526 CHAPTER 13 Physica l Properties of Solutions Think About It Ringer's lactate is isotonic with human plasma, which is often the case with fluids administered intravenousl y. Setup Because the volume of th e solution described is 1 L, thenumber of moles is also the molarity for each solute. R = 0.08206 L . at m/m ol . K, T = 310 K, and the van't H off factors for the solutes in Ringer's lactate are as follow s: NaCI(s) - _. Na+(aq) + Cl - (aq) i = 2 KCI (s) • K +(aq) + C1 -(aq) i = 2 CaCI 2 (s) • Ca 2+ (aq) + 2CI- (aq) i = 3 NaCH 3 CH 2 COO(s) • Na+(aq) + CH 3 CH 2 COO - (aq) i = 2 Solution The total concentration of ions in solution is the sum of the individual concentrations. total concentration = 2[NaCI] + 2[KCI] + 3[MgCI 2 ] + 2[NaCH 3 CH 2 COO] = 2(0.102 M) + 2(4 X 10- 3 M) + 3(1.5 X 10- 3 M) + 2(2 .8 X 10- 2 M) = 2.73 X 10- 1 M 7T = MRT = (0.273 M)(0.08206 L . atm/mol . K)(3 1O K) = 6.93 atm Practice Problem A Determine the osmotic pressure of a solution that is 0.200 M in glucose and 0.100 M in sodium chloride at 37.0° C. Assume no ion pairing. Practice Problem B Determine the concentration of an aqueous solution that has an osmotic pressure of 5.6 atm at 37°C if (a) the solute is glucose, and (b) the solute is sodium chloride. Assume . no IOn pamn g. To review: • Vapor-pressure lowering de pend s on concentration expressed as mole fraction, X. • Boiling-point ele va tion depend s on concentration expressed as molality, m. • Freezing-point depression depends on concentration expre ss ed as molalit y, m. • Os motic pre ss ure depend s on concentration expre ss ed as molarity, M. Checkpoint 13.5 Colligative Properties 13.5.1 13 .5.2 A solution contains 75.0 g of glucose (molar mass 180.2 g/mo!) in 425 g of water. Determine the vapor pressure of water over the solution at 35° C. (Plip = 42.2 mmHg at 35° C. ) a) 0.732 mmHg b) 42.9 mmHg c) 243 mmHg d) 41.5 mmHg e) 42.2 mmHg Determine the boiling point and the freezing point of a solution prepared by dissolving 678 g of glucose in 2.0 kg of water. For water, Kb = 0.52°Clm and K f = 1. 86°C lm . a) 101 °C and 3.5°C b) 99°C and -3.5 °C c) 101 °C and - 3SC d) 112°C and 6.2°C e) 88°C and - 6.2°C 13 .5.3 • 13.5.4 Calculate the osmotic pressure of a solution prepared by dissolving 65.0 g of Na2S04 in enough water to make 500 mL of solution at 20° C. (Assume no ion pairing.) a) 0.75 atm b) 66 atm c) 44 atm d) 1 X 10- 2 atm e) 22 atm A 1.00 m solution of HCI has a freezing point of - 3.30°C. Determine the experimental van't Hoff factor for HCI at this concentration. a) 1.77 b) 2.01 c) 1.90 d) 2 e) 1 - SECTION 13.6 Calculations Using Colligative Properties 527 Bringing Chemistry to Life Hemodialysis Osmosis refers to the movement of solvent through a membrane from the side where the solute concentration is lower to the side where the solute concentration is higher. Hemodialysis involves a more porous membrane, through which both solvent (water) and small solute particles can pass. The size of the membrane pores is such that only small waste products such as excess potas- sium ion, creatinine, urea, and extra fluid can pass through. Larger components in blood, such as blood cells and proteins, are too large to pass through the membrane. A solute will pass through the membrane from the side where its concentration is higher to the side where its concentration is lower. The composition of the dialysate ensures that the necessary solutes in the blood (e.g., sodium and calcium ions) are not removed. I Purified blood is pumped from the dialyzer back to the patient. Blood is pumped from the patient to the dialyzer. Pump -++ I Artificial membrane Blood out _ __ Dialysate in • /' o + Dialyzer '\ Blood in Dialysate out Because it is not normally found in blood, fluoride ion, if present in the dialysate, will flow across the membrane into the blood. In fact, this is true of any sufficiently small solute that is not normally found in blood necessitating requirements for the purity of water used to prepare dialy- sate solutions that far exceed those for drinking water. Calculations Using Colligative Properties The colligative properties of nonelectrolyte solutions provide a means of determining the molar mass of a solute. Although any of the four colligative properties can be used in theory for this pur- pose, only freezing-point depression and osmotic pressure are used in practice because they show the most pronounced, and therefore the most easily measured, changes. From the experimentally determined freezing-point depression or osmotic pressure, we can calculate the solution's molal- ity or molarity, respectively. Knowing the mass of dissolved solute, we can readily determine its molar mass. Sample Problems 13.10 and 13.11 illustrate this technique. . . . . T he se ca l cul a tions req ui re Equations 13 .7 an d 13. 