6/1/2018 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 6/1/2018 Enthalpy of Solution . Dissolution of Ionic Solids: • Enthalpy of solution (∆Hsoln) depends on: » Energies holding solute ions in crystal lattice » Attractive force holding solvent molecules together » Interactions between solute ions and solvent molecules • ∆Hsoln = ∆Hion-ion + ∆Hdipole-dipole + ∆H ion-dipole • When solvent is water: = » ∆Hsoln ∆Hion-ion + ∆Hhydration Enthalpy of Solution for NaCl 2 6/1/2018 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. 2+ - e.g. MgF2(s) Mg (g) + 2F (g) 푄 푥푄 Lattice energy (E) increases as Q 퐸 = 2.31 푥 10−19퐽 ∙ 푛푚 1 2 푒푙 푑 increases and/or as r decreases. Lattice Q1 is the charge on the cation Compound Energy (U) Q2 is the charge on the anion d is the distance between the ions MgF2 2957 Q= +2,-1 MgO 3938 Q= +2,-2 LiF 1036 radius F < LiCl 853 radius Cl 3 6/1/2018 Born-Haber Cycles – Calculating Ulattice Hess’ Law is used to calculate Ulattice The calculation is broken down into a series of steps (cycles) Steps: 1. sublimation of 1 mole of the metal = ΔHsub 2. if necessary, breaking bond of a diatomic = ΔHBE 3. ionization of the gas-phase metal = IE1 + IE2 4. electron affinity of the nonmetal = EA1 + EA2 etc 5. formation of 1 mole of the salt from ions(g) = ΔHlast = -Ulattice Calculating Ulattice for NaCl(s) 4 6/1/2018 ΔHlast = ΔHf − ΔHsub − ½ ΔHBE − IE1 − EA1 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 CaF2. ΔHsub,Ca = 168 kJ/mol BEF2 = 155 kJ/mol EAF = -328 kJ/mol IE1,Ca = 590 kJ/mol IE2,Ca = 1145 kJ/mol 5 6/1/2018 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 6 6/1/2018 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 7 6/1/2018 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 (wsystem < 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.” 8 6/1/2018 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) 9 6/1/2018 Clausius-Clapeyron Equation Vapor Pressure vs Temperature H vap 1 lnPvap C R T liquid solid gas Pressure Temperature Clausius-Clapeyron Equation Vapor Pressure vs Temperature Plot of ln(P) vs 1/T yields straight line: • Slope = −ΔHvap/R • Intercept = constant 10 6/1/2018 Clausius-Clapeyron Equation How to use when given (P1, T1) and (P2,T2) H vap 1 lnPvap C R T Sample Exercise 11.2: Calculating Vapor Pressure Using Hvap The compound with the common name isooctane has the structure o shown below. Its normal boiling point is 99 C, and Hvap = 35.2 kJ/mol. What is the vapor pressure at 25 oC? 11 6/1/2018 Practice Exercise, p. 473 Pentane (C5H12) 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 oC; its vapor pressure at 25 oC 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 12 6/1/2018 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 • Ptotal = 1P1 + 2P2 + 3P3 + … where i = mole fraction of component i, and Pi = 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 o of heptane (C7H16) in 87 g of octane (C8H18) at 25 C? 13 6/1/2018 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? 14 6/1/2018 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 15 6/1/2018 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 • Psolution = 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. 16 6/1/2018 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: n m solute kg of solvent – Preferred concentration unit for properties involving temperature changes because it is independent of temperature – For typical solutions: molality > molarity 17 6/1/2018 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? 18 6/1/2018 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 19 6/1/2018 Boiling-Point Elevation and Freezing-Point Depression . Boiling Point Elevation (ΔTb): • ΔTb = Kb∙m • Kb = boiling point elevation constant of solvent; m = molality. Freezing Point Depression (ΔTf): • ΔTf = Kf∙m • Kf = 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? o The freezing point depression constant of water, Kf = 1.86 C/m. Tf = Kf m Assume 1.00 L of each = 1000 mL 1.00 g 1 kg kg H2O (solvent) = 1000 mL x x =1.00 kg H O mL 1000 g 2 1.114 g x 1 mol mol E.G.