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CHAPTER 11

CHAPTER 11. SOLUTIONS AND THEIR PROPERTIES. Solution Formation 01. Saturated: Contains the maximum amount of solute that will dissolve in a given solvent. Unsaturated: Contains less solute than a solvent has the capacity to dissolve.

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CHAPTER 11

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  1. CHAPTER 11 SOLUTIONS AND THEIR PROPERTIES Chapter 11

  2. Solution Formation 01 • Saturated: Contains the maximum amount of solute that will dissolve in a given solvent. • 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. • Saturated: Contains the maximum amount of solute that will dissolve in a given solvent. • 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. Chapter 11

  3. Solution Formation 02 Chapter 11

  4. Solution Formation 02 Chapter 11

  5. Solution Formation 03 Chapter 11

  6. Solution Formation 04 • Exothermic ∆Hsoln: • The solute–solvent interactions are stronger than solute–solute or solvent–solvent. • Favorable process. Chapter 11

  7. Solution Formation 05 • Endothermic ∆Hsoln: • The solute–solvent interactions are weaker than solute–solute or solvent–solvent. • Unfavorable process. Chapter 11

  8. Solution Formation 06 • Solubility: A measure of how much solute will dissolve in a solvent at a specific temperature. • Miscible: Two (or more) liquids that are completely soluble in each other in all proportions. • Solvation: The process in which an ion or a molecule is surrounded by solvent molecules arranged in a specific manner. Chapter 11

  9. Solution Formation 07 • Predict the relative solubilities in the following cases: (a) Br2 in benzene (C6H6) and in water, (b) KCl in carbon tetrachloride and in liquid ammonia, (c) urea (NH2)2CO in carbon disulfide and in water. • Is iodine (I2) more soluble in water or in carbon disulfide (CS2)? • Which would have the largest (most negative) hydration energy and which should have the smallest? Al3+, Mg2+, Na+ Chapter 11

  10. Concentration Units 01 Chapter 11

  11. mass of solute % by mass of solute = 100 % ´ mass of solution mass of solution = mass of solute + mass of solvent Concentration Units 01 • Concentration:The amount of solute present in a given amount of solution. • Percent by Mass (weight percent):The ratio of the mass of a solute to the mass of a solution, multiplied by 100%. Chapter 11

  12. Concentration Units 02 • Parts per Million: • Parts per million (ppm) = = % mass x 104 • One ppm gives 1 gram of solute per 1,000,000 g or one mg per kg of solution. For dilute aqueous solutions this is about 1 mg per liter of solution. Chapter 11

  13. Concentration Units 03 • A sample of 0.892 g of potassium chloride (KCl) is dissolved in 54.6 g of water. What is the percent by mass of KCl in this solution? • An aqueous solution is 5.50% H2SO4. How many moles of sulfuric acid (MM = 98.08 g/mol) are dissolved in 250.0 g of the solution? Chapter 11

  14. Concentration Units 04 • Mole Fraction (X): • Molarity (M): • Molality (m): Chapter 11

  15. Units of Concentration Mass Percent (mass %) Mass % = (mass of component/total mass of SOLUTION) 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

  16. Concentration Units 06 • Molality from Mass: Calculate the molality of a sulfuric acid solution containing 24.4 g of sulfuric acid in 198 g of water. The molar mass of sulfuric acid is 98.08 g. • Molality from Molarity: Calculate the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/ml. Chapter 11

  17. Concentration Units 07 • Molality from Mass %: Assuming that seawater is a 3.50 mass % aqueous solution of NaCl, what is the molality of seawater? • Molarity from Molality: 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. What is the molarity of the solution? Chapter 11

  18. Concentration Units 08 • Mole Fraction from Molality: An aqueous solution is 0.258 m in glucose (MM = 180.2 g/mol). What is the mole fraction of the glucose? • Mass from Molality: What mass (in grams) of a 0.500 m aqueous solution of urea [(NH2)2CO, MM = 60.1 g/mol] would you use to obtain 0.150 mole of urea? Chapter 11

  19. Effect of Temperatureon Solubility 01 • Solids: Chapter 11

  20. Effect of Temperature on Solubility 02 • Gases: Chapter 11

  21. The Effect of Pressure on theSolubility of Gases 01 • Henry’s Law: • The solubility of a gas is proportional to the pressure of the gas over the solution. c  P c = k·P Chapter 11

  22. The Effect of Pressure on theSolubility of Gases 02 Flash Animation - Click to Continue Chapter 11

  23. The Effect of Pressure on theSolubility of Gases 02 • Calculate the molar concentration of O2 in water at 25°C for a partial pressure of 0.22 atm. The Henry’s law constant for O2 is 3.5 x 10–4 mol/(L·atm). • The solubility of CO2 in water is 3.2 x 10–2 M at 25°C and 1 atm pressure. What is the Henry’s law constant for CO2 in mol/(L·atm)? Chapter 11

  24. Colligative Propertiesof Nonvolatile Solutes 01 • Colligative Properties: Depend only on the number of solute particles in solution. These affect properties of the solvent. • There are four main colligative properties: • Vapor pressure lowering • Freezing point depression • Boiling point elevation • Osmotic pressure Chapter 11

