Chapter 11: Properties of Solutions - Their Concentrations and Colligative Properties. Chapter Outline

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Chapter 11: Properties of Solutions - Their Concentrations and Colligative Properties Chapter Outline 11.1 Energy Changes when Substances Dissolve 11.

1 Chapter 11: Properties of Solutions - Their Concentrations and Colligative Properties Chapter Outline 11.1 Energy Changes when Substances Dissolve 11.2 Vapor Pressure 11.3 Mixtures of Volatile Substances 11.4 Colligative Properties of Solutions 11.5 Osmosis and Osmotic Pressure 11.6 Using Colligative Properties to Determine Molar Mass 1

2 Calculating Lattice Energies Using the Born-Haber Cycle Lattice energy (U) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. It is always endothermic. e.g. MgF 2 (s) Mg 2+ (g) + 2F - (g) E el = 2.31 x J nm Q 1xQ 2 d Lattice energy (E) increases as Q increases and/or as r decreases. Q 1 is the charge on the cation Q 2 is the charge on the anion Compound Lattice Energy (U) d is the distance between the ions MgF Q= +2,-1 MgO 3938 Q= +2,-2 LiF LiCl radius F < radius Cl Born-Haber Cycles Calculating U lattice Hess Law is used to calculate U lattice The calculation is broken down into a series of steps (cycles) Steps: 1. sublimation of 1 mole of the metal = ΔH sub 2. if necessary, breaking bond of a diatomic = f x ΔH BE 3. ionization of the gas-phase metal = IE 1 + IE electron affinity of the nonmetal = EA 1 + EA 2 + formation of 1 mole of the salt from ions(g) = ΔH last = -U lattice 2

3 ΔH f = ΔH sub + f x ΔH BE + IE + EA + ΔH last M + (g) + X(g) M + (g) + X(g) IE 1 + IE 2 etc M + (g) + X - (g) EA 1 + EA 2 etc M(g) + X(g) M(g) + f X 2 (g) f ΔH BE ΔH last = -U lattice ΔH sub M(s) + n X 2 (g) ΔH f MX(s) Calculating U lattice for NaCl(s) 3

4 ΔH last = ΔH f ΔH sub ½ ΔH BE IE 1 EA 1 4

5 Sample Exercise 11.1 Calculating Lattice Energy Calcium fluoride occurs in nature as the mineral fluorite, which is the principle source of the world s supply of fluorine. Use the following data to calculate the lattice energy of CaF 2. ΔH sub,ca = 168 kj/mol BE F2 = 155 kj/mol EA F = -328 kj/mol IE 1,Ca = 590 kj/mol IE 2,Ca = 1145 kj/mol Chapter Outline 11.1 Energy Changes when Substances Dissolve 11.2 Vapor Pressure 11.3 Mixtures of Volatile Substances 11.4 Colligative Properties of Solutions 11.5 Osmosis and Osmotic Pressure 11.6 Using Colligative Properties to Determine Molar Mass 5

Vapor Pressure Pressure exerted by a gas in

this? Colligative Properties boiling point elevation,

Solutions Vapor Pressure: Pressure exerted by a gas in

6 Vapor Pressure Pressure exerted by a gas in equilibrium with its liquid Where are we going with this? Colligative Properties boiling point elevation, freezing point depression, osmosis Vapor Pressure of Solutions Vapor Pressure: Pressure exerted by a gas in equilibrium with liquid Rates of evaporation/condensation are equal 6

7 The tendency for a liquid to evaporate increases as - 1. the temperature rises 2. the surface area increases 3. the intermolecular forces decrease Boiling Point If you have a beaker of water open to the atmosphere, the mass of the atmosphere is pressing down on the surface. As heat is added, more and more water evaporates, pushing the molecules of the atmosphere aside (w system < 0). If enough heat is added, a temperature is eventually reached at which the vapor pressure of the liquid equals the atmospheric pressure, and the liquid boils. 7

Increasing the external atmospheric pressure increases the boiling point Decreasing the external

8 Normal boiling point the temperature at which the vapor pressure of a liquid is equal to the external atmospheric pressure of 1 atm. Increasing the external atmospheric pressure increases the boiling point Decreasing the external atmospheric pressure decreases the boiling point Intermolecular Forces and Vapor Pressure Diethyl ether - dipole-dipole interactions Water - hydrogen bonding (stronger) 8

9 Pressure Clausius-Clapeyron Equation Vapor Pressure vs Temperature ln H vap 1 P C vap R T solid liquid gas Temperature ln Clausius-Clapeyron Equation Vapor Pressure vs Temperature H vap 1 P C vap R T Plot of ln(p) vs 1/T yields straight line: Slope = ΔH vap /R Intercept = constant 9

