Colligative Properties

2. Colligative Properties

3. Colligative Properties • The properties of solvent that depend ONLY on the amount of solute added to it • NOT depend on the identity of solute.

4. Colligative Properties • VAPOR PRESSURE • BOILING POINT • FREEZING POINT • OSMOTIC PRESSURE

5. Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

6. 1. Vapor Pressure

7. Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

8. Vapor Pressure Lowering Figure 14.12

9. Understanding Colligative Properties VP of H2O over a solution depends on the number of H2O molecules per solute molecule. Psolventproportional to Xsolvent Psolvent = Xsolvent • Posolvent VP of solvent over solution = (Mol frac solvent)•(VP pure solvent) RAOULT’S LAW

10. Raoult’s Law PA = XA • PoA An ideal solution obeys Raoult’s law. mole fraction A (XA): always less than 1, PA is always less than PoA. The vapor pressure of solvent over a solution is always LOWERED!

11. Raoult’s Law A solution contains 62.1 g of glycol (1 mol) in 250. g of water. What is the vapor pressure of water over the solution at 30 oC? (The VP of pure H2O is 31.8 torr) Solution Mole fraction Xg = 0.0672 Know Xg + XH2O = 1.000 XH2O = 1.000 - 0.0672 = 0.933 PH2O = XH2O • PoH2O = (0.933)(31.8 torr) Pwater = 29.7 mm Hg

12. Changes in Freezing and Boiling Points of Solvent See Figure 14.12

13. Elevation of Boiling Point ∆TBP = KBP•m (where KBP is characteristic of solvent) ∆TBP

15. Change in Boiling Point What is the BP of the solution with 62.1 g of glycol (1.00 mol) dissolved in 250. g of water? KBP = +0.512 oC/molal for water Solution 1. Calculate solution molality = 4.00 m 2. ∆TBP = KBP • m ∆TBP = +0.512 oC/molal (4.00 molal) ∆TBP = +2.05 oC new BP = 100 + 2.05 oC = 102.05 oC

16. Boiling Point Elevation and Freezing Point Depression ∆T = K•m•i i= van’t Hoff factor = # of particles produced / formula unit. Compound Value of i glycol 1 NaCl 2 CaCl2 3

17. Boiling PointD6C32 • The addition of solute raises the boiling point. • The larger the “m” the greater the increase in temperature: NOT matter what you add

18. WHY?

19. 3. Freezing Point: • The addition of solute lowers the freezing point. • Salt is added to streets • ice cream freezes at different temp (depends on flavor, more you add the lower the temp)

20. Change in Freezing Point Ethylene glycol/water solution Pure water The freezing point of a solution is LOWERthan that of the pure solvent. FP depression = ∆TFP = KFP•m

22. Boiling Point Elevation and Freezing Point Depression ∆T = K•m•i i= van’t Hoff factor = # of particles produced / formula unit. Compound Value of i glycol 1 NaCl 2 CaCl2 3

23. Freezing Point Depression Calculate the FP of a 4.00 molal glycol/water solution. KFP = -1.86 oC/molal (Table 14.4) Solution ∆TFP = KFP • m ∆TFP = (-1.86 oC/molal) (4.00 m) ∆TFP = -7.44 oC FP= 0 - 7.44 oC = -7.44 oC ∆TBP = +2.05 ˚C for this solution.

24. Freezing Point Depression How much NaCl (g) must be dissolved in 1.00 kg of water to lower FP to -10.00 oC?. Calc. required molality ∆TFP = KFP • m -10.00 oC = (-1.86 oC/molal) • m Conc = 5.38 molal ions

25. Freezing Point Depression Conc req’d = 5.38 molal ions We need 5.38 mol dissolved particles / kg of solvent. NaCl(aq) --> Na+(aq) + Cl-(aq) need 5.38 mol / 2 = 2.69 mol NaCl / kg 2.69 mol NaCl / kg ---> 157 g NaCl / kg

26. Boiling Point Vapor Pressure Freezing Point

27. 4. Osmosis • The movement of solvent through a membrane from an area of higher to lower solvent concentration. • Semipermeable Membrane- barrier allowing solvent through, but block larger solute particles.

28. Osmosis at the Particulate Level Figure 14.15

29. 4. Osmosis • Osmotic Pressure- Pressure developed across a membrane that allows the diffusion of solvent particles. • Chemistry behind your kidneys and making pickles

30. Osmosis Driving force is entropy The semi-permeable 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.

31. Process of Osmosis

32. Osmotic pressure Osmosic Pressure, ∏ Equilibrium is reached when pressure — the OSMOTIC PRESSURE, ∏ — produced by extra solution counterbalances pressure of solvent molecules moving thru the membrane. ∏ = cRT (c is conc. in mol/L)

33. 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 torr at 25 ˚C. Calc. molar mass of hemoglobin. Solution (a) Calc. ∏ in atmospheres ∏ = 10.0 torr • (1 atm / 760 mmHg) = 0.0132 atm (b) Calc. concentration

34. Osmosis Calculating a Molar Mass Solution (b) Use ∏ = cRT c = 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 = 65,100 g/mol

35. Osmosis • 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 14.16

36. Osmosis and Living Cells

37. Reverse OsmosisWater Desalination

38. Colligative Properties On adding a solute to a solvent, the props. of the solvent are modified. • Vapor pressure decreases • Melting point decreases • Boiling point increases • Osmosis is possible (osmotic pressure) They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.