SOLUTIONS
SOLUTION – A homogeneous mixture
SOLVENT – The major component of a solution
SOLUTE – The minor component(s) of a solution
3G-1 (of 15)
LIKES DISSOLVE LIKES
Polar Component with a Nonpolar Component
Polar
Nonpolar
Polar molecules attract with LDF, DDA
Nonpolar molecules attract with LDF
To make a solution, polar and nonpolar molecules attract with LDF
These are too weak to overcome the polar molecules’ LDF, DDA
3G-2 (of 15)
LIKES DISSOLVE LIKES
Two Nonpolar Components
Nonpolar molecules attract with LDF
To make a solution, different nonpolar molecules attract with LDF
All molecules equally attract each other  a solution forms
3G-3 (of 15)
LIKES DISSOLVE LIKES
Two Polar Components
Polar molecules attract with LDF, DDA
To make a solution, different polar molecules attract with LDF, DDA
All molecules equally attract each other  a solution forms
Unless…
3G-4 (of 15)
LIKES DISSOLVE LIKES
Hydrogen-Bonding Solvent
The solute must be able to H-Bond with the solvent to dissolve
S  Cl

Cl
No H-Bonding
HO
HO
HO
H
H
H

H-Bonding
3G-5 (of 15)


LIKES DISSOLVE LIKES
Hydrogen-Bonding Solvent
The solute must be able to H-Bond with the solvent to dissolve
HNH

H
H

HNH
H-Bonding
H-Bonding
HO
HO
HO
H
H
H

H-Bonding
3G-6 (of 15)


LIKES DISSOLVE LIKES
Hydrogen-Bonding Solvent
The solute must be able to H-Bond with the solvent to dissolve
HCH

O
H-Bonding
HO
HO
HO
H
H
H

H-Bonding
3G-7 (of 15)


Solvent
Solute
Solubility
C6H6
C8H18
High
C6H6
CH2Br2
Low
CH2Cl2
C8H18
Low
CH2Cl2
CH2Br2
High
H2O
C8H18
Low
H2O
CH2Br2
Low (a little)
H2O
CH3OH
High
H2O
CH3CCH3
High
║
O
3G-8 (of 15)
When water-soluble polar molecules
dissolve in water, the MOLECULES separate
from each other, and exist as intact, neutral
molecules
These solutions do not conduct electricity
because no ions are formed
NONELECTROLYTE – A water-soluble
compound whose solution does not conduct
electricity
3G-9 (of 15)
Alcohols (CxHyOH) and sugars (Cx(H2O)y) are nonelectrolytes
Solute molecules surrounded by water molecules are said to be HYDRATED
H2O
C2H5OH (l)
→
C2H5OH (aq)
H2O
C6H12O6 (s) →
3G-10 (of 15)
C6H12O6 (aq)
When acid molecules dissolve in water, the
waters rip the acid molecules into IONS
These solutions conduct electricity because
ions are formed
ELECTROLYTE – A water-soluble compound
whose solution conducts electricity
Acids are electrolytes
3G-11 (of 15)
IONIZATION – The formation of ions during
the dissolving process
HCl (g)
H2O
→
H+ (aq) + Cl- (aq)
Strong acids produce many ions in solution,
and their solutions are good conductors
HCl, HBr, HI, and acids with at least 2 more
O’s than H’s are strong acids
Strong acids are STRONG ELECTROLYTES
3G-12 (of 15)
Weak acids produce few ions in solution,