  25. Colligative Propertiesof Nonvolatile Solutes 02 • 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. Chapter 11

  26. Colligative Propertiesof Nonvolatile Solutes 03 • Raoult’s Law:Psoln = P°solvXsolv • For a single solute solution, Xsolv= 1 – Xsolute, • We can obtain an expression for the change in vapor pressure of the solvent (the vapor pressure lowering). Psoln = P°solv – Psoln = P°solv – Xsolv P°solv = P°solv – (1 – Xsolute) P°solv ∆P = XsoluteP°solv Where superscript o is for pure substance. Chapter 11

  27. Van’t Hoff Factor • For incompletely dissociating ionic solids • Van’t Hoff Factor i = moles of particles in solution • moles of solute dissolved Chapter 11

  28. Colligative Propertiesof Nonvolatile Solutes 05 • The vapor pressure of a glucose (C6H12O6) solution is 17.01 mm Hg at 20°C, while that of pure water is 17.25 mm Hg at the same temperature. Estimate the molality of the solution. • How many grams of NaBr must be added to 250 g of water to lower the vapor pressure by 1.30 mm Hg at 40°C? The vapor pressure of water at 40°C is 55.3 mm Hg. Chapter 11

  29. Colligative Properties of a Mixture of Two Volatile Liquids 01 • What happens if both components are volatile(have measurable vapor pressures)? • The vapor pressure has a value intermediate between the vapor pressures of the two liquids. PT= PA + PB = XAP°A + XBP°B =XAP°A + (1 – XA)P°B PT = P°B + (P°A– P°B)XA Chapter 11

  30. Boiling-Point Elevation and Freezing-Point Depression 01 • Boiling-Point Elevation (∆Tb): The boiling point of the solution (Tb) minus the boiling point of the pure solvent (T°b):∆Tb = Tb – T°b ∆Tb is proportional to concentration:∆Tb = Kbm Kb= molal boiling-point elevation constant. Also for incompletely dissociating ionic solids ∆Tb = Kbm i Chapter 11

  31. Boiling-Point Elevation and Freezing-Point Depression 02 • Freezing-Point Depression (∆Tf):The freezing point of the pure solvent (T°f) minus the freezing point of the solution (Tf). ∆Tf = T°f – Tf ∆Tf is proportional to concentration:∆Tf = KfmKf= molal freezing-point depression constant. ∆Tb = Kbm i Chapter 11

  32. Boiling-Point Elevation and Freezing-Point Depression 04 Chapter 11

  33. Boiling-Point Elevation and Freezing-Point Depression 06 • van’t Hoff Factor, i: This factor equals the number of ions produced from each molecule of a compound upon dissolving. i = 1 for CH3OH i = 3 for CaCl2 i = 2 for NaCl i = 5 for Ca3(PO4)2 • For compounds that dissociate on dissolving, use: ∆Tb = iKbm∆Tf = iKf m ∆P = ix2P°1 Chapter 11

  34. Boiling-Point Elevation and Freezing-Point Depression 07 • How many grams of ethylene glycol antifreeze, CH2(OH)CH2(OH), must you dissolve in one liter of water to get a freezing point of –20.0°C. The molar mass of ethylene glycol is 62.01 g. For water, Kf = 1.86 (°C·kg)/mol. What will be the boiling point? Chapter 11

  35. Boiling-Point Elevation and Freezing-Point Depression 08 • What is the molality of an aqueous solution of KBr whose freezing point is –2.95°C? Kf for water is 1.86 (°C·kg)/mol. • What is the freezing point (in °C) of a solution prepared by dissolving 7.40 g of K2SO4 in 110 g of water? The value of Kf for water is 1.86 (°C·kg)/mol. Chapter 11

  36. Osmosis and Osmotic Pressure 01 Chapter 11

  37. Osmosis and Osmotic Pressure 01 • Osmosis: The selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. • Osmotic pressure (π or ∏): The pressure required to stop osmosis. π = iMRT R = 0.08206 (Latm)/(molK) Chapter 11

  38. Osmosis and Osmotic Pressure 02 Chapter 11

  39. Osmosis and Osmotic Pressure 03 Chapter 11

  40. Osmosis and Osmotic Pressure 04 • Isotonic: Solutions have equal concentration of solute, and so equal osmotic pressure. • Hypertonic: Solution with higher concentration of solute. • Hypotonic: Solution with lower concentration of solute. Chapter 11

  41. Osmosis and Osmotic Pressure 05 • The average osmotic pressure of seawater is about 30.0 atm at 25°C. Calculate the molar concentration of an aqueous solution of urea [(NH2)2CO] that is isotonic with seawater. • What is the osmotic pressure (in atm) of a 0.884 M sucrose solution at 16°C? Chapter 11

  42. Uses of Colligative Properties 01 • Desalination: Chapter 11

  43. Uses of Colligative Properties 02 • A 7.85 g sample of a compound with the empirical formula C5H4 is dissolved in 301 g of benzene. The freezing point of the solution is 1.05°C below that of pure benzene. What are the molar mass and molecular formula of this compound? Chapter 11

  44. Uses of Colligative Properties 03 • A 202 ml benzene solution containing 2.47 g of an organic polymer has an osmotic pressure of 8.63 mm Hg at 21°C. Calculate the molar mass of the polymer. • What is the molar mass of sucrose if a solution of 0.822 g of sucrose in 300.0 mL of water has an osmotic pressure of 149 mm Hg at 298 K? Chapter 11

  45. Uses of Colligative Properties 06 • Two miscible liquids, A and B, have vapor pressures of 250 mm Hg and 450 mm Hg, respectively. They were mixed in equal molar amounts. What is the total vapor pressure of the mixture and what are their mole fractions in the vapor phase? Chapter 11

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