First Mid-quarter Examination
MQ-1 on Monday, Jan. 29
at 6:30 pm
Covering Chapters 10, 11, and 13
room TBA
For students with a
scheduled-class conflict
with the first Mid-Quarter Exam:
(1) Send me an email today with a copy of your
class schedule, and
(2) Let me know when you want to take the
makeup exam:
a) during the last week of classes.
b) early (5:00-6:18 pm) on Mon, Jan 29.
c) late (8:00-9:18 pm) on Mon, Jan 29.
MQ-1 on Monday, Jan. 29
at 6:30 pm
Covering Chapters 10, 11, and 13
Help session on Saturday, Jan 27
at 4:00 – 6:00 pm
in MP1000
Week 4 Sections 13.1- 13.6—Properties of Solutions
13.1
13.2
13.3
The Solution Process
Energy Changes, Entropy, Rxns
Saturated Solutions and Solubility
Factors Affecting Solubility
Solute-Solvent Interactions
Pressure Effects
Temperature Effects
13.4
Ways of Expressing Concentration
Mass Percentage, ppm, and ppb
Mole Fraction, Molarity, and Molality
Conversion of Concentration Units
13.5
13.6
Colligative Properties
Colloids
endothermic
endothermic
exothermic
Figure 13.3
Figure 13.4
e.g. MgSO4
e.g. NH4NO3
Realize there is an inherent tendency for the two isolated materials to form solution,
regardless of the energetics!!! This represents an “entropy” factor.
Factors that FAVOR solubility:
1. Strong solute-solvent interactions
2. Weak solute-solute interactions
3. Weak solvent-solvent interactions
More often we’ll settle for the solute-solvent interactions
being similar to the solute-solute and solvent-solvent
interactions.
A general rule: Like dissolves like.
i.e. polar and polar
non-polar and non-polar
• Dissolution: solute + solvent  solution.
• Crystallization: solution  solute + solvent.
• Saturation: crystallization and dissolution are in
equilibrium.
• Solubility: amount of solute required to form a saturated
solution.
• Supersaturation: a solution formed when more solute is
dissolved than in a saturated solution.
Fig 13.12 Structure of glucose—note red O atoms in OH groups
which can interact nicely with water.
Consider gas solubilities in water at 20 oC with 1 atm gas pressure (~Table 13.2)
Solubility/M
He
0.40 x 10-3
N2
0.69 x 10-3
CO
1.04 x 10-3
O2
1.38 x 10-3
Ar
1.50 x 10-3
Kr
2.79 x 10-3
CO2
3.1 x 10-2
NH3
~ 53
Fig 13.14 Henry’s Law, Cg = k Pg
Note Table 13.2 again for k
Consider N2 dissolved in water at 4.0 atm.
Note k = 0.69 x 10-3 mol/L-atm
C g = k Pg
= (0.69 x 10-3 mol/L-atm)(4.0 atm)
= 2.76 x 10-3 mol/L
at normal atmospheric conditions, however, Pg = 0.78 atm
Cg = (0.69 x 10-3 mol/L-atm)(0.78 atm)
= 0.538 x 10-3 mol/L
Note that (2.76 - 0.54) x 10-3 mol/L = 2.22 x 10-3 mol/l
Thus for 1.0 L of water, 0.0022 mol of nitrogen would
be released = 0.0022 x 22.4L = 0.049 L = 49 mL !
To read about nitrogen narcosis, see
http://www.scuba-doc.com/narked.html and about the bends,
see http://www.diversalertnetwork.org/medical/articles/index.asp
Solubility of GASES as a
Function of Temperature
And of
SALTS
Are these exothermic or endothermic processes?
Ways of expressing concentration:
a) percent, ppm, ppb
usually mass/mass
b) mole fraction = XA , XB sum of Xi = 1
c) molarity = M or mol/Lsolution
depends on T and density of soln
preparation requires dilution
d) molality = m or mol/kgsolvent
independent of T
easily prepared
(a) Prepare 0.500 L of a 0.100 M aqueous solution of
KHCO3 (MW = 100.12 g)
Recall M=n/L or n = (M)(L)
therefore we need
5.01 g of KHCO3 dissolved and diluted to 0.500 L
(b) Use this solution as a ‘stock’ solution to prepare
a final solution of 0.0400 M concentration.
What is the final volume of this solution?
Since n = M V,
M1V1 = M2V2
and V2 = M1V1/M2 = (0.100 M)(0.500 L) / (0.0400 M)
= 1.25 L
Consider a solution prepared by dissolving 22.4 g MgCl2
in 0.200 L of water. Assume the density of water is
1.000 g/cm3 and the density of the solution is
1.089 g/cm3.
