1 / 44

Physical Properties of Solutions

Physical Properties of Solutions. Chapter 12. Solution Stoichiometry end of Chapter 4. Problems: 12.12, 12.15, 12.16, 12.17, 12.18, 12.21, 12.22, 12.28, 12.36, 12.38, 12.51, 12.54, 12.55, 12.57, 12.60, 12.65, 12.76, 12.77, 12.112 4.12, 4.60, 4.70, 4.72, 4.74, 4.77, 4.88.

travis-travis
Download Presentation

Physical Properties of Solutions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Physical Properties of Solutions Chapter 12 SolutionStoichiometry end of Chapter 4 Problems: 12.12, 12.15, 12.16, 12.17, 12.18, 12.21, 12.22, 12.28, 12.36, 12.38, 12.51, 12.54, 12.55, 12.57, 12.60, 12.65, 12.76, 12.77, 12.112 4.12, 4.60, 4.70, 4.72, 4.74, 4.77, 4.88

  2. A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount 12.1

  3. nonelectrolyte weak electrolyte strong electrolyte An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. 4.1

  4. A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. 12.1

  5. Three types of interactions in the solution process: • solvent-solvent interaction • solute-solute interaction • solvent-solute interaction DHsoln = DH1 + DH2 + DH3 12.2

  6. “like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. • non-polar molecules are soluble in non-polar solvents • CCl4 in C6H6 • polar molecules are soluble in polar solvents • C2H5OH in H2O • ionic compounds are more soluble in polar solvents • NaCl in H2O or NH3 (l) 12.2

  7. moles of A XA = sum of moles of all components x 100% mass of solute x 100% = mass of solution mass of solute mass of solute + mass of solvent Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass % by mass = Mole Fraction(X) 12.3

  8. moles of solute liters of solution moles of solute m = mass of solvent (kg) M = Concentration Units Continued Molarity(M) Molality(m) 12.3

  9. moles of solute moles of solute m= m= moles of solute M = mass of solvent (kg) mass of solvent (kg) liters of solution 5.86 moles C2H5OH = 0.657 kg solvent What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? • Assume 1 L of solution: • 5.86 moles ethanol = 270 g ethanol • 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg = 8.92 m 12.3

  10. Convert % mass to Molarity • What is the Molarity of a 95% acetic acid solution? (density = 1.049 g/mL) If you assume 1 L, that amount of solution = 1049 g 95% of the solution is acetic acid 1049 g solution x 0.95 = 997 g solute 997 g X 1 mol/60.05 g = 16.6 mol solute Since we assumed 1 L, that’s 16.6 mol / 1 L or 16.6 M

  11. solubility increases with increasing temperature solubility decreases with increasing temperature Temperature and Solubility Solid solubility and temperature 12.4

  12. Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have 90 g KNO3 contaminated with 10 g NaCl. • Fractional crystallization: • Dissolve sample in 100 mL of water at 600C • Cool solution to 00C • All NaCl will stay in solution (s = 34.2g/100g) • 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g 12.4

  13. Temperature and Solubility Gas solubility and temperature solubility usually decreases with increasing temperature 12.4

  14. low P high P low c high c Pressure and Solubility of Gases The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). c is the concentration (M) of the dissolved gas c = kP P is the pressure of the gas over the solution k is a constant (mol/L•atm) that depends only on temperature 12.5

  15. Chemistry In Action: The Killer Lake 8/21/86 CO2 Cloud Released 1700 Casualties • Trigger? • earthquake • landslide • strong Winds Lake Nyos, West Africa

  16. moles of solute M = molarity = What mass of KI is required to make 500. mL of a 2.80 M KI solution? liters of solution volume KI moles KI grams KI 1 L 2.80 mol KI 166 g KI x x x 1000 mL 1 L soln 1 mol KI Solution Stoichiometry (Chapter 4) The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M KI M KI 500. mL = 232 g KI 4.5

  17. 4.5

  18. Moles of solute before dilution (i) Moles of solute after dilution (f) = Dilution Add Solvent = MfVf MiVi Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. 4.5

  19. How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? MfVf 0.200 x 0.06 Vi = = 4.00 Mi MiVi = MfVf Mi = 4.00 Vi = ? L Mf = 0.200 Vf = 0.06 L = 0.003 L = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution 4.5

