Transcript Chem 1201 - LSU Department of Chemistry

Watkins
Chapter 13
Chapter 13
Solutions
Solubility
Concentration Units
Colligative Properties
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Chapter 13
Solvation
Dissolving: the solvent molecules separate
and then bond to the solute particles.
Polar solvent bonded to ionic solute by
ion-dipole force
Polar solvent bonded to polar solute
by dipole-dipole force
Non-polar solvent bonded to nonpolar solute by London force
Like Dissolves Like
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Chapter 13
Concentration
amount of solute
amount of solution
If “amount” = mass...
mass fraction =
g Su
g Su
=
g Sn
g Su + g Sv
g Su
mass pph =
× 102 (mass %)
g Sn
g Su
mass ppm =
× 106
g Sn
g Su
mass ppb =
× 109
g Sn
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Su = Solute
Sv = Solvent
Sn = Solution
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Chapter 13
Concentration
amount of solute
amount of solution
amount of solute
amount of solvent
If “amount” = moles...
mol Su
mole fraction =
= Xsu
mol Su + mol Sv
XSu + XSv = 1
The denominator may be different...
mol Su
molarity =
Volume Sn
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Su = Solute
Sv = Solvent
Sn = Solution
mol Su
molality =
Kg Sv
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Chapter 13
Concentration
1.0 g sucrose (C12H22O11, F.W. = 342.3 g/mol) is dissolved in
300 mL of ethanol (C2H5OH, density 0.79 g/cc). Calculate
the molarity and the molality of the solution.
1 g sucrose = 2.92 mmol. Assume the volume of the
solution is 300 mL.
2.92 mmol
M=
= 0.0097 molar
300 mL
The mass of solvent is 0.79 g/cc × 300 cc = 237 g
2.92 mmol
m=
= 0.0123 molal
237 g
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Chapter 13
Gases Dissolve in Liquids
The solubility of a gas in a liquid depends on (a) what the
gas is; (b) what the liquid is; (c) the gas pressure above
the liquid; (d) the temperature of the liquid.
Henry's Law: Conc = kPgas
k  T-1
1.38 mM/atm
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DT = 0
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Chapter 13
Gases Dissolve in Liquids
The solubility of a gas in a liquid depends on (a) what the
gas is; (b) what the liquid is; (c) the gas pressure above
the liquid; (d) the temperature of the liquid.
Henry's Law: Conc = kPgas
What is the concentration of dissolved oxygen in the ocean
if the water temperature is 20 oC?
1. k = 0.00138 M/atm
2. P(O2) = X(O2)Ptotal = (0.20)(1atm) = 0.20 atm
3. conc = kP = 0.00138 M/atm × 0.20 atm = 0.000276 M
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Chapter 13
Vapor Pressure of a Solution
The vapor pressure Po of a pure liquid depends on
the number of “hot” molecules at the surface.
Po
Su = Solute
Sv = Solvent
Sn = Solution
P
T
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Chapter 13
Vapor Pressure of a Solution
The vapor pressure of solvent above solution, PSn,
depends on the number of “hot” molecules at the
surface and on the relative number of solvent
molecules at the surface.
Po > PSn
Su = Solute
Sv = Solvent
Sn = Solution
is a nonvolatile solute.
P
T
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Chapter 13
Raoult's Law
n Sv
PSn =
Po
n Sv  n Su
PSn = XSvPo
Po > PSn
Su = Solute
Sv = Solvent
Sn = Solution
is a nonvolatile solute.
P
T
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Chapter 13
Raoult's Law
PSn = XSvPo
Three direct consequences of
Raoult's Law are the colligative
properties of solutions
Boiling Point Elevation
Freezing Point Depression
Osmotic Pressure
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Boiling Point Elevation
Su = Solute
Sv = Solvent
Sn = Solution
The normal boiling point of a solution, TSn, is
always greater than the normal boiling point
of pure solvent, TSv
DTb = |TSn – TSv| > 0 (b.p. elevation)
1 atm
PSn = XSvPo
P
T
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Chapter 13
Boiling Point Elevation
Su = Solute
Sv = Solvent
Sn = Solution
The normal boiling point of a solution, TSn, is
always greater than the normal boiling point
of pure solvent, TSv
DTb = |TSn – TSv| > 0 (b.p. elevation)
Boiling point elevation depends on the nature
of the solute, the concentration of the solution,
and what the solvent is:
DTb = |TSn – TSv| = imKb
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Chapter 13
Boiling Point Elevation
Su = Solute
Sv = Solvent
Sn = Solution
The normal boiling point of a solution, TSn, is
always greater than the normal boiling point
of pure solvent, TSv
DTb = |TSn – TSv| > 0 (b.p. elevation)
Concentration is expressed as molality
because molarity (volume) changes with
temperature.
DTb = |TSn – TSv| = imKb
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Chapter 13
Boiling Point Elevation
DTb = imKb
Su = Solute
Sv = Solvent
Sn = Solution
m is the molality of the solution
i is the number of solute particles per
formula unit (van't Hoff factor)
for molecular solutes, i = 1
For ionic solutes, i = total number of ions
per formula unit (e.g., i = 3 for MgCl2)
im is the “effective molality”
Kb is the boiling point elevation constant of
the solvent (Kb = 0.51 for water)
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Chapter 13
Boiling Point Elevation
DTb = imKb
Su = Solute
Sv = Solvent
Sn = Solution
Arrange the following aqueous solutions in order of
increasing boiling point:
0.10 m calcium nitrate
0.15 m trisodium phosphate
0.20 m sodium chloride
0.25 m sugar
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im = 0.30
im = 0.60
im = 0.40
im = 0.25
2
4 (highest)
3
1 (lowest)
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Chapter 13
Freezing Point Depression
Su = Solute
Sv = Solvent
Sn = Solution
The normal freezing point of a solution, TSn,
is always lower than the normal freezing
point of pure solvent, TSv
DTf = |TSv - TSn| > 0 (f.p. depression)
1 atm
PSn = XSvPo
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Chapter 13
Freezing Point Depression
Su = Solute
Sv = Solvent
Sn = Solution
The normal freezing point of a solution, TSn,
is always lower than the normal freezing
point of pure solvent, TSv
DTf = |TSv - TSn| > 0 (f.p. depression)
Freezing point depression depends on the
same factors as boiling point elevation
(Kf = 1.86 for water)
DTf = |TSv – TSn| = imKf
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Chapter 13
Freezing Point Depression
Su = Solute
Sv = Solvent
Sn = Solution
DTf = imKf
Into your car radiator, pour 10 L of water and 6 L of
antifreeze (ethylene glycol, C2H6O2, 62.1 g/mol,
1.11 g/cc). What is the freezing point of this
solution?
solvent (water ): 10 L = 10 kg
solute (antifreeze):
(6000 cc)(1.11 g/cc)/(62.1 g/mol) = 107.2 mol
m = (107.2 mol)/10 kg = 10.7 m
DTf = (1)(10.7)(1.86) = 20
Tf = -20 oC
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Chapter 13
Molecular Weight from
Colligative Properties
Su = Solute
Sv = Solvent
Sn = Solution
DTb = imKb
DTf = imKf
mol Su
g Su
molality 

