1. Solutions

2. Occur in all phases u The solvent does the dissolving. u The solute is dissolved. u There are examples of all types of solvents dissolving all types of solvent. u We will focus on aqueous solutions. u The components of a mixture are uniformly intermingled (the mixture is homogeneous).

3. Dissolving u (YDVD)

4. 11.1 Solution Composition u Molarity(M) = moles of solute liters of solution u Mass % = mass of solute x 100 mass of solution u Mole fraction of component A =  A = moles A moles A + moles B (total) u Molality( m ) = moles of solute kilograms of solvent

5. Practice u An aqueous antifreeze solution is 40% ethylene glycol (C 2 H 6 O 2 ) by mass. The density of the solution is 1.05 g/cm 3. Calculate the molality, molarity and mole fraction of the ethylene glycol. #26 u In lab you need to prepare at least 100 mL of each of the following solutions. Explain how you would proceed using the given information. #28 a. 2.0 m KCl in water (density of water=1.00 g/cm 3 ) b. 15% NaOH by mass in water (d=1.00 g/cm 3 ) c. 25% NaOH by mass in CH3OH (d=0.79 g/cm 3 ) d. 0.10 mole fraction of C 6 H 12 O 6 in water (d=1.00 g/cm 3 )

6. Practice u A bottle of wine contains 12.5% ethanol by volume. The density of ethanol (C 2 H 5 OH) is 0.789 g/cm 3. Calculate the concentration of ethanol in wine in terms of mass percent and molality. #30 u Calculate the molarity and mole fraction of acetone in a 1.00 m solution of acetone (CH 3 COCH 3 ) in ethanol (C 2 H 5 OH) (Density of acetone = 0.788g/cm 3 ; density of ethanol = 0.789 g/cm 3.) Assume that the volumes of acetone and ethanol add. #32

7. 11.2 Energy of Making Solutions  Heat of solution (  H soln ) is the energy change for making a solution. u Most easily understood if broken into steps. u Step 1 - Expanding (breaking up) the solute (endothermic). u Step 2 -Expanding the solvent (endothermic). u Step 3 -Allowing the solute and solvent to interact to form a solution (exothermic).

8.  H soln =  H step1 +  H step2 +  H step3 1. Break apart Solvent  Have to overcome attractive forces.  H 1 > 0 (Lattice Energy) 2. Break apart Solute.  Have to overcome attractive forces.  H 2 > 0

9. 3. Mixing solvent and solute   H 3 depends on what you are mixing.  Molecules can attract each other  H 3 is large and negative.  Molecules can’t attract -  H 3 is small and negative. u This explains the rule “Like dissolves Like”

10. EnergyEnergy Solute Solution H1H1 H2H2 H3H3 Solvent Solute and Solvent  Size of  H 3 determines whether a solution will form

11. Favoring the solution  Negative values for  H soln u Increase in entropy.  If  H soln is small and positive, a solution will still form because of entropy. u There are many more ways for components to become mixed than there is for them to stay separate.

12. Practice

13. 11.3 Structure and Solubility u Water soluble molecules must have dipole moments - polar bonds. u To be soluble in non polar solvents the molecules must be non polar. u Vitamin A and Vitamin C p.519 (YDVD)

14. Soap PO-O- CH 3 CH 2 O-O- O-O-

15. Soap u Hydrophobic non- polar end PO-O- CH 3 CH 2 O-O- O-O-

16. Soap u Hydrophilic polar end PO-O- CH 3 CH 2 O-O- O-O-

17. PO-O- CH 3 CH 2 O-O- O-O- _

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

19. Hydrophobic ends dissolve in grease

20. Hydrophilic ends dissolve in water

21. u Water molecules can surround and dissolve grease. u Helps get grease out of your way.

22. Pressure Effects u Changing the pressure doesn’t effect the amount of solid or liquid that dissolves. u They are incompressible. u It does effect gases.

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

24. u The gas is at equilibrium with the dissolved gas in this solution. u The equilibrium is dynamic.

25. u If you increase the pressure the gas molecules dissolve faster. u The equilibrium is disturbed.

26. u The system reaches a new equilibrium with more gas dissolved. u Henry’s Law. P= kC Pressure = constant x Concentration of dissolved gas

27. Temperature Effects u Increased temperature increases the rate at which a solid dissolves. u We can’t predict whether it will increase the amount of solid that dissolves. (Usually does) u We must read it from a graph of experimental data.

28. 20 406080 100

29. Gases are predictable u As temperature increases, solubility decreases. u Gas molecules can move fast enough to escape.

