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Role of Disorder in Solutions

Role of Disorder in Solutions. Ch 13: Solutions. Disorder (Entropy) is a factor Solutions mix to form maximum disorder. Two Ways to Form Solutions. 1. Physical Dissolving (Solvation) NaCl(s)  Na + (aq) + Cl - (aq) C 12 H 22 O 11 (s)  C 12 H 22 O 11 (aq)

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Role of Disorder in Solutions

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  1. Role of Disorder in Solutions Ch 13: Solutions • Disorder (Entropy) is a factor • Solutions mix to form maximum disorder

  2. Two Ways to Form Solutions 1. Physical Dissolving (Solvation) • NaCl(s)  Na+(aq) + Cl-(aq) • C12H22O11(s)  C12H22O11(aq) • Particles are surrounded by solvent molecules • Can evaporate water/solvent to get original compound back

  3. Types of Reactions

  4. Chemical reaction • Ni(s) + 2HCl(aq)  NiCl2(aq) + H2(g) • Evaporating solvent gives the products

  5. Solubility – Maximum amount of a solute that can dissolve in 100 mL of a solution Ex: NaCl 35.7 g/100mL • Saturated solution – Contains the max. amount of solute with some undissolved solid, • Unsaturated – more solute will dissolve.

  6. Supersaturated – More than the max is dissolved by heating and slowly cooling.

  7. Like Dissolves Like: Miscibility • Polar dissolves polar (dipole-dipole Forces) and ionic (ion:dipole) • Water and Ammonia • Non-Polar dissolves non-polar (London Forces) • Soap and grease

  8. :O: || CH3CCH3 Would acetone (shown below) dissolve in water? Acetone

  9. Using you knowledge of “like dissolves like”, explain the following trends in solubility.

  10. Pressure Effects • Solubilty of a gas increases with pressure of gas over the liquid (soda bottle) • Henry’s Law Sgas = kPgas

  11. Henry’s Law The Henry’s law constant for CO2 is 0.031 mol/L-atm. • Calculate the concentration of CO2 in a soda bottle pressurized to 4.00 atm of CO2. • After the bottle has been opened, the concentration drops to 9.3 X 10-6 M. Calculate the partial pressure of CO2 over the soda.

  12. Temperature Effects • Solubility of most solids increases with temperature • Solubility of most gases decreases with temperature (warm soda) • Warm water is deoxygenated • Problem with thermal pollution of lakes

  13. Ways of Expressing Concentration Mass % = mass of compound in soln X 100 total mass of soln Parts Per Million ppm = mass of component X 106 total mass of soln

  14. Concentration: Ex 1 13.5 g of C6H12O6 is dissolved in 0.100 kg of water. Calculate the mass percentage. mass % = 13.5 g X 100 = 11.9% (100 g + 13.5 g)

  15. Concentration: Ex 2 A 2.5 g sample of groundwater is found to contain 5.4 mg of Zn2+.What is the concentration of the Zn2+ ion in ppm.

  16. Concentration: Ex 2 A 2.5 g sample of groundwater is found to contain 5.4 mg of Zn2+.What is the concentration of the Zn2+ ion in ppm. 5.4 mg | 1X10-6g = 5.4 X 10-6 g | 1 mg ppm = mass of component X 106 total mass of soln ppm = 5.4 X 10-6 g X 106 = 2.2 ppm 2.5g

  17. Concentration: Ex 3 Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. ANS: 2.91 %

  18. Concentration: Ex 4 Bleach is 3.62 % NaOCl. What mass of NaOCl is contained in 2500 g of bleach? ANS: 90.5 g NaOCl

  19. Mole Fraction Mole Fraction X = moles of component total moles of all components What is the mole fraction of HCl if 36.5 grams is dissolved in 144 grams of water? ANS: 0.111

  20. Molality Molality = moles of solute kilograms of solvent Why not use Molarity? • Molarity varies with temperature • Total volume of a solution changes with temperature (liquid expands) • Mass does not change with temperature

  21. Molality: Ex 1 A solution is made by dissolving 4.35 grams of C6H12O6 in 25.0 mL of water. Calculate the molality of the glucose. 4.35 g 1 mol = 0.0241 mol 180.2 g Molality = 0.0241 mol = 0.964 m 0.0250 kg

  22. Molality: Ex 2 Calculate the molality of a solution made by dissolving 36.5g C10H8 in 425 grams of toluene (solvent). ANS: 0.671 m

  23. Molality: Ex 3 A solution of HCl contains 36 percent HCl by mass. Calculate the mole fraction and molality of HCl. Pretend 100 g 36 g HCl 64 g H2O

  24. 36 g HCl 1 mol HCl = 0.99 mol HCl 36.5 g HCl 64 g H2O 1 mol H2O = 3.6 mol HCl 18 g H2O XHCl = 0.99 mol HCl = 0.22 0.99 mol + 3.6mol Molality = 0.99 mol HCl = 15 m 0.064 kg H2O

  25. Molality: Ex 4 A commercial bleach solution contains 3.62 percent NaOCl by mass. Calculate the mole fraction and molality of NaOCl. ANS: XNaOCl = 0.00900, 0.505 m

  26. Molality: Ex 5 The density of a solution of 5.0 g of toluene (C7H8) and 225 g of benzene (C6H6) is 0.876 g/mL. Calculate the molality and molarity of the toluene.

