2. POINT > Define colligative properties POINT > Describe how solutes affect the equilibrium vapor pressure of a solution POINT > Describe boiling point elevation effects POINT > Describe freezing point depression effects POINT > Compare electrolyte and non-electrolyte effects POINT > Define effects on osmotic pressure

3. POINT > Define colligative properties Colligative properties are properties of solutions that depend on the concentration of solute, but not on the identity of the solute We use molal (m) concentration in calculations involving colligative properties Colligative properties include: vapor pressure effects, boiling point elevation, freezing point depression and changes in osmotic pressure

4. WB CHECK: Which would have a greater effect on the boiling point of a solution, a 1.0m concentration of glucose (molar mass = 180.0g) or a 1.0m concentration of sucrose (molar mass = 342.3g)? a) glucose b) sucrose c) same effect d) cannot be determined

5. POINT > Describe how solutes affect the equilibrium vapor pressure of a solution Solute-solvent intermolecular attractions make it harder for solvent particles to escape to the vapor phase Ex. the equilibrium vapor pressure of pure water is greater than that of salt water

6. POINT > Describe how solutes affect the equilibrium vapor pressure of a solution Since the vapor pressure of a solution is lower than that of the pure solvent, the boiling point of the solution is higher than that of the pure solvent. (Recall that boiling point is when vapor pressure = atm. pressure)

7. POINT > Describe boiling point elevation effects Solute-solvent interactions cause solutions to have higher boiling points and lower freezing points than the pure solvent

8. WB CHECK: Which would have the lowest equilibrium vapor pressure? a) a 1.0m aqueous solution of NaCl b) a 2.0m aqueous solution of NaCl c) they would be the same d) pure water

9. POINT > Define non-electrolyte boiling point elevation effects The change in boiling point,  T b, is proportional to the molality of the solution:  T b = K b  m where K b is the molal boiling point elevation constant, a colligative property of the solvent. K b = 0.51°C/m for water  T b is added to the normal boiling point of the solvent

10. POINT > Define non-electrolyte boiling point elevation effects Ex. What is the boiling point of a solution comprised of 45.0g glucose in 100.0g water?  T b = K b  m K b = 0.51°C/m for water  T b = 1.3°C Boiling point = 101.3°C

11. POINT > Define freezing point depression effects The change in freezing point can be found similarly:  T f = K f  m Here K f is the molal freezing point depression constant of the solvent. K f = -1.86°C/m for wa ter  T f is subtracted from the normal freezing point of the solvent.

12. POINT > Define colligative properties Note that in both equations,  T does not depend on what the solute is, but only on how many particles are dissolved  T b = K b  m  T f = K f  m

13. WB CHECK: Determine the freezing point of a 2.75m aqueous solution of glucose. -5.12°C Determine the boiling point of a 0.75m aqueous solution of sucrose. 100.38°C An aqueous solution of glucose has a boiling point of 100.62°C. What is the molality of the solution? 1.22m

14. POINT > Compare electrolyte and non-electrolyte effects Since colligative properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater colligative changes than those of non-electrolytes

15. POINT > Compare electrolyte and non-electrolyte effects The expected changes in boiling point or freezing point for NaCl would be double that for glucose The expected changes in boiling point or freezing point for CaCl 2 would be triple that for glucose...

16. POINT > Compare electrolyte and non-electrolyte effects To calculate expected changes in boiling or freezing point for electrolytes, factor in the number of particles:  T b = K b  m  #particles  T f = K f  m  #particles

17. WB CHECK: Determine the freezing point of a 1.50m aqueous solution of potassium chloride. -5.58°C Determine the boiling point of a 3.80m aqueous solution of sodium sulfide. 105.81°C An aqueous solution of sodium chloride has a freezing point of -1.62°C. What is the molality of the solution? 0.435m NaCl

18. POINT > Compare electrolyte and non-electrolyte effects However, a 1m solution of NaCl does not quite show twice the depression in freezing point that a 1m solution of glucose does. Why?...

19. POINT > Compare electrolyte and non-electrolyte effects One mole of NaCl in water does not quite give rise to two moles of ions The ions interact and are not completely solitary particles (except in very dilute solutions)

20. POINT > Compare electrolyte and non-electrolyte effects Na + and Cl − ions interact for short times, so the true concentration of particles is somewhat less than two times the concentration of NaCl The colligative effects are not quite doubled...

21. POINT > Compare electrolyte and non-electrolyte effects Ion interactions are more likely at higher concentration Therefore, the number of “colligative particles” present is concentration dependent

22. POINT > Define effects on osmotic pressure Semi-permeable membranes allow some particles to pass through, but block other particles In biological systems, membranes allow water to pass through, but some solutes are not free to do so Osmosis is the diffusion of water through a semi- permeable membrane

23. POINT > Define effects on osmotic pressure In osmosis, there is net movement of water from the area of higher water concentration (lower solute concentration) to the area of lower water concentration (higher solute concentration) Osmotic pressure is the pressure needed to oppose this diffusion of water

25. POINT > Define effects on osmotic pressure In living cells: If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic. Water will flow out of the cell, and crenation results (the cell shrinks)

26. POINT > Define effects on osmotic pressure If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic Water will flow into the cell, and hemolysis results (the cell swells/bursts)

27. WB CHECK: A semi-permeable membrane separates 2 aqueous solutions. Water can pass through the membrane but ions cannot. Solution 1 is 0.27m calcium chloride. Solution 2 is 0.27m sodium chloride. Which way will the net movement of water be? a)Toward Solution 1 b)Toward Solution 2 c) No net movement of water (be able to explain this)

28. Homework: Read 13.2 pages 422-432 Practice #1-4 page 426 Practice #1-4 page 427 Practice #1-3 page 430 Practice #1-2 page 433 FA #1-4 page 432