The Chemistry of Water and the Nature of Liquids Chapter 11 CHAPTER OUTLINE 11.2 I. The Structure of Water: An Introduction to Intermolecular Forces II. A Closer Look at Intermolecular lar Forces A. London Dispersion Forces : Induced Dipoles, B. Permanent Dipole : Dipole Forces, C. Hydrogen Bonds III. Impact of Intermolecular Forces on the Physical Properties of Water, I IV. Phase Diagrams V. Impact of Intermolecular Forces on the Physical Properties of Water, II A. Viscosity, it B. Surface Tension, C. Capillary Action VI. Water: The Universal Solvent A. Why Do So Many Substances Dissolve in Water? VII. Measures of Solution Concentration ti A. Measures Based on Moles, B. Measures Based on Mass VIII. The Effect of Temperature and Pressure on Solubility A. Temperature Effects, B. Pressure Effects IX. Colligative Properties A. Vapor Pressure Lowering, B. Boiling Point Elevation, C. Freezing Point Depression, D. Osmosis, E. Reverse Osmosis, F. Back to the Future 1 2
What we learn from chap. 11 11.3 Our approach to this chapter is different than typical texts because although we spend much of the chapter on water, we look at it as only one (albeit the key) type of liquid, and take an approach that covers liquids broadly. That is, we believe that the understanding of liquids includes water, rather than being distinct from water. The chapter is spilt into two parts, with the first part dealing with pure liquids and the second part primarily dealing with aqueous solutions. the worldwide use of water : majority of worldwide water use is for agriculture the structure and properties of water the need for clean water 1 3 Water Consumption (UNESCO ) 11.4 1 4
11.1 Structure of Water 11.5 Bent molecule H-O-H bond angle of 104.5 o Polar molecule permanent dipole In the liquid state, intermolecular forces cause between 3 and 6 molecules to aggregate. Water vs Methane O-H (BE 940kJ/mol) C-H (BE 1650kJ/mol) H vap =44kJ/mol H vap =9kJ/mol 1 5 Structure of Water 11.6 1 6
11.2 Intermolecular Forces 11.7 London dispersion forces : Induced dipole Permanent Dipole-Dipole Forces Hydrogen Bonds 1 7 11.8 London Dispersion Forces Induced dipoles polarizability The result of temporary dipoles. 1 8
London Dispersion Forces 11.9 1 9 Polarizability vs Distance 11.10 1 10
11.11 Permanent Dipole-Dipole Forces The result of the dipole in polar covalent molecules Approximately 1% as strong as a covalent bond 1 11 Dipole-Dipole Forces 11.12 1 12
Hydrogen Bonds 11.13 The attraction between a hydrogen, bonded to F, O, or N in a molecule, and the lone electrons of F, O, or N in another molecule. The strongest of all intermolecular forces, about 10% of a covalent bond. 1 13 Hydrogen Bonds 11.14 1 14
11.15 1 15 11.3 Phase Changes 11.16 Evaporation the process of molecules leaving the surface of the liquid phase and entering the vapor phase. Condensation the process of molecules leaving the vapor phase and entering the liquid phase. Sublimation the escape of molecules from the solid phase directly to the vapor phase. : Opposite behavior -- Deposition 1 16
Water -- Vapor 11.17 1 17 Vapor Pressure 11.18 1 18
Vapor Pressure 11.19 The pressure of a vapor over a liquid. Vapor pressure increases with temperature. Heavier molecules have lower vapor pressures than lighter molecules. Molecules with the strongest intermolecular forces will have the lowest vapor pressure. 1 19 Vapor Pressure 11.20 1 20
Heating Curves 11.21 1 21 Changes of State 11.22 Boiling point the pressure of a liquid s vapor is equal to the surrounding pressure. Normal boiling gpoint the boiling point of a liquid if the surrounding pressure is 1 atm. Melting point the temperature at which a solid changes to a liquid. 1 22
