Physical Properties
of Solutions
Chapter 12
Solution
Stoichiometry
end of Chapter 4
Problems: 12.12, 12.15, 12.16, 12.17,
12.18, 12.21, 12.22, 12.28, 12.36, 12.38,
12.51, 12.54, 12.55, 12.57, 12.60, 12.65,
12.76, 12.77, 12.112
4.12, 4.60, 4.70, 4.72, 4.74, 4.77, 4.88
A solution is a homogenous mixture of 2 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
12.1
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
4.1
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.
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate.
12.1
Three types of interactions in the solution process:
• solvent-solvent interaction
• solute-solute interaction
• solvent-solute interaction
ΔHsoln = ΔH1 + ΔH2 + ΔH3
12.2
“like dissolves like”
Two substances with similar intermolecular forces are likely
to be soluble in each other.
•
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)
12.2
Concentration Units
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
Percent by Mass
mass of solute
x 100%
% by mass =
mass of solute + mass of solvent
mass of solute x 100%
=
mass of solution
Mole Fraction (X)
moles of A
XA =
sum of moles of all components
12.3
Concentration Units Continued
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
12.3
What is the molality of a 5.86 M ethanol (C2H5OH)
solution whose density is 0.927 g/mL?
moles of solute
moles of solute
m =
M =
mass of solvent (kg)
liters of solution
Assume 1 L of solution:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
moles of solute
m =
mass of solvent (kg)
=
5.86 moles C2H5OH
= 8.92 m
0.657 kg solvent
12.3
Convert % mass to Molarity
• What is the Molarity of a 95% acetic acid
solution? (density = 1.049 g/mL)
If you assume 1 L, that amount of solution = 1049 g
95% of the solution is acetic acid
1049 g solution x 0.95 = 997 g solute
997 g X 1 mol/60.05 g = 16.6 mol solute
Since we assumed 1 L, that’s 16.6 mol / 1 L or
16.6 M
Temperature and Solubility
Solid solubility and temperature
solubility
solubility decreases
increases with
with
increasing temperature
12.4
Fractional crystallization is the separation of a mixture of
substances into pure components on the basis of their differing
solubilities.
Suppose you have 90 g KNO3
contaminated with 10 g NaCl.
Fractional crystallization:
1. Dissolve sample in 100 mL of
water at 600C
2. Cool solution to 00C
3. All NaCl will stay in solution
(s = 34.2g/100g)
4. 78 g of PURE KNO3 will
precipitate (s = 12 g/100g).
90 g – 12 g = 78 g
12.4
Temperature and Solubility
Gas solubility and temperature
solubility usually
decreases with
increasing temperature
12.4
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).
c is the concentration (M) of the dissolved gas
c = kP
P is the pressure of the gas over the solution
k is a constant (mol/L•atm) that depends only
on temperature
low P
high P
low c
high c
12.5
Chemistry In Action: The Killer Lake
8/21/86
CO2 Cloud Released
1700 Casualties
Trigger?
•
earthquake
•
landslide
•
strong Winds
Lake Nyos, West Africa
Solution Stoichiometry (Chapter 4)
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
What mass of KI is required to make 500. mL of
a 2.80 M KI solution?
M KI
volume KI
500. mL x
moles KI
1L
1000 mL
x
2.80 mol KI
1 L soln
M KI
x
grams KI
166 g KI
1 mol KI
= 232 g KI
4.5
4.5
Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
=
Moles of solute
after dilution (f)
MiVi
=
MfVf
4.5
How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Mi = 4.00
Vi =
Mf = 0.200
MfVf
Mi
Vf = 0.06 L
Vi = ? L
0.200 x 0.06
=
= 0.003 L = 3 mL
4.00
3 mL of acid + 57 mL of water = 60 mL of solution
4.5
Gravimetric Analysis
1. Dissolve unknown substance in water
2. React unknown with known substance to form a precipitate
3. Filter and dry precipitate
4. Weigh precipitate
5. Use chemical formula and mass of precipitate to determine
amount of unknown ion
4.6
Titrations
In a titration a solution of accurately known concentration is
added gradually added to another solution of unknown
concentration until the chemical reaction between the two
solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
4.7
What volume of a 1.420 M NaOH solution is
Required to titrate 25.00 mL of a 4.50 M H2SO4
solution?
