Successfully reported this slideshow.
Upcoming SlideShare
×

# Ch11 z5e solutions

2,253 views

Published on

Published in: Education
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

### Ch11 z5e solutions

1. 1. Ch11 Solutions pp
2. 2. 11.1 Solution Composition <ul><li>Solutions occur in all phases </li></ul><ul><li>The solvent does the dissolving. </li></ul><ul><li>The solute is dissolved. </li></ul><ul><li>There are examples of all types of solvents dissolving all types of solutes. </li></ul><ul><li>We will focus on aqueous solutions. </li></ul>
3. 3. Ways of Measuring <ul><li>Molarity = moles of solute Liters of solut ion </li></ul><ul><li>% mass = Mass of solute x 100% Mass of solut ion </li></ul><ul><li>Mole fraction of component A =  A = n A n A + n B </li></ul>
4. 4. <ul><li>Mol al ity = moles of solute Kilograms of solv ent </li></ul><ul><li>Molality is abbreviated m . </li></ul><ul><li>Normality - read for general info. </li></ul>Ways of Measuring
5. 5. <ul><li>Molarity </li></ul><ul><li>Mass % </li></ul><ul><li>Mole Fraction </li></ul><ul><li>Molality </li></ul><ul><li>Answers are . . . </li></ul>Given 1.00 g C 2 H 5 OH & 100. g H 2 O and final volume of 101 mL <ul><li>0.215 M </li></ul><ul><li>0.990% </li></ul><ul><li>0.00389 </li></ul><ul><li>0.217 m </li></ul>
6. 6. 11.2 The Energies of Solution Formation pp <ul><li>Heat of solution (  H soln ) is the energy change for making a solution. </li></ul><ul><li>Most easily understood if broken into 3 energy steps (that must be added). </li></ul><ul><li>1.Break apart sol vent </li></ul><ul><li>2.Break apart sol ute </li></ul><ul><li>3. Mixing solvent and solute </li></ul>
7. 7. 1. Break apart Solvent pp <ul><li>Have to overcome attractive forces. So,  H 1 > 0 (must put in energy). </li></ul><ul><li>2. Break apart Solute </li></ul><ul><li>Have to overcome attractive forces. So,  H 2 > 0 </li></ul>
8. 8. 3. Mixing solvent and solute pp <ul><li> H 3 depends on what you are mixing. </li></ul><ul><li>If molecules can attract each other then  H 3 is large and negative . </li></ul><ul><li>If molecules can’t attract then  H 3 is small and negative (negligible interactions). </li></ul><ul><li>This explains the rule “ Like dissolves Like ” </li></ul>
9. 9. Figure 11.1 The Steps in the Dissolving Process pp
10. 10. <ul><li>Size of  H 3 determines whether a solution will form pp </li></ul>Energy Solute Solution  H 1  H 2  H 3 Solvent Solute and Solvent  H 3 No solution
11. 11. Types of Solvent and solutes pp <ul><li>If  H soln is small and positive , a solution will still form because of entropy . </li></ul><ul><li>This is because there are many more ways for them to become mixed than there is for them to stay separate. </li></ul><ul><li>This why NaCl is so soluble in water even though its  H soln is + 3 kJ/mol </li></ul><ul><li>Memorize Table 11.3 p. 492 </li></ul>
12. 12. Table 11.3 p.521 pp
13. 13. Energetics of Solutions & Solubility pp <ul><li>The lattice energy of KCl is -715 kJ/mol, & the enthalpy of hydration is -684 kJ/mol. Calculate the enthalpy of solution per mole of solid KCl. Steps . . . </li></ul><ul><li>Lattice energy was defined in Chapter 8 as M + (g) + X - (g)  MX (s) (memorize this). </li></ul><ul><li>Use Hess’ law in two steps: 1st, take the solid and convert it into gaseous ions (opposite of lattice energy) to get . . . </li></ul>
14. 14. Energetics of Solutions & Solubility pp <ul><li>KCl (s)  K 1+ (g) + Cl 1- (g) whose ∆H is the (-) of ∆H lattice energy = -(-) 715 kJ/mol </li></ul><ul><li>Next, hydrate the ions to get a solution: </li></ul><ul><li>K 1+ (g) + Cl 1- (g)  K 1+ ( aq ) + Cl 1- ( aq ) , where ∆H hydration = -684 kJ/mol </li></ul>
15. 15. Energetics of Solutions & Solubility pp <ul><li>Add the energies to get . . . </li></ul><ul><li>KCl (s)  K 1+ (g) + Cl 1- (g) = -(-) 715 kJ/mol </li></ul><ul><li>K 1+ (g) + Cl 1- (g)  K 1+ ( aq ) + Cl 1- ( aq ) = -684 kJ/mol </li></ul><ul><li>KCl (s)  K 1+ ( aq ) + Cl 1- ( aq ) = 31 kJ/mol </li></ul><ul><li>∆H solution = 31 kJ/mol (endothermic). </li></ul><ul><li>You have a HW problem on this (#33). </li></ul><ul><li>You will have a test question on this!!! </li></ul>