and their solutions are poor conductors
All other acids are weak acids
Weak acids are WEAK ELECTROLYTES
HF (g)
H2O
→
3G-13 (of 15)
HF(aq)
When water-soluble ionic compounds
dissolve in water, the IONS separate from
each other
ION-DIPOLE ATTRACTIONS between the
ions and the water molecules pull the ions
into solution
These solutions conduct electricity because
ions are formed
Soluble ionic compounds are strong
electrolytes
3G-14 (of 15)
DISSOCIATION – The separation of ions
from an ionic crystal during the dissolving
process
NaCl (s)
H2O
→
3G-15 (of 15)
Na+ (aq) + Cl- (aq)
SOLUBILITY – The maximum amount of solute that will dissolve in a
specific amount of solvent
SATURATED – A solution that contains as much dissolved solute as
possible
Dissolving Rate
Crystallizing Rate
SOLUTION EQUILIBRIUM – When the rates of dissolving and crystallizing
are equal
A saturated solution is in solution equilibrium
3H-1 (of 8)
FACTORS AFFECTING SOLUBILITY
(1) Temperature
Increasing the temperature generally increases a solid’s solubility
Increasing the temperature decreases a gas’s solubility
(2) Pressure
No effect on a solid’s solubility
Increasing the pressure increases a gas’s solubility
HENRY’S LAW – The amount of gas dissolved in a liquid is directly
proportional to the pressure of the gas in contact with the liquid
pgas = kCgas
3H-2 (of 8)
(Cgas is concentration of dissolved gas)
CONCENTRATION UNITS
(1) MASS PERCENT – The mass of solute per mass of solution, times 100
Mass Percent
=
Mass Solute
___________________
x 100
Mass Solution
Find the mass percent of KCl in a solution made with 20.0 g KCl
dissolved in 80.0 g H2O.
20.0 g KCl
_____________________
100.0 g Solution
3H-3 (of 8)
x 100 = 20.0 %
(2) MOLE FRACTION (X) – The moles of solute per moles of solution
Find the mole fraction of KCl in a solution made with 20.0 g KCl
dissolved in 80.0 g H2O.
20.0 g KCl
x
mol KCl
_______________
= 0.2683 mol KCl
74.55 g KCl
80.0 g H2O
x
mol H2O
_______________
= 4.440 mol H2O
18.02 g H2O
0.2683 mol KCl
____________________________________
0.2683 + 4.440 mol Solution
3H-4 (of 8)
= 0.0570
(3) MOLARITY (M) – The moles of solute per liter of solution
Used to easily calculate moles of solute by only measuring the volume
of solution
M
=
n
n = quantity of matter (moles), V = volume (liters)
____
V
Find the molarity of a solution with 6.24 grams of magnesium chloride
dissolved in enough water to make 500.0 mL of solution.
6.24 g MgCl2 x
mol MgCl2
_________________
= 0.06554 mol MgCl2
95.21 g MgCl2
0.06554 mol MgCl2
_________________________
0.5000 L Solution
3H-5 (of 8)
= 0.131 M MgCl2
Electrolyte solutions really consist of individual ions
Each MgCl2 formula unit has 1 Mg2+ ion and 2 Cl- ions

