier is made of membranes that have a nonpolar interior region.) 12.132 12.133 12.134 12.135 12.136 12.137 QUESTIONS AND PROBLEMS 501 Given the ge neral properties of water and ammonia, comment on the problems that a biological system (as we know it) would have developing in an ammonia medium. Boiling point Melting point Molar heat capacity Molar heat of vaporization Molar heat of fusion Viscosity Dipole mom ent Phase at 300 K H 2 0 373. 15K 273.15 K 75.3 11 K· mol 40.79 kl lmol 6.0 kl lmol 0.001 N . s/m2 1. 82 D Liquid NH3 239.65 K 195.3 K 8. 53 11 K· mol 23.3 kl lmol 5.9 kl lmo! 0.0254 N . s/m2 (at 240 K) 1.46 D Gas A pressure cooker is a sealed container that allows steam to escape when it exceeds a predetermined pressur e. How does this device redu ce the time needed for cooking? A 1.20-g sample of water is injected into an evacuated 5.00-L flask at 65°C. What percentage of the water will be vapor when the syst em reaches eq uilibrium? Assume ideal beh av ior of water vapor and that the vo lume of liquid water is negligible. The vapor pressure of water at 65°C is 187.5 mmHg. What are the advantages of cooking the vegetable br occoli with steam instead of boiling it in water? A quantitative measure of how efficiently spheres pack into unit cells is called packing efficiency, which is the percentage of the cell space occupied by the spheres. Calculate the packing efficiencies of a simple cubic cell, a body-centered cubic cell, and a face-centered cubic cell. (Hint: Refer to Figure 12.24 and use the relationship that the volume of a sphere is l 1T ?, where r is the radius of the sphere.) The phase diagram of helium is show n. Helium is the only known substance that ha s two different liquid phases: helium-I and helium-II. (a) What is the maximum temperature at which helium-II can exist? (b) Wha t is the minimum pressure at which solid helium ca n exist? (c) What is the normal boiling point of helium-I? (d) Can solid helium sublime? 100 Solid 10 Liquid ~ >= - (helium-I) ~ 0:: ~ 1 8 Liquid ::l </) (helium-II) </) <l.l 0.1 0 Gas 0.01 1 2 3 4 5 6 Temperature (K) 502 CHAPTER 12 Intermole cular Forces and the Physical Properties of Liquids and Solids 12.138 The phase cliagram of sulfur is shown. (a) Ho w many triple points are there? (b) Which is the more stable allotrope under ordinary atmospheric conditions? (c) Describe what happens when sulfur at I atm is heated from 80°C to 200°C. 154°C 1288 atm 12.139 Provide an explanation for each of the following phenomena: (a) Solid argon (m.p. -189.2 °C; b.p. - 185.7°C) can be prepared by immer sing a flask containing argon gas in liquid nitrogen (b.p. -195.8 °C) until it liquefies and then connecting the fla sk to a vacuum pump. (b) The melting point of cyclohexane (C 6 H, z) increases with increasing pressure exerted on the solid cyclohexane. (c) Certain high-altitude clouds contain water droplets at -lO °C. (d) When a piece of dry ice is added to a beaker of water, fog forms above the water. 12.140 Argon crystallizes in the face-centered cubic arrangement at 40 K. Given that the atomic radius of argon is 191 pm, calculate the density of solid argon. 12.141 Given the phase diagram of carbon, answer the following questions: (a) How many triple points are there and what are the phases that can coexist at each triple point? (b) Which has a higher density, graphite or diamond? (c) Synthetic diamond can be made from graphite. Using the phase diagram, how would you go about making diamond? Diamond Graphite 3300 T (°C) Liquid Vapor 12.142 A chemistry instructor pelformed the following mystery demonstration. Just before the students arrived in class, she heated some water to boiling in an Erlenmeyer flask. She then removed the flask from the flame and closed the flask with a rubber stopper. After the class commenced, she held the fla sk in front of the students and announced that s he could make the water boil simply by rubbing an ice cube on the outside walls of the flask. To the amazement of everyone, it worked. Give an explanation for this phenomenon. 