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AP Chemistry
Chapter 11 Notes - Properties of Solutions
In a ___________________, or homogeneous mixture, the components of the mixture are uniformly
intermingled (on a molecular level).
State of solute
gas
liquid
Types of Solutions
State of solvent
State of solution
gas
gas
liquid
liquid
solid
gas
solid
solid
solid
liquid
liquid
gas
solid
liquid
liquid
solid
Examples
air, natural gas,
vodka in water,
antifreeze
brass, steel
soda water
seawater, koolaid
hydrogen in platinum
How do we express solution concentration?
Solution Composition:
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
1. Molarity (M) = 𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
2. Mass (weight) percent = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 x 100%
3. Mole Fraction (χA) =
𝑚𝑜𝑙𝑒𝑠 𝐴
𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
4. Molality (m) = 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Example: 1.00 g of ethanol (formula = _________) is dissolved in 100.0 mL of water.
Molarity Calculations:
Mass % Calculations:
Mole Fraction:
Molality Calculations:
Notice from previous calculations that in dilute solutions Molarity (M) and molality (m) are very similar.
Normality (N) =
𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Acid-Base Equivalents = (moles) (total (+) charge)
Redox Equivalents = (moles) (# e- transferred)
Molarity and Normality Comparisons:
Acid or Base
Molar Mass
HCl
36.5
H2SO4
98
H3PO4
98
NaOH
40
Ca(OH)2
74
Normality Calculations:
Equivalent Mass
36.5
98/2 = 49
98/3 = 32.6
40
74/2 = 37
Molarity and Normality
1M=1N
1M=2N
1M=3N
1M=1N
1M=2N
Concentration and Density Calculations: 11.2 on pages 513-514
The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density
of 1.230 g/mL. Calculate the mass percent, molality and normality of the sulfuric acid.
Steps in Solution Formation (see illustration below):
1. Expanding the solute (endothermic)
2. Expanding the solvent (endothermic)
3. Allowing the solute and solvent to interact to form a solution (exothermic)
ΔHsolution = ΔHstep 1 + ΔHstep 2 + ΔHstep 3
11_269
Step 1
Step 2
 H1
 H2
Solute
Expanded solute
Direct
formation
of solution
Step 3
Solvent
H soln
Expanded solvent
H3
Processes that require large amounts of energy tend not to occur. Solution processes are favored by an
increase in entropy.
Polar solvent,
polar solute
Polar solvent,
non-polar
solute
Non-polar
solvent, nonpolar solute
Non-polar
solvent, polar
solute
Energy Terms for Various Types of solutes and Solvents
ΔH1
ΔH2
ΔH3
ΔHsoln
large
large
large, negative small
outcome
solution forms
small
large
small
large, positive
no solution
forms
small
small
small
small
solution forms
large
small
small
large, positive
no solution
forms
Structure and Solubility:
 “Like dissolves like.” (see above)
 Hydrophobic (water fearing) – fat soluble vitamins (A, D, E, K)
 Hydrophillic (water loving) – water soluble vitamins (Bs, C)
 Hypervitaminosis – excessive build-up of A, D, E, K in the body
Henry’s Law
The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.
P = kC
where:
P=
C=
k=
partial pressure of gaseous solute above the solution
concentration of dissolved gas
a constant
Solubility Curve (solid salts):
11_274
300
Sugar
(C12 H 22 O 11 )
Solubility (g solute/100 g H2O)
260
KNO
3
220
180
140
NaNO 3
NaBr
100
KBr
Na 2 SO 4 KCl
60
20
0
Ce 2 (SO 4 )3
0
20
40
60
Temperature (C)
80
100
Solubility Curve (gases):
Raoult’s Law: The presence of a nonvolatile solute lowers the vapor pressure of a solvent.
Psoln = χsolvent P°solvent
where:
Psoln = vapor pressure of the solution
χsolvent = mole fraction of the solvent
P°solvent = vapor pressure of the pure solvent
Sample Exercise 11.6 on page 524.
Predict the vapor pressure of a solution prepared by mixing 35.0 g solid Na2SO4
(molar mass = 142.05 g/mol) with 175 g water at 25 ºC. The vapor pressure of
pure water at 25 ºC is 23.76 torr.
Na2SO4 forms 3 ions so the number of moles of solute is multiplied by three.
Apply a modified version of Raoult’s Law when both components of a liquid-liquid solution are volatile
(see below):
Vapor pressure for a solution of two volatile liquids.
a) Ideal(benzene & toluene) -- obeys Raoult’s Law,
b) Positive deviation (ethanol & hexane) from Raoult’s Law
c) Negative deviation (acetone & water). Negative deviation is due to hydrogen bonding.
