1. Why does a raw egg swell or shrink when placed in different solutions?

2. Some Definitions
A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase.
One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

3. Definitions Solutions can be classified as saturated or unsaturated.
A saturated solution contains the maximum quantity of solute that dissolves at that temperature.
SUPERSATURATED SOLUTIONS contain more than is possible and are unstable.

4. Dissolving An Ionic Solid
Active Figure 14.9

5. Energetics of the Solution Process
Figure 14.8

6. Energetics of the Solution Process
If the enthalpy of formation of the solution is more negative that that of the solvent and solute, the enthalpy of solution is negative.
The solution process is exothermic!

7. Supersaturated Sodium Acetate
One application of a supersaturated solution is the sodium acetate “heat pack.”
Sodium acetate has an ENDOthermic heat of solution.

8. Supersaturated Sodium Acetate
Sodium acetate has an ENDOthermic heat of solution.
NaCH3CO2 (s) + heat > Na+(aq) + CH3CO2-(aq)
Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC.
Na+(aq) + CH3CO2-(aq) ---> NaCH3CO2 (s) + heat

9. Colligative Properties
On adding a solute to a solvent, the properties of the solvent are modified.
Vapor pressure decreases
Melting point decreases
Boiling point increases
Osmosis is possible (osmotic pressure)
These changes are called COLLIGATIVE PROPERTIES.
They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

10. Concentration Units
An IDEAL SOLUTION is one where the properties depend only on the concentration of solute.
Need concentration units to tell us the number of solute particles per solvent particle.
The unit “molarity” does not do this!

11. Concentration Units MOLE FRACTION, X MOLALITY, m
For a mixture of A, B, and C
MOLALITY, m
WEIGHT % = grams solute per 100 g solution
(also ppm, ppb, which is like %, which is parts
per hundred)

12. Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate mol fraction, molality, and weight % of ethylene glycol.

13. Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol.
Calculate mole fraction
250. g H2O = 13.9 mol
X glycol =

14. Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol.
Calculate molality
Calculate weight %

15. Dissolving Gases & Henry’s Law
Sg Gas solubility (mol/L) = kH • Pgas
kH for O2 = x 10-6 M/mmHg
When Pgas drops, solubility drops.

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

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

18. Raoult’s Law PA = XA • PoA
An ideal solution is one that obeys Raoult’s law.
PA = XA • PoA
Because mole fraction of solvent, XA, is always less than 1, then PA is always less than PoA.
The vapor pressure of solvent over a solution is always LOWERED!

19. Vapor Pressure Lowering
Figure 14.14

20. Raoult’s Law Pwater = 29.7 mm Hg
Assume the solution containing 62.1 g of glycol in 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 oC? (The VP of pure H2O is 31.8 mm Hg)
Solution
Xglycol = and so Xwater = ?
Because Xglycol + Xwater = 1
Xwater = =
Pwater = Xwater • Powater = (0.9382)(31.8 mm Hg)
Pwater = 29.7 mm Hg

21. Elevation of Boiling Point
Elevation in BP = ∆TBP = KBP•m
(where KBP is characteristic of solvent)
The boiling point of a solution is higher than that of the pure solvent.

22. Change in Boiling Point
Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP of the solution?
KBP = oC/molal for water (see Table 14.3).
Solution
1. Calculate solution molality = 4.00 m
2. ∆TBP = KBP • m
∆TBP = oC/molal (4.00 molal)
∆TBP = oC
BP = oC

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

24. Lowering the Freezing Point
Water with and without antifreeze
When a solution freezes, the solid phase is pure water. The solution becomes more concentrated.

25. Freezing Point Depression
How much NaCl must be dissolved in kg of water to lower FP to oC?.
Solution
Calc. required molality (*This one is a little different because the solute is ionic, not molecular.)
∆TFP = KFP • m
oC = (-1.86 oC/molal) • Conc
Conc = molal

26. Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg of water to lower FP to oC?.
Solution
Conc req’d = 5.38 molal
This means we need 5.38 mol of dissolved particles per kg of solvent.
*Recognize that m represents the total concentration of all dissolved particles.
Recall that 1 mol NaCl(aq)--> 1 mol Na+(aq) + 1 mol Cl-(aq)

27. Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg of water to lower FP to oC?.
Solution
Conc req’d = 5.38 molal
We need 5.38 mol of dissolved particles per kg of solvent. Each mol of NaCl contributes 2 mol of dissolved particles.
NaCl(aq) --> Na+(aq) + Cl-(aq)
To get 5.38 mol/kg of particles we need
5.38 mol / 2 = mol NaCl / kg
2.69 mol NaCl / kg ---> 157 g NaCl / kg
(157 g NaCl / kg)•(4.00 kg) = 629 g NaCl

28. Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
A generally useful equation
i = van’t Hoff factor = number of particles produced per formula unit.
Compound Theoretical Value of i
glycol 1
NaCl 2
CaCl2 3

29. Osmosis Dissolving the shell in vinegar Egg in pure water
Egg in corn syrup

30. Osmosis Driving force is entropy
The semipermeable 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.
Driving force is entropy

31. Process of Osmosis

32. Osmosis at the Particulate Level
Figure 14.17

33. Osmosis at the Particulate Level
Figure 14.17

34. Osmosis
Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration.

35. Osmosis and Living Cells

36. Reverse Osmosis Water Desalination
Water desalination plant in Tampa

37. 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 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin.
Solution
(a) Calc. ∏ in atmospheres
∏ = mmHg • (1 atm / 760 mmHg)
= atm
(b) Calculate molarity

38. 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 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin.
Solution
(b) Calc. molarity from ∏ = iMRT
Conc = 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