12.143 Swimming coaches sometimes suggest that a drop of alcohol (ethanol) placed in an ear plugged with water "draws out the water." Exp lain this action from a molecular point of view. 12.144 Use the concept of intermolecular forces to explain why the far end of a walking cane rises when one raises the handle. 12.145 Why do citrus growers spray their trees with water to protect them from freezing? 12.1 46 What is the origin of dark spots on the inner glass walls of an old tungsten Iightbulb? What is the purpose of filling these lightbulbs with argon gas? 12.147 A student heated a beaker of cold water (on a tripod) with a Bunsen bumer. When the gas was ignited, she noticed that there was water condensed on the outside of the beaker. Explain what happened. The compound dichlorodifluoromethane (CClzF z ) ha s a normal boiling point of -30 °C, a critical temperature of 112°C, and a corresponcling critical pressure of 40 atm. If the ga s is compressed to 18 atm at 20 °C, will the gas condense? Your answer should be ba sed on a graphical interpretation. Iron crystallizes in a body-centered cubic lattice. The cell length as determined by X -r ay diffraction is 286.7 pm. Given that the density of iron is 7.874 g/cm 3 , calculate Avogadro's number. 12 .1 50 Sketch the cooling curves of waterfrom about 110°C to about - lO °C. How would you al so show the formation of supercooled liquid below O°C which then freezes to ice? The pressure is at 1 atrn throughout the process. The curves need not be drawn quantitatively. 12.151 The boiling point of methanol is 65.0°C, and the standard enthalpy of formation of methanol va por is -201.2 kJ/mo!. Calculate the vapor pressure of methanol (in mmHg) at 25°C. (Hint: See Appendix 2 for other thermodynamic data of methanol). , 12.152 A sample of water shows the following behavior as it is heated at a constant rate: 12.153 Heat added If twice the mass of water has the same amount of heat transferred to it, which of the following graphs best describes the temperature variation? Note that the scales for all the graphs are the same. ~ U ° '-' N V ~ U ° '-' Heat added (a) Heat added (c) Heat added (b) Heat added (d) A closed vessel of volume 9.6 L contains 2.0 g of water. Calculate the temperature (in 0C) at which only half of the water remains in the liquid pha se. (See Table 11.5 for vapor pressures of water at different temperatures.) ANSWERS TO IN-CHAPTER MATERIALS 503 PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES Silicon used in computer chips must have an impurity level below 10- 9 (that i s, fewer than one impurity atom for every 10 9 Si atoms). Silicon is prepared by the reduction of quartz (Si0 2 ) with coke (a form of carbon made by the destructive distillation of coal) at about 2000°C: Si0 2 (s) + 2C(s) -_. Si(l) + 2CO(g) Next, solid silicon is separated from other solid impurities by treatment with hydrogen chloride at 350°C to form gaseous trichlorosilane (S iCI 3 H): SiCs) + 3HCI(g) - _. SiCI 3 H(g) + H 2 (g) Finally, ultrapure Si can be obtained by reversing the preceding reaction at 1000°C: SiChH(g) + H 2 (g) - _. SiCs) + 3HCl(g) 1. The molar heat of vaporization of trichlorosilane is 28.8 kJ/mo!. Using this information and the equation determine the normal boiling point of trichlorosilane. a) -28 .0°C b) -276 °C c)30.1 °C d) 275°C 2. What kind(s) of intermolecular forces exist between trichlorosilane molecules? a) Dispersion, dipole-dipole, and hydrogen bonding b) Dispersion and dipole-dipole c) Ion-ion d) Dipole-dipole 3. Each cubic unit cell (edge length a = 543 pm ) contains eight Si atoms. If there are 1.0 X 1013 boron atoms per cubic centimeter in a sample of pure silicon, how many Si atoms are there for every B atom in the sample? a) 5 X 10 22 b) 2 X 10- 10 c) 2 X 10 - 22 d) 5 X 10 9 4. Calculate the density of pure silicon. a) 2.33 g/cm 3 b) 0.292 g!cm 3 c) 4.67 g!em 3 d) 3.72 X 10- 22 g/ cm 3 ANSWERS TO IN-CHAPTER MATERIALS Answers to Practice Problems 12.1 (a) Dispersion forces, (b) dipole-dipole forces, H bonding and dispersion forces, (c) dipole-dipole and dispersion forces, (d) dispersion forces. 