1. John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend http://academic.cengage.com/kotz Chapter 14 Solutions and Their Behavior

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3. 3 © 2009 Brooks/Cole - Cengage Solutions Chapter 14 Why does a raw egg swell or shrink when placed in different solutions? PLAY MOVIE

4. 4 © 2009 Brooks/Cole - Cengage 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. 5 © 2009 Brooks/Cole - Cengage Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. Definitions PLAY MOVIE

6. 6 © 2009 Brooks/Cole - Cengage 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. Definitions PLAY MOVIE

7. 7 © 2009 Brooks/Cole - Cengage Dissolving An Ionic Solid See Active Figure 14.9

8. 8 © 2009 Brooks/Cole - Cengage Energetics of the Solution Process See Figure 14.8

9. 9 © 2009 Brooks/Cole - Cengage 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. 10 © 2009 Brooks/Cole - Cengage Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.”One application of a supersaturated solution is the sodium acetate “heat pack.” Sodium acetate has an ENDOthermic heat of solution.Sodium acetate has an ENDOthermic heat of solution. PLAY MOVIE

11. 11 © 2009 Brooks/Cole - Cengage Supersaturated Sodium Acetate Sodium acetate has an ENDOthermic heat of solution. NaCH 3 CO 2 (s) + heat f Na + (aq) + CH 3 CO 2 - (aq) Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC. Na + (aq) + CH 3 CO 2 - (aq) f NaCH 3 CO 2 (s) + heat

12. 12 © 2009 Brooks/Cole - Cengage Colligative Properties On adding a solute to a solvent, the props. of the solvent are modified. Vapor pressure decreasesVapor pressure decreases Melting point decreasesMelting point decreases Boiling point increasesBoiling point increases Osmosis is possible (osmotic pressure)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. 13 © 2009 Brooks/Cole - Cengage 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! Concentration Units

14. 14 © 2009 Brooks/Cole - Cengage Concentration Units MOLE FRACTION, X For a mixture of A, B, and C MOLE FRACTION, X For a mixture of A, B, and C WEIGHT % = grams solute per 100 g solution MOLALITY, m

15. 15 © 2009 Brooks/Cole - Cengage Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate mol fraction, molality, and weight % of glycol.

16. 16 © 2009 Brooks/Cole - Cengage Calculating Concentrations 250. g H 2 O = 13.9 mol Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate X, m, and % of glycol. X glycol = 0.0672

17. 17 © 2009 Brooks/Cole - Cengage Calculating Concentrations Calculate molality Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate X, m, and % of glycol. Calculate weight %

18. 18 © 2009 Brooks/Cole - Cengage Dissolving Gases & Henry’s Law Gas solubility (mol/L) = k H · P gas k H for O 2 = 1.66 x 10 -6 M/mmHg When P gas drops, solubility drops. PLAY MOVIE

19. 19 © 2009 Brooks/Cole - Cengage Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

20. 20 © 2009 Brooks/Cole - Cengage Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution. PLAY MOVIE

21. 21 © 2009 Brooks/Cole - Cengage P solvent = X solvent · P o solvent Understanding Colligative Properties VP of H 2 O over a solution depends on the number of H 2 O molecules per solute molecule. P solvent proportional to X solvent P solvent proportional to X solvent VP of solvent over solution = (Mol frac solvent)(VP pure solvent) RAOULT’S LAW RAOULT’S LAW

22. 22 © 2009 Brooks/Cole - Cengage P A = X A · P o A Raoult’s Law An ideal solution is one that obeys Raoult’s law. Because mole fraction of solvent, X A, is always less than 1, then P A is always less than P o A. The vapor pressure of solvent over a solution is always LOWERED !

23. 23 © 2009 Brooks/Cole - Cengage 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 o C? (The VP of pure H 2 O is 31.8 mm Hg; see App. E.) Solution X glycol = 0.0672and so X water = ? Because X glycol + X water = 1 X water = 1.000 - 0.0672 = 0.9328 P water = X water · P o water = (0.9382)(31.8 mm Hg) P water = 29.7 mm Hg

24. 24 © 2009 Brooks/Cole - Cengage Raoult’s Law For a 2-component system where A is the solvent and B is the solute ∆P A = VP lowering = X B P o A ∆P A = VP lowering = X B P o A VP lowering is proportional to mol frac solute! For very dilute solutions, ∆P A = K·molality B where K is a proportionality constant. This helps explain changes in melting and boiling points.