kg Sv
MWSu kg Sv 
MWSu
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i K b g Su
i K f g Su
=

DTb kg Sv
DTf kg Sv
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Chapter 13
Osmosis
Su = Solute
Sv = Solvent
Sn = Solution
The vapor pressure above the solvent is greater than the
vapor pressure above the solution: PSn = XSvPSv
As a result, solvent vapor flows into the solution in an
effort to equalize the two pressures.
solvent vapor flow
PSv
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Osmosis: the flow of
PSn
solvent from a less
concentrated solution into
a more concentrated
solution.
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Chapter 13
Osmosis
Concentration differential
across a Semi-Permeable
Membrane (SPM)
More solvent molecules on
the dilute side collide with
the SPM, so there is a net
flow of solvent from dilute
to concentrated (osmosis).
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Osmosis: the flow of
solvent from a less
concentrated solution
into a more
concentrated solution.
rises
falls
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Chapter 13
Osmosis
 is the osmotic pressure
dhg
V = inRT
i is the van't Hoff factor of the
concentrated solution
M is the molarity of the
concentrated solution
T is the Kelvin temperature
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 = i MRT
concentrated
dilute
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Chapter 13
Osmosis
 is the osmotic pressure
dhg
V = inRT
 = i MRT
mol Su
g Su

M 
L Sn
MWSu L Sn
MWSu
i g Su R T
=
 L Sn
concentrated
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dilute
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Chapter 13
Molecular Weight from
Colligative Properties
DTb,f = imKb,f
MWSu
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 = iMRT
i K f g Su

D Tf kg Sv
Freezing point depression
i K b g Su
=
DTb kg Sv
Boiling point elevation
i g Su R T
=
 L Sn 
Osmotic Pressure
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