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
Enthalpy of Solution Dissolution of Ionic Solids: Enthalpy of solution ( H soln ) depends on:» Energies holding solute ions in crystal lattice» Attractive force holding solvent molecules together» Interactions between solute ions and solvent molecules H soln = H ion-ion + H dipole-dipole + H ion-dipole When solvent is water:» H soln = H ion-ion + H hydration Enthalpy of Solution for NaCl 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 10 19 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 d is the distance between the ions Compound Lattice Energy (U) MgF 2 2957 Q= +2,-1 MgO 3938 Q= +2,-2 LiF LiCl 1036 853 radius F < radius Cl 3
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. Sublimate the metal = ΔH sub 2. if necessary, breaking bond of a diatomic = ΔH BE 3. ionization of the metal = IE 1 + IE 2 etc 4. electron affinity of the nonmetal = EA 1 + EA 2 etc 5. formation of 1 mole of the salt from ions = ΔH last = -U lattice 4
In General for M(s) + n X 2 (g) MX 2 n (s) 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) + n X 2 (g) n ΔH BE ΔH last = -U lattice ΔH sub M(s) + n X 2 (g) ΔH f MX(s) Calculating U lattice ΔH f = ΔH sub + n ΔH BE + IE + EA + ΔH last Rearranging and solving for ΔH last - ΔH last = ΔH f ΔH sub n ΔH BE IE EA U lattice = -ΔH last 5
Calculating U lattice for NaCl(s) Na + (g) + Cl(g) Na + (g) + Cl(g) IE 1 Na + (g) + Cl - (g) EA 1 Na(g) + Cl(g) Na(g) + ½ Cl 2 (g) ½ ΔH BE ΔH last = -U lattice ΔH sub Na(s) + ½ Cl 2 (g) ΔH f NaCl(s) 6
ΔH last = ΔH f ΔH sub ½ ΔH BE IE 1 EA 1 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 7
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 8
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 9
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. 10
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 Location Elevation (ft) b.p H 2 O ( o C) San Francisco sea level 100.0 Salt Lake City 4400 95.6 Denver 5280 95.0 Mt. Everest 29,028 76.5 11
Pressure 5/31/2018 Intermolecular Forces and Vapor Pressure Diethyl ether - dipole-dipole interactions Water - hydrogen bonding (stronger) Clausius-Clapeyron Equation Vapor Pressure vs Temperature ln H vap 1 P C vap R T solid liquid gas Temperature 12
Clausius-Clapeyron Equation Vapor Pressure vs Temperature Plot of ln(p) vs 1/T yields straight line: Slope = ΔH vap /R Intercept = constant Sample Exercise 11.2: Calculating Vapor Pressure Using H vap The compound with the common name isooctane has the structure shown below. Its normal boiling point is 99 o C, and H vap = 35.2 kj/mol. What is the vapor pressure at 25 o C? 13
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? 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 14
Fractional Distillation A method of separating a mixture of compounds based on their different boiling points Vapor phase enriched in the more volatile component 15
Solutions of Volatile Compounds 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 1 + 2 P 2 + 3 P 3 + where i = mole fraction of component i, and P i = equilibrium vapor pressure of pure volatile component at a given temperature 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? 16
Real vs. Ideal Solutions Deviations from Raoult s Law: Due to differences in solute solvent and solvent solvent interactions (dashed lines = ideal behavior) Negative deviations Positive deviations Concept Test, p. 478 Which of the following solutions is least likely to follow Raoult s Law: (a) acetone/ethanol, (b) pentane/hexane, (c) pentanol/water? 17
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 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 18
Vapor Pressure of Solutions Raoult s Law: 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 Sample Exercise 11.4: Calculating the Vapor Pressure of a Solution of One or More Nonvolatile Substances Most of the world s supply of maple syrup is produced in quebec, Ontario, where the sap from maple trees is evaporated until the concentration of sugar (mostly sucrose) in the sap reaches at least 66% by weight. What is the vapor pressure in atm of a 66% aqueous solution of sucrose (MW=342) at 100 oc? Assume the solution obeys Raoults s Law. 19
Colligative Properties of Solutions Colligative Properties: Vapor pressure lowering Consequences: in b.p. and f.p. of solution relative to pure solvent 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 20
Calculating Molality Starting with: a) Mass of solute b) Volume of solvent, or c) Volume of solution Sample Exercise 11.5: Calculating the Molality of a Solution A popular recipe for preparing corned beef calls for preparing a seasoned brine (salt solution) that contains 0.50 lbs of kosher salt (NaCl) dissolved in 3.0 qt of water. Assuming the density of water = 1.00 g/ml, what is the molality of NaCl in this solution for corning beef? 21
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 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. 2012 by W. W. Norton & Company 22
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. 2012 by W. W. Norton & Company 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 1.114 g/ml and the density of water is 1.000 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 1.114 g x ml x 1 kg 1000 g =1.00 kg H 2 O 1 mol =17.95 mol E.G. 62.07 g 17.95 mol m = 1.00 kg = 17.95 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 = -33.4 o C 23
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. Values of van t Hoff Factors 2012 by W. W. Norton & Company 24
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) = mol NaCl (solute) = 910 g x 15.52 mol m = 7.57 kg = 2.05 m 7.6 x 103 ml x 1.00 g ml x 1 kg 1000 g = 7.57 kg H 2 O 1 mol = 15.52 mol NaCl 58.44 g T f = i K f m = (1.85)(1.86 o C/m)(2.05 m) = 7.0 o C 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 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 25
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 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 = 0.357 m Step 2: solve for moles moles solute = 0.357 mol kg benzene X 0.00100 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 26