8, respectively. 528 CHAPTER 13 Physical Properties of Solutions Think About It Check the re sult using the molecular formula of quinine: C2oH24N202 (324.4 g/mol). Multistep problems such as this one require careful tracking of units at each step. Think About It Biological molecules can have very high molar ma sses. Sample Problem 13.10 Quinine was the first drug widely used to treat malaria, and it remains the treatment of choice for severe cases. A solution prepared by dissolving 10.0 g of quinine in 50.0 mL of ethanol has a freezing poi nt 1.55°C below that of pure ethanol. Determine the molar mass of quinine. (The density of ethanol is 0.789 g/mL.) Assume that quinine is a nonelectrolyte. Strategy Use Equation 13.7 to determine the molal concentration of the solution. Use the density of ethanol to determine the mass of solvent. The molal concentration of quinine multiplied by the ma ss of ethanol (in kg) gives moles of quinine. The mass of quinine (in grams) divided by moles of quinine gives the molar mass. Setup ma ss of ethanol = 50.0 mL X 0.789 g/mL = 39.5 g or 3.95 X 10- 2 kg Kr for ethanol (f r om Table 13.2) is 1.99°C/m. Solution Solving Equation 13.7 for molal concentration, tJ.T r l.55 °C m = K f = 1.990 C/m = 0.779 m The solution is 0.779 m in quinine (i.e., 0.779 mol quinine/kg ethanol solvent). 0.779 mol quinine kg ethanol (3 .95 X 10- 2 kg ethanol) = 0.0308 mol quinine 10.0 g quinine molar mass of quinine = 0 0308 = 325 g/mol . mol qumme Practice Problem A Calculate the molar mass of naphthalene, the organic compound in "mothballs," if a solution prepared by dissolving 5.00 g of naphthalene in exactly 100 g of benzene has a freezing point 2.0°C below that of pure benzene. Practice Problem B What mass of naphthalene must be dissolved in 2.00 X 10 2 g of benzene to give a solution with a freezing point 2.50°C below that of pure benzene? A solution is prepared by dissolving 50.0 g of hemoglobin (Hb) in enough water to make 1.00 L of solution. The osmotic pressure of the solution is measured and found to be 14.3 mmHg at 25°C. Calculate the molar mass of hemoglobin. (Assume that there is no change in volume when the hemoglobin is added to the water.) Strategy Use Equation 13.8 to calculate the molarity of the solution. Because the solution volume is 1 L, the molarity is equal to the number of moles of hemoglobin. Dividing the mass of hemoglobin, which is given in the problem statement, by the number of moles gives the molar mass. Setup R = 0.08206 L . atm/mol . K, T = 298 K, and 7T = 14.3 mmHg/(760 mmHg/atm) = 1.88 X 10- 2 atm. Solution Rearranging Equation 13.8 to solve for molarity, we get M = 7T = 1.88 X 10 - 2 atm = 7.69 X 10- 4 M RT (0.08206 L . atm/mol . K)(298 K) Thus, the solution contains 7.69 X 10- 4 mole of hemoglobin. 50 g molar mass of hemoglobin = = 6.5 X 10 4 g/mol 7.69 X 10- 4 mol Practice Problem A A solution made by dissolving 25 mg of insulin in 5.0 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the molar mass of insulin. (Assume that there is no change in volume when the insulin is added to the water.) Practice Problem B What mass of insulin must be dissolved in 50.0 mL of water to produce a solution with an osmotic pressure of 16 .8 mmHg at 25°C? SECTION 13.6 Calculations Using (olligative Properties 529 The colligative properties of an electrolyte solution can be used to determine percent dis- sociatio n. Percent dissociation is the percentage of dissolved molecules (or formula unit s, in the case of an ionic compound) that separate into ions in solution. For a strong electrolyte such as NaC l, there should be complete, or 100 percent, dissociation. However, the data in Table 13.4 indi- cate that this is not