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CHAPTER 13 Physical Properties of Solutions

CHAPTER 13 Physical Properties of Solutions. Mixtures - Heterogeneous Homogeneous (solutions); examples Solvent; solute Concentration - Mass solute per mass solvent Molar solubility (molarity). Thermodynamics - Energy ( H) Entropy (S). nonpolar + nonpolar polar + polar

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CHAPTER 13 Physical Properties of Solutions

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  1. CHAPTER 13 Physical Properties of Solutions

  2. Mixtures - Heterogeneous Homogeneous (solutions); examples Solvent; solute Concentration - Mass solute per mass solvent Molar solubility (molarity)

  3. Thermodynamics - Energy (H) Entropy (S) nonpolar + nonpolar polar + polar nonpolar + polar “like dissolves like”

  4. Types of mixing - Mixing of gases Mixing of liquids (miscible and immiscible) Solid in liquid (soluble and insoluble) Polar molecular solid + water Nonpolar molecular solid + water Ionic solid + water

  5. Classification of solutions - Unsaturated; saturated; supersaturated Preparation of a saturated solution - 1) Add an excess amount of the solid substance being dissolved. 2) Stir the mixture. 3) Filter the excess undissolved solid.

  6. Effect of Temperature On Solubility

  7. Recrystallization

  8. Concentration units - Molarity (M) moles solute (mol/L) liters solution Molality (m) moles solute (mol/kg) kilograms solvent Mole fraction (XA) moles A total moles

  9. Percent by mass mass A x 100% total mass Parts per million by mass mass A x 106 (ppm) total mass Know - 1) Definitions for concentration units. 2) How to calculate concentration 3) How to convert between concentration units

  10. Example: A solution is prepared by dissolving 2.163 g of potassium hydroxide (KOH, MW = 56.11 g/mol) in water. The final volume of the solution is V = 250.0 mL. What is the molarity of the solution?

  11. Example: A solution is prepared by dissolving 2.1634 g of potassium hydroxide (KOH, MW = 56.11 g/mol) in water. The final volume of the solution is V = 250.0 mL. What is the molarity of the solution? M = moles solute liters solution moles KOH = 2.1634 g KOH 1 mol KOH = 0.03856 mol 56.11 g KOH liters solution = 0.2500 L M = 0.03856 mol KOH = 0.1542 mol/L 0.2500 L soln

  12. Example: A solution is prepared that is 40.0 % by mass ethylene glycol (C2H6O2, MW = 62.07 g/mol) in water (H2O, MW = 18.02 g/mol). The density of the solution is D = 1.0514 g/mL. What are the molarity, molality, and mole fraction ethylene glycol in the solution?

  13. Example: A solution is prepared that is 40.0 % by mass ethylene glycol (C2H6O2, MW = 62.07 g/mol) in water (H2O, MW = 18.02 g/mol). The density of the solution is D = 1.0514 g/mL. What are the molarity, molality, and mole fraction ethylene glycol in the solution? Assume that you have 100.0 g of solution. Then you have 40.00 g of ethylene glycol and 60.00 g water. nE = 40.00 g 1 mol = 0.6444 mol E 62.07 g nW = 60.00 g 1 mol = 3.3296 mol W 18.02 g XE = nE = (0.6444 mol) = 0.1622 nE + nW (0.6444 + 3.3296)mol

  14. The molality ethylene glycol in the solution is mE = mol E = 0.6444 mol = 10.74 mol/kg kg H2O 0.06000 kg The total mass of solution is m = 40.00 g + 60.00 g = 100.00 g. Since the density of the solution is D = 1.0514 g/mL, the volume of the solution is V = 100.00 g 1 mL = 95.11 mL 1.0514 g So ME = 0.6444 mol = 6.775 mol/L 0.09511 L

  15. Solubility of gases in liquids - Henry’s law [B] = k pB Example: The Henry’s law constant for oxygen dissolving in water is k = 2.02 x 10-3 mol/L.atm at T = 20 C. What is the concentration of dissolved oxygen in water in equilibrium with the Earth’s atmosphere (pO2 = 0.21 atm)?

