1. Chapter 11 Properties of Solutions

2. Copyright © Cengage Learning. All rights reserved 2 Solution – a homogeneous mixture. Solute – substance being dissolved. Solvent – liquid the substance is dissolved in. 11.1 Solution Composition

3. Solution Composition 11.1 Solution Composition Molarity (M) Mass percent (mass%) Mole fraction (X A ) Molality (m)

4. A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 in 100.0 mL of water. Calculate the mole fraction, molarity, and molality of H 3 PO 4. (Assume water has a density of 1.00 g/mL.) EXERCISE #1 = 0.0145 m == 0.816 m M = = 0.816 M n (H 3 PO 4 ) = n (H 2 O) = = 5.55 mol

5. Little effect on solubility of solids or liquids. Henry’s law: c = kP c = concentration of dissolved gas k = constant P = partial pressure of gas solute above the soln 1. Structure Effects Polarity – “like dissolves like”. 2. Pressure Effects (on gases) 11.3 Factors Affecting Solubility c ∝ P gas  [gas] increases with increasing P gas

6. Copyright © Cengage Learning. All rights reserved 6 Solubility of most solids in water increases with increasing temperature. The Solubilities of Several Solids as a Function of Temperature 3a. Temperature Effects (on solids) 11.3 Factors Affecting Solubility

7. 3b. Temperature Effects (on gases) 7 Solubility of a gas in water usually decreases with increasing temperature. 11.3 Factors Affecting Solubility The solubilities of several gases in water

8. Colligative Properties of Non-Electrolyte Solutions Depend only on the number of solute particles in solution and not on the nature of the solute particles.  Vapor-pressure  Boiling-point elevation  Freezing-point depression  Osmotic pressure Copyright © Cengage Learning. All rights reserved 8 11.4-11.6 Colligative Properties of Non-Electrolyte Solutions

9. Vapor-Pressure of Solutions Raoult’s law 11.4 The Vapor Pressures of Solutions Nonvolatile solute lowers the vapor pressure of a solvent.  P = vapor-pressure lowering

10. Boiling-Point Elevation 10 Nonvolatile solute elevates the boiling point of the solvent. 11.5 Boiling-Point Elevation and Freezing-Point Depression ΔT b = K b m solute ΔT b = boiling-point elevation K b = molal boiling-point elevation constant m solute = molality of solute

11. Freezing-Point Depression Copyright © Cengage Learning. All rights reserved 11 11.5 Boiling-Point Elevation and Freezing-Point Depression ΔT f = K f m solute Nonvolatile solute lowers the freezing point of the solvent. ΔT = freezing-point depression K f = molal freezing-point depression constant m solute = molality of solute

12. 11.6 Osmotic Pressure Osmotic pressure  = MRT  = osmotic pressure (atm) M = molarity of the solution R = gas law constant T = temperature (Kelvin) Osmosis – flow of solvent into the solution through a semipermeable membrane. Minimum pressure required to stop osmosis.

13. EXERCISE #2 The vapor pressure of water at 30 o C is 31.82 mmHg. Calculate the vapor pressure of a solution made by dissolving 218 g of glucose (molar mass = 180.2 g/mol) in 460 mL of water at 30 o C. The density of water at 30 o C is 1.00 g/mL. n (H 2 O) =25.5 mol n (glucose) =1.21 mol 0.955 P soln = (0.955)(31.82 mmHg) =30.4 mmHg

14. What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g/mol. K f (H 2 O) = 1.86 o C/m EXERCISE #3 ΔTfΔTf n (eth. gly.) =7.708 mol m = 2.407 m T f = 0oC0oC – (1.86 o C/m) (2.407 m) = –4.48 o C = K f m eth. gly.

15. V (H 2 O) = 10.0 mL = 0.01 L P soln = 558 torr = 0.7342 atm R = 0.08206 L∙atm/K∙mol mass (compound) = 33.4 mg = 0.0334 g When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25 o C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound. EXERCISE #4  = MRT n (compound) =3.0 x 10 −4 mol Molar mass = 111 g/mol T = 25 o C = 298 K

16. Colligative Properties of Nonelectrolyte Solutions Boiling-Point Elevation  T b = K b m Freezing-Point Depression  T f = K f m Osmotic Pressure (  )  = MRT Review 11.7 Colligative Properties of Electrolyte Solutions Vapor-pressure

17. 0.1 m NaCl solution 0.1 m Na + ions & 0.1 m Cl − ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution0.2 m ions in solution van’t Hoff factor nonelectrolytes NaCl CaCl 2 i should be 1 2 3 11.7 Colligative Properties of Electrolyte Solutions Colligative Properties of Electrolyte solutions

18. Boiling-Point Elevation  T b = i K b m Freezing-Point Depression  T f = i K f m Osmotic Pressure (  )  = i M R T 11.7 Colligative Properties of Electrolyte Solutions Colligative Properties of Electrolyte Solutions

19. Ion Pairing  Ion pairing is most important in concentrated solutions.  As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs.  Ion pairing is most important for highly charged ions. Copyright © Cengage Learning. All rights reserved 19 11.7 Colligative Properties of Electrolyte Solutions

20. 1.90 Which aqueous liquid will have the lowest freezing point? a)pure water b)aq. 0.6 m glucose c)aq. 0.7 m fructose d)aq. 0.25 m FeCl 3 The osmotic pressure of a 0.010 M potassium iodide (KI) solution at 25 o C is 0.465 atm. Calculate the van’t Hoff factor for KI at this concentration. EXERCISE #5 EXERCISE #6 a) − b) i = 1 c) i = 1 d) i = 4 i = R = 0.08206 L∙atm/K∙mol  = iMRT  T f = = i K f m T f = 0oC0oC− i K f m (1 x 0.6 m) = 0.6 m (1 x 0.7 m) = 0.7 m (4 x 0.25 m) = 1.0 m i x m

21. END OF CHAPTER 11