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Chapter 12: Solutions and Their Properties

Chapter 12: Solutions and Their Properties. Renee Y. Becker Valencia Community College. Introduction 1. A mixture is any intimate combination of two or more pure substances 2. Can be classified as heterogeneous or homogeneous Heterogeneous

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Chapter 12: Solutions and Their Properties

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  1. Chapter 12: Solutions and Their Properties Renee Y. Becker Valencia Community College

  2. Introduction 1. A mixture is any intimate combination of two or more pure substances 2. Can be classified as heterogeneous or homogeneous Heterogeneous -The mixing of components is visually nonuniform and have regions of different composition Homogenous -Mixing is uniform, same composition throughout -Can be classified according to the size of their particles as either solutions or colloids

  3. Solutions Solution 1. Homogeneous mixtures 2. Contain particles with diameters in the range of 0.1–2 nm 3. Transparent but may be colored 4. Do not separate on standing Colloids 1. Milk & fog 2. Diameters 2-500 nm 3. Do not separate on standing

  4. Types of Solutions

  5. Solution Formation Solute • Dissolved substance, or smaller quantity substance Solvent • Liquid dissolved in, larger quantity substance Saturated solution • Contains the maximum amount of solute that will dissolve in a given solvent.

  6. Solution Formation Unsaturated Contains less solute than a solvent has the capacity to dissolve. Supersaturated Contains more solute than would be present in a saturated solution. Crystallization The process in which dissolved solute comes out of the solution and forms crystals.

  7. Energy Changes and the Solution Process Three Types of interactions 1.      Solvent-solvent 2.      Solvent-solute 3.      Solute-solute “Like dissolves like” solutions will form when three types of interactions are similar in kind and magnitude

  8. Energy Changes and the Solution Process Example NaCl and water: Ionic solid NaCl dissolve in polar solvents like water because the strong ion-dipole attractions between Na+ and Cl- ions and polar water molecules are similar in magnitude to the strong dipole-dipole attractions between water molecules and to the strong ion-ion attractions between Na+ and Cl- ions Example Oil and water Oil does not dissolve in water because the two liquids have different kinds of intermolecular forces. Oil is not polar or an ionic solvent

  9. Energy Changes and the Solution Process NaCl in H2O 1. Ions that are less tightly held because of their position at a corner or an edge of the crystal are exposed to water molecules 2. Water molecules will collide with the NaCl until an ion breaks free 3. More water molecules then cluster around the ion, stabilizing it by ion-dipole attractions 4. The water molecules attack the weak part of the crystal until it is dissolved 5. Ions in solution are said to be solvated they are surrounded and stabilized by an ordered shell of solvent molecules

  10. Energy Changes and the Solution Process G, Free energy change 1. If G is negative the process is spontaneous, and the substance is dissolved 2. If G is positive the process is non-spontaneous, the substance is not dissolved 3. G = H -TS H, enthalpy, heat flow in or out of the system, Hsoln heat of solution S, entropy, disorder, Ssoln entropy of solution

  11. Energy Changes and the Solution Process

  12. Energy Changes and the Solution Process Ssoln Entropy of Solution Usually a positive number because when you dissolve something you are increasing disorder Hsoln Heat of Solution 1. Harder to predict because it could be exothermic (-Hsoln) or endothermic (+Hsoln) 2. The value of the heat of solution for a substance results from an interplay of the three kinds of interactions

  13. Energy Changes and the Solution Process • Solvent-solvent interactions: Energy is required (endothermic) to overcome intermolecular forces between solvent molecules because the molecules must be separated and pushed apart to make room for solute particles • Solute-solute interactions: Energy is required (endothermic) to overcome interactions holding solute particles together in a crystal. For an ionic solid, this is the lattice energy. Substances with higher lattice energies therefore tend to be less soluble than substances with lower lattice energies.

  14. Energy Changes and the Solution Process 3) Solvent-solute interactions: Energy is released (exothermic) when solvent molecules cluster around solute particles and solvate them. For ionic substances in water, the amount of hydration energy released is generally greater for smaller cations than for larger ones because water molecules can approach the positive nuclei of smaller ions more closely and thus bind more tightly. Hydration energy generally increases as the charge on the ion increases.

  15. Energy Changes and the Solution Process Exothermic -Hsoln The solute–solvent interactions are stronger than solute–solute or solvent–solvent. Favorable process Exothermic rxn.

