1. 1 © 2006 Brooks/Cole - Thomson Chemistry and Chemical Reactivity 6th Edition John C. Kotz Paul M. Treichel Gabriela C. Weaver CHAPTER 14 Solutions and Their Behavior © 2006 Brooks/Cole Thomson Lectures written by John Kotz

2. 2 © 2006 Brooks/Cole - Thomson Solutions Chapter 14 Why does a raw egg swell or shrink when placed in different solutions?

3. 3 © 2006 Brooks/Cole - Thomson 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.

4. 4 © 2006 Brooks/Cole - Thomson Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. Definitions

5. 5 © 2006 Brooks/Cole - Thomson Solutions can be classified as unsaturated or saturated. A saturated solution contains the maximum quantity of solute that dissovles at that temperature. SUPERSATURATED SOLUTIONS contain more than is possible and are unstable. Definitions

6. 6 © 2006 Brooks/Cole - Thomson Dissolving An Ionic Solid Active Figure 14.9

7. 7 © 2006 Brooks/Cole - Thomson Energetics of the Solution Process Figure 14.8

8. 8 © 2006 Brooks/Cole - Thomson 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 !

9. 9 © 2006 Brooks/Cole - Thomson 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.

10. 10 © 2006 Brooks/Cole - Thomson Supersaturated Sodium Acetate Sodium acetate has an ENDOthermic heat of solution. NaCH 3 CO 2 (s) + heat ----> 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) ---> NaCH 3 CO 2 (s) + heat

11. 11 © 2006 Brooks/Cole - Thomson 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.

12. 12 © 2006 Brooks/Cole - Thomson 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

13. 13 © 2006 Brooks/Cole - Thomson 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

14. 14 © 2006 Brooks/Cole - Thomson 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.

15. 15 © 2006 Brooks/Cole - Thomson 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

16. 16 © 2006 Brooks/Cole - Thomson 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 %

17. 17 © 2006 Brooks/Cole - Thomson 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.

18. 18 © 2006 Brooks/Cole - Thomson Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

19. 19 © 2006 Brooks/Cole - Thomson Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

20. 20 © 2006 Brooks/Cole - Thomson 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

21. 21 © 2006 Brooks/Cole - Thomson 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 !

22. 22 © 2006 Brooks/Cole - Thomson 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

23. 23 © 2006 Brooks/Cole - Thomson 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 = Kmolality B where K is a proportionality constant. This helps explain changes in melting and boiling points.

24. 24 © 2006 Brooks/Cole - Thomson Changes in Freezing and Boiling Points of Solvent See Figure 14.14

25. 25 © 2006 Brooks/Cole - Thomson Vapor Pressure Lowering Figure 14.14

26. 26 © 2006 Brooks/Cole - Thomson The boiling point of a solution is higher than that of the pure solvent.

27. 27 © 2006 Brooks/Cole - Thomson Elevation of Boiling Point Elevation in BP = ∆T BP = K BP m (where K BP is characteristic of solvent)

28. 28 © 2006 Brooks/Cole - Thomson 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

29. 29 © 2006 Brooks/Cole - Thomson 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

30. 30 © 2006 Brooks/Cole - Thomson Lowering the Freezing Point Water with and without antifreezeWhen a solution freezes, the solid phase is pure water. The solution becomes more concentrated.

31. 31 © 2006 Brooks/Cole - Thomson Calculate the FP of a 4.00 molal glycol/water solution. K FP = -1.86 o C/molal (Table 14.4) 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

32. 32 © 2006 Brooks/Cole - Thomson 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

33. 33 © 2006 Brooks/Cole - Thomson 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) --> 1 mol Na + (aq) + 1 mol Cl - (aq) Freezing Point Depression

34. 34 © 2006 Brooks/Cole - Thomson 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) --> Na + (aq) + Cl - (aq) NaCl(aq) --> 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 ---> 157 g NaCl / kg (157 g NaCl / kg)(4.00 kg) = 629 g NaCl Freezing Point Depression

35. 35 © 2006 Brooks/Cole - Thomson Boiling Point Elevation and Freezing Point Depression ∆T = Kmi ∆T = Kmi 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

36. 36 © 2006 Brooks/Cole - Thomson Osmosis Dissolving the shell in vinegar Egg in corn syrupEgg in pure water

37. 37 © 2006 Brooks/Cole - Thomson 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

38. 38 © 2006 Brooks/Cole - Thomson Process of Osmosis

39. 39 © 2006 Brooks/Cole - Thomson Osmosic 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

40. 40 © 2006 Brooks/Cole - Thomson Osmosis

41. 41 © 2006 Brooks/Cole - Thomson Osmosis at the Particulate Level Figure 14.17

42. 42 © 2006 Brooks/Cole - Thomson 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. Osmotic pressure in living systems: FIGURE 14.18Osmotic pressure in living systems: FIGURE 14.18

43. 43 © 2006 Brooks/Cole - Thomson Osmosis and Living Cells

44. 44 © 2006 Brooks/Cole - Thomson Reverse Osmosis Water Desalination Water desalination plant in Tampa

45. 45 © 2006 Brooks/Cole - Thomson 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

46. 46 © 2006 Brooks/Cole - Thomson 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