  20. Gravimetric Analysis • Dissolve unknown substance in water • React unknown with known substance to form a precipitate • Filter and dry precipitate • Weigh precipitate • Use chemical formula and mass of precipitate to determine amount of unknown ion 4.6

  21. Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 4.7

  22. What volume of a 1.420 M NaOH solution is Required to titrate 25.00 mL of a 4.50 M H2SO4 solution? WRITE THE CHEMICAL EQUATION! H2SO4 + 2NaOH 2H2O + Na2SO4 M M rx volume acid moles acid moles base volume base base acid coef. 4.50 mol H2SO4 2 mol NaOH 1000 ml soln x x x 1000 mL soln 1 mol H2SO4 1.420 mol NaOH 25.00 mL = 158 mL 4.7

  23. CaCO3 (s) CaO (s) + CO2 (g) - CaO (s) + H2O (l) Ca2+ (aq) + 2OH (aq) - Mg2+ (aq) + 2OH (aq) Mg(OH)2(s) Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l) Mg2+ + 2e- Mg 2Cl- Cl2 + 2e- MgCl2 (l) Mg (l) + Cl2 (g) Chemistry in Action: Metals from the Sea Now back to Chapter 12…

  24. = vapor pressure of pure solvent 0 P1 = X1 P 1 0 0 0 P 1 P 1 P 1 - P1 = DP = X2 Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering X1= mole fraction of the solvent Raoult’s law If the solution contains only one solute: X1 = 1 – X2 X2= mole fraction of the solute 12.6

  25. Fractional Distillation Apparatus 12.6

  26. 0 DTb = Tb – T b 0 T b is the boiling point of the pure solvent 0 Tb > T b DTb = Kbm Boiling-Point Elevation T b is the boiling point of the solution DTb > 0 m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) 12.6

  27. 0 DTf = T f – Tf 0 T f is the freezing point of the pure solvent 0 T f > Tf DTf = Kfm Freezing-Point Depression T f is the freezing point of the solution DTf > 0 m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) 12.6

  28. 12.6

  29. 0 DTf = T f – Tf moles of solute m= mass of solvent (kg) = 3.202 kg solvent 1 mol 62.01 g 478 g x 0 Tf = T f – DTf What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g. DTf = Kfm Kf water = 1.86 0C/m = 2.41 m DTf = Kfm = 1.86 0C/m x 2.41 m = 4.48 0C = 0.00 0C – 4.48 0C = -4.48 0C 12.6

  30. Vapor-Pressure Lowering Boiling-Point Elevation DTb = Kbm 0 P1 = X1 P 1 Freezing-Point Depression DTf = Kfm p = MRT Osmotic Pressure (p) Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 12.6

  31. actual number of particles in soln after dissociation van’t Hoff factor (i) = number of formula units initially dissolved in soln Colligative Properties of Electrolyte Solutions 0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution 0.2 m ions in solution i should be 1 nonelectrolytes 2 NaCl CaCl2 3 12.7

  32. Change in Freezing Point • Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision? • sand, SiO2 • Rock salt, NaCl • Ice Melt, CaCl2

  33. Boiling-Point Elevation DTb = iKbm Freezing-Point Depression DTf = i Kfm p = iMRT Osmotic Pressure (p) Colligative Properties of Electrolyte Solutions 12.7

  34. Change in Freezing Point Common Applications of Freezing Point Depression Ethylene glycol – deadly to small animals Propylene glycol

  35. Change in Boiling Point Common Applications of Boiling Point Elevation

  36. Freezing Point Depression At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆TFP = Kf • m • i ∆TFP = (1.86 oC/molal) • 5.4 m • 2 ∆TFP = 20.1oC FP = 0 – 20.1 = -20.1oC

  37. The Cleansing Action of Soap 12.8

  38. Osmotic Pressure (p) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (p) is the pressure required to stop osmosis. more concentrated dilute 12.6

  39. Osmotic Pressure (p) High P Low P p = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K) 12.6

  40. A cell in an: isotonic solution hypotonic solution hypertonic solution 12.6

  41. Chemistry In Action: Desalination

  42. A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. • Colloid versus solution • collodial particles are much larger than solute molecules • collodial suspension is not as homogeneous as a solution 12.8

  43. Colloids • Brownian motion • Tyndall Effect

  44. Suspensions • These are mixed, but not dissolved in each other • Will settle over time • Particles are bigger than 1 micrometer (larger than colloid) • Examples: dust in air, muddy water

More Related