Changes of State 11.23 Heat of Fusion ( fus H=334 J/g for ice) the amount of heat necessary to convert a solid to a liquid at its melting point and constant pressure. Heat of Vaporization ( vap H=2.44 kj/g for water) the amount of heat needed to convert a liquid to a vapor at its normal boiling point. 1 23 Sample Problem 11.24 How much heat is necessary to bring 10.0 g of ice at 10.0 o C to a temperature of 50.0 o C? The specific heat of ice is 2.05 J/g o C, the heat of fusion of water is 334 J/g, the specific heat of water is 4.184 J/g o C. : There are three steps: 1. warming the ice; 2. melting the ice; and 3. warming the water. q m SH t warming ice ice o o o 10.0 g2.05 J/g C 0.0 C 10.0 C 205 J 1 24
Sample Problem (cont) 11.25 10.0 g334 J/g q = m H melting ice fus 3340 J q m SH t heating water water o o o 10. 0 g4. 184 J/g C50. 0 C- 0.0 C 2092 J qtotal qwarming qmelting qheating 205 J + 3340 J + 2092 J 3 = 5.64 10 J 1 25 11.4 Phase Diagram 11.26 (4.6torr) Phase diagram of CO 2 1 26
Phase Diagram 11.27 Triple point the point representing the temperature and pressure at which the three phases coexist in equilibrium. Critical temperature the temperature above which the liquid state can no longer exist at any pressure. Critical pressure the vapor pressure at the critical temperature. Critical point the point defined by the critical temperature and the critical pressure. 1 27 11.5 Properties of Water 11.28 Viscosity the resistance of a liquid to flow. Water ae : 0890 0.890 mpa.s(25 C) 0.378 as( 03 mpa.s(75 C) as( C 5 H 12 : 0.224 mpa.s, C 8 H 18 : 0.508 mpa.s intermolecular force, temperature (average kinetic energy) Surface tension a measure of the energy per area on the surface of a liquid To maximize the # of hydrogen bond (minimum energy) Water forms a shape with minimum i surface area Capillary action the upward rise of a liquid in a small diameter tube caused by the adhesion of molecules to the surface of the tube. H water interact with the O glass : adhesion Cohesive force 1 28
Cohesive force & Adhesion 11.29 meniscus 1 29 11.30 11.6 Water : The Universal Solvent Solution solvent, solute aqueous solution universal solvent Process of Dissolving Dissolving can be described as occurring in three steps: Solute separation. Solvent separation. Electrostatic t ti interaction ti between solvent and solute. 1 30
Solution Process 11.31 1 31 Solution Process 11.32 Ion-dipole interaction cf) solvation, hydration 1 32
Solution Process 11.33 Solution formation can be exothermic or endothermic. Heat of solution is: H sol = H solute + H solvent + H solvation Like dissolves like polar molecules dissolve in polar liquids. nonpolar molecules dissolve in nonpolar liquids. 1 33 Solution Process 11.34 1 34
Sample Problem 11.35 Which of the following substances are soluble in water? KBr, ethanol (CH 3 CH 2 OH), hexane (C 6 H 12 ) KBr is an ionic compound it will dissolve in water. ethanol is polar it will dissolve in polar water. hexane is nonpolar, it will not dissolve in water. 1 35 Solubility 11.36 g solute solubility 100 ml solvent 1 36
10.7 Molarity 11.37 Molarity (M) is the number of moles of solutes per volume of solution in liters. Molarity ( M) = mol of solute L solution 1 37 Sample Problem 11.38 Calculate the molarity of a solution of 24.0g of HCl made up to a volume of 500. ml. M mol of solute = L solution 1 mol HCl 24.0g HCl 36.5g HCl 0.500 L 1.32 M 1 38
Molality 11.39 Molality (m) is the number of moles of solute per kilogram of solvent. Molality ( m) = mol of solute kg solvent 1 39 Sample Problem 11.40 Calculate the molality of a solution of 13.5g of KF dissolved in 250. g of water. mol of solute m = kg solvent 1 mol KF 13.5g 58.1 0.250 kg 0.929 m 1 40
Mole Fraction 11.41 Mole oefraction (χ) is the ratio of the number of moles of a substance over the total number of moles of substances in solution. i number of moles of i total t number of mol es n i = nt 1 41 Sample Problem -Conversions between units- 11.42 ex) What is the molality of a 0.200 M aluminum nitrate solution (d = 1.012g/mL)? Work with 1 liter of solution. mass = 1012 g mass Al(NO 3 ) 3 = 0.200 mol 213.01 g/mol = 42.6 g ; mass water = 1012 g -43 g = 969 g Molality 0.200mol 0969 0.969kgk 0.206 mol / kg 1 42