necessarily the case. An experimentally determined van 't Hoff factor smaller than the corresponding calculated value indicates less than 100 percent dissociation. As the experi- mentally determined van't Hoff factors for NaCl indicate, dissociation of a strong electrolyte is . . " . . more complete at lower concentration. The percent ionization of a weak electrolyte, such as a weak acid, also depends on the concentration of the solution. Sample Problem 13 .12 shows how to use colligative propelties to determine the percent dis- sociation of a weak electrolyte. Sample Problem 13.12 A solution that is 0.100 M in hydr ofl uoric acid ( HF ) has an osmotic pressure of 2.64 atm at 25°C. Calculate the percent ionization of HF at this concentration. Strategy Use the osmotic pressure and Equation 13.8 to determine the molar con ce ntration of the particles in solution. Compar e the concentration of particles to the nominal concentration (0.100 M) to determine what percentage of the or iginal HF molecules are ionized. Setup R = 0.08206 L . atm/mol . K, and T = 298 K. Solution Rearranging Equation 13 .8 to solve for molarity, M = 7T = 2.64 atm = 0.108 M RT (0.08206 L . atm/mo1 . K)(298 K) The concentration of dissolved particles is 0.108 M. Consider the ionization ofHF [ ~ Section 4.3] : According to this equation, if x HF molecules ionize, we get x H+ ions and x F- ion s. Thus, the total concentration of particles in sol ution will be the or iginal concentration of HF minus x, which gives the concentration of intact HF molec ul es, plus 2x, which is the concentration of ions (H+ and F- ): (0.100 - x) + 2x = 0.100 + x Th erefore, 0.108 = 0.100 + x and x = 0.008. Because we earlier defined x as the amount of HF ionized, the percent ionization is given by . . . 0.008 M 100 0/, 80/, percent 10 Dl zat lOn = 0.100 M X 0 = 0 At this concentration HF is 8 per ce nt io ni zed. Practice Problem A An aqueous so lu tion that is 0.0100 M in acetic acid (HC 2 H 3 0 2 ) has an osmotic pressure of 0.255 atm at 25°C. Calculate the percent ionization of acetic acid at this concentration. Practice Problem B An aqueous solution that is 0.015 M in acetic acid ( HC 2 H 3 0 2 ) is 3.5 percent ionized at 25°C. Calculate the osm ot ic pressure of this solution. Checkpoint 13.6 Calculations Using Colligative Properties 13.6.1 A solution made by di ssolving 14.2 g of sucrose in 100 g of water exhibits a freezing point depression of O.77 °C. Ca lculate the molar mass of sucrose. a) 34 gl mol b) ? 3.4 X 10- gl mol c) 2.4 glmol d) 1.8 X 10 2 glmol e) 68 g/mo] 13.6.2 A 0.01 a M solution of the weak electrolyte HA has an osmotic pressure of 0.27 atm at 25°C. What is the percent ionization of the electrolyte at this con ce ntration? a) 27% b) 10% c) 15% d) 81% e) 90% Recall that the term dissociation is used for ionic electrolytes and the term ionization is used for molecular electrolytes. In this context, they mean essentially the same th ing. Think About It For weak acids, the lower the concentration, the greater the percent ionization. A 0.010 M solution of HF has an osmotic pressure of 0.30 atm, co rresponding to 23 percent io ni zation. A 0.0010 M solution of HF has an osmotic pressure of 3.8 X 10- 2 atm, corresponding to 56 percent ionization. 530 CHAPTER 13 Physical Properties of So l utions The substan ce disp e rsed is called the dispersed phase; the s ubstan ce in w hich it is dispersed is c alled the di spe rsi ng medi um. Styrofoam is a regi s te r ed trademark of the Dow Che m ical Compan y. It refers specif i call y to extruded p oly styr e ne used for i nsu l at i on in home const ru ctio n. " Styrofoam " cups , co ol er s, and packing peanu ts are not reall y made of Styrof oa m. Figure 13.13 The Tyndall effect. Light is scattered by colloidal particles (right) but not by dissolved particles (left). Colloids The solutions discussed so far in this chapter are true homogeneous mixtures. Now consider what happens if we add fine sand to a beaker of water and stir. The sand particles are suspended at first but gradually settle to the bottom of the beaker. This is an example of a heterogeneous mixture. Between the two extremes of homogeneous and heterogeneous mixtures is an intermediate state called a colloidal suspension, or