1. Chapter 11 Properties of Solutions

2. Important Vocabulary Homogeneous means there is only one phase (compositions do not vary) Ex: Kool Aid, air, steel Solute: Gets dissolved Solvent: Does the dissolving Solution: Homogeneous mixture consisting of a solute and solvent

3. Dilute vs. Concentrated Can’t be used in calculations Molarity, mass percent, and mole fraction can be used to show solution concentrations

4. Molarity Moles of solute/liters of solution Represented by M Example: A solution was prepared by adding 5.84 g of formaldehyde, H 2 CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the molarity. Answer: 1.87 M H 2 CO

5. Mass Percent Percent by mass of the solute in the solution Mass Percent = (mass of solute/mass of solution) X 100% Example: A solution was prepared by adding 5.84 g of formaldehyde, H 2 CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the mass percent. Answer: 5.52 % H 2 CO, 94.48% H 2 O

6. Mole Fraction Represented by X Moles of part/moles of solution X 100% Mole Frac. A = X A = n A /(n A +n B ) Example: A solution was prepared by adding 5.84 g of formaldehyde, H 2 CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the mole fraction. Answer: X H 2 CO = 0.0338, X H 2 O = 0.9662

7. Molality Represented by m Moles of solute per kilogram of solvent Molality = moles of solute/kilogram of solvent Example: A solution was prepared by adding 5.84 g of formaldehyde, H 2 CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the molality. Answer: 1.94 m H 2 CO

8. Solubility Shows what will dissolve in what “Like dissolves like” = polar solvents will dissolve polar/ionic solutes and nonpolar solvents will dissolve nonpolar solutes

9. Factors Affecting Solubility 1. Structure 2. Pressure 3. Temperature

10. 1. Structure Effects Polarity of solute/solvent (like dissolves like) Example: vitamins are fat-soluble and water- soluble –Fat-soluble = nonpolar, hydrophobic (water-fearing), build up/stored in fatty tissue, too much = hypervitaminosis –Water-soluble = polar, hydrophilic (water-loving), extra are excreted by the body

11. 2. Pressure Effects Doesn’t affect liquids/solids, but has a large affect on gases Gas solubility increases as the partial pressure of the gas above the solution increases

12. Henry’s Law Shows relationship between gas pressure and concentration of dissolved gas: C = kP C = concentration of dissolved gas K = constant for particular solution P = partial pressure of gas above solution Works best with gases that don’t dissociate in/react with solvent

13. Henry’s Law Example The solubility of O 2 is 2.2 X 10 -4 M at 0C and 0.10 atm. Calculate the solubility of O 2 at 0C and 0.35 atm. Answer: 7.7 X 10 -4 M O 2

14. 3. Temperature Effects For most solids, solubility increases as temperature increases For most gases, solubility decreases as temperature increases –Thermal Pollution in lakes: increase in temp. lowers dissolved oxygen concentrations

15. Vapor Pressure of Solutions If a solution contains a nonvolitile (not easily vaporized) solute, its vapor pressure is LOWER than the pure solvent. Shells of water solvation make it so it’s harder for the solvent to vaporize

16. Molecules that do not dissociate (break up) in water (solvent) have higher vapor pressures than ionic compounds that do dissociate The decrease in a solution’s vapor pressure is proportional to the number of particles the solute makes in solution.

17. Answer This… Which compound affects the vapor pressure of a solution the least: glucose, sodium chloride, or calcium chloride? Solutions with covalent compounds > Solutions with ionic compounds

18. Raoult’s Law Calculates the expected vapor pressure of a solution based on the solute/solvent P soln = X solvent P 0 solvent P soln = observed vapor pressure of solution X solvent = mole fraction of solvent P 0 solvent = vapor pressure of the pure solvent

19. Example Glycerin, C 3 H 8 O 3, is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g of glycerin to 338 mL of H 2 O at 39.8C? The vapor pressure of pure water at 39.8C is 54.74 torr and its density is 0.992 g/mL. Answer: 50.0 torr

20. Example #2 What is the vapor pressure of a solution made by adding 52.9 g of CuCl 2, a strong electrolyte, to 800.0 mL of water at 52.0C? The vapor pressure of water is 102.1 torr, and its density is 0.987 g/mL. Answer: 99.4 torr

21. Colligative Properties Depend on the number of solute particles, NOT their identity in an ideal solution 1. Boiling-Point Elevation 2. Freezing-Point Depression 3. Osmotic Pressure

22. 1. Boiling-Point Elevation Review boiling point definition: when vapor pressure = atmospheric pressure When solute is added to solvent, it lowers the vapor pressure. More kinetic energy must be added to bring the solution to boiling Boiling point is HIGHER in solutions than in pure solvents

23. Antifreeze in car engines (solute) makes it so car engines don’t boil in high temperatures The more solute particles dissolved, the higher the boiling point (identity doesn’t matter)

24. Boiling-Point Elevation Calculation Change in boiling point ∆T b is the difference between the boiling point of the solution and the pure solvent Unit: °C/m Calculated using ∆T b = K b X m solute K b is a molal boiling-point elevation constant of the solvent found on pg. 517

25. Example What is the boiling point of a solution containing 96.7g of sucrose (C 12 H 22 O 11 ) in 250.0g water at 1atm? Answer: 100.579°C

26. 2. Freezing-Point Depression When solute is present, the normal molecular freezing pattern is disrupted This makes it so the solution has to lose more kinetic energy (get colder) in order to solidify Freezing point of the solution is LOWER than that of the pure solvent

27. The more solute particles dissolved, the more the freezing point decreases (identity doesn’t matter) Sidewalk salt and car antifreeze work this way

28. Freezing-Point Depression Calculation Change in freezing point ∆T f is the difference between the freezing point of the solution and the pure solvent Unit: °C/m Calculated using ∆T f = K f X m solute K f is a molal freezing-point depression constant of the solvent found on pg. 517

29. Example Determine the freezing point of a solution made by adding 27.5 g of methanol (CH 3 OH) to 250.0 g of water. Answer: -6.39°C

30. Example Find the boiling point of a 1.50m solution of calcium chloride, CaCl 2 and water. Answer: 2.30°C so the new boiling point would be 102.30°C.