  16. Example: The Henry’s law constant for oxygen dissolving in water is kH = 2.02 x 10-3 mol/L.atm at T = 20 C. What is the concentration of dissolved oxygen in water in equilibrium with the Earth’s atmosphere (pO2 = 0.21 atm)? [O2] = k pO2 = (2.02 x 10-3 mol/L.atm) (0.21 atm) = 4.2 x 10-4 mol/L (or 0.0135 g/L) While small, this is enough dissolved oxygen to support fish life.

  17. Liquids - Volatile liquid Vapor pressure Raoult’s law pA = XA pAo Ideal liquid solution Example: At 20C the vapor pressure of pure benzene is 57. torr and the vapor pressure of pure toluene is 21. torr. Consider a solution with XB = 0.40. Assuming that benzene and toluene form an ideal solution, find the partial pressure of benzene and toluene and the total pressure above the solution.

  18. Example: At 20C the vapor pressure of pure benzene (B) is 57. torr and the vapor pressure of pure toluene (T) is 21. torr. Consider a solution with XB = 0.40. Assuming that benzene and toluene form an ideal solution, find the partial pressure of benzene and toluene and the total pressure above the solution. From Raoult’s law, pA = XApA XB + XT = 1, so XT = 1 – XB = 1 – 0.40 = 0.60 So pB = (0.40)(57. torr) = 22.8 torr pT = (0.60)(21. torr) = 12.6 torr ptotal = psoln = pB + pT = 22.8 torr + 12.6 torr = 35.4 torr

  19. Ideal and Nonideal Solution For an ideal solution a plot of pressure vs mole fraction will be linear. If such a plot is not linear, then we have a nonideal solution. A = CH3COCH3; B = CS2

  20. Colligative property Definition Vapor pressure lowering p = XB pAo Boiling point elevation Tb = Kb mB Freezing point depression Tf = Kf mB Osmotic pressure  = [B]RT

  21. Example: A solution is prepared by dissolving 1.00 mol glucose (C6H12O6) in 2500. g of water. What are the normal freezing point and normal boiling point for this solution?

  22. Example: A solution is prepared by dissolving 1.00 mol glucose (C6H12O6) in 2500. g of water. What are the normal freezing point and normal boiling point for this solution? glucose(s)  glucose(aq) Tb = 100.00 C Tf = 0.00 C Kb = 0.52 C/m Kf = 1.86 C/m molality glucose = 1.00 mol glucose = 0.400 mol particles = 0.400 m 2.500 kg water kg water So Tb = (0.52 C/m) (0.400 m) = 0.208 C Tf = (1.86 C/m) (0.400 m) = 0.744 C Tb = 100.00 C + 0.208 C = 100.21 C Tf = 0.00 C - 0.744 C = - 0.74 C

  23. Phase Diagram For Solutions

  24. Example: What is the osmotic pressure for a 0.0100 M solution of naphthalene (C10H8) in benzene (C6H6) at T = 25. C?

  25. Example: What is the osmotic pressure for a 0.0100 M solution of naphthalene (C10H8) in benzene (C6H6) at T = 25. C?  = MBRT , where MB molarity of solute particles (mol/L) For naphthalene, MB = 0.0100 M, T = 273. + 25. = 298. K (naph.) = (0.0100 mol/L)(0.08206 L.atm/mol.K)(298 K) = 0.245 atm ( = 186. Torr) Note that we use absolute temperature (Kelvin) in the equation for osmotic pressure.

  26. Hypotonic - Lower molarity than plasma. Isotonic - Same molarity as plasma. Hypertonic - Higher molarity than plasma.

  27. Reverse Osmosis

  28. Reverse Osmosis

  29. Colligative properties for ionic compounds van’t Hoff factor Example: Consider the following three substances C6H12O6 (sugar) NaCl (sodium chloride) K2SO4 (potassium sulfate) When 1.000 moles of each of the above sub-stances are dissolved in water how many moles of particles will form? What is the van’t Hoff factor?

