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Solutions

Solutions. Occur in all phases. The solvent does the dissolving. The solute is dissolved. There are examples of all types of solvents dissolving all types of solvent. We will focus on aqueous solutions. Energy of Making Solutions.

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Solutions

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  1. Solutions

  2. Occur in all phases • The solvent does the dissolving. • The solute is dissolved. • There are examples of all types of solvents dissolving all types of solvent. • We will focus on aqueous solutions.

  3. Energy of Making Solutions • Heat of solution ( DHsoln ) is the energy change for making a solution. • Most easily understood if broken into steps. • 1.Break apart solvent • 2.Break apart solute • 3. Mixing solvent and solute

  4. 1. Break apart Solvent • Have to overcome attractive forces. DH1 >0 2. Break apart Solute. • Have to overcome attractive forces. DH2 >0

  5. 3. Mixing solvent and solute • DH3 depends on what you are mixing. • If molecules can attract each other DH3 is large and negative. • Molecules can’t attract- DH3 issmall and negative. • This explains the rule “Like dissolves Like”

  6. Solute and Solvent DH3 DH2 Solvent DH1 DH3 Solution Solution • Size of DH3 helps determine whether a solution will form Energy Reactants

  7. Ways of Measuring • Molarity = moles of solute Liters of solution • % mass = Mass of solute x 100 Mass of solution • Mole fraction of component AcA = nA nA + nB

  8. Ways of Measuring • Molality = moles of solute Kilograms of solvent • Molality is abbreviated m • Normality - read but don’t focus on it. • It is molarity x number of active pieces

  9. Vapor Pressure of Solutions • A nonvolatile solute lowers the vapor pressure of the solution. • The molecules of the solventmust overcome the force of both the other solvent molecules and the solute molecules.

  10. Raoult’s Law: • Psoln = csolvent x Psolvent • Vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent • Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

  11. Water has a higher vapor pressure than a solution Aqueous Solution Pure water

  12. Water evaporates faster from for water than solution Aqueous Solution Pure water

  13. The water condenses faster in the solution so it should all end up there. Aqueous Solution Pure water

  14. Practice Problem • A solution of cyclopentane with a nonvolatile compound has vapor pressure of 211 torr. If vapor pressure of the pure liquid is 313 torr, what is the mole fraction of the cyclopentane? • Psoln = XcpPcp • 211 torr = Xcp (313 torr) • .674

  15. Try one on your own • Determine the vapor pressure of a solution at 25 C that has 45 grams of C6H12O6, glucose, dissolved in 72 grams of H2O. The vapor pressure of pure water at 25 C is 23.8 torr. • Psolution= Xsolvent Psolvent • Psolution = .941(23.8 torr) • Psolution = 22.4 torr

  16. Ideal solutions • Liquid-liquid solutions where both are volatile. • Modify Raoult’s Law to • Ptotal= PA + PB = cAPA0 + cBPB0 • Ptotal= vapor pressure of mixture • PA0= vapor pressure of pure A • If this equation works then the solution is ideal.

  17. Vapor Pressure of solution P of pure A P of pure B Vapor Pressure χA χb

  18. Deviations • If solvent has a strong affinity for solute (H bonding). • Lowers solvent’s ability to escape. • Lower vapor pressure than expected. • Negative deviation from Raoult’s law. • DHsoln is large and negative exothermic. • Endothermic DHsoln indicates positive deviation.

  19. Positive deviations- Weak attraction between solute and solvent Positive ΔHsoln Vapor Pressure χA χb

  20. Negative deviations- Strong attraction between solute and solvent Negative ΔHsoln Vapor Pressure χA χb

  21. Problem #1 • The vapor pressure of a solution containing 53.6g of glycerin C3H8O3 in133.7g ethanol C2H5OH is 113 torr at 40C. Calculate the vapor pressure of pure ethanol at 40C assuming that the glycerin is a non volatile, nonelectrolyte solute in ethanol.

  22. Answer to #1 Psoln = Xeth Peth 113torr = 2.90mol/3.48mol (Peth) 135.6 torr = Peth

  23. Problem #2 • At a certain temperature, the vapor pressure of pure benzene C6H6 is 0.930atm. A solution was prepared by dissolving 10.0g of a nondissociating, nonvolatile solute in 78.11g of benzene at that temperature. The vapor pressure was found to be 0.900atm. Assuming the solution behave ideally, determine the molar mass of the solute.