simply, a colloid. A colloid is a dispersion of particles of one . . . . . . . . , . . . substance throughout another substance. Colloidal particles are much larger than the normal sol- ute molecules; they range from I X 10 3 pm to I X 10 6 pm. Also, a colloidal suspension lacks the homogeneity of a true solution. Colloids can be further categorized as aerosols (liquid or solid dispersed in gas), foams (gas dispersed in liquid or solid), emulsions (liquid dispersed in another liquid), sols (solid dispersed in liquid or in another solid), and gels (liquid dispersed in a solid). Table 13.5 lists the different types of colloids and gives one or more examples of each. One way to distinguish a solution from a colloid is by the Tyndall 4 effect. When a beam of light passes through a colloid, it is scattered by the dispersed phase (Figure 13 .13). No such scat- tering is observed with true solutions because the solute molecules are too small to interact with visible light. Another demonstration of the Tyndall effect is the scattering of light from automobile headlights in fog (Figure 13.14). Among the most important colloids are those in which the dispersing medium is water. Such colloids can be categorized as hydrophilic (water loving) or hydrophobic (water fearing). Hydro- philic colloids contain extremely large molecules such as proteins. In the aqueous phase, a protein like hemoglobin folds in such a way that the hydrophilic parts of the molecule, the parts that can interact favorably with water molecules by ion-dipole forces or hydrogen-bond formation, are on the outside surface (Figure 13.15). Dispersing Dispersed Medium Phase Name Example Gas Liquid Aerosol Fog, mist Gas Solid Aerosol Smoke Liquid Gas Foam Whipped cream, meringue Liquid Liquid Emulsion Mayonnaise Liquid Solid Sol Milk of magnesia • •• • •••• . •• •• • • • • •• •• •• • • • • • • ••• Solid Gas Foam Styrofoam Solid Liquid Gel Jelly, butter . Solid Solid Solid sol Alloys such as steel, gemstones (glass with dispersed metal) 4. John Tyndall (1820-1893). Irish phys ici s t. Tyndall did important work in magnetism and also explained glacier motion. o -0 0 ,,~ c I Protein I c /~ -0 0 Repulsion • A hydrophobic colloid normally would not be stable in water, and the particles would clump together, like droplets of oil in water merging to form a film at the water's surface. They can be . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . stabilized, however, by the adsorption of ions on their surface (Figure 13.16). Material that col- lects on the surface is adsorbed, whereas material that passes to the interior is absorbed. The adsorbed ions are hydrophilic and can interact with water to stabilize the colloid. In addition, because adsorption of ions leaves the colloid particles charge d, electrostatic repulsion prevents them from clumping together. Soil particles in rivers and streams are hydrophobic particles that are stabilized in this way. When river water enters the sea, the charges on the dispersed particles are neutralized by the high-salt medium. With the charges on their surfaces neutralized, the par- ticles no longer repel one another and they clump together to form the silt that is seen at the mouth of the river. Another way hydrophobic colloids can be stabilized is by the presence of other hydrophilic groups on their surfaces. Consider sodium stearate, a soap molecule that has a polar group at one SECTION 13.7 Colloids 531 Figure 13.14 A familiar example of the Tyndall effect: headlights illuminating fog. Figure 13.15 Hydrophilic groups on the surface of a large molecule such as a protein stabilize the molecule in water. Note that all the hydrophilic groups can form hydrogen bonds with water. Figure 13.16 Diagram showing the stabilization of hydrophobic colloids . Negative ions are adsorbed onto the surface, and the repulsion between like charges prevents aggregation of the particles . A hy drophob ic co lloi d must be stab iliz ed in ord er to remain suspe nded in water. 