0.131 M x 1 = 0.131 M Mg2+
0.131 M x 2 = 0.262 M ClFind the molarities of each ion in a 0.10 M Al2(SO4)3 solution
3H-6 (of 8)
Find the mass of potassium nitrate needed to prepare 250. mL of a 0.200 M
potassium nitrate solution.
M
= n
___
V
MV = n
0.200 mol KNO3
_____________________
x 0.250 L solution = 0.05000 mol KNO3
L solution
0.05000 mol KNO3
x 101.11 g KNO3
___________________
mol KNO3
3H-7 (of 8)
=
5.06 g KNO3
(4) MOLALITY (m) – The moles of solute per kilogram of solvent
Used because it does not change when the temperature changes
m
=
mol solute
_______________
kg solvent
Find the molality of a solution with 11.8 grams of glucose (C6H12O6,
m
_ = 180.16 g/mol) dissolved in 150.0 grams of water.
11.8 g C6H12O6
x
mol C6H12O6
______________________
= 0.06550 mol C6H12O6
180.16 g C6H12O6
0.06550 mol C6H12O6
___________________________
0.1500 kg H2O
3H-8 (of 8)
= 0.437 m C6H12O6
COLLIGATIVE PROPERTIES OF SOLUTIONS
COLLIGATIVE PROPERTIES – Properties that depend only on the
number of dissolved solute particles, not what they are
VOLATILE – A substance (or solute) that evaporates easily
NONVOLATILE – A substance (or solute) that does not evaporate
3I-1 (of 11)
(1) Vapor Pressure Lowering
A liquid’s equilibrium vapor pressure is lowered by having a
nonvolatile solute dissolved in it
The pressure exerted by a vapor in
equilibrium with its liquid is called the
EQUILIBRIUM VAPOR PRESSURE
With solute particles dissolved in the liquid,
there are less solvent molecules on the
surface, therefore less solvent molecules
can evaporate, so the rate of evaporation
decreases
Now vapor molecules condense faster than
liquid molecules evaporate, so the amount of
vapor decreases
When the two rates are again equal, there are
less vapor molecules than before, so a lower
equilibrium vapor pressure
3I-2 (of 11)
Temp (ºC)
10
EVP of pure
H2O (torr)
9.2
17.5
31.8
20
EVP of a dilute salt
water solution (torr)
9.1
17.4
31.6
30
EVP
Pure Water
Water Solution
(Temp °C)
3I-3 (of 11)
1882 FRANÇOIS-MARIE RAOULT
RAOULT’S LAW – The equilibrium vapor pressure of a
solution is proportional to the mole fraction of the
solvent in the solution
psolution = Xsolventpºsolvent
psolution = equilibrium vapor pressure of the solution
Xsolvent = mole fraction of the solvent in the solution
pºsolvent = equilibrium vapor pressure of the pure solvent
A solution will obey Raoult’s Law if
(1) the solution is dilute
(2) the solvent and solute particles are about the same size
(3) the solvent and solute particles have about the same attractive forces
3I-4 (of 11)
Find the equilibrium vapor pressure of a solution at 20ºC that is prepared
with 0.500 moles of sucrose dissolved in 35.00 moles of water, if the
equilibrium vapor pressure of water at 20ºC is 17.5 torr.
35.00 mol H2O
____________________________________
= 0.9859
35.00 + 0.500 mol solution
psoln = Xsolvpºsolv = (0.9859)(17.5 torr)
3I-5 (of 11)
= 17.3 torr
When a solute is an electrolyte (a salt or acid), the fact that it dissociates
or ionizes must be considered
Find the EVP of a solution at 20ºC that is prepared with 0.250 moles of
Na2SO4 dissolved in 10.00 moles of water, if the EVP of water at 20ºC is
17.5 torr.
nsolute = 3 x 0.250 mol = 0.750 mol
10.00 mol H2O
__________________________________
= 0.9302
10.00 + 0.750 mol solution
psoln = Xsolvpºsolv
3I-6 (of 11)
= (0.9302)(17.5 torr)
= 16.3 torr
Find the EVP of a solution at 23ºC that is prepared with 0.300 moles of KCl
dissolved in 15.00 moles of water, if the EVP of water at 23ºC is 21.1 torr.
nsolute = 2 x 0.300 mol = 0.600 mol
15.00 mol H2O
__________________________________
= 0.9615
15.00 + 0.600 mol solution
psoln = Xsolvpºsolv
3I-7 (of 11)
= (0.9615)(21.1 torr)
= 20.3 torr
In a solution, if both the solvent and solute are volatile, vapor pressure
is produced from both components in the solution
psolution = psolvent + psolute
psolution = Xsolventpºsolvent + Xsolutepºsolute
3I-8 (of 11)
Find the EVP of a solution prepared with 0.300 moles of acetone and
0.100 moles of chloroform if the EVP of acetone is 293 torr and the EVP of
chloroform is 345 torr.
psolution = Xsolventpºsolvent + Xsolutepºsolute
0.300 mol acetone
=
0.7500
0.100 mol chloroform =
0.2500