  30. C6H12O6 (glucose) (i = 1) C6H12O6(s)  C6H12O6(aq) mol particles = 1.00 mole glu. 1 mole particle = 1.00 mol particles 1 mol glu. NaCl (sodium chloride) (i = 2) NaCl(s)  Na+(aq) + Cl-(aq) mol particles = 1.00 mol NaCl 2 mol particles = 2.00 mol particles 1 mol NaCl K2SO4 (potassium sulfate) (i = 3) K2SO4(s)  2 K+(aq) + SO42-(aq) mol particles = 1.00 mol K2SO43 mol particles = 3.00 mol particles 1 mol K2SO4

  31. Colligative Properties With Several Solutes Example: Consider the following two solutions: A) 1.000 L of an aqueous solution containing 0.200 moles of Ca(NO3)2. B) 1.000 L of an aqueous solution containing 0.200 moles of Ca(NO3)2, and 0.200 moles of glucose. What is the osmotic pressure for each of the above solutions at T = 37. C?

  32. Example: Consider the following two solutions: A) 1.000 L of an aqueous solution containing 0.200 moles of Ca(NO3)2. B) 1.000 L of an aqueous solution containing 0.200 moles of Ca(NO3)2, and 0.200 moles of glucose. What is the osmotic pressure for each of the above solutions at T = 37. C? For solution A (Ca(NO3)2(s)  Ca2+(aq) + 2 NO3-(aq) ; i = 3) mol particles = 0.200 mol Ca(NO3)2 .3 mol particles 1 mol Ca(NO3)2. = 0.600 mole particles For solution B mol particles = (0.600 + 0.200) = 0.800 mol particles

  33. For solution A MsolA = 0.600 mol particles = 0.600 M 1.000 L soln  = MsolART = (0.600 mol/L)(0.08206 L.atm/mol.K)(310. K) = 15.3 atm For solution B MsolB = 0.800 mol particles = 0.800 M 1.000 L soln  = MsolBRT = (0.800 mol/L)(0.08206 L.atm/mol.K)(310. K) = 20.4 atm

  34. Experimental Values for theVan’t Hoff Factor The value for the van't Hoff factor can be found experimentally. For dilute solutions the experimental value for the van’t Hoff factor is close to the value predicted from theory. However, for more concentrated solutions the experimental value is lower than expected. For example, for NaCl NaCl(s)  Na+(aq) + Cl-(aq) Molality NaCl itheory iexperimental 0.0001 2.00 1.99 0.001 2.00 1.97 0.01 2.00 1.94 0.1 2.00 1.87 Unless otherwise stated we will use the theoretical value for i, which corresponds to ideal behavior.

  35. Applications to Molecular Mass Determination Example: A solution is prepared by dissolving 0.1438 g of a nonvolatile and nonionizing solute in water. The final volume of the solution is V = 25.00 mL. The osmotic pressure of the solution, measured at T = 37.0 C, is 21.6 torr. What is the molecular mass of the solute?

  36. Example: A solution is prepared by dissolving 0.1438 g of a nonvolatile and nonionizing solute in water. The final volume of the solution is V = 25.00 mL. The osmotic pressure of the solution, measured at T = 37.0 C, is 21.6 torr. What is the molecular mass of the solute? MW = m/n = grams/moles  = MBRT = 21.6 torr (1 atm/760 torr) = 0.02842 atm MB =  = 0.02842 atm = 1.117 x 10-3 mol/L RT (0.08206 L.atm/mol.K) (310. K) n = MBV = (1.117 x 10-3 mol/L) (0.02500 L) = 2.792 x 10-5 mol M = 0.1438 g = 5150. g/mol 2.792 x 10-5 mol

  37. Colloid

  38. End of Chapter 13 “If you're not part of the solution, you're part of the precipitate.” - unknown “How can you distinguish science from junk? Science posits hypothesis and tests them. Pseudoscience assumes conclusions and finds evidence to back them up.” - Wendy Kaminer, Sleeping With Extra-Terrestrials

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