Sample Problem 11.43 Calculate the mole fraction of 10.0g of NaCl dissolved in 100. g of water. NaCl mol of NaCl mol of NaCl + mol H O 2 1 mol NaCl 10.0g 58.5g NaCl 1molH 0 1 mol NaCl 2O 10.0g 100.g H2O 58.5g NaCl 18.0g H 2O 0.0299 1 43 Conc. Based on mass 11.44 gsolute mass % or weight % 100 g solution ppm g solute 10 gsolution g solute ppb 10 gsolution ppt gsolute 10 gsolution 9 12 6 1 44
ppm, ppb, ppt in dilute aqueous solution 11.45 ppm ppb ppt 6 10 10 9 1gsolute 1mgsolute g of solution L solution 1gsolute 1g solute g of solution L solution 1g solute 1ng solute 12 10 g of solution L solution 1 45 11.46 11.8 Effect of Temperature on Solubility The solubility of a gas decreases with temperature. 1 46
Solubility of O 2 11.47 1 47 Effect of Temperature on Solubility 11.48 The solubility of an ionic solid generally increases with temperature. 1 48
Effect of Pressure on Solubility 11.49 Henry s Law P gas = k gas C gas P gas = pressure of the gas above the solution C gas = concentration of the gas k gas = Henry s law constant Henry s law holds best for gases O 2 and N 2, does not hold HCl 1 49 Sample Problem 11.50 A liter of water dissolves 0.0404 g of oxygen at 25 o C at a pressure of 760. torr. What would be the concentration ti of oxygen (in g/l) if the pressure were increased to 1880 torr at the same temperature? t P C P = C 1 2 1 2 0.04040404 g/l C = 2 760. torr 1880 torr C = 0.0999 g/l 2 g 1 50
11.9 Colligative Properties 11.51 Vapor pressure lowering the vapor pressure of a solvent is lowered by the addition of a nonvolatile solute. cf) volatile solute 1 51 11.52 1 52
Raoult s Law 11.53 P P o solution solvent solvent P o solvent = vapor pressure of the pure solvent 1 53 Sample Problem 11.54 What will be the vapor pressure of a solution made by dissolving 6.25g of glucose, C 6 H 12 O 6, in 50.0g of water at 25 o C? How much was the vapor pressure of the pure water lowered? The vapor pressure of water at 25 o C is 23.8 torr 1 mol mol glucose = 6.25 g 0.0347 mol glucose 180. g 1 mol mol water = 50.0 g 2.78 mol water 18.0g 2.78 water 0.988 0.0347+2.78 o P P 0.988 23.8 torr 23.5 torr soln water water vapor pressure lowering = 23.8 torr - 23.5 torr = 0.2 torr 1 54
Colligative Properties 11.55 Boiling point elevation the change in the boiling point is: T b = ik b m i = sum of the coefficients of the ions (i = 1 for molecular compounds) K b = boiling point elevation constant m = molality 1 55 Colligative Properties 11.56 Freezing point depression the change in the freezing point is: T f = ik f m i=sum of the coefficients of the ions (i = 1 for molecular compounds) K f = freezing gpoint depression constant m = molality 1 56
Antifreeze e solution o 11.57 1 57 Sample Problem 11.58 Calculate the boiling point elevation and the freezing point depression of a solution made by dissolving 12.2g of KCl in 45.0g of water. K b = 0.512 o C/m and K f = 1.86 o C/m i = 2 for KCl K + + Cl 1 mol mol KCl = 12.2g 0.164 mol 74.6g mol KCl 0.164 mol mkcl 3.64 m kg water 0.045 kg o o T ik m 2 0.512 C/ m 3.64 m 3.73 C b b o i m m m o T f i K fm 2 1.86 C/ m 3.64 m 13.5 C 1 58
Colligative Properties 11.59 Osmotic pressure П = imrt i = sum of the coefficients of the ions (i = 1 for molecular compounds) M = molarity R = gas constant (0.0821 L atm/mol K) T = temperature in Kelvin 1 59 Osmosis 11.60 1 60
Osmosis 11.61 1 61 Osmosis 11.62 1 62
11.63 1 63 Sample Problem 11.64 What is the osmotic pressure of a 100. ml solution containing 9.50 g of glucose, C 6 H 12 O 6, at 20.0 o C? 1 mol mol glucose = 9.50 g =0.0528 mol 180. g 0.05280528 mol M glucose= = 0.528 mol/l 0.100 L L atm π = (1) 0.528 mol/l 0.08206 08206 [20.0+273] 0+273] K = 127atm 12.7 Kmol 1 64
Problems 11.65 4,8,36,47,48,66,72,84,98,106, 112 1 65