12.lB (a) Dipole-dipole and dispersion forces, (b) dipole-dipole forces, H bonding and dispersion forces, (c) dipole-dipole forces, H bonding and dispersion forces, (d) dispersion forces. 12.2A 265 mrnHg. 12.2B 29°C. 12.3A 315 pm. 12.3B 24.2°. 12.4A 10.5 g/cm 3 . 12.4B Body-centered cubic. 12.SA 4 Ca, 8 F. 12.5B 1 Cs, 1 C!. 12.6A 2.65 g/cm 3 12.6B 421 pm. 12.7A 2. 72 g/cm 3 . 12.7B 361 pm. 12.8A 984 kJ. 12.8B 100°C, liquid and vapor in equilibrium. 12.9A (a) - 110°C, 10°C, (b) liquid. 12.9B 2.0 ~ S L 8 ~ '" ~ " 1.0 ~ '" G '" '" " ~ 0- a a 100 200 300 Temperature (DC) Answers to Checkpoints 12.1.1 a. 12.1.2 d. 12.2.1 a. 12.2.2 e. 12.3.1 d. 12.3.2 e. 12.6.1 a. 12.6.2 c. 12.7.1 a. 12.7.2 e. Answers to Applying What You've Learned a) Both molecules exhibit dipole-dipole forces (including hydrogen bonding) and dispersion forces. b) 1.9758 A. c) 4. d) 1.526 g/cm 3 e) 4.5 X 10 2 kJ, l.5 X 10 3 kJ. • • SIca erties • o utionS Dialysis In 1993, nine patients who had undergone routine hemodialysis treatment at the Univer- sity of Chicago Hospitals became seriously ill, and three of them died. The illnesses and deaths were attributed to fluoride poisoning, which occurred when the equipment meant to remove fluoride from the water failed. Although hemodialysis is supposed to remove impurities from the blood, the inadvertent use of fluoridated water in the process actually added a toxin to the patients' blood. Hemodialysis, often called simply dialysis, is the cleansing of toxins from the blood of patients whose kidneys have failed. It works by routing a patient's blood temporarily through a special filter called a dialyzer. Inside the dialyzer, the blood is separated (by an artificial porous membrane) from an aqueous solution called the dialysate. The dialysate contains a variety of dissolved substances, typically including sodium chloride, sodium bicarbonate or sodium acetate, calcium chloride, potassium chloride, magnesium chlo- ride, and sometimes glucose. Its composition mimics that of blood plasma. When the two solutions, blood and dialysate, are separated by a porous membrane, the smallest of the dissolved solutes pass through the membrane from the side where the concentra- tion is high to the side where the concentration is low. Because the dialysate contains vital components of blood in concentrations equal to those in blood, no net passage of these substances occurs through the membrane. However, the harmful substances that accumulate in the blood of patients whose kidneys do not function properly pass through the membrane into the dialysate, in which their concentration is initially zero, and are thereby removed from the blood. Properly done, hemodialysis therapy can add years to the life of a patient with kidney failure. The development and refinement of medical procedures such as hemodialysis require an understanding of the properties of solutions. In This Chapter, You Will Learn about the energy changes associated with solution formation, how the concentration of a solution is expressed, and how concentration determines certain properties of a solution. Before you begin, you should review • Intermolecular forces. [ ~~ Sect ion 12.1] • Molecular geometry and polarity [ ~~ Sect i on 9.2 ] • Solution stoichiometry [ ~~ Sectio n 4.5] Dialysis uses the properties of solutions to remove harmful substances from the blood. , s ,'. Media Player/ MPEG Content Chapter in Re vi ew 505 506 CHAPTER 13 Physical Properties of Solutions The term solubility was also defined in Section 4. 