1 / 37

Chapter 11c: Solutions and Their Properties

Chapter 11c: Solutions and Their Properties. 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

alissa
Télécharger la présentation

Chapter 11c: Solutions and Their Properties

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 11c: Solutions and Their Properties

  2. 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

  3. 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

  4. 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

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

  6. 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

  7. Physical Behavior of Solutions: Colligative Properties o • H2O b.p. 100.0o C f.p. 0.0o C • 1.00 m NaCl b.p. 101.0o C f.p. -3.7o C o Colligative properties: Properties that depend on the amount of a dissolved solute but not on its chemical identity There are four main colligative properties: 1. Vapor pressure lowering 2. Freezing point depression 3. Boiling point elevation 4. Osmosis:

  8. The vapor pressure of a solution is different from the vapor of the pure solvent. Two different cases: 1. solute is non-volatile solute has no vapor pressure of its own example: dissolving a solid vapor pressure of the solution is always lower than that of the pure solvent 2. solute is volatile solute has its own vapor pressure example: mixing 2 liquids vapor pressure of the mixture is intermediate between the vapor pressures of the two pure liquids

  9. Vapor-pressure Lowering of Solutions: Raoult’s Law • Solutions with a Nonvolatile Solute If the solute is nonvolatile and has no appreciable vapor pressure of its own (solid dissolved) the vapor pressure of the solution is always lowerthat that of the pure solvent.

  10. 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 Vapor pressure lowering is a colligative property (only dependent on amount of solute and not on its chemical identity!) For ionic substances calculate the total moles of solute particles, 1 mol NaCl will result in 1 mol Na+ and 1 mol Cl- = 2 moles of particles 1 mol Na2SO4 will give 3 moles of particles

  11. Raoult’s Law applies to only Ideal solutions 1. Law works best when solute concentrations are low an d when solute and solvent particles have similar intermolecular forces. 2. Further complication is that at higher concentrations ionic compounds are not 100% dissociated. Example 1 mol NaCl is only 90% dissociated 10% is undissociated resulting in less particles in solution than expected

  12. Example 9 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

  13. Psoln = Psolv· Xsolv MM C7H6O2 = 122.12 g/mol Psolv = 100.5 mm Hg MM C2H6O = 46.07 g/mol 1mol a) 5 g C7H6O2 x ---------- = 0.0409 mol 122.12g b) 100 g C2H6O x 1 mol/ 46.07 g = 2.17 mol c) Xsolv = 2.17 mol / (0.0409 + 2.17 mol) = 0.982 Psoln = Psolv· Xsolv = 100.5 mm Hg · 0.982 = 98.7 mm Hg

  14. 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?

  15. Reason for vapor pressure lowering DG = DHvap - T DS DHvap = positive, disfavored DS = positive, favored DHvap is (nearly) the same for a pure solvent and a solvent in a solution DS is different solvent in a solution has more disorder than pure solvent entropy of a solution is higher than the pure solvent entropy of the vapor in both cases the same Entropy increase for vaporization from a solution is smaller than vaporization from a pure solvent Less entropy increase means less favored

  16. Solutions with a Volatile Solute!! For a mixture of 2 volatile liquids A and B the overall vapor pressure is the sum of the vapor pressure of the 2 components (Dalton’s law) Ptotal = PA + PB The vapor pressure for each component is calculated by Raoult’s law: vapor pressure is equal to the mole fraction of A times the vapor pressure of pure A Ptotal = PA + PB = (P0A · XA) + (P0B · 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

  17. Solutions with a Volatile Solute Close-up view of part of the vapor pressure curves for two pure liquids and a mixture of the two. Which curves represent the pure liquids, and which the mixture?

  18. Ptotal should be intermediate to A & B Raoult’s law applies only to ideal solutions Most real solutions show deviations

  19. Example 10 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

  20. Example 10 P°H2O = 23.8 mm Hg P°C2H5OH = 61.2 mm Hg XH2O = mole fraction of A 25 g C2H5OH x 1 mol / 46.07 g = 0.543 mol C2H5OH 100.0 g H2O x 1 mol/ 18 g = 5.56 mol H2O XH2O = 5.56 /(5.56 + 0.543) = 0.91 XC2H5OH = mole fraction of B XC2H5OH = 0.543 / (0.543 + 5.56) = 0.09 Ptot = (23.8 x 0.91) + (61.2 x 0.09) = 27.2 mm Hg