532 CHAPTER 13 Physical Properties of Solutions Figure 13.17 (a)Asodium stearate molecule. (b) The simplified representation of the molecule that shows a hydrophilic head and a hydrophobic tail. Figure 13.18 The mechanism by which soap removes grease. (a) Grease (oily substance) is not soluble in water. (b) When soap is added to water, the nonpolar tails of soap molecules dissolve in grease. (c) The grease can be washed away when the polar heads of the soap molecules stabilize it in water. Figure 13.19 Structure of sodium glycocholate. The hydrophobic tail of sodium glycocholate dissolves in ingested fats, stabilizing them on the aqueous medium of the digestive system. Itis being nonpolar th at makes some vitamins so l uble in fat. Remember the ax i om "like dissolves like." o II /C~Z/C~ 2 /C~ Z /C~2/C~Z/C~ z /C~z/C~ z /C~ CH 3 CH z CH 2 CH z CH z CH z CH z CH 2 CH 2 O-Na + Sodium stearate (C 17 H 35 COO-Na +) (a) Hydrophilic head Hydrophobic tail (b) ~- - " ~ (a) HO" " , ~~d'<; _ ;r~ ~ " . . - (b) OH , H " , "OH o NH O~ V' (c) end, often called the "head," and a long hydrocarbon "tail" that is nonpolar (Figure 13.17). The cleansing action of soap is due to the dual nature of the hydrophobic tail and the hydrophilic head. The hydrocarbon tail is readily soluble in oily substances, which are also nonpolar, while the ionic -COO- group remains outside the oily surface. When enough soap molecules have surrounded an oi 1 droplet, as shown in Figure 13.18, the entire system becomes stabilized in water because the exterior portion is now largely hydrophilic. This is how greasy substances are removed by the action of soap. In general, the process of stabilizing a colloid that would otherwise not stay dis- persed is called emulsification, and a substance used for such stabilization is called an emulsifier or emulsifying agent. A mechanism similar to that involving sodium stearate makes it possible for us to digest dietary fat. When we ingest fat, the gallbladder excretes a substance known as bile. Bile contains a variety of substances including bile salts. A bile salt is a derivative of cholesterol with an attached amino acid. Like sodium stearate, a bile salt has both a hydrophobic end and a hydrophilic end. (Figure 13.19 shows the bile salt sodium glycocholate.) The bile salts surround fat particles with their hydrophobic ends oriented toward the fat and their hydrophilic ends facing the water, emul- sifying the fat in the aqueous medium of the digestive system. This process allows fats to be . , . digested and other nonpolar substances such as fat-soluble vitamins to be absorbed through the wall of the small intestine. APPLYING WHAT YOU'VE LEARNED 533 Applying What You've Learned Despite the use of fluoride in municipal water supplies and many topical dental products, acute fluoride poisoning is relatively rare. Fluoride is now routinely removed from the water used to prepare dialysate solutions. However, water-supply fluoridation became common during the 1960s and ]970s just when hemodialysis was first being made widely available to patients. This unfortunate coincidence resulted in large numbers of early dialysis patients suffering the effects of fluoride poisoning, before the danger of introducing fluoride via dialysis was recognized. The level of fluoride considered safe for the drinking water supply is based on the presumed ingestion by a healthy person of 14 L of water per week. Many dialysis patients routinely are exposed to as much as 50 times that volume, putting them at significantly increased risk of absorbing toxic amounts of fluoride. Problems: a) Early fluoridation of municipal water supplies was done by dissolving enough sodium fluoride to achieve a I-ppm concentration of fluoride ion. Convert 1.0 ppm F- to percent by mass F- and molality ofNaF. [ ~~ Sample Problem l3.3] b) Calculate the boiling point and freezing point of a solution made by dissolving 4.10 g NaF in 100 mL H 2 0. (For water, d = 1 g/mL.) [ ~~ Sample Problem l3.7] c) Many fluoridation facilities now use flu oro silicic acid instead of sodium fluoride. Fluorosilicic acid typically is distributed as a 23 % ( 1.596 m) aqueous solution. Calculate the van't Hoff factor of fluorosilicic acid given that a ?3 % solution has a freezing point of -IS.S ° C. [ ~~ Sample Problem l3 .8] d) The density of 23 % fiuorosilicic acid is 1.19 g/mL. Given that the osmotic pressure of this solution at 25°C is 242 atm, calculate the molar mass of fluorosilicic acid. [ ~~ Sample Problems 13.9 and 13.10] e) Hydrofluoric