________________________
0.400 mol solution
___________________________
0.400 mol solution
(0.7500)(293 torr) + (0.2500)(345 torr)
219.8 torr
3I-9 (of 11)
+
86.25 torr
=
306.05 torr
=
306 torr
If the EVP of a solution equals the sum of the EVP’s of the 2 components,
the solution obeys Raoult’s Law
IDEAL SOLUTION – A solution that obeys Raoult’s Law
3I-10 (of 11)
POSITIVE DEVIATION – Occurs
when the attractions between
solvent and solute molecules are
WEAKER than the attractions in
either pure component
NEGATIVE DEVIATION – Occurs
when the attractions between
solvent and solute molecules are
STRONGER than the attractions in
either pure component
Solution formation is endothermic
Solution formation is exothermic
3I-11 (of 11)
(2) Boiling Point Elevation
A liquid’s boiling point is raised by having a nonvolatile solute
dissolved in it
A solvent boils when its EVP equals the
prevailing atmospheric pressure
Atmospheric Pressure
With solute particles dissolved in the
solvent, its EVP is lowered
Only when the temperature increases will the
EVP again equal atmospheric pressure
 the boiling point of the solution is now
higher than the boiling point of the solvent
102ºC
101ºC
100ºC
3J-1 (of 16)
EVP
Pure Water
Water Solution
100
(Temp °C)
3J-2 (of 16)
To determine the increase in boiling point from the solvent to the solution:
ΔTb = Kbmi
ΔTb = increase in the boiling point
Kb
= molal boiling point constant of the solvent
m
= molality of the solution
i
= van’t Hoff factor of the solute = moles of particles in solution
_____________________________________
moles of dissolved solute
i=
3J-3 (of 16)
C6H12O6
MgBr2
HCl
Al2(SO4)3
1
3
2
5
Find the boiling point of a solution that is prepared with 0.150 moles of
sodium chloride dissolved in 90.0 grams of water.
The boiling point of pure water is 100.00ºC and molal boiling point
constant for water is 0.51 Cºkg H2O/mol solute.
ΔTb = Kbmi
0.150 mol NaCl
____________________
= 1.667 m NaCl
0.0900 kg H2O
ΔTb = (0.51 Cºkg H2O/mol solute)(1.667 mol solute/kg H2O)(2) = 1.7 Cº
100.00ºC + 1.7 Cº
3J-4 (of 16)
= 101.7ºC
All colligative properties can be used to calculate the molar mass of a
nonvolatile solute
3J-5 (of 16)
Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with
4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon
disulfide has a boiling point of 47.5ºC.
molar mass X
=
grams X
_____________
moles X
molar mass N.E. =
4.80 grams N.E.
____________________
? moles N.E.
3J-6 (of 16)
Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with
4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon
disulfide has a boiling point of 47.5ºC.
Pure carbon disulfide has a boiling point of 46.3ºC and molal boiling point
constant of 2.37 Cºkg CS2/mol solute.
ΔTb = Kbmi
ΔTb = 47.5ºC – 46.3ºC = 1.2 Cº
ΔTb = Kb (mol solute) i
_______________
(kg solvent)
(kg solvent) ΔTb = (mol solute)
_____________________
Kb i
=
(0.1500 kg CS2)(1.2 Cº)
_____________________________________
(2.37 Cºkg CS2/mol solute)(1)
= 0.0759 mol solute
3J-7 (of 16)
Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with
4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon
disulfide has a boiling point of 47.5ºC.
Pure carbon disulfide has a boiling point of 46.3ºC and molal boiling point
constant of 2.37 Cºkg CS2/mol solute.
4.80 grams N.E.
_______________________
0.0759 moles N.E.
3J-8 (of 16)
=
63 g/mol
(3) Freezing Point Depression
A liquid’s freezing point is lowered by having a nonvolatile solute
dissolved in it
A solvent freezes when the
EVP of its liquid equals the
EVP of its solid
With solute particles
dissolved in the liquid
solvent, its EVP is lowered
Only when the temperature
decreases will the EVP of
the liquid solution equal the
EVP of the solid solvent
-2ºC
-1ºC
0ºC
 the freezing point of the solution is now lower than the freezing
point of the solvent
3J-9 (of 16)
EVP
Pure Water
Ice
Water Solution
0
(Temp °C)
3J-10 (of 16)
To determine the decrease in freezing point from the solvent to the solution:
ΔTf = Kfmi
ΔTf = decrease in the freezing point
Kf
= molal freezing point constant of the solvent
m
= molality of the solution
i
= van’t Hoff factor of the solute
3J-11 (of 16)
Find the freezing point of a solution that is prepared with 0.150 moles of
