3. Solutions pg 453 n Solution - n Solution - homogeneous mixture of pure substances. Solvent Solvent – Medium used to dissolve, present in greater amounts in mixture Solute Solute - substance being dissolved

4. B. Solvation n Solvation – n Solvation – the process of dissolving solute particles are separated and pulled into solution solute particles are surrounded by solvent particles ent particles

5. The process of dissolution is favored by: 1) A decrease in the energy of the system (exothermic) 2) An increase in the disorder of the system (entropy)

6. Liquids Dissolving in Liquids n Liquids that are soluble in one another (“mix”) are MISCIBLE. –“LIKE dissolves LIKE” n POLAR liquids are generally soluble in other POLAR liquids. n NONPOLAR liquids are generally soluble in other NONPOLAR liquids.

7. Factors affecting rate of dissolution: think iced tea vs. hot tea & the type of sugar you use: cubes or granulated 1) Surface area / particle size –Greater surface area, faster it dissolves 2) Temperature –Most solids dissolve faster @ higher temps 3) Agitation –Stirring/shaking will speed up dissolution

8. Solution Concentration: pg. 462 n Concentration refers to the amount of solute dissolved in a solution. n If something is a concentrate we usually say it is dissolved into something else n Ex. Orange juice concentrate. We mix it with water to drink it!

9. Saturation: a solid solute dissolves in a solvent until the soln is SATURATED n Unsaturated solution – is able to dissolve more solute n Saturated solution – has dissolved the maximum amount of solute n Supersaturated solution – has dissolved excess solute (at a higher temperature). Solid crystals generally form when this solution is cooled.

10. C. Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

11. SOLUBILITY n Solubility = the amount of solute that will dissolve in a given amount of solvent n We can use a solubility chart like on page 458 to figure this out!

12. C. Solubility n Solubility Curve –shows the dependence of solubility on temperature

13. Factors Affecting Solubility n The nature of the solute and solvent: different substances have different solubilities n Temperature: many solids substances become more soluble as the temp of a solvent increases; however, gases are less soluble in liquids at higher temps. n Pressure: Only affects the solubility of gases. As pressure increases, the solubility of gases increases. (Henry’s Law)

14. n Solids are more soluble at... –high temperatures.  Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: nitrogen narcosis, the “bends,” soda

15. Gases: solubility n Temp and Pressure (think: flat soda) n Have you ever seen mentos in Diet Coke? –Nucleation site: the following factors that contribute to the bubble formation: Diet coke –carbon dioxide is what makes the bubbles form in the first place –in synthetic mixtures aspartam, caffeine and potassium benzoate where shown give better fountains Mentos –the most important property is the rough surface which provides plenty of nucleation sites for bubble formation –the density makes them sink which is ideal as the bubbles formed at the bottom of the bottle help expel much more soda –mentos contains gelatin and gum arabic which could also reduce surface tension

16. Molarity 2M HCl What does this mean?

17. Molarity Calculations molar mass (g/mol) 6.02  10 23 (particles/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molarity (mol/L)

18. B. Molarity Calculations n How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L 0.25 mol 1 L = 7.3 g NaCl 58.44 g 1 mol

19. B. Molarity Calculations n Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g 1 mol 41.99 g = 0.238 mol NaF 0.238 mol 0.25 L M = = 0.95M NaF

20. MOLARITY BY DILUTION n When you dilute a solution, you can use this equation:

21. Example: Describe how you would prepare 2.50 L of 0.665 M Na 2 SO 4 solution starting with: a) 5.00 M Na 2 SO 4 Add 0.333 L of Na 2 SO 4 to 2.17 L of water.

22. Example: Describe how you would prepare 2.50 L of 0.665 M Na 2 SO 4 solution starting with: solid Na 2 SO 4. Dissolve 236 g of Na 2 SO 4 in enough water to create 2.50 L of solution.

23. MASS PERCENT

24. n Example: What is the percent of NaCl in a solution made by dissolving 24 g of NaCl in 152 g of water?

25. *MOLALITY

26. MOLALITY n Example: What is the molality of a solution that contains 12.8 g of C 6 H 12 O 6 in 187.5 g water?

27. MOLALITY n Example: How many grams of H 2 O must be used to dissolve 50.0 g of sucrose to prepare a 1.25 m solution of sucrose, C 12 H 22 O 11 ?