acid (HF) is another compound that can be used in the fluoridation of water. HF is a weak acid that only partially ionizes in solution. If an aqueous solution that is 0.15 Min HF has an osmotic pressure of 3.9 atm at 25°C, what is the percent ionization of HF at this concentration? [ ~~ Sample Problem 13.11] Elevated l evels of fluor ide are associated with osteomalacia, a condition marked by deb il itating bone pain and mu s cle weakn e ss. 534 CHAPTER 13 Physical Properties of Solutions CHAPTER SUMMARY Section 13.1 • Solutions are homogeneous mixtures of two or more substance s, which may be solids, liquids, or ga ses. • Saturated solutions contain the maximum possible amount of dissolved solute. • The amount of solute dissolved in a saturated solution is the solubility of the solute in the specified sol vent at the specified temperature. • Unsaturated solutions contain less than the ma ximum possible amount of solute. • Supersaturated solutions contain more solute than specified by the solubility. Section 13.2 • Substances with similar intermolecular forces tend to be soluble in one another. "Like dissolves like." Two liquids that are soluble in each other are called miscible. • Solution formation may be endothermic or exothermic overall. An increase in entropy is the driving force for solution formation. Solute particles are surrounded by solvent molecules in a process called solvation. , Section 13.3 • In addition to molarity (M) and mole fraction ex) , molality (m) and percent by mass are used to express the concentrations of solutions. • Molality is defined as the number of moles of solute per kilogram of solvent. Percent by mass is defined as the mass of solute di vided by the total ma ss of the solution, all multiplied by 100 percent. • Molality and percent by ma ss have the advantage of being temperature independent. Conversion among molarity, molality, and percent by mass requires solution density. • The units of concentration used depend on the type of problem to be solved. Section 13.4 • Increasing the temperature increases the solubility of mo st solids in water and decreases the solubility of most gases in water. • Increasing the pressure increases the solubility of gases in water but does not affect the solubility of solids. • According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the solution: c = kP • The proportionality constant k is the Henry's law constant. Henry's law constants are specific to the ga s and solvent, and they are temperature dependent. !(EyWORDS Colligative properties, 517 Hydrophilic, 530 Colloid, 530 Hydrophobic, 530 Entropy, 508 Hypertonic, 525 Henry's law, 516 Hypotonic, 525 Henry's law constant (k), 516 Ideal solution, 519 Section 13.S • Colligative properties depend on the number (but not on the type) of dissolved particles. The colligative properties are vapor-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure. • A volatile substance is one that has a measurable vapor pressure. A nonvolatile s ub stance is one that does not have a measurable vapor pressure. • According to Raoult's law, the partial pressure of a substance over a solution is equal to the mole fraction ex ) of the substance times its pure vapor pressure (PO ). An ideal solution is one that obeys Raoult's law. • Osmosis is the flow of solvent through a semipermeable membrane, one that allows solvent molecules but not solute particles to pass, from a more dilute solution to a more concentrated one. • Osmotic pressure (rr) is the pressure required to prevent osmosis from . occurrlllg. • Two solutions with the same osmotic pressure are called isotonic. Hypotonic refers to a solution with a lower osmotic pressure. Hypertonic refers to a solution with a higher osmotic pressure. These terms are often us ed in reference to human plasma, which has an osmotic pressure of 7.6 atm. • In electrolyte solutions, the number of dissolved particles is increased by dissociation or ionization. The magnitudes of colligative properties are increas ed by the van't Hofffactor (i), which indicates the degree of dissociation or ionization. • The experimentally determined van't Hoff factor is generally smaller than the calculated value due to the formation of ion pairs especially at high