16. 16. 11.3 Factors Affecting Solubility Structure and Solubility <ul><li>Water soluble molecules must have dipole moments; i.e. , polar bonds. </li></ul><ul><li>To be soluble in non polar solvents the molecules must be non polar. </li></ul><ul><li>Read Vitamins A and C discussion on figure 11.4 pg. 493 on your own. </li></ul>
17. 17. Soap P O - CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O - O -
18. 18. Soap <ul><li>Hydro phobic non polar end </li></ul>P O - CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O - O -
19. 19. Soap <ul><li>Hydro philic polar end </li></ul>P O - CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O - O -
20. 20. Skeletal structure of Soap _ P O - CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O - O -
21. 21. <ul><li>A drop of grease in water </li></ul><ul><li>Grease is non polar </li></ul><ul><li>Water is polar </li></ul><ul><li>Soap lets you dissolve the nonpolar in the polar. </li></ul>
22. 22. <ul><li>Hydro phobic ends dissolve in grease </li></ul>
23. 23. <ul><li>Hydro philic ends dissolve in water </li></ul>
24. 24. <ul><li>Water molecules can surround and dissolve grease. </li></ul><ul><li>Helps get grease out of your way. </li></ul>
25. 25. Figure 23.29 Soap Micelles
26. 26. Pressure effects <ul><li>Changing the pressure doesn’t affect much the amount of solid or liquid that dissolves </li></ul><ul><li>They are basically incompressible. </li></ul><ul><li>But, changing pressure does affect gases (including those dissolved in liquids). </li></ul>
27. 27. Dissolving Gases <ul><li>Pressure affects the amount of gas that can dissolve in a liquid. </li></ul><ul><li>The dissolved gas is at equilibrium with the gas above the liquid . </li></ul><ul><li>(Mentos video and http://eepybird.com/ ) </li></ul>
28. 28. <ul><li>The gas is at equilibrium with the dissolved gas in this solution. </li></ul><ul><li>The equilibrium is dynamic . </li></ul>
29. 29. <ul><li>If you increase the pressure the gas molecules dissolve faster than they escape. </li></ul><ul><li>The equilibrium is disturbed . </li></ul>
30. 30. <ul><li>The system reaches a new equilibrium with more gas dissolved. </li></ul><ul><li>This is Henry’s Law. </li></ul><ul><li>P= kC </li></ul><ul><li>Pressure = constant x Concentration of gas </li></ul>
31. 31. Calculations using Henry’s Law pp <ul><li>A bottle of soft drink at 25 o Celsius contains CO 2 at a pressure of 5.0 atm over the liquid. </li></ul><ul><li>Assuming that the partial pressure of CO 2 in the atmosphere (that is, in an open system) is 4.0 x 10 -4 atm, </li></ul><ul><li>Calculate the equilibrium concentrations of CO 2 both before and after the bottle is opened. Henry’s constant here is 32 L atm/mol. . . </li></ul>
32. 32. Calculations using Henry’s Law pp <ul><li>Henry’s Law is P = kC </li></ul><ul><li>For this problem it is P CO2 = k CO2 C CO2 </li></ul><ul><li>In the unopened bottle, solve for C CO2 to get . . . </li></ul><ul><li>C CO2 = P CO2 /k CO2 = 5.0 atm/32 L atm/mol = 0.16 mol/L. </li></ul>
33. 33. Calculations using Henry’s Law pp <ul><li>In the opened bottle the gas will reach equilibrium with the atmosphere </li></ul><ul><li>So P CO2 = the partial pressure = 4.0 x 10 -4 atm. </li></ul><ul><li>C CO2 = P CO2 /k CO2 = 4.0 x 10 -4 atm /32 L atm/mol = 1.2 x 10 -5 mol/L . </li></ul><ul><li>This large change in concentration (from 0.16 mol/L to 1.2 x 10 -5 mol/L) explains why soda goes flat. </li></ul>
34. 34. Temperature Effects <ul><li>Increased temperature increases the rate at which a solid dissolves. </li></ul><ul><li>We can not predict whether it will increase the amount of solid that dissolves (only the rate). </li></ul><ul><li>We must read it from a graph of experimental data. </li></ul>
35. 35. 20 40 60 80 100
36. 36. Gases are predictable <ul><li>As temperature increases, solubility decreases. </li></ul><ul><li>So, gas molecules can move fast enough to escape. </li></ul><ul><li>Causes thermal pollution. </li></ul>
37. 37. Figure 11.7 The Solubilities of Several Gases in Water