Calculate
mole fraction
molarity
molality
MQ-1 on Monday, Jan. 29
at 6:30 pm
Covering Chapters 10, 11, and 13
Help session on Saturday, Jan 27
at 4:00 – 6:00 pm
in MP1000
For students with a
scheduled-class conflict
with the first Mid-Quarter Exam:
(1) Send me an email today with a copy of your
class schedule, and
(2) Let me know when you want to take the
makeup exam:
a) during the last week of classes.
b) early (5:00-6:18 pm) on Mon, Jan 29.
c) late (8:00-9:18 pm) on Mon, Jan 29.
A 9.386 M solution of H2SO4 has a density of 1.509 g/cm3.
Calculate
molality
% by mass
mole fraction of H2SO4
Week 4 Sections 13.1- 13.6
13.3
Factors Affecting Solubility
13.4
Ways of Expressing Concentration
Mass Percentage, ppm, and ppb
Mole Fraction, Molarity, and Molality
Conversion of Concentration Units
13.5
Colligative Properties
Lowering the Vapor Pressure
Boiling-Point Elevation
Freezing-Point Depression
Osmosis
Determination of Molar Mass
Colloids
13.6
Colligative Properties
Solution properties that depend only on the
total # of ‘particles’ present.
Vapor Pressure
Boiling Point
Freezing Point
Osmotic Pressure
Note that VP of a solution is lower than that of pure solvent.
A fascinating and somewhat surprising observation:
Raoult’s Law
PA = XA PAo
PA = vapor pressure over solution
XA = mole fraction of component A (solvent)
PAo = vapor pressure of pure
component A (solvent)
also PA = (1 – XB) PAo
where XB = mol fraction of B (solute)
(Recall also Dalton’s Law: PA = XA Ptotal )
At first, we consider only nonvolatile solutes.
At 20 oC, the vapor pressure of benzene (cmpd A, MW=78)
is 0.1252 atm,i.e. PAo = 0.1252 atm. If 6.40 g of naphthalene
(cmpd B, C10H8, 128.17 g/mol) is dissolved in 78.0 g of
benzene calculate the vapor pressure of benzene over
the solution.
As shown on page 549, we also can consider
solutions with two volatile components.
Consider a liquid soln containing 1.0 mol benzene
and 2.0 mol of toluene at 20 oC. This yields
Xbenzene = 0.33 and Xtoluene = 0.67
This can be coupled with the fact that
Pobenzene = 75 torr
Potoluene = 22 torr
Apply Raoult’s law to each separately, to obtain
Pbenzene = XbenzenePobenzene = 25 torr
Ptoluene = Xtoluene Potoluene = 15 torr
and PT = Pb + Pt = 40 torr
But with Pb = 25 torr and Pt = 15 torr
We also can calculate the concentrations
of the two in the gas phase!
Xbgas = 25/40 = 0.63
and Xtgas = 15/40 = 0.37
Boiling Point Elevation and Freezing Point Depression
(directly related to Raoult’s Law)
Boiling Point Elevation
ΔTb = Tb final– Tb initial= Kb m => + quantity
where m is the molal concentration
Freezing Point Depression
ΔTf = Tffinal – Tfinitial = - Kf m => - quantity
where m is the molal concentration.
[note the definition and the negative sign!!!]
for H2O, Kb = 0.052 oC/m and Kf = 1.86 oC/m
Consider a water solution which has 0.500 mol
of sucrose in 1.000 kg of water. Therefore it
has a concentration of 0.500 molal or 0.500 mol/kg.
recall Kb = 0.52 oC/m and Kf = 1.86 oC/m
What is the boiling point and freezing point
of this solution?
(a) When 5.50 g of biphenyl (C12H10, 154.2 g/mol) is
dissolved in 100 g benzene (C6H6, 78.0 g/mol), the
BP increases by 0.903 oC. Calculate Kb for benzene.
(a) When 5.50 g of biphenyl (C12H10, 154.2 g/mol) is
dissolved in 100 g benzene (C6H6, 78.0 g/mol), the
BP increases by 0.903 oC. Calculate Kb for benzene.
 Kb = 2.53 K kg/mol
(b) When 6.30 g of an unknown hydrocarbon is dissolved
in 150.0 g of benzene, the BP of the solution increases
by 0.597 oC.
What is the MW of the unknown substance?
But recall we said Colligative Properties depend on the
total concentration of ‘species’.
A sample of sea water contains the following in 1.000 L
of solution. Estimate the freezing point of this solution.
Na+ = 4.58 mol
ClMg2+ = 0.052 mol
SO42Ca2+ = 0.010 mol
HCO3K+
= 0.010
Brneutral species = 0.001 mol
Sum of species = 1.095 mol
= 0.533 mol
= 0.028 mol
= 0.002 mol
= 0.001 mol
List the following aqueous solutions in increasing order
of their expected freezing points.
0.050 m CaCl2  0.05 m Ca+2 and 0.10 m Cl-1
0.15 m NaCl  0.30 m total
0.10 m HCl  0.20 m total
0.050 m HOAc  between 0.05 and 0.10 m total
0.10 m C12H22O11  0.10 m total
List the following aqueous solutions in increasing order
of their expected freezing points.