28. Colligative Properties of Solutions (page 471) Colligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind of particle.

29. Colligative Properties n Lowering vapor pressure n Raising boiling point n Lowering freezing point n Generating an osmotic pressure

30. 2 things to focus on… n Raising boiling point n Lowering freezing point

31. Boiling Point Elevation n a solution that contains a nonvolatile solute has a higher boiling pt than the pure solvent; the boiling pt elevation is proportional to the # of moles of solute dissolved in a given mass of solvent. Like when adding salt to a pot of boiling water to make pasta

32. Boiling Point Elevation n  T b = k b m where:  T b = elevation of boiling pt m = molality of solute ( mol solute/kg solvent ) k b = the molal boiling pt elevation constant n k b values are constants; see table 15.4 pg. 472 n k b for water = 0.52 °C/m

33. Ex: What is the normal boiling pt of a 2.50 m glucose, C 6 H 12 O 6, solution? n “normal” implies 1 atm of pressure n  T b = k b m n  T b = (0.52  C/m)(2.50 m) n  T b = 1.3  C n T b = 100.0  C + 1.3  C = 101.3  C

34. Freezing/Melting Point Depression n The freezing point of a solution is always lower than that of the pure solvent. Like when salting roads in snowy places so the roads don’t ice over or when making ice cream

35. Freezing/Melting Point Depression n  T f = k f m where:  T f = lowering of freezing point m = molality of solute k f = the freezing pt depression constant n k f for water = 1.86 °C/m n k f values are constants; see table 15.4 pg. 472

36. Ex: Calculate the freezing pt of a 2.50 m glucose solution. n  T f = k f m n  T f = (1.86  C/m)(2.50 m) n  T f = 4.65  C n T f = 0.00  C - 4.65  C = -4.65  C

37. Example Calculate the freezing-point depression ( ΔT f ) of a benzene solution containing 400. g of benzene and 200. g of acetone, C 3 H 6 O (solute). K f for benzene is 5.12 °C/m

38. Answer n ΔT f = Kf x m n ΔT f = (5.12 °C/m) x (m) n m = So, ΔT f = (5.12 °C/m) x (8.61 m) = 44.1 °C

39. Ex: When 15.0 g of ethyl alcohol, C 2 H 5 OH, is dissolved in 750 grams of formic acid, the freezing pt of the solution is 7.20°C. The freezing pt of pure formic acid is 8.40°C. Determine K f for formic acid.  T f = k f m 1.20  C= (k f )( 0.4340 m) k f = 2.76  C/m  T f = k f m

40. Electrolytes and Colligative Properties Colligative properties depend on the # of particles present in solution. Because ionic solutes dissociate into ions, they have a greater effect on freezing pt and boiling pt than molecular solids of the same molal conc.

41. Electrolytes and Colligative Properties n For example, the freezing pt of water is lowered by 1.86°C with the addition of any molecular solute at a concentration of 1 m. –Such as C 6 H 12 O 6, or any other covalent compound n However, a 1 m NaCl solution contains 2 molal conc. of IONS. Thus, the freezing pt depression for NaCl is 3.72°C…double that of a molecular solute. –NaCl  Na + + Cl - (2 particles)

42. Electrolytes - Boiling Point Elevation and Freezing Point Depression The relationships are given by the following equations: n  T f = k f ·m·n or  T b = k b ·m·n  T f/b = f.p. depression/elevation of b.p. m = molality of solute k f/b = b.p. elevation/f.p depression constant n = # particles formed from the dissociation of each formula unit of the solute

43. Ex: What is the freezing pt of a 1.15 m sodium chloride solution? n NaCl  Na + + Cl - n=2 n  T f = k f ·m·n n  T f = (1.86  C/m)(1.15 m)(2) n  T f = 4.28  C n T f = 0.00  C - 4.28  C = -4.28  C

44. How to determine n???? n MgCl 2 = 1 Mg +2 and 2Cl - n=3 n Ca 3 (PO 4 ) 2 = 3 Ca +2 and 2 PO 4 -3 n= 5