  16. Energy Changes and the Solution Process Endothermic +Hsoln The solute–solvent interactions are weaker than solute–solute or solvent–solvent. Unfavorable process. Endothermic rxn

  17. Energy Changes and the Solution Process • Hydration • The attraction of ions for water molecules • Hydration Energy • The energy associated with the attraction between ions and water molecules • Lattice energy • The energy holding ions together in a crystal lattice

  18. Example 1 Arrange the following in order of their expected increasing solubility in water: Br2, KBr, C7H8

  19. Na+ Cs+ Li+ 4. Rb+ Mg2+ Na+ Li+ Example 2 Which would you expect to have the larger (more negative) hydration energy?

  20. Units of Concentration Molarity (M) M = mole of solute / Liter of solution Molality (m) m = moles of solute/mass of solvent (kg) Mole Fraction (x) X = mole of component / total moles

  21. Units of Concentration Mass Percent (mass %) Mass % = (mass of component / total mass of sol’n) x 100% Parts per million, ppm = (mass of component / total mass of solution) x 106 Parts per billion, ppb = (mass of component / total mass of solution) x 109

  22. Example 3 What is the mass % concentration of a saline sol’n prepared by dissolving 1.00 mol of NaCl in 1.00 L of water? DensityH2O=1.00 g/mL MMNaCl = 58.443 g/mol

  23. Example 4 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 % • Assume 1 L to make easier, 1000 mL

  24. Example 5 What is the molality of a solution prepared by dissolving 0.385 g of cholesterol, C27H46O in 40.0 g of chloroform, CHCl3? What is the mole fraction of cholesterol in the solution?

  25. Example 6 What mass in grams of a 0.500 m solution of sodium acetate, CH3CO2Na, in water would you use to obtain 0.150 mol of sodium acetate?

  26. Example 7 The density at 20°C of a 0.258 m solution of glucose in water is 1.0173 g/mL and the molar mass of glucose is 180.2 g/mol. What is the molarity of the solution? • Assume 1 kg

  27. Example 8 The density at 20°C of a 0.500 M solution of acetic acid in water is 1.0042 g/mL. What is the concentration of this solution in molality? The molar mass of acetic acid, CH3CO2H, is 60.05 g/mol. • Assume 1 L

  28. Some Factors Affecting Solubility Solubility The amount of solute per unit of solvent needed to form a saturated solution Miscible Mutually soluble in all proportions Effect of Temperature on Solubility 1. Most solid substances become more soluble as temperature rises 2. Most gases become less soluble as temperature rises

  29. Some Factors Affecting Solubility Effect of Pressure on Solubility 1. No effect on liquids or solids 2. The solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas over the solution, @ 25°C Henry’s Law solubility = k x P k = constant characteristic of specific gas, mol/Latm P = partial pressure of the gas over the sol’n

  30. Some Factors Affecting Solubility a) Equal numbers of gas molecules escaping liquid and returning to liquid b)   Increase pressure, increase # of gas molecules returning to liquid, solubility increases c) A new equilibrium is reached, where the #’s of escaping = # of returning

  31. Example 9 Which of the following will become less soluble in water as the temperature is increased? • NaOH(s) • CO2(g)

  32. Example 10 The solubility of CO2 in water is 3.2 x 10-2 M @ 25°C and 1 atm pressure. What is the Henry’s-Law constant for CO2 in mol/L atm? solubility = k x P

  33. Physical Behavior of Solutions: Colligative Properties Colligative properties Properties that depend on the amount of a dissolved solute but not its chemical identity There are four main colligative properties: 1. Vapor pressure lowering 2. Freezing point depression 3. Boiling point elevation 4. Osmotic pressure

  34. Physical Behavior of Solutions: Colligative Properties In comparing the properties of a pure solvent with those of a solution… 1. Vapor pressure of sol’n is lower 2. Boiling point of sol’n is higher 3. Freezing point of sol’n is lower 4. Osmosis, the migration of solvent molecules through a semipermeable membrane, occurs when solvent and solution are separated by the membrane

  35. Vapor-pressure Lowering of Solutions: Raoult’s Law 1.   A liquid in a closed container is in equilibrium with its vapor and that the amount of pressure exerted by the vapor is called the vapor pressure. 2.      When you compare the vapor pressure of a pure solvent with that of a solution at the same temperature the two values are different. 3.      If the solute is nonvolatile and has no appreciable vapor pressure of its own (solid dissolved) the vapor pressure of the solution is always lower that that of the pure solvent. 4. If the solute is volatile and has a significant vapor pressure (2 liquids) the vapor pressure of the mixture is intermediate between the vapor pressures of the two pure liquids.

