advertisement

John E. McMurry • Robert C. Fay General Chemistry: Atoms First Chapter 11 Solutions and Their Properties Lecture Notes Alan D. Earhart Southeast Community College • Lincoln, NE Copyright © 2010 Pearson Prentice Hall, Inc. Solutions Solution: A homogeneous mixture. Solvent: The major component. Solute: A minor component. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/2 Solutions Chapter 11/3 Energy Changes and the Solution Process Energy Changes and the Solution Process The sodium and chloride ions are hydrated. Energy Changes and the Solution Process There is an entropy change for the solution process. Energy Changes and the Solution Process There is an entropy change for the solution process. Solution #1 Solution #2 Chapter 11/7 Energy Changes and the Solution Process ∆G = ∆H - T∆S endothermic: +∆H exothermic: -∆H spontaneous: -∆G nonspontaneous: +∆G Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/8 Energy Changes and the Solution Process Units of Concentration Moles of solute Molarity (M) = Liters of solution Moles of component Mole Fraction (X) = Total moles making up solution Mass of component Mass Percent = x 100% Total mass of solution Moles of solute Molality (m) = Mass of solvent (kg) Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/10 Units of Concentration Assuming that seawater is an aqueous solution of NaCl, what is its molarity? The density of seawater is 1.025 g/mL at 20 °C, and the NaCl concentration is 3.50 mass %. Assuming 100.00 g of solution, calculate the volume: 100.00 g solution 1 mL 1L x x = 0.09756 L 1.025 g 1000 mL Convert the mass of NaCl to moles: 3.50 g NaCl 1 mol x 58.4 g = 0.0599 mol Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/11 Units of Concentration Then, calculate the molarity: 0.0599 mol 0.09756 L = 0.614 M Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/12 Units of Concentration What is the molality of a solution prepared by dissolving 0.385 g of cholesterol, C27H46O, in 40.0 g chloroform, CHCl3? Convert the mass of cholesterol to moles: 0.385 g 1 mol x = 0.000 997 mol 386.0 g Calculate the mass of chloroform in kg: 40.0 g 1 kg x = 0.0400 kg 1000 g Calculate the molality of the solution: 0.000 997 mol = 0.0249 m 0.0400 kg Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/13 Units of Concentration Chapter 11/14 Some Factors Affecting Solubility Saturated Solution: A solution containing the maximum possible amount of dissolved solute at equilibrium. dissolve Solute + Solvent Solution crystallize Supersaturated Solution: A solution containing a greater-than-equilibrium amount of solute. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/15 Some Factors Affecting Solubility Chapter 11/17 Some Factors Affecting Solubility Some Factors Affecting Solubility Henry’s Law Solubility = k P Chapter 11/19 Physical Behavior of Solutions: Colligative Properties Colligative Properties: Properties that depend on the amount of a dissolved solute but not on its chemical identity. • • • • Vapor-Pressure Lowering Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/20 Vapor-Pressure Lowering of Solutions: Raoult’s Law Raoult’s Law Psoln = Psolv Xsolv Vapor-Pressure Lowering of Solutions: Raoult’s Law The vapor pressure of pure water at 25 °C is 23.76 mm Hg. What is the vapor pressure of a solution made from 1.00 mol glucose in 15.0 mol of water at 25 °C? Glucose is a nonvolatile solute. Psoln = Psolv Xsolv = 23.76 mm Hg x 15.0 mol = 22.3 mm Hg 1.00 mol + 15.0 mol Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/22 Vapor-Pressure Lowering of Solutions: Raoult’s Law Solutions of ionic substances often have a vapor pressure significantly lower than predicted, because the ion-dipole forces between the dissolved ions and polar water molecules are so strong. van’t Hoff Factor: i = NaCl(aq) moles of particles in solution moles of solute dissolved Na1+(aq) + Cl1-(aq) For sodium chloride, the predicted value of i is 2. For a 0.05 m solution of sodium chloride, the experimental value for i is 1.9. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/23 Chapter 11/24 Vapor-Pressure Lowering of Solutions: Raoult’s Law Ptotal = PA + PB = (P°A XA) + (P°B XB) Boiling-Point Elevation and Freezing-Point Depression Chapter 11/26 Boiling-Point Elevation and Freezing-Point Depression Nonelectrolytes Electrolytes ∆Tb = Kb m ∆Tb = Kb m i ∆Tf = Kf m ∆Tf = Kf m i Chapter 11/28 Chapter 11/29 Boiling-Point Elevation and Freezing-Point Depression What is the freezing point (in °C) of a solution prepared by dissolving 7.40 g of MgCl2 in 110 g of water? The van’t Hoff factor for MgCl2 is i = 2.7. Calculate the moles of MgCl2: 7.40 g x 1 mol = 0.0776 mol 95.3 g Calculate the molality of the solution: 0.0776 mol 110 g x 1000 g 1 kg = 0.71 mol kg Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/30 Boiling-Point Elevation and Freezing-Point Depression Calculate the freezing point of the solution: ∆Tf = Kf m i = 1.86 °C kg mol x 0.71 mol kg x 2.7 = 3.6 °C Tf = 0.0 °C - 3.6 °C = -3.6 °C Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/31 Osmosis and Osmotic Pressure Osmosis: The passage of solvent through a semipermeable membrane from the less concentrated side to the more concentrated side. Osmotic Pressure (): The amount of pressure necessary to cause osmosis to stop. Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/32 Chapter 11/33 = MRT Chapter 11/34 Osmosis and Osmotic Pressure Calculate the osmotic pressure of a 1.00 M glucose solution in water at 300 K. = MRT = 1.00 mol L x 0.08206 L atm x 300 K = 24.6 atm K mol Copyright © 2010 Pearson Prentice Hall, Inc. Chapter 11/35 Some Uses of Colligative Properties Reverse Osmosis Fractional Distillation of Liquid Mixtures Chapter 11/38