31. 11.4 Vapor Pressure of Solutions u A nonvolatile solvent lowers the vapor pressure of the solution. u The molecules of the solvent must overcome the force of both the other solvent molecules and the solute molecules.

32. Raoult’s Law:  P soln =  solvent x P solvent u Vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent. u Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure. u Ionic compounds have nearly 2, 3 or more times the VP lowering of nonionic solutes. Why?

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

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

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

36. 1. What is the vapor pressure, at 25ºC, for a solution of 158.0 g of sucrose (C 12 H 22 O 11 ) in 643.5 g of water? The vapor pressure of water at 25ºC is 23.8 torr. Ex 11.5 2. A solution of NaCl in water has a vapor pressure of 19.6 torr at 25ºC. What is the mole fraction of NaCl in this solution? #48 Review Questions

37. u A solution is prepared by mixing 0.0300mol CH 2 Cl 2 and 0.0500 mol CH 2 Br 2 at 25°C. Assuming the solution is ideal, calculate the composition of the vapor(in terms of mole fractions) at 25°C. At 25°, the vapor pressures of pure CH 2 Cl 2 and pure CH 2 Br 2 are 133 and 11.4 torr, respectively. #50

38. 11.5 Colligative Properties u Because dissolved particles affect vapor pressure - they affect phase changes. u Colligative properties depend only on the number - not the kind of solute particles present. 4 Boiling point elevation 4 Freezing point depression  Osmotic pressure u Useful for determining molar mass

39. Boiling Point Elevation u Because a non-volatile solute lowers the vapor pressure it raises the boiling point.  The equation is:  T = K b m solute   T is the change in the boiling point. u K b is a constant determined by the solvent. u m solute is the molality of the solute. u (BDVD)

40. Freezing Point Depression u Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point.  The equation is:  T = K f m solute   T is the change in the freezing point. u K f is a constant determined by the solvent. u m solute is the molality of the solute. u (BDVD)

41. 1 atm Vapor Pressure of solution Vapor Pressure of pure water

42. 1 atm Freezing and boiling points of water

43. 1 atm Freezing and boiling points of solution

44. 1 atm TfTf TbTb

45. Practice u A 2.00-g sample of a large biomolecule was dissolved in 15.0g of carbon tetrachloride. The boiling point of this solution was determined to be 77.85 ° C. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling-point constant is 5.03 0 Ckg/mol., and the boling point of pure carbon tetrachloride is 76.50 ° C. #58 u The freezing point of t-butanol is 25.50°C and K f of 9.10Ckg/mol. Usually t-butanol absorbs water on exposure to air. If the freezing point of a 10.0g sample of t-butanol is 24.59 0 C, how many grams of water are present in the sample? #60

46. Practice u What volume of ethylene glycol (C 2 H 6 O 2 ), a nonelectrolyte, must be added to 15.0 L of water to produce an antifreeze solution with a freezing point of -30.0 °C? What is the boiling point of this solution? (The density of ethylene glycol is 1.11 g/cm 3, and the density of water is 1.00 g/cm 3.) #62 u Water K f = 1.86 °Ckg/mol, K b = 0.51 °Ckg/mol

47. 11.7 Electrolytes in Solution u Since colligative properties only depend on the number of molecules. u Ionic compounds should have a bigger effect. u When they dissolve they dissociate. u Individual Na and Cl ions fall apart. u 1 mole of NaCl makes 2 moles of ions. u 1mole Al(NO 3 ) 3 makes 4 moles ions.

48. u Electrolytes have a bigger impact on on melting and freezing points per mole because they make more pieces. u Relationship is expressed using the van’t Hoff factor i i = Moles of particles in solution Moles of solute dissolved u The expected value can be determined from the formula. i is an integer greater than 1.

49. u The actual value is usually less because at any given instant some of the ions in solution will be paired. u Ion pairing increases with concentration. u i decreases with increasing concentration. u We can change our formulas to  H = i K m

50. Practice u From the Following: pure water, solution of C 12 H 22 O 11 (m=.01), solution of NaCl(m=.01), and CaCl 2 (m=.01)(all are in water). Choose one with the: a)Highest freezing point b)Lowest freezing point c)Highest boiling point d)Lowest boiling point e)Highest osmotic pressure u Calculate the freezing point and the boiling point of each of the following aqueous solutions.(Assume complete dissociation.) a).050 m MgCl 2 b).050 m FeCl 3

51. u Colloidal Dispersion (colloid): A suspension of tiny particles in some medium. aerosols, foams, emulsions, sols See Table 11.7 u Coagulation: The addition of an electrolyte or heat, causing destruction of a colloid. Colloids