  36. Solutions with a Nonvolatile Solute When solute molecules displace solvent molecules at the surface, the vapor pressure drops since fewer gas molecules are needed to equalize the escape rate and capture rates at the liquid surface.

  37. Solutions with a Nonvolatile Solute!!! Raoult’s Law Psoln = Psolv· Xsolv Psoln = vapor pressure of the solution Psolv = vapor pressure of the pure solvent Xsolv = mole fraction of the solvent in the solution

  38. Raoult’s Law applies to only Ideal solutions 1.    Law works best when solute concentrations are low and when solute and solvent particles have similar intermolecular forces. 2.      If intermolecular forces between solute particles and solvent molecules are weaker than solvent molecules alone, solvent molecules are less tightly held, vapor pressure is higher than Raoult predicts 3.      If intermolecular forces between solute and solvent are stronger than solvent alone, solvent molecules are more tightly held and the vapor pressure is lower than predicted 4.      No Van’t Hoff factor!!!

  39. Example 11 What is the vapor pressure (in mm Hg) of a solution prepared by dissolving 5.00 g of benzoic acid (C7H6O2) in 100.00 g of ethyl alcohol (C2H6O) at 35°C? The vapor pressure of the pure ethyl alcohol at 35°C is 100.5 mm Hg

  40. Solutions with a Nonvolatile Solute Close-up view of part of the vapor pressure curve for a pure solvent and a solution of a nonvolatile solute. Which curve represents the pure solvent, and which the solution? Why?

  41. The lower vapor pressure of a sol’n relative to that of a pure solvent is due to the difference in their entropies of vaporization, Svap. Because the entropy of the solvent in a sol’n is higher to begin with, Svap is smaller for the sol’n than for the pure solvent. As a result vaporization of the solvent from the sol’n is less favored (less negative Gvap), and the vapor pressure of the solution is lower.

  42. Solutions with a Volatile Solute!! Ptotal = PA + PB Ptotal = (P°A · XA) + (P°B · XB) P°A = vapor pressure of pure A XA = mole fraction of A P°B = vapor pressure of pure B XB = mole fraction of B Ptotal should be intermediate to A & B

  43. Close-up view of part of the vapor pressure curves for two pure liquids and a mixture of the two. Which curves represent the mixture? • Red • Green • blue

  44. Example 12 What is the vapor pressure ( in mm Hg) of a sol’n prepared by dissolving 25.0 g of ethyl alcohol (C2H5OH) in 100.0 g of water at 25°C? The vapor pressure of pure water is 23.8 mm Hg and the vapor pressure of ethyl alcohol is 61.2 mm Hg at 25°C

  45. Example 13 The following phase diagram shows part of the vapor pressure curves for a pure liquid (green curve) and a solution (red curve) of the first liquid with a second volatile liquid (not shown) a)      Is the boiling point of the second pure liquid higher or lower than that of the first liquid? b) On the diagram where is the approximate position of the second pure liquid?

  46. Boiling Point Elevation and Freezing Point Depression of Solutions

  47. Boiling Point Elevation and Freezing Point Depression of Solutions 1. Red line is pure solvent 2. Green line solution of nonvolatile solute 3. Vapor pressure of sol’n is lower 4. Temp at which vapor pressure = 1 atm for sol’n is higher 5. Boiling point of sol’n is higher by Tb 6. Liquid/vapor phase transition line is lower for sol’n 7. Triple point temp is lower for sol’n 8. Solid/liquid phase transition has shifted to a lower temp. 9. The freezing point of the sol’n is lower by Tf

  48. Boiling Point Elevation and Freezing Point Depression of Solutions • Tb = Kb · m • Tf = Kf · m Kb = molal boiling-point elevation constant Kf = molal freezing-point depression constant m = molality NO Van’t Hoff Factor!!

  49. Boiling Point Elevation and Freezing Point Depression of Solutions The higher boiling point of a solution relative to that of a pure solvent is due to a difference in their entropies of vaporization, Svap. Because the solvent in a solution has a higher entropy to begin with, Svap is smaller for the solution than for the pure solvent. As a result, the boiling point of the solution Tb is higher than that of the pure solvent.

  50. Boiling Point Elevation and Freezing Point Depression of Solutions The lower freezing point of a solution relative to that of a pure solvent is due to a difference in their entropies of fusion, Sfusion. Because the solvent in a solution has a higher entropy level to begin with, Sfusion is larger for the solution than for the pure solvent. As a result the freezing point of the solution Tf is lower than that of the pure solvent.

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