  27. Molality: Ex 5 The density of a solution of 5.0 g of toluene (C7H8) and 225 g of benzene (C6H6) is 0.876 g/mL. Calculate the molality and molarity of the solution. Molality 5.0 g 1 mol = 0.054 mol 92.0 g m = 0.054 mol/ 0.225 kg = 0.24 m

  28. Molarity D = mass/V V = mass/D V = 230 g = 263 mL 0.876 g/ml M = 0.054 mol = 0.21 M 0.263 L

  29. Molality: Ex 6 A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/mL. Calculate • molality (10.9 m) • mole fraction (XC3H8O3 = 0.163) • molarity of glycerol in the solution (5.97 M)

  30. Colligative Properties: Vapor Pressure Lowering Non-volatile solutes lower the vapor pressure of the solvent Raoult’s law PA = XAPoA PA = Vapor pressure XA = Mole fraction of solvent PoA = Pressure of pure solvent

  31. Raoult’s Law: Ex 1 What is the vapor pressure of a solution made by adding 50.0 mL of glycerin (C3H8O3) to 500.0 mL of water? The density of glycerin is 1.26 g/mL and the vapor pressure of pure water is 23.8 torr. MassC3H8O3 = (50.0 mL)(1.26 g/mL ) = 63.0 g MolesC3H8O3 = 63.0 g/92.1 g/mol = 0.684 mol MolesH2O = 500.0 g/18 g/mol = 27.8 mol

  32. XH2O = 27.8 mol = 0.976 (27.8 mol + 0.684 mol) PA = XAPoA PA = (0.976)(23.8 torr) = 23.2 torr

  33. Raoult’s Law: Ex 2 The vapor pressure of water at 110oC is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1 atm at 110oC. What is the mole fraction of ethylene glycol in the solution? ANS: 0.290

  34. Colligative Properties: Boiling Point Elevation • Non-volatile solute raises the boiling point of a solution • Shifts the phase diagram • The pressure of the solution reaches atmospheric pressure at a higher temp. DTb = iKbm

  35. Colligative Properties: Freezing Point Depression • Solutions freeze at a lower temperature than pure solvent • Salt water freezes lower (-2oC) than distilled water (0oC) DTf = iKfm i = Van’t Hoff factor m = molality of the nonvolatile solute

  36. The more ions produced, the greater the freezing point depression or boiling point elevation • C12H22O11 (i=1) • NaCl = Produces two ions (i=2) • CaCl2 = Produces three ions (i=3)

  37. Colligative: Ex 1 Ethylene Glycol, C2H6O2, is used in antifreeze. What will be the freezing and boiling point of a 25.0 mass percent solution of ethylene glycol and water? Pretend 100 grams of solution 25 grams of C2H6O2 75 grams of H2O (0.075 kg)

  38. 25 grams of C2H6O2 = 0.403 moles m = 0.403 moles = 5.37 m 0.075 kg H2O DTb = iKbm= (1)(0.52oC/m)(5.37 m) = 2.8oC DTf = iKfm = (1)( 1.86oC/m)(5.37 m) =10.0oC Boiling Point = 102.8oC Freezing Point = -10.0oC o o o

  39. Colligative: Ex 2 Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of C10H18O. Kf for CHCl3 is 4.68oC/m and the normal freezing point is -63.5 oC. ANS: -65.6oC

  40. Colligative: Ex 3 Rank the following aqueous solutions in order of their expected freezing points: 0.050 m CaCl2 0.15 m NaCl 0.10 m HCl 0.050 m HC2H3O2 (acetic acid) 0.10 m C12H22O11 (sugar)

  41. 0.050 m CaCl2 (0.15 m in particles) 0.15 m NaCl (0.30 m in particles) 0.10 m HCl (0.20 m in particles) 0.050 m HC2H3O2 (just above 0.05 m) 0.10 m C12H22O11 (0.10 m in particles) Lowest FP Highest FP NaCl < HCl < CaCl2 < C12H22O11 < HC2H3O2

  42. Colligative: Ex 4 Rank the following in order of the increase in boiling point that they will produce in 1 kg of water 1 mol Co(NO3)2 2 mol KCl 3 mol C2H6O2 (a very, very weak electrolyte(acidic))

  43. 1 mol Co(NO3)2 (3 mol particles) 2 mol KCl (4 mol of particles) 3 mol C2H6O2 (3+ mol of particles) Lowest BP Highest BP Co(NO3)2 < C2H6O2 < KCl

  44. Freezing Pt Depression: Ex 5 What would be the molality of salt water if it freezes at 0 oF? Kf= 1.86 oC/m. ANS: 4.78 m

  45. Colligative Properties: Osmotic Pressure • Osmosis – movement of solvent from high concentration to low concentration • semipermeable membrane – allows to passage of some particles but not others Cucumber Skin cell after in salt water soaking in a tub

  46. Note that solvent moves both ways • Solute too large to pass through membrane • Net movement is to try to dilute the side with solutes

  47. Osmotic Pressure (p) – pressure required to prevent osmosis PV = inRT • V = inRT p = inRT V p = iMRT M = molarity

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