  24. Answer #2 Psoln = XbenzenePbenzene .900atm = Xbenzene (.930atm) Xbenzene = .9677 Xsolute = .0323 MM = 10.0g/.0323mol = 310g/mol

  25. Problem #3 A solution of NaCl in water has a vapor pressure of 19.6 torr at 25C. What is the mol fraction of solute particle in this solution if the vapor pressure of water is 23.8 torr at 25C?

  26. Answer #3 Psoln = XwaterPwater 19.6torr = Xwater(23.8torr) .824 = Xwater therefore Xsolute = .176

  27. Problem #3 For the same problem as #3: What is the vapor pressure of the solution at 45C if the vapor pressure of water is 71.9 torr at 45C?

  28. Answer #3 Psoln = .824(71.9torr) Psoln = 59.2 torr

  29. Problem #4 A solution is made from 0.0300mol CH2Cl2 and 0.0500mol CH2Br2 at 25C. Assuming that the solution is ideal, calculate the % composition of the vapor at 25C. PCH2Cl2 = 133 torr PCH2Br2 = 11.4 torr

  30. Answer #4 Psoln = XCH2Cl2 P + XCH2Br2 P Psoln = (.03/.08)(133torr) + (.05/.08)(11.4 torr) Psoln = 57.0 torr XCH2Cl2 = 49.9 / 57 = .875 = 87.5% XCH2Br2 = 7.13 / 57 = .125 = 12.5%

  31. Colligative Properties • Because dissolved particles affect vapor pressure - they affect phase changes. • Colligative properties depend only on the number - not the kind of solute particles present • Useful for determining molar mass

  32. Water

  33. 1 atm Vapor Pressure of pure water Vapor Pressure of solution

  34. 1 atm Freezing and boiling points of solvent

  35. 1 atm Freezing and boiling points of solution

  36. 1 atm DTb DTf

  37. Boiling point Elevation • Because a non-volatile solute lowers the vapor pressure it raises the boiling point. • The equation is: DT = Kbmsolute • DT is the change in the boiling point • Kb is a constant determined by the solvent. • msolute is the molality of the solute

  38. Freezing point Depression • Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point. Freezing occurs when Pv (l) = Pv(s). Because the Pv(soln) is reduced, the temp must be lowered to form a solid. • The equation is: DT = -Kfmsolute • DT is the change in the freezing point • Kf is a constant determined by the solvent • msolute is the molality of the solute

  39. Electrolytes in solution • Since colligative properties only depend on the number of molecules. • Ionic compounds should have a bigger effect. • When they dissolve they dissociate. • Individual Na and Cl ions fall apart. • 1 mole of NaCl makes 2 moles of ions. • 1mole Al(NO3)3 makes 4 moles ions.

  40. Electrolytes have a bigger impact on on melting and freezing points per mole because they make more pieces. • Relationship is expressed using the van’t Hoff factor i i = Moles of particles in solution Moles of solute dissolved • The expected value can be determined from the formula of the compound.

  41. The actual value is usually less because • At any given instant some of the ions in solution will be paired up. • Ion pairing increases with concentration. • i decreases with increasing concentration. • We can change our formulas to DT = iKm

  42. Problem #1 A 2.00 gram sample of a large biomolecule was dissolved in 15.0 grams of CCl4 . The boiling point of this solution was found to be 77.85C. Calculate the molar mass of the biomolecule. Kb = 5.03C kg / mol BPCCl4 = 76.5C

  43. Answer #1 DT = Kbm (77.85 – 76.5C) = 5.03C Kg/mol x m .268mol/Kg = m .268 mol/Kg = Xmol / .015kg = .00402 mol 2.00g / .00402mol = 498g/mol

  44. Problem #2 The freezing point of t-butanol is 25.5C and Kf = 9.1C Kg/mol. Usually t-butanol absorbs water on exposure to air. If the freezing point of a 10.0 gram sample of t-butanol is 24.59C, how many grams of water are present in the sample?

  45. Answer #2 DT = -Kf m (24.59 – 25.5C) = -9.1 C Kg/mol x m .1 mol/Kg = m .1 m = X mol H2O /.01kg = .001mol H2O .018g H2O

  46. Problem #3 Calculate the freezing point and the boiling point for each: .050m MgCl2 .050m FeCl3 Kf = 1.86C Kg/mol Kb = 0.51C Kg/mol Predict which will have higher BP and which will have the lower FP

  47. Answer #3 Dissociate each: MgCl2 Mg2+ + 2Cl-i= 2.7 FeCl3 Fe3+ + 3Cl-i= 3.4 MgCl2 Tf = -.25C Tb = 100.069C FeCl3 Tf = -.32C Tb = 100.087C

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