38. 38. 11.4 The Vapor Pressures of Solutions <ul><li>A non volatile sol ute lowers the vapor pressure of the sol vent . </li></ul><ul><li>The molecules of the solvent must overcome the force of both the other solvent molecules and the solute molecules. </li></ul>
39. 39. Raoult’s Law: <ul><li>P solu tion = P o sol vent x  sol vent </li></ul><ul><li>Form of y = mx + b, plot of P soln and  sol vent is straight line with slope P o </li></ul><ul><li>Vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent </li></ul><ul><li>Applies only to an ideal solution where the solute itself doesn’t contribute to the vapor pressure ( i.e ., the solute is non volatile. </li></ul>
40. 40. <ul><li>Pure water has a higher vapor pressure than a solution </li></ul>Aqueous Solution Pure water
41. 41. <ul><li>Water evaporates faster from pure water than from a solution </li></ul>Aqueous Solution Pure water
42. 42. <ul><li>The water condenses faster in the solution so it should all end up there. </li></ul>Aqueous Solution Pure water
43. 43. <ul><li>What is the percent composition of a pentane-hexane solution (not the gas) that has a vapor pressure of 350 torr at 25ºC ? The vapor pressures at 25ºC are </li></ul><ul><ul><li>pentane 511 torr </li></ul></ul><ul><ul><li>hexane 150 torr. </li></ul></ul><ul><li>Try first, then see hints next slide: </li></ul>Review Question pp
44. 44. <ul><li>% Comp. Of pentane-hexane solution with V p soln = 350 and V p pen = 511, V p hex 150? </li></ul><ul><li>All are molecules, so no colligative property problem. </li></ul><ul><li>Use  can do w/ P since proportional to moles) </li></ul><ul><li>Answers are . . . </li></ul><ul><li>Hexane - 22.7% (150/(511+150) x 100%) </li></ul><ul><li>Pentane - 77.3% (100 - 22.7%) </li></ul>Review Question pp
45. 45. <ul><li>What is the composition (hexane & pentane pressures) in torr of the vapor ? </li></ul><ul><li>Remember, % Comp. of pentane-hexane is hexane 22.7%, pentane 77.3% and V p solution is 350 torr. Use mole fraction. </li></ul><ul><li>Answers are . . . </li></ul><ul><li>Hexane 79 torr and pentane 271 torr Vapor composition hexane = (.227)(350) = 79 torr, so  that of pentane = 271 </li></ul>Review Question pp
46. 46. <ul><li>Liquid-liquid solutions where both are volatile. </li></ul><ul><li>Modify Raoult’s Law to . . . </li></ul><ul><li>P total = P A + P B =  A P A 0 +  B P B 0 </li></ul><ul><li>P total = vapor pressure of mixture </li></ul><ul><li>P A 0 = vapor pressure of pure A (etc.) </li></ul><ul><li>If this equation works then the solution is ideal. </li></ul><ul><li>Solvent and solute are alike (as are the interactions between all species).. </li></ul>Ideal/Non-ideal solutions
47. 47. Test-type Question (will be on our test!) pp <ul><li>Pentane (C 5 H 12 ) & Hexane (C 6 H 14 ) form an ideal solution with V p 25° of 511 & 150. Torr respectively. A solution is made from 25 ml pentane (d = 0.63 g/mL) & 45 mL hexane (d = 0.66 g/mL) </li></ul><ul><li>(a) What is V p of the resulting solution ? </li></ul><ul><li>(b) What is composition by mole fraction of pentane in the vapor that is in equilibrium with this solution? </li></ul><ul><li>Reference: Z7e p. 521, # 49 </li></ul>
48. 48. Test-type Question pp <ul><li>Pentane (C 5 H 12 ) & Hexane (C 6 H 14 ) form an ideal solution with V p 25° of 511 & 150. Torr respectively. A solution is made from 25 ml pen (d = 0.63 g/mL) & 45 mL hex (d = 0.66 g/ML </li></ul><ul><li>(a) What is V p of the resulting solution ? </li></ul><ul><li>Use density to get number of moles. </li></ul><ul><li>Then find  of liquid pentane & hexane. </li></ul><ul><li>Then use Raoult’s law to get P in torr of both partial and total pressure </li></ul>
49. 49. Test-type Question rf. Z7e #49 pp <ul><li>25 ml Pen x 0.63g/ml x 1 mol/72.15 g = 0.22 mol Pen (and 0.34 mol Hex) </li></ul><ul><li> pen liq = 0.22/0.56 = 0.39 and  hex liq = 1.00 - 0.39 = 0.61 </li></ul><ul><li>P pen = (  pen liq )(P pen o ) = (0.39)(511 torr) = 2.0 x 10 2 torr ; </li></ul><ul><li>ditto for P hex = 92 torr </li></ul><ul><li>P total = 292 torr = 290 torr (sig figs) </li></ul>
