Chapter 16 Colligative properties of solutions
In the last chapter we dealt with pure solids and liquids. In this chapter we will deal with
what happens to a liquid when there is some material dissolved in it. While the
emphasis will be on aqueous solutions, the concepts and equation you learn will apply
to any solution aqueous or nonaqueous.
In the last chapter you learned about how all liquids have a vapor pressure and how this
vapor pressure varies with temperature. Most if not all of the properties we deal with in
this chapter will be based on the vapor pressure of the solvent
16-1 Molality and Mole Fraction
So, we just finished with colloids, let’s get back to what happens in an aqueous
solution
Ions interact strongly with water, so get hydrated ions, these hydrated ions are
uniformly dispersed at the molecular level, so there are no clumps or particles
like we had for colloids
Polar nonelectrolytes also interact strongly with water, so again they get
uniformly dispersed throughout the solutions
It is these strong interactions between the solute and the solvent that affect
many of the bulk properties of the solvent. Interestingly, it is not the interaction
per se that has the effect, but simply the ratio of solute to solvent molecules
The ratio of solvent to solute molecules is a measure of concentration so let’s
review our measures of concentration:
Molarity (M) = moles solute/liter solution
mass % = g solute/g solution x 100%
mole fraction = n1/(n1 + n2 +...)
Note: we had mole fraction in gases, but the same idea applies to solutions
While you are the most used to using molarity as a measure of concentration ,
there are some problems with it. 1.) You need the special volumetric glassware
to make it in the lab. And 2.) As the temperature changes, the volume of the
solution will expand and contract, so as T8 V8 and M9
For the average chemist doing simple stoichiometry these changes are so small
that they are insignificant, but they can be important in some of these colligative
properties we will talk about here.
Since colligative properties depend on the ratio of solvent to solute molecules,
mole fraction is a better way to measure this kind of concentration. The only
problem is that most of the solutions we will be dealing with are mostly water and
very little solute, so we will have water something like .999999 and solute maybe
.000001 mole fraction. And this is not really convenient
So I will introduce yet another measure of concentration
Key equation/definition:
Molality (m) = moles solute/kg solvent
Practice problems
EXAMPLE
Let’s dissolve 5 grams of acetic acid (CH3COOH) in 100 grams of water. If the
resulting solution has a total volume of 105 mls, calculate the following:
Molarity (M)
Moles/liter
Need mole of acetic acid = 5/((12.01x2)+(16.00x2)+(1.008x4) = 60.05
= .0833
Need liters 105 mL x 1L/1000mL = .105 L
M = mol/L = .0833mol/.105L =0.793M
Mass Percent
= mass solute/mass/solution x 100%
= 5/(100+5) x100% = 4.76%
Mole fraction
Mole acetic acid/total number of moles
Mol acetic acid (from above) = 5/60.05 = .0833
Mol water =100 g/ 18 = 5.56 mol
= .0833/(5.56+.0833) = .0148
Molality(m)
= mol solute / kg of solvent
mol solute (from above) = .0833
kg of solvent = 100 g x 1kg/1000g = .1 kg
= .0833mol/.1kg = .833m
Note molality and molarity not the same!
One additional detail
The colligative properties we will deal with depend on the number of solute
particles in the solution. How many moles of particles do you get from 1 mole of
CH3CH2OH? 1 mole of NaCl? 1 mole of FeCl3?
To take into account the number of particles(ions) released from an ionic
compound we will use the term colligatve molality
Key term
Colligative molality mc = im
where m is moles of the unionized electrolyte
and i is the number of particles (or ions) produced where than electrolyte
dissolves in solutions
Practice problem (Clicker question)
What is the colligative molality of a .1m solution Magnesium phosphate
16-2 Raoult’s Law
When a nonvolatile solute is dissolved in a volatile solvent, the vapr pressure of
the solvent decreases
Last chapter we saw that the vapor pressure was an equilibrium between the
rate of evaporation of the solvent and the rate of condensation of the solvent
from the vapor
How does this change when you add a solute? The rate of evaporation of the
solvent is directly proportional to the number of solvent molecules on the surface
of the liquid. When you add solute molecules you lower the concentration of
solvent molecules at the surface of the liquid, and this decreases the rate of
evaporation. By decreasing the number of molecules getting into the vapor, you
lower the vapor pressure of the solvent (Figure 16.1)
We can account for this decrease in pressure using an equation called Raoult’s
law
Key equation:
Raoult’s Law: VP solution =÷solvent VP o solvent
The amount that the vapor pressure of the solvent has decreased is called the
vapor pressure lowering = VP0pure solvent - VPsolution
=VP pure solvent - ÷ pure solvent x VP pure solvent
= VP Pure solvent (1-÷ pure solvent)
but, for a 2 component system ÷ solvent + ÷ solute = 1; so 1-÷solvent = ÷ solute
so:
Key equation:
Vapor pressure lowering = ÷solute VPsolvent
Practice problem:
The vapor pressure of water at 80oC is 355 Torr
Calculate the vapor pressure of an aqueous solution made by dissolving 50
grams of ethylene glycol (C2H6O2) in 50 grams of water. What is the vapor
pressure lowering of water in this solution?