  21. Boiling Point Elevation and Freezing Point Depression of Solutions A solution has a lower vapor pressure than the pure liquid. To reach the atmospheric pressure (boiling point) the temperature must be higher. • Tb = Kb · m Boiling point elevation • Tf = Kf · m Freezing point depression Kb = molal boiling-point elevation constant Kf = molal freezing-point depression constant m = molality

  22. Example 11 What is the normal boiling point in °C of a solution prepared by dissolving 1.50 g of aspirin (C9H8O4) in 75.00 g of chloroform (CHCl3)? The normal boiling point of chloroform is 61.7 °C and Kb of chloroform is 3.63 °C kg/mol

  23. Example 11 Tb = Kb · m m = mole solute / kg solvent MM C9H8O4 = 180.16 g/mol 1.50 g C9H8O4 x 1 mol / 180.16 g = 0.00833 mol C9H8O4 75.00 g CHCl3 = 0.07500 kg CHCl3 m =0 .00833 mol C9H8O4 / 0.07500 kg CHCl3 = 0.111 m Tb = 3.63 °C kg/mol · 0.111 mol/kg = 0.403 °C Boiling point = 0.403 °C + 61.7 °C = 62.1 °C

  24. Osmosis and Osmotic Pressure Membranes are semipermeable materials They allow water and other small molecules to pass through, but they block the passage of larger molecules or ions.

  25. All living cells contain membranes and osmosis is important in biological systems Osmosis provides the primary means by which water is transported into and out of cells Osmosis is responsible for the ability of plant roots to suck up water from the soil

  26. Osmosis

  27. Thermodynamic explanation Every system wants to balance out the concentration One side pure solvent Other side solution ordered system The system tries to get into a more disordered more randomness state The entropy will increase Osmosis is similar to diffusion

  28. Osmosis Pressure 1. The amount of pressure necessary to achieve equilibrium 2.  = MRT  = osmotic pressure M = molarity R = gas constant, .08206 L atm/K mol T = temperature in Kelvin

  29. Isotonic sodium chloride solution The total concentration of dissolved particles inside red blood cells is 0.30 M. What is the osmotic pressure at body temp (310 k) ?

  30. Isotonic sodium chloride solution The total concentration of dissolved particles inside red blood cells is 0.30 M. What is the osmotic pressure at body temp (310 k) ? • = MRT • = 0.30 mol/L x 0.08206 L atm/K mol x 310 K • = 7.63 atm

  31. All medical infusions must have the same osmotic pressure Otherwise the blood cells would burst! Therefore isotonic NaCl solutions are injected Q: What is the mass% of an isotonic NaCl solution? Since the molarity in blood cells is 0.3 M, we need 0.15 M NaCl (0.15 M Na+ and 0.15 M Cl-) Na = 23.0 amu Cl = 35.5 amu NaCl = 58.5 amu 0.15 M NaCl = 58.5 x 0.15 = 9 g/L ; 0.9 mass%

  32. Example 12 What osmotic pressure in atm would you expect for a solution of 0.125 M C6H12O6 that is separated from pure water by a semipermeable membrane at 310 K?  = MRT = (0.125 mol/L)(.08206 L atm/K mol)(310 K) = 3.18 atm

  33. Example 13 A solution of unknown substance in water at 300 K gives rise to an osmotic pressure of 3.85 atm. What is the molarity of the solution?  = MRT M = /RT M = 3.85 atm / [(.08206 L atm/K mol)(300 K)] M = .156 mol/L

  34. Some uses of colligative properties 1 Freezing-point depression - sprinkling of salt to melt snow - antifreeze in automobile cooling system - de-icing of airplane wings 2 Osmosis - desalination of seawater with reverse osmosis 3 Molar mass determination can use any four colligative properties most accurate is osmotic pressure, since the magnitude of osmosis effect is so great

  35. Example 14 • What is the molar mass of sucrose if a solution prepared by dissolving 0.822 g of sucrose in water and diluting to a volume of 300.0 mL has an osmotic pressure of 149 mm Hg at 298 K?

  36.  = MRT 149 mm Hg x 1 atm / 760 mm Hg = 0 .196 atm M =  /RT = 0.196 atm / [(0.08206 L atm/K mol)(298 K)] = 0.00802 mol/L 0.00802 mol/L x 1 L/1000 mL x 300 mL = 0.00241 mol MM = mass of sucrose / moles of sucrose = 0.822 g / 0.00241 mol = 341.08 g/mol

  37. Summary

More Related