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Chapter 14 Solutions and Their Behavior

Chapter 14 Solutions and Their Behavior. Important – Read Before Using Slides in Class

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Chapter 14 Solutions and Their Behavior

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  1. Chapter 14Solutions and Their Behavior

  2. Important – Read Before Using Slides in Class Instructor: This PowerPoint presentation contains photos and figures from the text, as well as selected animations and videos. For animations and videos to run properly, we recommend that you run this PowerPoint presentation from the PowerLecture disc inserted in your computer. Also, for the mathematical symbols to display properly, you must install the supplied font called “Symb_chm,” supplied as a cross-platform TrueType font in the “Font_for_Lectures” folder in the "Media" folder on this disc. If you prefer to customize the presentation or run it without the PowerLecture disc inserted, the animations and videos will only run properly if you also copy the associated animation and video files for each chapter onto your computer. Follow these steps: 1. Go to the disc drive directory containing the PowerLecture disc, and then to the “Media” folder, and then to the “PowerPoint_Lectures” folder. 2. In the “PowerPoint_Lectures” folder, copy the entire chapter folder to your computer. Chapter folders are named “chapter1”, “chapter2”, etc. Each chapter folder contains the PowerPoint Lecture file as well as the animation and video files. For assistance with installing the fonts or copying the animations and video files, please visit our Technical Support at http://academic.cengage.com/support or call (800) 423-0563. Thank you.

  3. SolutionsChapter 14 Why does a raw egg swell or shrink when placed in different solutions? PLAY MOVIE

  4. Some Definitions A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

  5. Definitions Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. PLAY MOVIE

  6. Definitions Solutions can be classified as unsaturated or saturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. SUPERSATURATED SOLUTIONS contain more than is possible and are unstable. PLAY MOVIE

  7. Dissolving An Ionic Solid See Active Figure 14.9

  8. Energetics of the Solution Process See Figure 14.8

  9. Energetics of the Solution Process If the enthalpy of formation of the solution is more negative that that of the solvent and solute, the enthalpy of solution is negative. The solution process is exothermic! PLAY MOVIE

  10. Supersaturated Sodium Acetate • One application of a supersaturated solution is the sodium acetate “heat pack.” • Sodium acetate has an ENDOthermic heat of solution. PLAY MOVIE

  11. Supersaturated Sodium Acetate Sodium acetate has an ENDOthermic heat of solution. NaCH3CO2 (s) + heatf Na+(aq) + CH3CO2-(aq) Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC. Na+(aq) + CH3CO2-(aq) f NaCH3CO2 (s) + heat

  12. Colligative Properties On adding a solute to a solvent, the props. of the solvent are modified. • Vapor pressure decreases • Melting point decreases • Boiling point increases • Osmosis is possible (osmotic pressure) These changes are called COLLIGATIVE PROPERTIES. They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

  13. Concentration Units An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need conc. units to tell us the number of solute particles per solvent particle. The unit “molarity” does not do this!

  14. Concentration Units MOLE FRACTION, X For a mixture of A, B, and C MOLALITY, m WEIGHT % = grams solute per 100 g solution

  15. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate mol fraction, molality, and weight % of glycol.

  16. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol. 250. g H2O = 13.9 mol X glycol = 0.0672

  17. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol. Calculate molality Calculate weight %

  18. Dissolving Gases & Henry’s Law PLAY MOVIE Gas solubility (mol/L) = kH· Pgas kH for O2 = 1.66 x 10-6 M/mmHg When Pgas drops, solubility drops.

  19. Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

  20. Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution. PLAY MOVIE PLAY MOVIE

  21. Understanding Colligative Properties VP of H2O over a solution depends on the number of H2O molecules per solute molecule. Psolventproportional to Xsolvent Psolvent = Xsolvent · Posolvent VP of solvent over solution = (Mol frac solvent)•(VP pure solvent) RAOULT’S LAW

  22. Raoult’s Law An ideal solution is one that obeys Raoult’s law. PA = XA · PoA Because mole fraction of solvent, XA, is always less than 1, then PA is always less than PoA. The vapor pressure of solvent over a solution is always LOWERED!