WRITE THE CHEMICAL EQUATION!
H2SO4 + 2NaOH
M
volume acid
25.00 mL x
acid
2H2O + Na2SO4
rx
moles acid
4.50 mol H2SO4
1000 mL soln
x
coef.
M
moles base
2 mol NaOH
1 mol H2SO4
x
base
volume base
1000 ml soln
1.420 mol NaOH
= 158 mL
4.7
Chemistry in Action: Metals from the Sea
CaCO3 (s)
CaO (s) + CO2 (g)
CaO (s) + H2O (l)
Ca2+ (aq) + 2OH- (aq)
Mg2+ (aq) + 2OH - (aq)
Mg(OH)2 (s) + 2HCl (aq)
Mg2+ + 2e2ClMgCl2 (l)
Mg
Cl2 + 2eMg (l) + Cl2 (g)
Now back to Chapter 12…
Mg(OH)2 (s)
MgCl2 (aq) + 2H2O (l)
Colligative Properties of Nonelectrolyte Solutions
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
0
1
Raoult’s law
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
If the solution contains only one solute:
X1 = 1 – X2
P 10 - P1 = DP = X2 P 10
X2 = mole fraction of the solute
12.6
Fractional Distillation Apparatus
12.6
Boiling-Point Elevation
ΔTb = Tb – T b0
T b0 is the boiling point of
the pure solvent
T b is the boiling point of
the solution
Tb > T b0
ΔTb > 0
ΔTb = Kb m i
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
12.6
Freezing-Point Depression
ΔTf = T 0f – Tf
T
0
Tf
f
is the freezing point of
the pure solvent
is the freezing point of
the solution
T 0f > Tf
DTf > 0
ΔTf = Kf mi
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
i is van’t hoffs factor for
12.6
electrolyte solutions
12.6
What is the freezing point of a solution containing 478 g
of ethylene glycol (antifreeze) in 3202 g of water? The
molar mass of ethylene glycol is 62.01 g.
ΔTf = Kf m
Kf water = 1.86 0C/m
moles of solute
m =
mass of solvent (kg)
478 g x
1 mol
62.01 g
=
= 2.41 m
3.202 kg solvent
ΔTf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
ΔTf = T 0f – Tf
Tf = T 0f – ΔTf = 0.00 0C – 4.48 0C = -4.48 0C
12.6
Colligative Properties of Nonelectrolyte Solutions
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 10
Boiling-Point Elevation
ΔTb = Kb mi
Freezing-Point Depression
ΔTf = Kf mi
Osmotic Pressure (p)
p= MRT
T is in Kelvin
12.6
Colligative Properties of Electrolyte Solutions
0.1 m Na+ ions & 0.1 m Cl- ions
0.1 m NaCl solution
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
van’t Hoff factor (i) =
0.2 m ions in solution
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be
nonelectrolytes
NaCl
CaCl2
1
2
3
12.7
Change in Freezing Point
Which would you use for the streets of
Bloomington to lower the freezing point
of ice and why? Would the temperature
make any difference in your decision?
a)
sand, SiO2
b)
Rock salt, NaCl
c)
Ice Melt, CaCl2
Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
ΔTb = i Kb m
Freezing-Point Depression
ΔTf = i Kf m
Osmotic Pressure (p)
p = iMRT
12.7
Change in Freezing Point
Common Applications
of Freezing Point
Depression
Propylene glycol
Ethylene
glycol –
deadly to
small
animals
Change in Boiling Point
Common Applications of
Boiling Point Elevation
Freezing Point Depression
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 = 20.1 oC
FP = 0 – 20.1 = -20.1 oC
The Cleansing Action of Soap
12.8
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.
dilute
more
concentrated
12.6
Osmotic Pressure (p)
High
P
Low
P
p = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
12.6
A cell in an:
isotonic
solution
hypotonic
solution
hypertonic
solution
12.6
Chemistry In Action:
Desalination
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
12.8
Colloids
• Brownian motion
• Tyndall Effect
Suspensions
• These are mixed, but not dissolved in each
other
• Will settle over time
• Particles are bigger than 1 micrometer
(larger than colloid)
• Examples: dust in air, muddy water