Gases & colligative
properties
Ch.14
Gases dissolving in liquids

Pressure and temperature influence gas
solubility

Solubility directly proportional to gas pressure

Henry’s Law:
Sg = kHPg
Sg = gas solubility (M = mol/L)
kH = Henry’s law constant (unique to each gas;
M/mm Hg)
Pg = partial pressure of gaseous solute (mm Hg)




Increase partial pressure 
increase solubility
Example


27.0 g of acetylene gas
dissolves in 1.00 L of
acetone at 1.00 atm
partial pressure of
acetylene.
If the partial pressure of
acetylene is increased to
6.00 atm, what is the
solubility of acetylene in
acetone in mol/L? MW of
acetylene = 26.037 g/mol
1.
2.
3.
27.0 g x (mol/26.037 g)
x (1/1.00 L) = 1.04 M
Sg = kHPg
1.04 M = kH x 1.00 atm
4.
1. kH = 1.04 M/atm
Sg = (1.04 M/atm) x 6.00 atm
1. = 6.24 M

Could also solve this by:
– (Sg1/Pg1) = (Sg2/Pg2)
– How did I come up with
this?
Problem

The partial pressure of oxygen gas, O2, in
air at sea level is 0.21 atm.
– Using Henry’s Law, calculate the molar
concentration of oxygen gas in the surface
water (at 20°C) of a lake saturated with air
given that the solubility of O2 at 20°C and 1.0
atm pressure is 1.38•10-3 M.
Solution
3
S2
1.38 10 M

1.0atm
0.21atm
4
S2  2.9 10 M
They call it “pop” in the Midwest

Drinks carbonated under high pressure
– Above 90 atm
– Under CO2 atmosphere

Once bottle opened, partial pressure of gas
above soda plummets
– CO2 solubility decreases drastically
– Gas bubbles out of soln

Once the fizz is gone, it can never be regained
– Truly, one of the existential tragedies of this universe
The bends

Deeper diving has higher pressures
– Must use breathing tank
– If it contains N2 then higher pressure forces N2 to
dissolve in higher amounts in blood

If ascension too fast, lower pressure causes N2
to start bubbling out of blood too quickly
– Rupturing of arteries
 Excruciatingly painful death
– Must be rushed to hyperbaric chamber

Tanks now don’t use N2, but He
– Why?
Effects of temp on solubility
Obviously, as temp
increases, solubility
decreases
 Since increasing heat
causes gases to
dissolve out
(endothermic)

–  dissolving gases is
an exothermic process
Another look at gas solubility: Le
Châtelier’s Principle
Explains temperature relevance of solubility
 For systems in equilibrium, change in one side
causes system to counteract on other side:
Gas + liquid solvent  sat. soln + heat

– So add heat, rxn goes to left by kicking out gas
– Add gas, rxn goes to right by saturating soln & giving
off heat
Solubility of solids based on
temperature

In general, solubility
increases w/
increasing temp
– But exceptions
– No general behavior
pattern noted
Crystals

One can separate
impure dissolved salts
by reducing
temperature
– Impurity or desired
product crystallizes out
at specific temp as
solubility collapses
Colligative properties

Vapor and osmotic pressures, bp, and mp
are colligative properties
– Depend on relative # of solute and solvent
particles
Vapor Pressure

Remember:
– Equilibrium vapor pressure
 Pressure of vapor when liq and vapor in equilibrium at
specific temp
Vapor pressure of soln lower than pure solvent
vapor pressure
 Vapor pressure of solvent  relative # of
solvent molecules in soln

– i.e., solvent vapor pressure  solvent mole fraction
Raoult’s Law

Psolution = Xsolvent  P°solvent

So if 75% of molecules in soln are solvent
molecules (0.75 = Xsolvent)
– Vapor pressure of solvent (Psolvent) = 75% of
P°solvent
Problem

The vapor pressure of pure acetone
(CH3COCH3) at 30°C is 0.3270 atm.
Suppose 15.0 g of benzophenone, C13H10O
(MW = 182.217 g/mol), is dissolved in
50.0 g of acetone (MW = 58.09 g/mol).
– Calculate the vapor pressure of acetone above
the resulting solution.
Solution
mol
solute :15.0g 
 0.0823mol
182.217g
mol
solvent : 50.0g 
 0.861mol
58.09g
0.861mol
X solvent 
 0.913
0.861mol  0.0823mol

Psolution  X solvent  P solvent  0.913  0.3270atm  0.2986atm
Problem

The vapor pressure of pure liquid CS2 is
0.3914 atm at 20°C. When 40.0 g of
rhombic sulfur (a naturally occurring form
of sulfur) is dissolved in 1.00 kg of CS2,
the vapor pressure falls to 0.3868 atm.
– Determine the molecular formula of rhombic
sulfur.
Solution
Psolution  X solvent  P  solvent
0.3868atm  X solvent  0.3914atm
X solvent  0.9882
mol
solvent :1.00kg  1.00 10 g 
 13.1mol
76.143g
13.1mol
0.9882 
13.1mol  mol rhombic sulfur
3
mol rhombic sulfur  0.156
40.0g
256g

