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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. Ways of Measuring. Molarity = moles of solute Liters of solvent

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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. Ways of Measuring • Molarity = moles of solute Liters of solvent • % mass = Mass of solute x 100 Mass of solution • Mole fraction of component A cA = NANA + NB

  4. Ways of Measuring • Molality = moles of solute Kilograms of solvent • Molality is abbreviated m • Normality - read but don’t focus on it (unless you plan on being an “Old School Biologist”)

  5. 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

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

  7. 3. Mixing solvent and solute • DH3 depends on what you are mixing. • Molecules can attract each other DH3 is large and negative. • Molecules can’t attract- DH3 issmall and negative. Solutions which form are typically exothermic overall! • This explains the rule “Like dissolves Like”

  8. Solute and Solvent DH3 DH2 Solvent DH1 DH3 Solution Solution • Size of DH3 determines whether a solution will form Energy Reactants

  9. Types of Solvent and solutes • If DHsoln is small and positive, a solution will still form because of entropy. • Typically, greater forms of disorder are favored over ordered matter.

  10. Structure and Solubility • Water soluble molecules must have dipole moments -polar bonds. • To be soluble in non polar solvents the molecules must be non polar. • I2 is very slightly soluble in water, but becomes more soluble in a concentrated KI solution…Explain.

  11. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 Soap

  12. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 Soap • Hydrophobic non-polar end

  13. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 Soap • Hydrophilic polar end

  14. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 _

  15. A drop of grease in water • Grease is non-polar • Water is polar • Soap lets you dissolve the non-polar in the polar.

  16. Hydrophobic ends dissolve in grease

  17. Hydrophilic ends dissolve in water

  18. Water molecules can surround and dissolve grease. • “Helps get grease out of your way.”

  19. Pressure effects • Changing the pressure doesn’t effect the amount of solid or liquid that dissolves • They are incompressible. • It does effect gases.

  20. Dissolving Gases • Pressure effects the amount of gas that can dissolve in a liquid. • The dissolved gas is at equilibrium with the gas above the liquid.

  21. The gas is at equilibrium with the dissolved gas in this solution. • The equilibrium is dynamic.

  22. If you increase the pressure the gas molecules dissolve faster. • The equilibrium is disturbed.

  23. The system reaches a new equilibrium with more gas dissolved. • Henry’s Law. P= kC Pressure = constant x Concentration of gas Examples: • Soda Can • Lake Nyos Tragedy

  24. Lake Nyos Tragedy • Suffocated 1,700 people within 25 kilometres (16 mi) of the lake, mostly rural villagers, as well as 3,500 livestock. • 300 ft fountain of water created an 82 ft wall of water which scoured the lake shore.

  25. Degassing Lake Nyos

  26. Temperature Effects • In general: Increased temperature increases the rate at which a solid dissolves. • However, we can’t always predict whether it will increase the amount of solid that dissolves. • Consult a graph of experimental data for the best results!

  27. 100 40 60 80 20

  28. Gases are predictable • As temperature increases, solubility decreases. • Gas molecules can move fast enough to escape the solution. • Thermal pollution of lakes.

  29. Vapor Pressure of Solutions • A nonvolatile solvent 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.

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

  31. Raoult’s Law Example • Determine the vapor pressure of a solution made by dissolving 125 g of sucrose into 455 mL of water at 25.0ºC. • The vapor pressure of water at this temp is 23.76 torr. (Density of water at this temp is 0.9971 g/cm3)

  32. Solution (get it?) • 125g sucrose = .365 mol • 155 mL of water = 8.58 mol • c of water = .959 • Pressure = 22.8 torr (down from 23.76 torr)

  33. Ionic vs Covalent Solute? • What if we used the same moles of salt rather than sucrose? • Same result? • I don’t think so…you get twice the number of solute particles = twice the effective mole fraction of solute! • c of water drops to .921 from .959…so the vapor pressure drops as well!

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

  35. Water evaporates faster from the water than the solution… Aqueous Solution Pure water

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

  37. Review Question • What is the composition of a pentane-hexane solution that has a vapor pressure of 350 torr at 25ºC ? • The vapor pressures at 25ºC are • pentane 511 torr • hexane 150 torr. • Nonideal solution: both can evaporate • What are the component vapor pressures?

  38. Review Question • Nonideal Solutions: Ptotal=PA + PB Psoln = csolA x PsolA + csolB x PsolB 350 torr = c A511 torr + c B 150 torr (A + B = 1…so B = 1-A now you may substitute…) 350 torr = 511 A+ 150 (1-A) A = .554 (so B = 1- .554…) B = .446 Check it in the original equation!

  39. 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

  40. 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

  41. Freezing point Depression • Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point. • 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

  42. Freezing/Boiling Point • Addition of a nonvolatile solute extends the liquid phase! • Decreasing the freezing point of the solution… • Increasing the boiling point of the solution!

  43. 1 atm Vapor Pressure of pure water Vapor Pressure of solution

  44. 1 atm Freezing and boiling points of water

  45. 1 atm Freezing and boiling points of solution

  46. 1 atm DTb DTf

  47. Electrolytes in solution • Since colligative properties only depend on the number of molecules. • Ionic compounds should have a greater 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.

  48. Electrolytes have a more significant impact on on melting and freezing points per mole because they generate more particles. • 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.

  49. The actual value is usually less because… …at any given instant some of the ions in solution will be paired. …ion pairing increases with concentration. …i decreases with increasing concentrations. • We can change our formulas to DT = imK

  50. Ideal solutions • Liquid-liquid solutions where both are volatile. • Modify Raoult’s Law to • Ptotal= PA + PB = cAPA0 + cAPB0 • Ptotal= vapor pressure of mixture • PA0= vapor pressure of pure A • If this equation works then the solution is ideal. • Solvent and solute are alike.

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