0.050 m CaCl2
0.15 m NaCl
0.10 m HCl
0.050 m HOAc
0.10 m C12H22O11
x
x
x
x
x
3
2
2
1
1
=
=
=
=
=
0.150
0.30
0.20
0.050
0.10
The van’t Hoft “i factor”
These calculations assume total dissociation of the
salts and zero dissociation of the last two.
This effect of the dissociation of electrolytes is usually
taken into account through the van’t Hoff i factor,
which can be stated formally as
ΔTb = i Kb m
Note that i may be defined as
i=
ΔTf(actual)
Kf meffective
meffective
-------------- = ------------- = -----------ΔTf(ideal)
Kf mideal
mideal
In real systems, these i factors are NOT integers,
but rather fractions whose values depend on
concentration.
Focus, for example on NaCl. Notice the limiting value
as well as the values at higher concentrations.
But how can i be less than 2.00 for NaCl?
Through this
partial
association.
This slide from earlier offers a useful intro to
Recall this observation:
Osmotic Pressure.
Osmotic
Pressure:
a fascinating
behavior.
Yet it is the
result of a
very simple
tendency to
equalize the
concentrations
of solutions.
With the addition
of the
semipermeable
membrane—
which permits only
solvent particles
to move from one
side to the other.
The critical part is the membrane!!!
A VERY practical application/consequence
of Osmotic Pressure.
Hypertonic soln
Hypotonic soln
crenation
(shrivels)
hemolysis
(bursts)
Examples:
– Cucumber placed in NaCl solution loses water to
shrivel up and become a pickle.
– Limp carrot placed in water becomes firm because
water enters via osmosis.
– Salty food causes retention of water and swelling of
tissues (edema).
– Water moves into plants through osmosis.
– Salt added to meat or sugar to fruit prevents bacterial
infection (a bacterium placed on the salt or honey will
lose water through osmosis and die).
Osmotic Pressure:
or π = (n/V) R T
or
(or
πV=nRT
π= MRT
π = ρ g h,
where ρ = density of solution
g = 9.807 m s-2
h = height of column
but be careful of the units with this form.)
π = ρ g h, where ρ = density of solution
g = 9.807 m s-2
h = height of column
If h = 0.17 meters of a dilute aqueous soln with ρ = 1.00 g/cm3
π = [(1.00g/cm3)(10-3 kg/g)(106 cm3/m3)](9.807 m/s2)(0.17 m)
= 1.7 x 103 kg m-1 s-1 = 1.7 x 103 Pa
or = (1.7 x 103 Pa) / (1.013 x 105 Pa / atm) = 0.016 atm
Sample Exercise 13.13
A chemist dissolves 3.50 mg of protein in water
to make 5.00 mL of solution.
water. The observed osmotic pressure is 1.54 torr
at 25 oC. What is the MW of the protein?
M = pi/RT = [(1.54 torr)(1 atm/760 torr)]/
(0.0821 L-atm/mol-K)(298K)
= 8.28 x 10-5 mol/L
 4.14 x 10-7 moles of protein
or a MW of 8.45 x 103 g/mol
Colloids
•
•
•
•
•
Hydrophilic and Hydrophobic Colloids
Focus on colloids in water.
“Water loving” colloids: hydrophilic.
“Water hating” colloids: hydrophobic.
Molecules arrange themselves so that hydrophobic
portions are oriented towards each other.
If a large hydrophobic macromolecule (giant molecule)
needs to exist in water (e.g. in a cell), hydrophobic
molecules embed themselves into the macromolecule
leaving the hydrophilic ends to interact with water.
•
•
•
•
•
•
Hydrophilic and Hydrophobic Colloids
Typical hydrophilic groups are polar (containing C-O, OH, N-H bonds) or charged.
Hydrophobic colloids need to be stabilized in water.
Adsorption: when something sticks to a surface we say
that it is adsorbed.
If ions are adsorbed onto the surface of a colloid, the
colloids appears hydrophilic and is stabilized in water.
Consider a small drop of oil in water.
Add to the water sodium stearate.
Colloids
Hydrophilic and Hydrophobic Colloids
Hydrophilic and Hydrophobic Colloids
• Sodium stearate has a long hydrophobic tail
(CH3(CH2)16-) and a small hydrophobic head (-CO2-Na+).
• The hydrophobic tail can be absorbed into the oil drop,
leaving the hydrophilic head on the surface.
• The hydrophilic heads then interact with the water and
the oil drop is stabilized in water.
Removal of Colloidal Particles. . . . . . .
• Colloid particles are too small to be separated by
physical means (e.g. filtration).
• Colloid particles are coagulated (enlarged) until they
can be removed by filtration.
• Methods of coagulation:
– heating (colloid particles move and are attracted to each other
when they collide);
– adding an electrolyte (neutralize the surface charges on the
colloid particles).
– Dialysis: using a semipermeable membranes separate ions
from colloidal particles