concentrations. Ion pairs are oppositely charged ions that are attracted to each other and effectively become a single "particle" in solution. Section 13.6 • Experimentally determined colligative properties can be used to calculate the molar ma ss of a nonelectrolyte or the percent dissociation ( or percent ionization) of a weak electrolyte. Section 13.7 • A colloid is a dispersion of particles (about I X 10 3 pm to 1 X 10 6 pm ) of one substance in another substance. • Colloids can be distinguished from true solutions by the Tyndall effect, which is the scattering of visible light by colloidal particles. • Colloids are classified either as hydrophilic (water loving) or hydrophobic (water fearing). • Hydrophobic colloids can be stabilized in water by surface interactions with ions or polar molecules. Ion pair, 523 Osmosis, 522 Isotonic, 523 Osmotic pressure (7T), 522 Miscible, 508 Percent by mass, 512 Molality (m), 512 Percent dissociation, 529 Nonvolatile, 517 • Raoult's law, 517 Saturated solution, 506 Semipermeable membrane, 522 Solubility, 506 KEY EQUATIONS Solvation, 507 Supersaturated solution, 506 Tyndall effect, 530 Unsaturated solution, 506 QUESTIONS AND PROBLEMS 535 v an't Hoff factor (i), 523 Volatile, 519 13.1 I l 't moles of solute mo a 1 y = m = ma ss of solvent (in kg) 13.2 ma ss of solute percent by mass = X 100% ma ss of solute + ma ss of solvent 13.3 c = kP 13.4 PI = XIP ) 13.5 P I - P j = t:.P = X2P I 13.6 t:.Tb = Kb m 13.7 t:. Tr = Krm 13.8 7T = MRT 13.9 t:.Tf = iKfm 13.10 t:.Tb = iK b m 13.11 7T = i MRT UESTIONS AND PROBLEMS Section 13.1: Types of Solutions Review Questions 13.1 13.2 Distinguish between an unsaturated solution, a saturated solution, and a supersaturated solution. Describe the different types of solutions that can be formed by the combination of solid s, liquids, and gase s. Give examples of each ty pe of solution. Section 13.2: A Molecular View of the Solution Process Review Questions 13.3 Briefly describe the solution process at the molecular level. Use the dissolution of a solid in a liquid as an example. 13.4 Basing your answer on intermolecular force considerations, explain what "like dissolves like" mean s. 13.5 13.6 13.7 13.8 What is solvation? What factors influence the extent to which solvation occurs? Give two examples of solvation; include one that involves ion-dipole interaction and one in which dispersion forces come into play. As you know, some solution processes are endothermic and others are exothermic. Provide a molecular interpretation for the difference. Explain why dissolving a solid almost always leads to an increase in disorder. Describe the factors that affect the solubility of a solid in a liquid. What does it mean to say that two liquids are miscible? • Problems 13.9 Wh y is naphthalene (C lO H 8 ) more soluble than CsF in benzene? 13.10 Explain why ethanol (C 2 H s OH ) is not soluble in cyclohexane (C 6 H d · 13.11 Arran ge the following compounds in order of increasing solubility in water: O 2 , LiCI, Br2, methanol (CH 3 0H). 13.12 Explain the variations in solubility in water of the listed alcohols: Compound Solubility in Water (g/100 g) at 20°C CH 3 0H 00 CH 3 CH 2 0H 00 CH 3 CH 2 CH 2 0H 00 CH 3 CH 2 CH 2 CH 2 0H 9 CH3 CH 2CH2 CH 2CH2 0H 2. 7 Note: 00 means that the alcohol and water are completely miscible in all proportion s. Section 13.3: Concentration Units Review Questions 13.13 Define the following concentration terms and give their units: percent by ma ss, mole fraction, molarity, molality. Compare their advantages and disadvantages. 13.14 Outline the steps required for conversion between molarity, molality, and percent by ma ss. Problems 13.15 Calculate the percent by ma ss of the solute in each of the fo ll owing aqueous solutions: (a) 5.75 g of NaBr in 67.9 g of solution, (b) 24.6 g ofKCI in 114 g of water, (c ) 4.8 g of toluene in 39 g of benz en e . [...]