10 Clausius-Clapeyron Equation How to use when given (P 1, T 1 ) and (P 2,T 2 ) ln H vap 1 P C vap R T Practice Exercise, p. 473 Pentane (C 5 H 12 ) gas is used to blow the bubbles in molten polystyrene that turn it into Styrofoam, which is used in coffee cups and other products that have good thermal insulation properties. The normal boiling point of pentane is 36 o C; its vapor pressure at 25 o C is 505 torr. What is the enthalpy of vaporization of pentane? 10

11 Chapter Outline 11.1 Energy Changes when Substances Dissolve 11.2 Vapor Pressure 11.3 Mixtures of Volatile Substances 11.4 Colligative Properties of Solutions 11.5 Osmosis and Osmotic Pressure 11.6 Using Colligative Properties to Determine Molar Mass Mixtures of Volatile Substances Raoult s Law: Total vapor pressure of an ideal solution depends on how much the partial pressure of each volatile component contributes to total vapor pressure of solution P total = 1 P P P 3 + where i = mole fraction of component i, and P i = equilibrium vapor pressure of pure volatile component at a given temperature 11

12 Sample Exercise 11.3 (simplified): Calculating the Vapor Pressure of a Solution What is the vapor pressure of a solution prepared by dissolving 13 g of heptane (C 7 H 16 ) in 87 g of octane (C 8 H 18 ) at 25 o C? Chapter Outline 11.1 Energy Changes when Substances Dissolve 11.2 Vapor Pressure 11.3 Mixtures of Volatile Substances 11.4 Colligative Properties of Solutions 11.5 Osmosis and Osmotic Pressure 11.6 Using Colligative Properties to Determine Molar Mass 12

13 Colligative Properties of Solutions Colligative Properties: Set of properties of a solution relative to the pure solvent Due to solute solvent interactions Depend on concentration of solute particles, not the identity of particles Include lowering of vapor pressure, boiling point elevation, freezing point depression, osmosis and osmotic pressure Vapor Pressure of Solutions Raoult s Law for Solutions: Vapor pressure of solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of solvent in the solution P solution = solvent P solvent Vapor pressure lowering: A colligative property of solutions (Section 11.5) Ideal Solution: One that obeys Raoult s law 13

$20 moles of glucose. Sucrose and glucose are nonvolatile. What is the mole fraction of water in the solution?$

14 Homework Problem 11.38: Calculating the Vapor Pressure of a Solution of One or More Nonvolatile Substances A solution contains 4.5 moles of H 2 O, 0.30 moles of sucrose (C 12 H 22 O 11 ), and 0.20 moles of glucose. Sucrose and glucose are nonvolatile. What is the mole fraction of water in the solution? What is the vapor pressure of the solution at 35 o C given that the vapor pressure of pure water at 35 o C is 42.2 Torr? χ H2O = χ H2O = mol H 2 O total moles 4.5 moles moles = 0.90 P solution = solvent P solvent P solution = H2O P H2O P solution = (0.90)(42.2 Torr) = 38 Torr Colligative Properties of Solutions Colligative Properties: Vapor pressure lowering Consequences: in b.p. and f.p. of solution relative to pure solvent 14

Solute Concentration: Molality Changes in boiling point/freezing point of solutions depends on molality: nsolute m kg of solvent Preferred concentration unit for properties involving temperature

15 Solute Concentration: Molality Changes in boiling point/freezing point of solutions depends on molality: nsolute m kg of solvent Preferred concentration unit for properties involving temperature changes because it is independent of temperature For typical solutions: molality > molarity Calculating Molality Starting with: a) Mass of solute b) Volume of solvent, or c) Volume of solution 15

16 Practice Exercise, p. 483: Calculating the Molality of a Solution What is the molality of a solution prepared by dissolving 78.2 g of ethylene glycol, HOCH 2 CH 2 OH (MW = 62.07), in 1.50 L of water? Assume that the density of the water is 1.00 g/ml m = mol solute kg solvent mol solute = 78.2 g x 1 mol = mol g kg solvent = 1.50 L x 1000 ml L x 1.0 g ml 1.0 kg x = 1.50 kg 1000 g m = mol = m 1.50 kg Chapter Outline 11.1 Energy Changes when Substances Dissolve 11.2 Vapor Pressure 11.3 Mixtures of Volatile Substances 11.4 Colligative Properties of Solutions 11.5 Osmosis and Osmotic Pressure 11.6 Using Colligative Properties to Determine Molar Mass 16

Colligative Properties of Solutions Colligative Properties: Solution properties that depend on concentration of solute particles, not the identity