1 . Types of Solutions As we noted in Section l.2 , a solution is a homogeneous mixture of two or more substances. Recall that a solution con sists of a solvent and one or more solutes [ ~~ Section 4.1] . Although many of the mo st familiar solutions are those in which a solid is dissolved in a liquid (e.g., salt- water or sugar water), the components of a solution may be solid, liquid, or gas. The possible combinations give rise to seven distinct types of solutions, which we classify by the original states of the solution components. Table 13.1 gives an example of each type. In this chapter, we will focus on solutions in which the solvent is a liquid; and the liquid solvent we will encounter mo st often is water. Recall that solutions in which water is the solvent are called aqueous solutions [ ~~ Section 4.1] . Solutions can also be classified by the amount of solute dissolved relative to the maximum amount that can be dissolved. A saturated solution is one that contains the maximum amount of a solute that will dissolve in a solvent at a specific temperature. The amount of solute dissolved in a . . . . . . . . . . . . . . . . . . . given volume of a saturated solution is called the solubility. It is important to realize that solubility refers to a specific solute, a specific solvent, and a specific temperature. For example, the solubil- ity of NaCI in water at 20 0 e is 36 g per 100 mL. The solubility of NaCI at another temperature, or in another solvent, would be different. An unsaturated solution is one that contains less solute than it has the capacity to dissolve. A supersaturated solution, on the ot her hand, contains more dissolved solute than is pre sent in a sa turated solution (Figure 13.1). It is generally not stable, and eventually the dissolved so lute will come out of solution. An example of this phenomenon is shown in Figure 13.2. (b) Solute Gas Gas Gas Liquid Liquid Solid Solid Solvent Gas Liquid Solid Liquid Solid Liquid Solid State of Resulting Solution Ga s· Liquid Solid Liquid Solid Liquid Solid ·Gaseous solutions can only contain gaseous solutes . (£-G- . - (c) (d) Example Air Carbonated water H2 gas in palladium Ethanol in water Mercury in silver Saltwater Brass (Cu/Zn) (e) (a) Figure 13.1 (a) Many solutions consist of a solid di ssolved in water. (b) When all the so lid dis so lves, the so lution is un saturated. (c) If more so lid is added than will di ssolve, the so lution is saturated. (d) A saturated solution i s, by definition, in contact with undi ss olved so lid. (e) Some saturated solutions can be made into supersaturated solutions by heating to dissolve more solid, and cooling carefully to prevent crystallization. SECTION 13.2 A Molecular View of the Solution Process 507 (a) (b) (c) (d) (e) Figure 13.2 In a supersaturated solution, (a) addition of a tiny seed crystal initiates crystallization of excess solute. (b)-(e) Crystallization proceeds rapidly to give a saturated solution and the crystallized solid. A Molecular View of the Solution Process In Chapter 4, we learned guidelines that helped us predict whether or not an ionic solid is soluble in water. We now take a more general look at the factors that determine solubility. This discussion will enable us to understand why so many ionic substances are soluble in water, which is a polar solvent; and it will help us to predict the solubility of ionic and molecular compounds in both polar and nonpolar solvents. The Importance of Intermolecular Forces The intermolecular forces that hold molecules together in liquids and solids playa central role in the solution process. When the solute dissolves in the solvent, molecules of the solute disperse throughout the solvent. They are, in effect, separated from one another and each solute molecule is surrounded by solvent molecules a process known as solvation. The ease with which solute molecules are separated from one another and surrounded by solvent molecules depends on the relative strengths of three types of interactions: 1. Solute-solute interactions 2. Solvent-solvent interactions 3. Solute-solvent interactions For simplicity, we can imagine the solution process taking place in the three distinct steps shown in Figure 13.3. Step 1 is the separation of solute molecules from one another, and step 2 is the separation of solvent