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Solution Properties

Solution Properties. 11.1 Solution Composition 11.2 The Energies of Solution Formation 11.3 Factors Affecting Solubility 11.4 The Vapor Pressures of Solutions 11.5 Boiling-Point Elevation and Freezing-Point Depression 11.6 Osmotic Pressure

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Solution Properties

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  1. Solution Properties 11.1 Solution Composition 11.2 The Energies of Solution Formation 11.3 Factors Affecting Solubility 11.4 The Vapor Pressures of Solutions 11.5 Boiling-Point Elevation and Freezing-Point Depression 11.6 Osmotic Pressure 11.7 Colligative Properties of Electrolyte Solutions 11.8 Colloids

  2. Various Types of Solutions

  3. Solution Composition

  4. Molarity

  5. Exercise #1 You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity. 8.00 M

  6. Exercise #2 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

  7. Exercise #3 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

  8. Mass Percent

  9. Exercise #4 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%

  10. Mole Fraction

  11. Exercise #5 A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.) 0.0145

  12. Molality

  13. Exercise #6 A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 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

  14. Formation of a Liquid Solution • Separating the solute into its individual components (expanding the solute). • Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). • Allowing the solute and solvent to interact to form the solution.

  15. Steps in the Dissolving Process

  16. 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.

  17. Enthalpy (Heat) of Solution • Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔHsoln = ΔH1 + ΔH2 + ΔH3 • ΔHsoln may have a positive sign (energy absorbed) or a negative sign (energy released).

  18. Enthalpy (Heat) of Solution

  19. 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.

  20. The Energy Terms for Various Types of Solutes and Solvents

  21. 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”.

  22. Factors Affecting Solubility • Structural Effects: • Polarity – “like dissolves like” • Pressure Effects: • Henry’s law – for solubility of gases • Temperature Effects: • Affecting aqueous solutions

  23. Pressure Effects • 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 partial pressure of gas above the solution.

  24. A Gaseous Solute

  25. 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.

  26. The Solubilities of Several Solids as a Function of Temperature

  27. The Solubilities of Several Gases in Water

  28. Ideal Solution: One that obeys Raoult’s Law

  29. Ideal Solutions Consisting of Two Volatile Liquids • Two volatile Liquids form ideal solution if: • they are structurally very similar, and • molecular interactions between nonidentical molecules were relatively similar to identical molecules. • The vapor of each liquid obeys Raoult’s Law: PA = XAPoA; PB = XBPoB PT = PA + PB = XAPoA +XBPoB (X : mole fraction; Po : vapor pressure of pure liquid)

  30. Summary of the Behavior of Various Types of Solutions of Two Volatile Liquids

  31. Vapor Pressure for a Solution of Two Volatile Liquids

  32. Laboratory Fractional Distillation Apparatus

  33. Fractional Distillation Towers in Oil Refinaries

  34. Refined Crude Oil Mixtures

  35. 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? • Hexane (C6H14) and chloroform (CHCl3) • Ethyl alcohol (C2H5OH) and water • Hexane (C6H14) and octane (C8H18)

  36. Exercise #7 • A solution of benzene (C6H6) and toluene (C7H8) contains 50.0% benzene by mass. The vapor pressures of benzene and pure toluene at 25oC are 94.2 torr and 28.4 torr, respectively. Assuming ideal behavior, calculate the following: (a) The mole fractions of benzene and toluene; (b) The vapor pressure of each component in the mixture, and the total vapor pressure above the solution. (c) The composition of the vapor in mole percent.

  37. Exercise #8 • A solution composed of 24.3 g acetone (CH3COCH3) and 39.5 g of carbon disilfide (CS2) has a measured vapor pressure of 645 torr at 35oC. (a) Is the solution ideal or nonideal? (b) If not, does it deviate positively or negatively from Raoult’s law? (c) What can you say about the relative strength of carbon disulfide-acetone interactions compared to the acetone-acetone and carbon disulfide-carbon disulfide interaction? (Vapor pressures at 35oC of pure acetone and pure carbon disulfide are 332 torr and 515 torr, respectively.)

  38. An Aqueous Solution and Pure Water in a Closed Environment

  39. Liquid/Vapor Equilibrium

  40. Vapor Pressure Lowering: Addition of a Solute

  41. Vapor Pressures of Solutions of Nonvolatile Solutes • Nonvolatile solute lowers the vapor pressure of solvent. • Raoult’s Law: Psoln= vapor pressure of solution solv= mole fraction of solvent = vapor pressure of pure solvent

  42. Colligative Properties • Lowering of solvent vapor pressure • Freezing-point depression • Boiling-point elevation • Osmotic pressure • Colligative properties depend only on the number, not on the identity, of the solute particles in an ideal solution.

  43. Lowering of Solvent Vapor Pressure • The presence of nonvolatile solute particles lowers the number of solvent molecules in the vapor that is in equilibrium with the solution. • The solvent vapor pressure is lowered; • Assuming ideal behavior, the lowering of vapor pressure is proportional to the mole fraction of solute: DP = Xsolute.Posolvent (for nonelectrolytes) = iXsolute.Posolvent (for electrolytes) (iis the van’t Hoff’s factor, which approximately relates to the number of ions per formula unit of the compound)

  44. Changes in Boiling Point and Freezing Point of Water

  45. 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 = Kfmsolute(for nonelectrolytes) = iKfmsolute(for electrolytes) ΔT = freezing-point depression Kf= freezing-point depression constant msolute= molality of solute

  46. Freezing Point Depression: Solid/Liquid Equilibrium

  47. Freezing Point Depression: Addition of a Solute

  48. Freezing Point Depression: Solid/Solution Equilibrium

  49. Boiling-Point Elevation • Nonvolatile solute elevates the boiling point of the solvent. • ΔT = Kbmsolute ΔT = boiling-point elevation Kb= boiling-point elevation constant msolute = molality of solute

  50. Boiling Point Elevation: Liquid/Vapor Equilibrium

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