W. Norton & Company Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation (ΔT b ): ΔT b = K b m K b = boiling point

17 Colligative Properties of Solutions Colligative Properties: Solution properties that depend on concentration of solute particles, not the identity of particles. Previous example: vapor pressure lowering. Consequences: change in b.p. and f.p. of solution by W. W. Norton & Company Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation (ΔT b ): ΔT b = K b m K b = boiling point elevation constant of solvent; m = molality. Freezing Point Depression (ΔT f ): ΔT f = K f m K f = freezing-point depression constant; m = molality by W. W. Norton & Company 17

Sample Exercise 11.7: Calculating the Freezing Point Depression of a Solution What is the freezing point of an automobile radiator fluid prepared by mixing equal volumes of ethylene glycol (MW=62.

18 Sample Exercise 11.7: Calculating the Freezing Point Depression of a Solution What is the freezing point of an automobile radiator fluid prepared by mixing equal volumes of ethylene glycol (MW=62.07) and water at a temperature where the density of ethylene glycol is g/ml and the density of water is g/ml? The freezing point depression constant of water, K f = 1.86 o C/m. T f = K f m Assume 1.00 L of each = 1000 ml kg H 2 O (solvent) = mol E.G. (solute) = 1000 ml x 1.00 g ml 1000 ml x g x ml x 1 kg 1000 g =1.00 kg H 2 O 1 mol =17.95 mol E.G g mol m = 1.00 kg = m T f = K f m = (1.86 o C/m)(17.95 m ) = 33.4 o C New f.p. = 0.0 o C 33.4 o C = o C The van t Hoff Factor Solutions of Electrolytes: Need to correct for number of particles formed when ionic substance dissolves. van t Hoff Factor (i): number of ions in formula unit. e.g., NaCl, i = 2 ΔT b = i K b m & ΔT f = i K f m Deviations from theoretical value due to ion pair formation. 18

19 Values of van t Hoff Factors 2012 by W. W. Norton & Company Sample Exercise 11.9: Using the van t Hoff Factor Some Thanksgiving dinner chefs brine their turkeys prior to roasting them to help the birds retain moisture and to season the meat. Brining involves completely immersing a turkey for about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl, MW=58.44) in 7.6 L of water (d H2O = 1.00 g/ml). Suppose a turkey soaking in such a solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 o C. Are the brine (and turkey) in danger of freezing? K f = 1.86 o C/m; assume that i = 1.85 for this NaCl solution. T f = i K f m kg H 2 O (solvent) = 7.6 x 103 ml x 1.00 g ml mol NaCl (solute) = 910 g x 1 mol = 15.6 mol NaCl g 15.6 mol m = 7.6 kg = 2.05 m T f = i K f m = (1.85)(1.86 o C/m)(2.05 m) = 7.0 o C x 1 kg 1000 g = 7.6 kg H 2 O New f.p. = 0.0 o C 7.0 o C = -7.0 o C BUT the air T is -8 o C so the brine will freeze 19

20 Chapter Outline 11.1 Energy Changes when Substances Dissolve 11.2 Vapor Pressure 11.3 Mixtures of Volatile Substances 11.4 Colligative Properties of Solutions 11.5 Osmosis and Osmotic Pressure 11.6 Using Colligative Properties to Determine Molar Mass Molar Mass from Colligative Properties ΔT f = K f m and ΔT b = K b m Step 1: solve for molality ΔT f or ΔT b m = ΔT f /K f Step 2: solve for moles molality kg of solvent moles Step 3: calc. MW MW = g sample/moles 20

21 Sample Exercise 11.12: Using Freezing Point Depression to Determine Molar Mass Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to enhance recovery of petroleum and natural gas, a process sometimes called fracking. The freezing point of a solution prepared by dissolving 100 mg of eicosene in 1.00 g of benzene is 1.75 o C lower that the freezing point of pure benzene. What is the molar mass of eicosene? (K f for benzene is 4.90 o C/m) Step 1: solve for molality m = T K f = 1.75 o C 4.90 o C/m = m Step 2: solve for moles moles solute = mol kg benzene X kg benzene = 3.57 x 10-4 mol eicosene Step 3: calc. MW MW = 100 x 10 3 g eicosene 3.57 x 10 4 mol = 280 g/mol 21

Chapter 11: Properties of Solutions - Their Concentrations and Colligative Properties. Chapter Outline

Chapter 11: Properties of Solutions - Their Concentrations and Colligative Properties Chapter Outline 11.1 Energy Changes when Substances Dissolve 11.2 Vapor Pressure 11.3 Mixtures of Volatile Substances