0.156mol mol
256g
mol

 7.98  8
mol 32.066g sulfur
S8
Limitations of Raoult’s Law



Doesn’t take into consideration
attractive forces in solns
For ideal soln (to right), forces
between solute/solvent
molecules = forces w/in pure
solvent
– Thus, Ptot = PA + PB
– Like graph to right
Fine for similarly constructed
molecules (hydrocarbons)
– London dispersion forces
are weakest
Solute-solvent > solv-solv

Decreases vapor
pressure
– decreased volatility
Get lower vapor
pressure than
calculated
 Ex:

– CHCl3 & C2H5OC2H5
 H on former H-bonds to
latter
– Does it increase or
decrease the latter’s
IMF?
Solute-solvent < solv-solv

Increases vapor
pressure
– increased volatility
Get higher vapor
pressure than
calculated
 Ex:

– C2H5OH and H2O
– Former disrupts Hbonding of latter
 Does it increase or
decrease the latter’s
IMF?
Nonvolatile solute added to
solvent

Salts
– Lower vapor pressure of solvent
– Make solvent less volatile
Nonvolatile solute added to
solvent


Raises bp
Lowers mp
– Why?

Adding more nonvolatile
solute or increasing
solute molality
– decreases vapor pressure
even more

Phase diagram to right
– Pure water (black)
– Adulterated water (pink)
Bp and molality relationship
Tbp = Kbp  msolute
 Kbp = molal boiling pt elevation constant
for solvent (°C/m)
 Bp elevation, Tbp, directly
proportional to solute molality

Antifreeze

Propylene glycol
– 1,2-propanediol
– Formerly used ethylene glycol
 Phased out
 Poisonous
– Lowers melting pt
– Increases boiling pt
– Reduces risk of radiator
“boiling over”
– Appreciated during the
summer months in the desert
Example
Pure toluene (C7H8)
has a normal boiling
point of 110.60°C.
 A solution of 7.80 g of
anthracene (C14H10) in
100.0 g of toluene
has a boiling point of
112.06°C.
– Calculate Kb for
toluene.

1.
2.
3.
4.
5.
Tbp = Kbp  msolute
Tbp = 112.06°C 110.60°C = 1.46°C
7.80g x (mol/178.23g)
= 4.38 x 10-2 mol
(4.38 x 10-2 mol/0.1000
kg) = 0.438 m
1.46°C/0.438 m =
3.33°C/m
Freezing point depression
Similarly, Tfp = Kfp  msolute
 Kfp = molal fp depression constant (°C/m)
 Antifreeze & CaCl2

Problem
Barium chloride has a freezing point of
962°C and a Kf of 108 °C/m.
 A solution of 12.0 g of an unknown
substance dissolved in 562 g of barium
chloride gives a freezing point of 937°C.

– Determine the molecular weight of the
unknown substance.
Solution
T  962 C  937 C  25 C



108
C

T  25 C 
m
m
m  0.23
moles solute
0.23 
0.562g solvent
moles solute  0.13
12.0g
92g

0.13mol mol

Solutions containing ions: their
colligative properties


Colligative properties based on amount of solute/solvent
Molality of ions depend on number of constituents in


1.
Different for ionic vs. covalent cmpds
Ex:
NaCl ionizes into two ions
2.
Benzene doesn’t ionize

Using equation w/out above factor will lead to values
that are off
cmpd
– So 0.5 m NaCl has 0.5 x 2 m = 1 mtot
– So 0.5 m benzene = 0.5 mtot
How to correct for it: the van’t
Hoff factor

i = the number of solute particles after dissolving

Colligative properties are larger for electrolytes
than for nonelectrolytes of the same molality
– Why? (Hint: solve the below)

Give the i-values for: methanol, CaSO4, BaCl2

Tfp (measured) = Kfp  m  i
Problem
How many grams of Al(NO3)3 must be
added to 1.00 kg of water to raise the
boiling point to 105.0°C
 Kb = 0.51 °C/m
 MW = 212.9962 g/mol

solution
T  105.0 C - 100.0 C  5.0 C

0.51
C

5.0 C 
 m 4
m
m  2.5
moles solute
2.5 
1.00kg solvent
moles solute  2.5
212.9962g
2.5mol 
 530g Al(NO 3 )3 needed
mol
Osmosis

Net movement of water (solvent) from
area of lower solute concentration to area
of higher solute concentration across a
semi-permeable membrane
– Bio101
More…
Pressure of column of soln = pressure of water moving
through membrane
 Osmotic pressure = pressure made by column of soln
= diff of heights
  = cRT
 c = mol/L = M
 R = 0.08206 L  atm/(mol  K)  ideal gas law
 T = in Kelvin
  = atm
 Useful for measuring MM of biochemical macromolecules

– Proteins and carbs