1. Solutions Solution: homogeneous mixture
components are uniformly intermingled on a molecular level
Solutions can be
solid:
brass (zinc in copper)
liquid:
salt water, sugar water, etc
gas:
Air (oxygen & others in nitrogen)

2. Solutions Unsaturated solution:
a solution that is capable of dissolving more solute
Saturated solution:
a solution that is in equilibrium with undissolved solid
Supersaturated solution:
a solution that contains more dissolved solute than is needed to form a saturated solution

3. Solubility
Example: Which of the following solutes would you expect to be soluble in water:
CH3CH2CH3:
CH3CH2OH:
HCl:
Vitamin A Vitamin C
Remember: Substances with similar intermolecular forces tend to dissolve in each other

4. Solubility of Gases
The solubility of a gas in a solvent depends on the nature of the solute and solvent, the temperature, and the pressure.
In general, the solubility of gases in water increases with increasing molar mass.
Larger LDF

5. Solubility of Gases
The solubility of a gas in a solvent increases as the pressure of the gas over the solvent increases.
Henry’s Law: The solubility of a gas in a solvent is directly proportional to its partial pressure above the solution.
Cg = kPg
where Cg = solubility of the gas in the solution phase
Pg = partial pressure of the gas
k = proportionality constant (value depends on solute, solvent, and temperature

6. Solubility
Example: Calculate the concentration of CO2 in a soft drink that was bottled with a partial pressure of carbon dioxide of 3.5 atm over the liquid at 25oC. (k = 3.1 x 10-2 mol/L.atm)

7. Solubility of Gases
Example: Why does a bottle of soda bubble when the cap is first removed?
Carbonated beverages like soda are bottled under a carbon dioxide pressure slightly greater than 1 atm.
Opening the bottles, reduces the partial pressure of CO2 above the soda.
Solubility of CO2 decreases so CO2 bubbles out of the solution.

8. Solubility of Gases
The solubility of solid solutes generally increases with increasing temperature.

9. Solubility of Gases
The solubility of a gas in a solution decreases with increasing temperature.
Gas molecules have greater KE and can escape from the solution more easily.

10. Solubility of Gases
Example: A warm bottle of soda tends to taste “flat” compared to a cold bottle of soda. Explain why.

11. Concentration
Several different units can be used to express the concentration of a solute in a solution:
mass (weight) percent
parts per million (ppm)
parts per billion (ppb)
mole fraction
Molality
Molarity
Varies with temperature
Independent of temperature

12. Concentration Mass Percent = mass of component x 100
total mass of sol’n
Example: A solution is prepared by dissolving 6.8 g of NaCl in g of water. What is the mass percent of the solute?
Mass % NaCl = ___g NaCl____ x 100
g NaCl + g H2O
= ___6.8 g____ x 100 = 0.90 %
6.8 g g

13. Concentration ppm = mass of component x 106 total mass of sol’n
Example: A g sample of lake water contains 1.28 x 10-2 mg of arsenic. What is the concentration of arsenic in ppm?
ppm As = mg As x __1 g As_ x 106
10.25 g water 103 mg As
ppm As = 1.25 ppm

14. Concentration ppb = mass of component x 109 total mass of sol’n
Example: A 225 g sample of lake water contains 1.2 mg of pesticide. What is the concentration of pestcide in ppb?
ppb = 1.2 mg pest. x _1 g _ x 109
225 g water 106 mg
ppb = 5.3 ppb

15. Concentration
Mole fraction = moles of component total moles of all components
Example: Calculate the mole fraction HCl present in a solution prepared by dissolving 0.25 mol HCl in 9.50 mol H2O.
XHCl = ___moles HCl____
mol HCl + mol H2O
XHCl = ___0.25 mol ______ = 0.026
0.25 mol mol

16. Concentration Molality = m = moles of solute kg solvent
Example: Calculate the molality of a solution prepared by dissolving 1.25 g of sodium hydroxide in 250 g of water.
m = _moles NaCl_
kg H2O
m = 1.25 g NaCl x 1 mol NaCl x 103 g H2O
250 g H2O 58.5 g NaCl 1 kg H2O
m = m

17. Concentration Molarity = M = moles of solute L solution
Example: Calculate the molarity of a solution that contains 73.0 g of HCl per 250 mL of solution.
M = _moles HCl_
L soln
M = 73.0 g HCl x 1 mol HCl x 103 mL
250 mL soln 36.5 g HCl 1 L
M = 8.0 M

18. Concentration Why does molarity vary with temperature??
You must be able to interconvert between the different concentration units.

19. Concentration
Example: An aqueous solution of sodium hydroxide contains 4.4% NaOH by mass. Calculate the mole fraction and molality of the solution.
To find mole fraction:
Given: 4.4 g NaOH per
100.0 g solution
Find: XNaOH
XNaOH = mol NaOH
total mol

20. Concentration mol NaOH = 4.4 g NaOH x 1 mol = 0.11 mol 40.0 g
grams H2O = g – 4.4 g = 95.6 g
mol H2O = 95.6 g H2O x 1 mol = 5.31 mol
18.0 g
XNaOH = ___0.11 mol______ =
0.11 mol mol

21. Concentration To find molality: Given: 4.4 g NaOH per 100.0 g solution
m = mol NaOH
kg solvent
m = 0.11 mol NaOH x g H2O = 1.2 m
95.6 g H2O 1 kg H2O

22. Concentration
Example: Calculate the molarity of a 1.50 m solution of toluene (C7H8) in benzene if the solution has a density of g/mL.
Given: 1.50 mol toluene
per 1 (exact) kg benzene
d = g/mL
Find: M = mol toluene
L solution

23. Concentration
Solution:
Answer: M

24. Concentration The dilution equation is used to calculate either
the new concentration of a solution prepared by diluting a stock solution Or
The volume of a stock solution needed to prepare a known volume of a more dilute solution
C1V1 = C2V2

25. Concentration
Example: Calculate the molarity of a solution prepared by diluting 225 mL of 1.5 M KMnO4 to a total volume of mL.
Given:
Find:
Solution:
Answer: M

26. Concentration
Example: Describe in detail how you would prepare mL of 0.60 M HCl from a 12.0 M HCl stock solution?

27. Concentration
Example: A solution is prepared by dissolving 1.50 g of sodium chloride in enough water to give g of solution. A 25.0 g aliquot of this solution was then diluted with water to a total mass of g. Calculate the weight percent sodium chloride present in the final solution.