calcium chloride dissolved in 97.0 grams of water.
The freezing point of pure water is 0.00ºC and molal freezing point
constant for water is 1.86 Cºkg H2O/mol solute.
ΔTf = Kfmi
0.150 mol CaCl2
_______________________
= 1.546 m CaCl2
0.09700 kg H2O
ΔTf = (1.86 Cºkg H2O/mol solute)(1.546 mol solute/kg H2O)(3)
0.00ºC - 2.88 Cº = -8.83ºC
3J-12 (of 16)
= 8.63 Cº
Find the molar mass of a nonelectrolyte if a solution prepared with 10.0
grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a
freezing point of 3.51ºC.
molar mass X
=
grams X
_____________
moles X
molar mass N.E. =
10.00 grams N.E.
____________________
? moles N.E.
3J-13 (of 16)
Find the molar mass of a nonelectrolyte if a solution prepared with 10.0
grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a
freezing point of 3.51ºC.
Pure benzene has a freezing point of 5.48ºC, a molal freezing point
constant of 5.12 Cºkg benzene/mol solute, and a density of 0.780 g/mL.
ΔTf = Kfmi
ΔTf = 5.48ºC – 3.51ºC = 1.97 Cº
ΔTf = Kf (mol solute) i
_______________
(kg solvent)
100.0 mL benzene x 0.780 g benzene
_____________________
= 78.00 g benzene
mL benzene
= 0.07800 kg benzene
3J-14 (of 16)
Find the molar mass of a nonelectrolyte if a solution prepared with 10.0
grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a
freezing point of 3.51ºC.
Pure benzene has a freezing point of 5.48ºC, a molal freezing point
constant of 5.12 Cºkg benzene/mol solute, and a density of 0.780 g/mL.
ΔTf = Kfmi
ΔTf = 5.48ºC – 3.51ºC = 1.97 Cº
ΔTf = Kf (mol solute) i
_______________
(kg solvent)
(kg solvent) ΔTb = (mol solute)
_____________________
Kb i
=
(0.07800 kg C6H6)(1.97 Cº)
______________________________________
(5.12 Cºkg C6H6/mol solute)(1)
= 0.03001 mol solute
3J-15 (of 16)
Find the molar mass of a nonelectrolyte if a solution prepared with 10.0
grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a
freezing point of 3.51ºC.
Pure benzene has a freezing point of 5.48ºC, a molal freezing point
constant of 5.12 Cºkg benzene/mol solute, and a density of 0.780 g/mL.
10.0 grams N.E.
_________________________
0.03001 moles N.E.
3J-16 (of 16)
=
333 g/mol
(4) OSMOTIC PRESSURE
Cells in an ISOTONIC SOLUTION
have equal flow rates of water in and out
Cells in a HYPERTONIC SOLUTION
have water flowing out faster than flowing in
Cells in a HYPOTONIC SOLUTION
have water flowing in faster than flowing out
3K-1 (of 6)
SEMIPERMEABLE MEMBRANE – A membrane that will allow small
molecules to pass through but not big ones
water / not hydrated salt ions
water, salt ions / not proteins
Because the membrane blocks salt
ions, water goes out of the tube
slower than it goes in
The extra mass of salt water builds
up pressure on the membrane,
forcing water through faster
OSMOTIC PRESSURE (π) – The pressure on a semipermeable membrane
needed to equalize the passage of water across the membrane between
two solutions of different concentrations
3K-2 (of 6)
To determine the osmotic pressure in a solution:
πV = inRT
π
= Osmotic Pressure (atm)
V
= Volume of Solution (L)
i
= van’t Hoff factor
n
= Difference in Quantity of Solute between the 2 solutions (mol)
R
= Universal Gas Constant (0.08206 Latm/molK)
T
= Temperature (K)
3K-3 (of 6)
To determine the osmotic pressure in a solution:
πV = inRT
π
= inRT
_______
V
π
= iMRT
3K-4 (of 6)
n/V is molarity
Find the osmotic pressure that would develop in a solution that is
prepared at 22ºC with 0.0300 moles of glucose dissolved in enough water
to make 100. mL of solution.
π = inRT
_______
V
22ºC + 273.2 K = 295.2 K
π = (1)(0.0300 mol)(0.08206 Latm/molK)(295.2 K)
__________________________________________________________
0.100 L
3K-5 (of 6)
= 7.27 atm
Find the molar mass of a protein if a solution is prepared by dissolving
1.00 x 10-3 grams of the protein in enough water to make 1.00 milliliter of
solution, and the osmotic pressure at 25ºC is 1.12 torr.
π = inRT
_______
V
25ºC + 273.2 = 298.2 K
1.12 torr x
1 atm
_____________
= 0.001474 atm
760.0 torr
πV = n =
(0.001474 atm)(0.00100 L)
_____
_________________________________________
iRT
(1)(0.08206 Latm/molK)(298.2 K)
1.00 x 10-3 g protein
__________________________________
6. 024 x 10-8 moles protein
3K-6 (of 6)
=
16,600 g/mol
=
6.024 x 10-8 mol