11_279
Vapor pressure
of solution
Vapor pressure
Vapor pressure of pure B
eB
tial
Par
ssu
pre
re A
A
B
(a)
Vapor pressure
of solution
Va
Vapor pressure
Pa por pr
ess of pure A
rtia
ure
lp
of s
re
olu
ss
tion
ur
A
B
(b)
A
B
(c)
Liquid-Liquid Solutions:
Ptotal = PA + PB = χA P°A + χB P°B
Sample Exercise 11.7 on page 526.
A solution os prepared by mixing 5.81 g acetone (C3H6O, molar mass = 58.1 g/mol) and 11.9 g chloroform (HCCl3,
molar mass = 119.4 g/mol). At 35 ºC, this solution has a total vapor pressure of 260. Torr. Is this an ideal solution?
The vapor pressures of pure acetone and pure chloroform at 35 ºC are 345 and 293 torr, respectively.
Colligative Proerties:
Depend only on the number, not on the identity, of the solute particles in an ideal solution.
 Boiling point elevation
 Freezing point depression
 Osmotic pressure
Phase diagrams for pure water and for an aqueous solution containing a nonvolatile solute -- liquid
range is extended for the solution.
11_280
atm
Pressure (atm)
Vapor pressure
of pure water
Vapor pressure
of solution
Freezing point
of water
Freezing
point of
solution
Boiling point
of water
Tf
Boiling point
of solution
T b
Temperature (C)
Boiling Point Elevation:
A nonvolatile solute elevates the boiling point of the solvent. The solute lowers the vapor
pressure of the solution.
ΔT = Kb msolutei
where:
Kb = molal boiling point elevation constant
m = molality of the solute
i = van’t Hoff factor ( # ions formed)
Boiling Point Calculations:
Sample Exercise 11.8 on page 528: Calculate the Molar mass by Boiling Point Elevation:
A solution was prepared by dissolving 18.00 g glucose in 150.0 g water. The resulting solution was found to have a
boiling point of 100.34 ºC. Calculate the molar mass of glucose. Glucose is a molecular solid that is present as
individual molecules in solution.
A nonvolatile solute depresses the freezing point of the solvent. The solute interferes with crystal
formation.
ΔT = Kf msolutei
where:
Kf = molal boiling point elevation constant
m = molality of the solute
i = van’t Hoff factor ( # ions formed)
11.10 on pg 530 Calculate the Molar mass by Freezing Point Depression:
A chemist is trying to identify a human hormone that controls metabolism by determining its molar mass. A
sample weighing 0.546 g was dissolved in 15.0 g benzene, and the freezing point depression was determined to be
0.240 ºC. Calculate the molar mass of the hormone.
Osmotic Pressure:
Osmosis: The flow of solvent into the solution through the semi-permeable membrane.
Osmotic Pressure: The excess hydrostatic pressure on the solution compared to the pure
solvent.
Due to osmotic pressure, the solution is diluted by water transferred through the semi-permeable
membrane. The diluted solution travels up the thistle tube until the osmotic pressure is balanced by the
gravitational pull.
Osmosis: The solute particles interfere with the passage of the solvent, so the rate of transfer is slower
from the solution to the solvent than in the reverse direction.
a) The pure solvent travels at a greater rate into the solution than solvent molecules can travel
in the reverse direction.
b) At equilibrium, the rate of travel of solvent molecules in both directions is equal.
Osmotic Pressure:
where:
 = osmotic pressure (atm)
M = Molarity of solution
R = 0.08206
T = Kelvin temperature
11.11 on p. 532
 = MRT
Crenation and Lysis:
Crenation-solution in which cell is bathed is hypertonic (more concentrated)-cell shrinks. Pickle, hands
after swimming in ocean. Meat is salted to kill bacteria and fruits are placed in sugar solution.
Lysis-solution in which cell is bathed is hypotonic (less concentrated)-cell expands. Intravenous solution
that is hypotonic to the body instead of isotonic.
 If the external pressure is larger than the osmotic pressure, reverse osmosis occurs.
 One application is desalination of seawater.
Colligative Properties of Electrolyte Solutons:
 van’t Hoff factor, “i”, relates to the number of ions per formula unit.
NaCl: i = 2
K2SO4: i = 3
i=
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑒𝑑
ΔT = mKi
Π = MRTi
The value of i is never quite what is expected due to ion-pairing. Some ions stay linked together--this
phenomenon is most noticeable in concentrated solutions.
11.13 on p. 537-538
Colloids:
 Colloidal Dispersion (colloid): A suspension of tiny particles in some medium.
aerosols, foams, emulsions, sols
 Coagulation: The addition of an electrolyte, causing destruction of a colloid.
Examples are electrostatic precipitators and river deltas.