25. 25 © 2009 Brooks/Cole - Cengage Changes in Freezing and Boiling Points of Solvent See Figure 14.13

26. 26 © 2009 Brooks/Cole - Cengage Vapor Pressure Lowering See Figure 14.13

27. 27 © 2009 Brooks/Cole - Cengage The boiling point of a solution is higher than that of the pure solvent. PLAY MOVIE

28. 28 © 2009 Brooks/Cole - Cengage Elevation of Boiling Point Elevation in BP = ∆T BP = K BP ·m (where K BP is characteristic of solvent)

29. 29 © 2009 Brooks/Cole - Cengage 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? K BP = +0.512 o C/molal for water (see Table 14.3). Solution 1.Calculate solution molality = 4.00 m 2.∆T BP = K BP · m ∆T BP = +0.512 o C/molal (4.00 molal) ∆T BP = +0.512 o C/molal (4.00 molal) ∆T BP = +2.05 o C BP = 102.05 o C

30. 30 © 2009 Brooks/Cole - Cengage Change in Freezing Point The freezing point of a solution is LOWER than that of the pure solvent. FP depression = ∆T FP = K FP ·m Pure water Ethylene glycol/water solution PLAY MOVIE

31. 31 © 2009 Brooks/Cole - Cengage Lowering the Freezing Point Water with and without antifreezeWhen a solution freezes, the solid phase is pure water. The solution becomes more concentrated.

32. 32 © 2009 Brooks/Cole - Cengage Calculate the FP of a 4.00 molal glycol/water solution. K FP = -1.86 o C/molal (Table 14.3) Solution ∆T FP = K FP · m = (-1.86 o C/molal)(4.00 m) = (-1.86 o C/molal)(4.00 m) ∆T FP = -7.44 o C Recall that ∆T BP = +2.05 ˚C for this solution. Freezing Point Depression

33. 33 © 2009 Brooks/Cole - Cengage How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 o C?. Solution Calc. required molality ∆T FP = K FP · m -10.00 o C = (-1.86 o C/molal)(Conc) -10.00 o C = (-1.86 o C/molal)(Conc) Conc = 5.38 molal Conc = 5.38 molal Freezing Point Depression

34. 34 © 2009 Brooks/Cole - Cengage How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 o C?. 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) Freezing Point Depression

35. 35 © 2009 Brooks/Cole - Cengage How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 o C?. 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) 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 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 Freezing Point Depression

36. 36 © 2009 Brooks/Cole - Cengage Boiling Point Elevation and Freezing Point Depression ∆T = K·m·i ∆T = K·m·i A generally useful equation i = van’t Hoff factor = number of particles produced per formula unit. CompoundTheoretical Value of i glycol1 NaCl2 CaCl 2 3

37. 37 © 2009 Brooks/Cole - Cengage Osmosis Dissolving the shell in vinegar Egg in corn syrupEgg in pure water PLAY MOVIE

38. 38 © 2009 Brooks/Cole - Cengage 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. 39 © 2009 Brooks/Cole - Cengage Process of Osmosis

40. 40 © 2009 Brooks/Cole - Cengage Osmotic Pressure, ∏ Equilibrium is reached when pressure — the OSMOTIC PRESSURE, ∏ — produced by extra solution counterbalances pressure of solvent molecules moving thru the membrane. ∏ = cRT ∏ = cRT (c is conc. in mol/L) Osmotic pressure

41. 41 © 2009 Brooks/Cole - Cengage Osmosis PLAY MOVIE

42. 42 © 2009 Brooks/Cole - Cengage Osmosis at the Particulate Level See Figure 14.17

43. 43 © 2009 Brooks/Cole - Cengage Osmosis Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration.Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration.

44. 44 © 2009 Brooks/Cole - Cengage Osmosis and Living Cells

45. 45 © 2009 Brooks/Cole - Cengage Reverse Osmosis Water Desalination Water desalination plant in Tampa

46. 46 © 2009 Brooks/Cole - Cengage 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) ∏ = (10.0 mmHg)(1 atm / 760 mmHg) = 0.0132 atm (b)Calc. concentration

47. 47 © 2009 Brooks/Cole - Cengage Osmosis Calculating a Molar Mass Conc = 5.39 x 10 -4 mol/L 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 = 35.0 g / 5.39 x 10 -4 mol/L Molar mass = 65,100 g/mol Molar mass = 65,100 g/mol 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