1. Solutions & Colligative Properties
AP Chemistry Notes
Solutions & Colligative Properties

2. Solutions…
A solution is a homogenous mixture of two or more substances.
The solute is(are) the substance(s) present in the smaller amount(s).
The solvent is the substance present in the larger amount.

3. Solutions…
An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.
nonelectrolyte
weak electrolyte
strong electrolyte

4. Solutions…
A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.
An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature.
A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.

5. Solutions… DHsoln = DH1 + DH2 + DH3
Three types of interactions in the solution process:
solvent-solvent; solute-solute; and solvent-solute
DHsoln = DH1 + DH2 + DH3

6. Solutions…
Two substances with similar intermolecular forces are likely to be soluble in each other.
‘LIKE DISSOLVES LIKE’
Non-polar molecules are soluble in non-polar solvents
CCl4 in C6H6
Polar molecules are soluble in polar solvents
C2H5OH in H2O
Ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)

7. Concentration…
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
Percent by Mass: =
x 100%
mass of solute
mass of solution

8. Concentration… Mole Fraction (X) moles of A XA =
sum of moles of all components

9. Concentration… Molarity: Molality: moles of solute M =
liters of solution
m =
moles of solute
mass of solvent (kg)

10. Concentration…
What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is g/mL?
m =
moles of solute
mass of solvent (kg)
M =
moles of solute
liters of solution
Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = kg
m =
moles of solute
mass of solvent (kg) =
5.86 moles C2H5OH
0.657 kg solvent
= 8.92 m

11. Solid solubility and temperature

12. Gas solubility and temperature

13. Solubility… Pressure and Solubility of Gases
The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).
Mathematically expressed: c = kP
c is the concentration (M) of the dissolved gas
P is the pressure of the gas over the solution
k is a constant (mol/L•atm) that depends only on temperature

14. Henry’s Law:
The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).
high P
low P
high c
low c

15. Colligative Properties
NON-Electrolytic Solutions

16. Colligative Properties…
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
Vapor-Pressure Lowering
Boiling Point Elevation
Freezing Point Depression
Osmosis

17. Vapor Pressure Lowering…
Raoult’s law:
P1 = X1 P 1
P 1
= vapor pressure of pure solvent
X1 = mole fraction of the solvent
If the solution contains only one solute:
X1 = 1 – X2
P 1
- P1 = DP = X2
X2 = mole fraction of the solute

18. Vapor Pressure Lowering…
At a given temperature water has a vapor pressure of mmHg. Calculate the vapor pressure above a solution of g of sucrose (C12H22O11) in mL of water, assuming the water to have a density of g/mL.

19. Vapor Pressure Lowering…
90.40 g C12H22O11  0.26 mol
350.0 ml H2O  19.4 mol
X = (19.4/( )) =
VP = XP
VP = (0.9868)(22.80)
VP = mmHg

20. Vapor Pressure Lowering…
23.00 g of an unknown substance was added to g of water. The vapor pressure above the solution was found to be mmHg. Given that the vapor pressure of pure water at this temperature is mmHg, calculate the Molar Mass of the unknown.

21. Vapor Pressure Lowering…
23.0 g X
120.0 g H2O  6.65 mol H2O
VP = 21.34
P = 22.96
VP = XP
21.34 = X (22.96)
X = 0.929
X = mol H2O / total mol
0.929 = 6.65/( x)
X mol = 0.506
M = g/mol = 23.0 / 0.506
M = g/mol

22. Vapor Pressure Lowering…
PA = XA P A
Ideal Solution
PB = XB P B
PT = PA + PB
PT = XA P A
+ XB P B

23. Vapor Pressure Lowering…
At 20.0 oC the vapor pressures of methanol (CH3OH)and ethanol (C2H5OH) are 95.0 and 45.0 mmHg respectively. An ideal solution contains 16.1 g of methanol and 92.1 g of ethanol. Calculate the vapor pressure.

24. Vapor Pressure Lowering…
16.1 g CH3OH  0.5 mol
92.1 g C2H5OH  2 mol
Total mol = 2.5
VP = XaPa + XbPb
VP = (0.5/2.5)(95) + (2/2.5)(45)
VP = 55 mmHg

25. Boiling Point Elevation…
DTb = Tb – T b
T b is the boiling point of
the solution
T b is the boiling point of
the pure solvent
Tb > T b
DTb > 0
DTb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)

26. Freezing Point Depression…
DTf = T f – Tf
T f is the freezing point of
the solution
T f is the freezing point of
the pure solvent
T f > Tf
DTf > 0
DTf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)

27. Constants…

28. DTf = Kf m Kf water = 1.86 0C/m DTf = Kf m
What is the freezing point of a solution containing 478g of ethylene glycol (antifreeze) in 3202 g of water? GIVEN: The molar mass of ethylene glycol is g.
DTf = Kf m
Kf water = C/m =
3.202 kg solvent
478 g x
1 mol
62.01 g
m =
moles of solute
mass of solvent (kg)
= 2.41 m
DTf = Kf m
= C/m x 2.41 m = C
DTf = T f – Tf
Tf = T f – DTf
= C – C = C

29. Summary…
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 1
Boiling-Point Elevation
DTb = Kb m
Freezing-Point Depression
DTf = Kf m
Osmotic Pressure (p)
p = MRT

30. Colligative Properties
Electrolytic Solutions

31. Colligative Properties…
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 solution m Na+ ions & 0.1 m Cl- ions
0.1 m NaCl solution 0.2 m ions in solution
van’t Hoff factor (i) =
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be 1 2 3
Nonelectrolytes
NaCl
CaCl2

32. Colligative Properties…
Boiling-Point Elevation
DTb = i Kb m
Freezing-Point Depression
DTf = i Kf m
Osmotic Pressure (p)
p = iMRT

33. Colligative Properties…
At what temperature will a 5.4 molal solution of NaCl freeze?
Solution:
∆TFP = Kf • m • i
∆TFP = (1.86 oC/molal) • 5.4 m • 2
∆TFP = oC
FP = 0 – 20.1 = oC

34. Colligative Properties…
Osmotic Pressure (p):
Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.
more
concentrated
dilute

35. Colligative Properties…
High P
Low P
p = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)

36. Colloids…
A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.
Colloid versus solution:
Collodial particles are much larger than solute molecules
Collodial suspension is not as homogeneous as a solution