Chapter 11 Properties of Solutions

1. Chapter 11 Properties of Solutions • Solutions – homogeneous mixtures that could be gasses, liquids, or solids. • Let’s remember the terms… • Dilute – relatively little solute present • Concentrated – relatively large amount of solute.

2. More precisely… • We talk about solution composition in terms of Molarity – moles of solute per liter of solution. • And Mass Percent -

3. And even… • The mole fraction – the ratio of the number of moles of a component to the total number of moles of a solution. • And molality – number of moles of solute per kilogram of solvent.

4. Let’s see how they work • A solution is prepared by mixing 1.00g ethanol (C2H5OH) with 100.0g water to give a final volume 101 mL. Calculate the molarity, mass percent, mole fraction and molality of ethanol.

5. Molarity =0.215M

6. Mass percent =0.990% C2H5OH

7. Mole fraction

8. Molality

9. How about Normality • Symbolized by N • It is defined as the number of equivalents per liter of solution. Where the definition of equivalent depends on the reaction taking place in the solution. • THIS IS NOT ON THE AP TEST!!

10. You try one… • The electrolyte in automobile lead storage batteries is a 3.75M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent and molality of the sulfuric acid.

11. 11.2 The energies of solution formation. • Solubility is important! • Real life examples… • Vitamin solubility is correct in determining dosages. • Barium is not soluble, which is why it is used to improve the x-rays of the gratrointestinal tract (although Ba is toxic)

12. The cardinal rule of solubility… LIKE DISSOLVES LIKE Polar solvents dissolve polar or ionic solutes Nonpolar sovents dissolve nonpolar solutes BUT WHY?

13. Assume the formation of a liquid solution occurs in 3 distinct steps Separating the solute into its individual components (expanding the solute) Overcoming the intermolecular forces in the solvent to make room for the solute (expanding the solvent) Allowing the solute and solvent to interact to form the solution. (mixing) What does this all mean????

14. Energy of making solutions Heat of solution ( DHsoln ) is the energy change for making a solution. The energy comes from the 3 steps since forces must be overcome to expand the solute and the solvent.

15. Here’s how… Break apart the solvent Have to overcome attractive forces DH1 >0 (endothermic) Break apart the solute Have to overcome attractive forces DH2>0 (endothermic)

16. 3. Mixing solvent and solute DH3 depends on what you are mixing. If molecules can attract each other DH3 is large and negative. Molecules can’t attract- DH3 issmall and negative. This explains the rule “Like dissolves Like”

18. Heat of solutionendo or exothermic?

19. Generally speaking…Processes that required LARGE amounts of energy tend NOT to occur.

20. Types of Solvent and solutes If DHsoln is small and positive, a solution will still form because of entropy. There are many more ways for them to become mixed than there is for them to stay separate.

21. 11.3 Factors affecting solubility Structure Since molecular structure is what determines polarity, there is a definite connection between structure and solubility Hydrophobic: water-fearing…nonpolar materials Hydrophilic: water-loving…polar substances

22. Let’s think Why does dish detergent (soap) help dissolve grease in water?

23. Soap O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2

24. Soap Hydrophobic non-polar end O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2

25. Soap Hydrophilic polar end O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2

26. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 _

27. A drop of grease in water Grease is non-polar Water is polar Soap lets you dissolve the non-polar in the polar.

28. Hydrophobic ends dissolve in grease

29. Hydrophilic ends dissolve in water

30. Water molecules can surround and dissolve grease. Helps get grease out of your way.

31. Pressure effects Changing the pressure doesn’t affect the amount of solid or liquid that dissolves because liquids and solids are incompressible. However, pressure DOES affect gases.

32. Dissolving Gases Pressure affects the amount of gas that can dissolve in a liquid. The dissolved gas is at equilibrium with the gas above the liquid.

33. The gas is at equilibrium with the dissolved gas in this solution. The equilibrium is dynamic.

34. If you increase the pressure the gas molecules dissolve faster. The equilibrium is disturbed. More gas enters the solution at a higher rate than it leaves.

35. The system reaches a new equilibrium with more gas dissolved. Henry’s Law. C=kP Concentration of gas = constant times pressure. Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.

36. Temperature Effects It is true that an Increased temperature will increase the rate at which a solid dissolves. BUT We can’t predict whether it will increase the amount of solid that dissolves.

37. Solubility vs Temperature Graph

38. Predicting solid solubility with respect to temperature This is NOT someething that is easy to too. The ONLY SURE WAY to determine the temperature dependence of a solid’s solubility is by experiment!!

39. Gasses on the other hand… Are much less complex.. As temperature increases, solubility decreases because the gas molecules can move fast enough to escape.

40. 11.4 Vapor pressures of solutions. Remember from chapter 10 that liquids with high vapor pressures are said to be volatile. A nonvolatile solvent lowers the vapor pressure of the solution

41. Water has a higher vapor pressure than a solution.. So VP to achieve equilibrium with the water > than that required to achieve equilibrium with the solution. Aqueous Solution Pure water

42. Water emits vapor to attempt to reach equilibrium and solution absorbs vapor to lower the VP towards equilibrium. Aqueous Solution Pure water

43. The water condenses faster in the solution. Thus the solution takes on water. This will happen and equilibrium VP will be reached only when all the water is transferred to the solution. Aqueous Solution

44. Raoult’s Law • Psoln = csolvent● P0solvent • Vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent • Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

45. Try this problem. • A solution of cyclopentane with a nonvolatile compound has vapor pressure of 211 torr. If vapor pressure of the pure liquid is 313 torr, what is the mole fraction of the cyclopentane?

46. Another? • Determine the vapor pressure of a solution at 25 C that has 45 grams of C6H12O6, glucose, dissolved in 72 grams of H2O. The vapor pressure of pure water at 25 C is 23.8 torr.

47. Vapor Pressure Lowering: Addition of a Solute

48. By lowering the VP, we now have a way to count molecules and then determine molar masses. Additionally, we can use VP to characterize solutions. For example, NaCl lowers vapor pressure almost twice as much as expected due to the fact that there are 2 ions per formula unit which separate when dissolved. Thus, lowering of vapor pressure depends on the number of solute particle present in the solution.

49. What if... …the solution is a liquid liquid solution and they are both volatile? We need to modify Raoult’s law to account for the partial pressures of both liquids. Ptotal= PA + PB = cAPA0 + cBPB0

50. Ideal solutions If the liquid-liquid solution obey’s Raoult’s law, it is called an ideal solution. As with gasses, ideal solutions are rarely achieved, but often closely approached. Nearly ideal solutions occur when the solute and solvent are very similar and the solute basically just dilutes the solvent.