50. 50. Test-type Question pp <ul><li>Pentane (C 5 H 12 ) & Hexane (C 6 H 14 ) form an ideal solution with V p 25° of 511 & 150. Torr respectively. A solution is made from 25 ml pen (d = 0.63 g/mL) & 45 mL hex (d = 0.66 g/ML </li></ul><ul><li>(b) What is composition by mole fraction of pentane in the vapor that is in equilibrium with this solution? </li></ul><ul><li>Hint: Since partial pressure of gas is proportional to number of moles of gas present, use mol pentane in vapor ÷ total mol vapor to get  </li></ul>
51. 51. Test-type Question pp <ul><li>Pentane (C 5 H 12 ) & Hexane (C 6 H 14 ) form an ideal solution with V p 25° of 511 & 150. Torr respectively. A solution is made from 25 ml pen (d = 0.63 g/mL) & 45 mL hex (d = 0.66 g/ML </li></ul><ul><li>(b) What is composition by mole fraction of pentane in the vapor that is in equilibrium with this solution? </li></ul><ul><li>The answer is . . . </li></ul><ul><li> pen v = P pen /P total = 2.0 x 10 2 torr/290 torr = 0.69. </li></ul>
52. 52. Deviations to non-ideality <ul><li>If Solvent has a strong affinity for solute (H bonding). </li></ul><ul><li>Lowers solvent’s ability to escape. </li></ul><ul><li>Lower vapor pressure than expected. </li></ul><ul><li>If (-) deviation from Raoult’s law then . . . </li></ul><ul><li> H soln is large and (-), so exothermic . </li></ul><ul><li>An Endothermic  H soln indicates positive deviation. </li></ul>
53. 53. Colligative Properties <ul><li>Dissolved particles affect vapor pressure, so they affect phase changes. </li></ul><ul><li>Colligative properties depend only on the number - not the kind of solute particles present </li></ul><ul><li>The 3 colligative properties are Bp , Fp and osmotic pressure </li></ul><ul><li>Useful to determine molar mass (Lab 11) </li></ul><ul><li>M = (K fp x g solute )/(kg solvent x ∆T fp ) </li></ul>
54. 54. Colligative Properties pp <ul><li>See SE 11.6 p. 500 ( this type will be on test) </li></ul><ul><li>Predict the V p of a solution of 35.0 g Na 2 SO 4 (M = 142 g/mol) & 175 g H 2 0, where V p H 2 O = 23.76 torr at 25 o C. . . </li></ul><ul><li>Strategy: use Raoult’s Law P soln =  H 2 O P o H 2 O </li></ul><ul><li> H 2 O = n H2O /(n H2O + n solute ) </li></ul><ul><li>But . . . Since the solute falls apart into 3 particles, the n solute has to be the moles of Na 2 SO 4 x 3 . </li></ul>
55. 55. Colligative Properties pp <ul><li>See SE 11.6 p. 532 ( this type will be on test) </li></ul><ul><li>n H2O = 175 g/18.0 g/mol = 9.72 mol H 2 O </li></ul><ul><li>n Na2SO4 = 35.0 g/142 g/mol = 0.246 </li></ul><ul><li>n solute = 3 x n Na2SO4 = 0.738 </li></ul><ul><li> H2O = 9.72/(0.738 + 9.72) = 0.929 </li></ul><ul><li>P soln =  H2O P o H2O = (0.929)(23.76 torr) = 22.1 torr </li></ul>
56. 56. 11.5 Boiling point Elevation & Freezing Point Depression <ul><li>Because a nonvolatile solute lowers the vapor pressure it raises the boiling point. </li></ul><ul><li>Stays liquid longer because intermolecular forces between solute/solvent help hold the liquid together. </li></ul><ul><li>The equation is:  T = K b m solute (for a molecular solute; otherwise use van’t Hoff factor, i , for electrolyte solutions). </li></ul><ul><li> T is the change in the boiling point </li></ul><ul><li>K b is a constant determined by the solvent (look up in a table). </li></ul><ul><li>m solute is the molality of the solute </li></ul>
57. 57. Freezing point Depression <ul><li>Because a nonvolatile solute lowers the vapor pressure of the solution it lowers the freezing point. </li></ul><ul><li>Stays liquid longer (before freezing) because IMF of solute/solution prevents forming orderly crystal </li></ul><ul><li>The equation is:  T = K f m solute </li></ul><ul><li> T is the change in the freezing point </li></ul><ul><li>K f is a constant determined by the solvent </li></ul><ul><li>m solute is the molality of the solute </li></ul>
58. 58. 1 atm Vapor Pressure of solution Vapor Pressure of pure water
59. 59. 1 atm Freezing and boiling points of water
60. 60. 1 atm Freezing and boiling points of solution
61. 61. 1 atm  T f  T b
62. 62. Figure 11.14 p. 536 The Development of Osmotic Pressure pp AP requires you to diagram the change when adding solute to a solvent The effect of adding a solute is to extend the range of a solvent.