... and 1 atm is 0.034 mollL What is its solubility under atmospheric conditions? (The partial pressure of CO 2 in air is 0.0003 atm.) Assume that CO 2 obeys Henry's law 13.38 The solubility of N2 in blood at 37°C and at a partial pressure of 0.80 arm is 5.6 X 10- 4 mollL A deep-sea diver breathes compressed air with the partial pressure of N2 equal to 4.0 atm Assume that the total volume of blood in the... amount of N2 gas released (in liters at 37°C and 1 atm) when the diver returns to the sUliace of the water, where the partial pressure of N2 is 0.80 atm 13.39 The Henry's law constant of oxygen in water at 25°C is 1.3 X 10- 3 mollL atm Calculate the molarity of oxygen in water under a partial pressure of 0.20 atm Assuming that the solubility in blood at 37°C is roughly the same as that in water at 25°C,... Calculate the mole fraction of each component in the co ndensed liquid 4 0.50, XB = 0.50 0.5 2, XB = 0.48 0.72, XB = 0 28 1.0, XB = 0.86 Calculate the partial pressures of the components above the condensed liquid at 55 °C a) P A b) P A c) PA d) PA Calculate the partial press ures of A and B over the solution at 55°C = 98 mmHg, P B = 42 mmHg PB = 27 mmHg = 7 1 mmHg, PB = 12 mmHg = 72 mmHg, PB = 28 mmHg =... consumed Thus, Brappears at twice the rate that Br2 disappears To avoid the potential ambiguity of reporting the rate of disappearance or appearance of a specific chemical species, we report the rate of reaction We , determine the rate of reaction such that the result is the same regardless of which species we monitor For the hypothetical reaction , A ·2B the rate of reaction can be written as either... applied to seawater at 25 °C in order for reverse osmosis to occur? (Treat seawater as a 0.70 M NaCI solution.) 13.106 What masses of sodium chloride, magnesium chloride, sodium sulfate, calcium chloride, potassium chloride, and sodium bicarbonate are needed to produce 1 L of artificial seawater for an aquarium? The required ionic concentrations are [Na +] = 2.56 M, [K + ] = 0.0090 M, [Mg2+] = 0.054 M,... is -0.203°C, calculate the percent of the acid that has undergone ionization 0.01 M NaCI 13.130 Aluminum sulfate [Al z(S04)3] is sometimes used in municipal water treatment plants to remove undesirable particles Explain how this process works 0.1 M NaCI 13.119 Concentrated hydrochloric acid is usually available at a concentration of 37.7 percent by mass What is its molar concentration? (The density of... molality of this solution (b) Calculate its molarity (c) What volume of the solution would contain 0.250 mole of ethanol? 13.25 Fish breathe the dissolved air in water through their gills Assuming the partial pressures of oxygen and nitrogen in air to be 0.20 and 0.80 atm, respectively, calculate the mole fractions of oxygen and nitrogen in water at 298 K The solubilities of O2 and N2 in water at 298... Step Experimental Support for Reaction Mechanisms Catalysis Heterogeneous Catalysis Homogeneous Catalysis Enzymes: Biological Catalysts , • • Inetlcs • Methanol Poisoning Although methanol itself is not particularly harmful, accidental or intentional ingestion of methanol can cause headache, nausea, blindness, seizures, and even death In the liver, methanol is metabolized by the enzyme alcohol dehydrogenase... r , Media Player/ MPEG Content Chapter in R eview Methanol is used to power some mass-transit vehicles When ingested, it is converted to formic acid, the substance responsible for methanol's toxic and potentially deadly effects 543 544 CHAPTER 14 Chemical Kinetics Reaction Rates Chemical kinetics is the study of how fast reactions take place Many familiar reactions, such as the initial steps in vision... the change in pressure over time with a manometer Electrical conductance measurement can be used to monitor the progress if ionic species are consumed or produced Average Reaction Rate Consider the hypothetical reaction represented by A •B in which A molecules are converted to B molecules Figure 14.1 shows the progress of this reaction as a function of time The decrease in the number of A molecules . of ions leaves the colloid particles charge d, electrostatic repulsion prevents them from clumping together. Soil particles in rivers and streams are hydrophobic particles that are stabilized. and Equation 13.8 to determine the molar con ce ntration of the particles in solution. Compar e the concentration of particles to the nominal concentration (0.100 M) to determine what. Styrof oa m. Figure 13.13 The Tyndall effect. Light is scattered by colloidal particles (right) but not by dissolved particles (left). Colloids The solutions discussed so far in this chapter