Chapter 11. Properties of Solutions

Chapter 11 Properties of Solutions

Section 11.1 Solution Composition Solution Composition moles of solute Molarity ( M) = liters of solution mass of solute Mass (weight) percent = 100% mass of solution Mole fraction ( moles A A ) = total moles of solution moles of solute Molality ( m) = kilogram of s olvent Copyright Cengage Learning. All rights reserved 3

Section 11.1 Solution Composition EXERCISE! You have 1.00 mol of sugar in 125.0 ml of solution. Calculate the concentration in units of molarity. 8.00 M Copyright Cengage Learning. All rights reserved 5

Section 11.1 Solution Composition EXERCISE! You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? 0.200 L Copyright Cengage Learning. All rights reserved 6

Section 11.1 Solution Composition EXERCISE! Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 ml of solution. Calculate the concentration of each solution in units of molarity. 10.0 M NaOH 5.37 M KCl Copyright Cengage Learning. All rights reserved 7

Section 11.1 Solution Composition EXERCISE! What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water? 6.6% Copyright Cengage Learning. All rights reserved 9

Section 11.1 Solution Composition EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 in 100.0 ml of water. Calculate the mole fraction of H 3 PO 4. (Assume water has a density of 1.00 g/ml.) 0.0145 Copyright Cengage Learning. All rights reserved 11

Section 11.1 Solution Composition EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 in 100.0 ml of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/ml.) 0.816 m Copyright Cengage Learning. All rights reserved 13

Section 11.2 The Energies of Solution Formation Formation of a Liquid Solution 1. Separating the solute into its individual components (expanding the solute). 2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). 3. Allowing the solute and solvent to interact to form the solution. Copyright Cengage Learning. All rights reserved 14

Section 11.2 The Energies of Solution Formation Steps in the Dissolving Process Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent. Step 3 usually releases energy. Steps 1 and 2 are endothermic, and step 3 is often exothermic. Copyright Cengage Learning. All rights reserved 16

Section 11.2 The Energies of Solution Formation Enthalpy (Heat) of Solution Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔH soln = ΔH 1 + ΔH 2 + ΔH 3 ΔH soln may have a positive sign (energy absorbed) or a negative sign (energy released). Copyright Cengage Learning. All rights reserved 17

Section 11.2 The Energies of Solution Formation CONCEPT CHECK! Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role. Copyright Cengage Learning. All rights reserved 19

Section 11.2 The Energies of Solution Formation The Energy Terms for Various Types of Solutes and Solvents ΔH 1 ΔH 2 ΔH 3 ΔH soln Outcome Polar solute, polar solvent Large Large Large, negative Small Solution forms Nonpolar solute, polar solvent Small Large Small Large, positive No solution forms Nonpolar solute, nonpolar solvent Small Small Small Small Solution forms Polar solute, nonpolar solvent Large Small Small Large, positive No solution forms Copyright Cengage Learning. All rights reserved 20

Section 11.2 The Energies of Solution Formation In General One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed. Processes that require large amounts of energy tend not to occur. Overall, remember that like dissolves like. Copyright Cengage Learning. All rights reserved 21

Section 11.3 Factors Affecting Solubility Structure Effects Hydrophobic (water fearing) Non-polar substances Hydrophilic (water loving) Polar substances Copyright Cengage Learning. All rights reserved 23

Section 11.3 Factors Affecting Solubility Pressure Effects Little effect on solubility of solids or liquids Henry s law: C = kp C = concentration of dissolved gas k = constant P = partial pressure of gas solute above the solution Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. Copyright Cengage Learning. All rights reserved 24

Section 11.3 Factors Affecting Solubility Temperature Effects (for Aqueous Solutions) Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature. Predicting temperature dependence of solubility is very difficult. Solubility of a gas in solvent typically decreases with increasing temperature. Copyright Cengage Learning. All rights reserved 26

Section 11.4 The Vapor Pressures of Solutions Liquid/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 30

Section 11.4 The Vapor Pressures of Solutions Vapor Pressure Lowering: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 31

Section 11.4 The Vapor Pressures of Solutions Vapor Pressures of Solutions Nonvolatile solute lowers the vapor pressure of a solvent. Raoult s Law: P soln = observed vapor pressure of solution solv = mole fraction of solvent = vapor pressure of pure solvent P solv P = P soln solv solv Copyright Cengage Learning. All rights reserved 32