63. 63. 11.6 Osmotic Pressure <ul><li>Osmosis : The flow of solvent into the solution through a semi permeable membrane. </li></ul><ul><li>Osmotic Pressure : The excess hydrostatic pressure on the solution compared to the pure solvent. </li></ul><ul><li>Osmotic Pressure π = M RT </li></ul><ul><li>Watch units for R! </li></ul>
64. 64. Figure 11.16 Osmotic Pressure. Net transfer of solvent molecules until hydrostatic pressure equalizes the solvent flow in both directions
65. 65. Figure 11.18 Osmosis at Equilibrium
66. 66. <ul><li>If the external pressure is larger than the osmotic pressure, reverse osmosis occurs. </li></ul><ul><li>One application is desalination of seawater . </li></ul>
67. 67. Figure 11.17 Osmosis When pressure exceeds this then reverse osmosis occurs.
68. 68. Review Problems Bp, Fp, π <ul><li>Try SE 11.8 p. 537 (& Text # 60) </li></ul><ul><li>Try SE 11.10 p. 539 (& Text # 65) </li></ul><ul><li>Try SE 11.11 p. 541 (& Text # 68) </li></ul><ul><li>Try SE 11.12 (& Text # 70) </li></ul><ul><li>#60 = 498 g/mol </li></ul><ul><li>#65 = 456 g/mol </li></ul><ul><li>#68 = 27 000 g/mol </li></ul><ul><li>#70 = Dissolve 18 g in 1 L of solution </li></ul>
69. 69. 11.7 Colligative Properties of Electrolyte Solutions <ul><li>Since colligative properties only depend on the number of particles: </li></ul><ul><li>Ionic compounds should have a bigger effect because . . . </li></ul><ul><li>When they dissolve they dissociate. </li></ul><ul><li>Individual Na and Cl ions fall apart. </li></ul><ul><li>1 mole of NaCl makes 2 moles of ions. </li></ul><ul><li>1 mole Al(NO 3 ) 3 makes 4 moles ions. </li></ul>
70. 70. <ul><li>Electrolytes have a bigger impact on on melting and freezing points per mole because they make more pieces. </li></ul><ul><li>Relationship is expressed using the van’t Hoff factor i </li></ul><ul><li>i = Moles of particles in solution </li></ul><ul><li>Moles of solute dissolved </li></ul><ul><li>The expected value can be determined from the above formula. </li></ul>
71. 71. <ul><li>The actual value is usually less because: </li></ul><ul><li>At any given instant some of the ions in solution will be paired. </li></ul><ul><li>Ion pairing increases with concentration. </li></ul><ul><li>i decreases with increased concentration. </li></ul><ul><li>We can change our generic formulae to </li></ul><ul><li> H = i K m </li></ul>
72. 72. Colligative Properties of Electrolytes <ul><li> T = i K b m solute = Bp elevation </li></ul><ul><li> T = i K f m solute = Fp depression </li></ul><ul><li>π = iM RT = Osmotic Pressure </li></ul><ul><li>11.8 - Tyndall effect = the scattering of light by particles (read on own). </li></ul><ul><li>End, Chapter 11. </li></ul>
73. 73. <ul><li>Label your solutions, in the flasks and the ice cube trays. </li></ul><ul><li>Final conclusion will be to compare the actual freezing point depression to the theoretical. </li></ul><ul><li>Give reasons for any differences. </li></ul>