molecules from one another. Both of these steps require an input of energy to overcome intermolecular attractions, so they are endothermic. In step 3 the solvent and solute . .' . . . . . . . . . . . . " . molecules mix. This process is usually exothermic. The enthalpy change for the overall process, t:: H so ln , is given by t:: H so ln = t:: H I + t:: H 2 + t:: H 3 The overall solution-formation process is exothermic (t:: H so ln < 0) when the heat given off in step 3 is greater than the sum of heat required for steps 1 and 2. The overall process is endothermic (t:: H s oln > 0) when the heat given off in step 3 is less than the total required for steps 1 and 2. (Fig- ure 13.3 depicts a solution formation that is endothermic overall.) The saying "like dissolves like" is useful in predicting the solubility of a substance in a given solvent. What this expression means is that two substances with intermolecular forces of similar type and magnitude are likely to be soluble in each other. For example, both carbon tetra- chloride (CCI 4 ) and benzene (C6H6) are nonpolar liquids. The only intermolecular forces present in these substances are dispersion forces [ ~~ Section 12.1]. When these two liquids are mixed, Like the formati on of chemical bonds [ ~~ Section 8.9), the formation of i nterm ole cular attracti ons is exotherm ic. If that is n't intuitiv el y o bv iou s, thi nk of it this w ay: It w ould r equire energy to separate molecules that are attra cte d to ea ch other. The r ev erse process, the comb in ation of mo le cu les that attra ct e ac h other, wo ul d give off an equal amount of energ y [ ~ Section 5.3). 508 CHAPTER 13 Physical Properties of So l ut ions Figure 13.3 A molecular view of the solution process portrayed as taking place in three steps: First the solute and solvent molecules are separted (steps 1 and 2, respectively-both endothermic). Then the solvent and solute molecules mix (step 3 exothermic). So lv ati on ref ers in a gene ra l way to s ol ut e pa rticl es b ei ng su rr o und ed by so lv en t mo lecu les. When the s olv ent is water, we use the mo re s peci fic term hy dra tion [ ~~ Section 4.1]. • • • • • • • • • • • • • • • Separated solute •• • • •• • • • • • • • • • Separated solute . , . . . +. . +. Solute '" J , , - J c,;/ J:') } '" ) } ) } J J } J ) J) ) Separated solvent Solvent Solvent Step 2 I1H 2 > 0 Step I I1H J > 0 Step 3 I1H 3 < 0 I1H so Jn > 0 Solution they readily dissolve in each other, because the attraction between CCl 4 and C6H6 molecules is comparable in magnitude to the forces between molecules in pure CCl 4 and those between mol- ecules in pure C 6 H 6 . Two liquids are said to be miscible if they are completely soluble in each other in all proportions. Alcohols such as methanol, ethanol, and 1,2-ethylene glycol are miscible with water because they can form hydrogen bonds with water molecules: H I H-C-O - H I H Methanol H H I I H-C-C-O-H I I H H Ethanol H H I I H- O-C-C-O - H I I H H 1,2-Ethylene glycol The guidelines listed in Tables 4.2 and 4.3 enable us to predict the solubility of a particular ionic compound in water. When sodium chloride dissolves in water, the ions are stabilized in solu- . . . . . . tion by hydration, which involves ion-dipole interactions. In general, ionic compounds are much more soluble in polar solvents, such as water, liquid ammonia, and liquid hydrogen fluoride, than in nonpolar solvents. Because the molecules of nonpolar solvents, such as benzene and carbon tet- rachloride, do not have a dipole moment, they cannot effectively solvate the Na + and CI- ions. The predominant intermolecular interaction between ions and nonpolar compounds is an ion-induced dipole interaction, which typically is much weaker than ion-dipole interactions. Consequently, ionic compounds usually have extremely low solubility in nonpolar solvents. Energy and Entropy in Solution Formation Although exothermic processes are generally more energetically favorable than endothermic pro- cesses, solutions