Section 11.4 The Vapor Pressures of Solutions Nonideal Solutions Liquid-liquid solutions where both components are volatile. Modified Raoult s Law: PTotal = APA + BPB Nonideal solutions behave ideally as the mole fractions approach 0 and 1. Copyright Cengage Learning. All rights reserved 34

Section 11.4 The Vapor Pressures of Solutions Summary of the Behavior of Various Types of Solutions Interactive Forces Between Solute (A) and Solvent (B) Particles ΔH soln ΔT for Solution Formation Deviation from Raoult s Law Example A A, B B A B Zero Zero None (ideal solution) Benzenetoluene A A, B B < A B Negative (exothermic) Positive Negative Acetonewater A A, B B > A B Positive (endothermic) Negative Positive Ethanolhexane Copyright Cengage Learning. All rights reserved 36

Section 11.4 The Vapor Pressures of Solutions CONCEPT CHECK! For each of the following solutions, would you expect it to be relatively ideal (with respect to Raoult s Law), show a positive deviation, or show a negative deviation? a) Hexane (C 6 H 14 ) and chloroform (CHCl 3 ) b) Ethyl alcohol (C 2 H 5 OH) and water c) Hexane (C 6 H 14 ) and octane (C 8 H 18 ) Copyright Cengage Learning. All rights reserved 37

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Colligative Properties Depend only on the number, not on the identity, of the solute particles in an ideal solution: Boiling-point elevation Freezing-point depression Osmotic pressure Copyright Cengage Learning. All rights reserved 38

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling-Point Elevation Nonvolatile solute elevates the boiling point of the solvent. ΔT = K b m solute ΔT = boiling-point elevation K b = molal boiling-point elevation constant m solute = molality of solute Copyright Cengage Learning. All rights reserved 39

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Liquid/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 40

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 41

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Solution/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 42

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing-Point Depression When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. ΔT = K f m solute ΔT K f = freezing-point depression = molal freezing-point depression constant m solute = molality of solute Copyright Cengage Learning. All rights reserved 43

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Solid/Liquid Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 44

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 45

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Solid/Solution Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 46

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression EXERCISE! A solution was prepared by dissolving 25.00 g of glucose in 200.0 g water. The molar mass of glucose is 180.16 g/mol. What is the boiling point of the resulting solution (in C)? Glucose is a molecular solid that is present as individual molecules in solution. 100.35 C Copyright Cengage Learning. All rights reserved 48

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression EXERCISE! You take 20.0 g of a sucrose (C 12 H 22 O 11 ) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be -0.426 C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture. 72.8% sucrose and 27.2% sodium chloride; mole fraction of the sucrose is 0.313 Copyright Cengage Learning. All rights reserved 49

Section 11.5 Boiling-Point Elevation and Freezing-Point Depression EXERCISE! A plant cell has a natural concentration of 0.25 m. You immerse it in an aqueous solution with a freezing point of 0.246 C. Will the cell explode, shrivel, or do nothing? Copyright Cengage Learning. All rights reserved 50

Section 11.6 Osmotic Pressure Osmosis flow of solvent into the solution through a semipermeable membrane. = MRT = osmotic pressure (atm) M = molarity of the solution R = gas law constant T = temperature (Kelvin) Copyright Cengage Learning. All rights reserved 51

Section 11.6 Osmotic Pressure Osmosis To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 53

Section 11.6 Osmotic Pressure EXERCISE! When 33.4 mg of a compound is dissolved in 10.0 ml of water at 25 C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound. 111 g/mol Copyright Cengage Learning. All rights reserved 55

Section 11.7 Colligative Properties of Electrolyte Solutions van t Hoff Factor, i The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as: i = moles of particles in solution moles of solute dissolved Copyright Cengage Learning. All rights reserved 56

Section 11.7 Colligative Properties of Electrolyte Solutions Ion Pairing At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle. Copyright Cengage Learning. All rights reserved 57

Section 11.7 Colligative Properties of Electrolyte Solutions Examples The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur). NaCl i = 2 KNO 3 i = 2 Na 3 PO 4 i = 4 Copyright Cengage Learning. All rights reserved 58

Section 11.7 Colligative Properties of Electrolyte Solutions Ion Pairing Ion pairing is most important in concentrated solutions. As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs. Ion pairing occurs to some extent in all electrolyte solutions. Ion pairing is most important for highly charged ions. Copyright Cengage Learning. All rights reserved 59

Section 11.8 Colloids A suspension of tiny particles in some medium. Tyndall effect scattering of light by particles. Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm. Copyright Cengage Learning. All rights reserved 61