  23. Raoult’s Law Assume the solution containing 62.1 g of glycol in 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 oC? (The VP of pure H2O is 31.8 mm Hg; see App. E.) Solution Xglycol = 0.0672 and so Xwater = ? Because Xglycol + Xwater = 1 Xwater = 1.000 - 0.0672 = 0.9328 Pwater = Xwater· Powater = (0.9382)(31.8 mm Hg) Pwater = 29.7 mm Hg

  24. Raoult’s Law For a 2-component system where A is the solvent and B is the solute ∆PA = VP lowering = XBPoA VP lowering is proportional to mol frac solute! For very dilute solutions, ∆PA = K·molalityB where K is a proportionality constant. This helps explain changes in melting and boiling points.

  25. Changes in Freezing and Boiling Points of Solvent See Figure 14.13

  26. Vapor Pressure Lowering See Figure 14.13

  27. The boiling point of a solution is higher than that of the pure solvent. PLAY MOVIE

  28. Elevation of Boiling Point Elevation in BP = ∆TBP = KBP·m (where KBP is characteristic of solvent)

  29. Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP of the solution? KBP = +0.512 oC/molal for water (see Table 14.3). Solution 1. Calculate solution molality = 4.00 m 2. ∆TBP = KBP· m ∆TBP = +0.512 oC/molal (4.00 molal) ∆TBP = +2.05 oC BP = 102.05 oC

  30. Change in Freezing Point Ethylene glycol/water solution Pure water The freezing point of a solution is LOWERthan that of the pure solvent. FP depression = ∆TFP = KFP·m PLAY MOVIE PLAY MOVIE

  31. Lowering the Freezing Point Water with and without antifreeze When a solution freezes, the solid phase is pure water. The solution becomes more concentrated.

  32. Freezing Point Depression Calculate the FP of a 4.00 molal glycol/water solution. KFP = -1.86 oC/molal (Table 14.3) Solution ∆TFP = KFP· m = (-1.86 oC/molal)(4.00 m) ∆TFP = -7.44 oC Recall that ∆TBP = +2.05 ˚C for this solution.

  33. Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?. Solution Calc. required molality ∆TFP = KFP· m -10.00 oC = (-1.86 oC/molal)(Conc) Conc = 5.38 molal

  34. Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?. Solution Conc req’d = 5.38 molal This means we need 5.38 mol of dissolved particles per kg of solvent. Recognize that m represents the total concentration of all dissolved particles. Recall that 1 mol NaCl(aq) f 1 mol Na+(aq) + 1 mol Cl-(aq)

  35. Freezing Point Depression How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?. Solution Conc req’d = 5.38 molal We need 5.38 mol of dissolved particles per kg of solvent. NaCl(aq) f Na+(aq) + Cl-(aq) To get 5.38 mol/kg of particles we need 5.38 mol / 2 = 2.69 mol NaCl / kg 2.69 mol NaCl / kg f 157 g NaCl / kg (157 g NaCl / kg)(4.00 kg) = 629 g NaCl

  36. Boiling Point Elevation and Freezing Point Depression ∆T = K·m·i A generally useful equation i = van’t Hoff factor = number of particles produced per formula unit. Compound Theoretical Value of i glycol 1 NaCl 2 CaCl2 3

  37. Osmosis PLAY MOVIE Dissolving the shell in vinegar Egg in pure water Egg in corn syrup

  38. Osmosis The semipermeable membrane allows only the movement of solvent molecules. Solvent molecules move from pure solvent to solution in an attempt to make both have the same concentration of solute. Driving force is entropy

  39. Process of Osmosis

  40. Osmotic pressure Osmotic Pressure, ∏ Equilibrium is reached when pressure — the OSMOTIC PRESSURE, ∏ — produced by extra solution counterbalances pressure of solvent molecules moving thru the membrane. ∏ = cRT (c is conc. in mol/L)

  41. Osmosis PLAY MOVIE PLAY MOVIE

  42. Osmosis at the Particulate Level See Figure 14.17

  43. Osmosis • Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC— they have the same concentration.

  44. Osmosis and Living Cells

  45. Reverse OsmosisWater Desalination Water desalination plant in Tampa

  46. Osmosis Calculating a Molar Mass Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. ∏ measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin. Solution (a) Calc. ∏ in atmospheres ∏ = (10.0 mmHg)(1 atm / 760 mmHg) = 0.0132 atm (b) Calc. concentration

  47. Osmosis Calculating a Molar Mass Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. ∏ measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin. Solution (b) Calc. concentration from ∏ = cRT Conc = 5.39 x 10-4 mol/L (c) Calc. molar mass Molar mass = 35.0 g / 5.39 x 10-4 mol/L Molar mass = 65,100 g/mol

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