do form in which the overall process is endothermic. The solution process, like all physical and chemical processes, is governed by two factors: energy and entropy. Entropy will be defined and examined in some detail in Chapter 19, but for now you can think of it as random- ness or disorder. There is an inherent tendency for entropy to increase in all natural events. In their pure states, the solvent and solute each possess a fair degree of order, characterized by the more or less regular arrangement of atoms, molecules, or ions in three-dimensional space, This order is disrupted when the solute dissolves in the solvent (Figure 13,3), Therefore, the solution process is accompanied by an increase in entropy. It is the increase in entropy of the system that favors the dissolution of any substance, even if the solution process is endothermic, Sample Problem 13,1 shows how to predict solubility based on the overall polarity of the solute and the solvent. SECTION 13.2 A Molecular View of the Solution Process 509 Determine for each solute whether the solubility will be greater in water, which is polar , or in benzene (C6H6)' which is nonpolar: (a) bromine (Br2)' (b) sodium iodide (Nal), (c) carbon tetrachloride (CCI 4 ), and (d) formaldehyde (CH 2 0). Strategy Consider the structure of each solute to determine whether or not it is polar. For molecular solutes, start with a Lewis structure and apply the VSEPR theory [ ~~ Section 9.1]. We expect polar solutes, including ionic compounds , to be more soluble in water. Nonpolar solutes will be more soluble in benzene. Setup (a) Bromine is a homonuclear diatomic molecule and is nonpolar. (b) Sodium iodide is ionic. (c) Carbon tetrachloride has the following Lewis structure: :CI: . . I :CI-C-CI: .• I •• :Cl: • • With four electron domains around the central atom, we expect a tetrahedral arrangement. A symmetrical arrangement of identical bonds results in a nonpolar molecule. (d) Formaldehyde has the following Lewis structure: • • "0" 111 c~ HY ""H Crossed arrows can be used to represent the individual bond dipoles [ ~~ Section 9.2]. This molecule is polar and can form hydrogen bonds with water. Solution (a) Bromine is more soluble in benzene. (b) Sodium iodide is more soluble in water. .(c) Carbon tetrachloride is more soluble in benzene. (d) Formaldehyde is more soluble in water. Practice Problem A Predict whether iodine (I2) is more soluble in liquid ammonia ( NH 3 ) or in carbon disulfide (CS 2 ). Practice Problem B Which of the following should you expect to be more soluble in benz ene than in water: C 3 H s , HCI, 1 2 , CS 2 ? Bringing Chemistry to Life Vitamin Solubility Vitamins can be categorized as either water soluble or fat soluble. An excess of a fat-soluble vita- min can be stored in the body, whereas excesses of most water-soluble vitamins are eliminated in the urine. The difference between water-soluble and fat-soluble vitamins is their molecular structures. Water-soluble vitamins, such as vitamin C (C 6 H s 06), have multiple polar groups that can interact with water to form hydrogen bonds: •• :OH • • HO • • • • HO >===0: • • • .' •• CH2QH Vitamin C Think About It Remember that molecular formula alone is not sufficient to determine the shape or the polarity of a polyatomic molecule. It must be determined by starting with a correct Lewis structure and applying the VSEPR theory. [...]... 4 + mass 0 water 142.0 g 1142.0 g X 100% = 12.4% The term percent literally means "parts per hundred." If we were to use Equation 13.2 but multiply by 1000 instead of 100, we would get "parts per thousand"; mUltiplying by 1,000,000 would give "parts per million" or ppm; and so on Parts per million, parts per billion, parts per trillion, and so fOlth, usually are used to express very low concentrations,... 10- 3 M , Colligative Properties Colligative properties are properties that depend on the number of solute particles in solution but do not depend on the nature of the solute particles That is , colligative properties depend on the concentration of solute particles regardless of whether those particles are atoms , molecules , or ions The colligative properties are vapor-pressure lowering, boiling-point... sum of the individual partial pressures exerted by the solution components Raoult's law holds equally well in thi s case: PA=XAP A P B = XBP E where P A and P B are the partial pressures over the solution for components A and B, P Aand P E are the vapor pressures of the pure substances A and B, and XA and XB are their mole fraction s The total pressure is given by Dalton's law of partial pressures [... molecules through a porous membrane from a more dilute solution to a more concentrated one Figure 13.10 illustrates osmosis The left compartment of the apparatus contains pure solvent; the right compartment contains a solution made with the same solvent The two compartments are separated by a semipermeable membrane, which allows the passage of solvent molecules but blocks the passage of solute molecules... chemists of his time, van't Hoff did significant work in thermodynamics, molecular structure and optical activity, and solution chemistry In 1901 he received the first Nobel Prize in Chemistry (a) Free ions and (b) ion pairs in solution Ion pairing reduces the number of dissolved particles in a solution, causing a decrease in the observed colligative properties Furthermore, an ion pair bears no net charge... (b) c = (3.1 X = (3.1 X 10- 2 mollL atm) (5.0 atm ) 10- 2 mollL atm) (O.0003 atm) = 9.3 = 1.6 X X 10- 1 mollL 10- 6 mollL Think About It With a pressure approximately 15,000 times smaller in part (b) than in part (a), we expect the concentration of COz to be approximately 15 ,000 times smaller-and it is Practice Problem A Calculate the concentration of CO 2 in water at 25 °C when the pressure of CO... pressure of 1 atm is 6.8 X 10- 4 mollL Calculate the concentration of di ssolved N2 in water under atmospheric conditions where the partial pressure of N2 is 0.78 atm 13.4.2 Calculate the molar concentration of O2 in water at 25 °C under atmospheric conditions where the partial pressure of O 2 is 0.22 atm The Henry's law constant for O 2 is 1.3 X 10- 3 mol/L atm a) 2 9 X 10- 4 M a) 6.8 X 10- 4 M b)... pressure is directly proportional to the concentration of the solution This is what we would expect, though, because all colligative properties depend only on the number of solute particles in solution, not on the identity of the solute particles Two solutions of equal concentration have the same osmotic pressure and are said to be isotonic to each other Electrolyte Solutions So far we have discussed the colligative... get three ions: one Ca + ion and two CI- ions Thus, for every mole of CaCl2 dissolved, we get three moles of ions in solution Colligative properties depend only on the number of dissolved particles-not on the type of particles This means that a 0.1 m solution of NaCI will exhibit a freezing point depression twice that of a 0.1 m solution of a nonelectrolyte, such as sucrose Similarly, we expect a 0.1... greater than the partial pressure of CO 2 in the air Thus, when a can or bottle of soda is opened, the CO 2 dissolved under highpressure conditions comes out of solution resulting in the bubbles that make carbonated drinks appealing Sample Problem 13.5 illustrates the use of Henry 's law Sample Problem 13.5 Calculate the concentration of carbon dioxide in a soft drink that was bottled under a partial pressure . mUltiplying by 1,000,000 would give "parts per million" or ppm; and so on. Parts per million, part s per billion, parts per trillion, and so fOlth, us ually. number of solute particles in solution but do not depend on the nature of the solute particles. That is, colligative properties depend on the concentration of solute particles regardless. mainly in solution properties and electrochemistry. Think About It With a pressure approximately 15 ,000 times smaller in part (b) than in part (a), we expect the concentration of