VP solition = ÷ solvent x VP solvent
÷ solvent = nsolvent /(nsolvent + nsolute)
nsolvent = 50 g water /18g/mole = 2.78 moles
nsolute = 50 g /[(2(12)+6(1) + 2(16)] = 50/62 = .806 moles
÷solvent = 2.78/(2.78+.806) = .775
VP solution = .775(355) = 275 Torr
Pressure lowering = ÷ solute VP solvent
÷ solute = .806/(.806+2.78) = .225
VP lowering = .225 x 355 Torr = 80 Torr
Check 355 - 275 = 80 Torr Answer checks
16-3 Boiling Point Elevation
In the last chapter we found that the boiling point of a solution is the point
temperature at which the vapor pressure of the solution = the atmospheric
pressure.
So what happens to the boiling point of the solution we just calculated in the
previous problem? Since we have lowered the VP of the solution, you will have
to heat the solution to a higher temperature to get it to boil.
Key definition:
The boiling point elevation is the amount by which the boiling point of a solution
exceeds the boiling point of the pure solvent = Tb -T o b
Sketch right half of Figure 16.4 on board
In solutions where mc # 1.0 m, the mole fraction of the solute is directly
proportional to the ÷ of the solute
So if VP lowering is proportional to ÷, then it is also proportional to mc
Figure 16.6
At 1 atm the vapor pressure lowering is directly proportional to the change in the
boiling point so
Tb - Tb o is proportional to mc solute as well. So
Key equation
Boiling point elevation equation: Tb - Tb o = Kbmc solute
where Kb is called the boiling point elevation constant
Kb for various solvents are shown in table 16.2
Practice problem
Let’s calculate the boiling point for that ethylene glycol solution we worked with in
the previous example
Tb -Tb o = Kbmc solute
X-100 = .513 oC@ kg/mol x mc solute
mc solute = moles solute /kg solvent
In the previous example we had 50 grams of ethylene glycol (C2H6O2) in 50
grams of water, so kg of solvent = .05 kg water
EG had a molar mass of 62 so 50/62 = .806 moles
m = .806/.05 = 16.1 mol/kg
i=1 (nonelectrolyte), so mc = 1(16.1) = 16.1
X-100 = .513 oC@ kg/mol x 16.1 mol/kg = 8.27 oC
X-100 = 8.27
X = 108.27oC
16-4 Freezing Point Depression
So if the boiling point goes up, why does the freezing point go down?
Lets think a little more about the vapor pressure of a solid and a liquid and what
happens when something melts or freezes.
Sketch like 10.45 of Zumdahl on board
Let’s look at the VP of water and ice just above and below the MP, 0
Notice that the VP of solid is lower <0 and the VP of the liquid is lower >0.
And they equal each other at zero so our definition of normal melting point where
solid and liquid have equal vapor pressures makes sense.
Now try this thought experiment with a set up like 10.46 from Zumdahl
if Temp < this point, then solid has lower vp, and liquid will move or solid and freeze.
If Temp > this point, then liquid has lower VP and solid will move to liquid and melt
if Temp = this point, then in equilibrium and liquid and solid can coexist.
Now return to sketch like 10.45 -Z
What happens when the VP of the solution get lower?
The liquid curve drops
Where is the intersection of solid and liquid so they can be in equilibrium?
At a lower T!
If you want you can now return to Mcquarry figure 16.4 and point out region at
bottom of liquid phase you have been discussion. If you do this be sure to
emphasize that the line between solid and liquid in this figure has a visible slope
to the left, which is correct, but the line is more nearly vertical so the ÄT you get
at the bottom tip of the liquid phase is ~ same as the ÄTnormal @ 1 atm
Using the same chain of logic we used for the boiling point elevation, you can
find that the freezing point depression is also proportional to the m c of the solute
so:
Key Equation
Freezing point elevation equation:
where Kf is called the Freezing point depression constant
Notice that this equation closely resembles the boiling point elevation equation,
but the two T’s have been switched around.
Practice problem 1
Let’s calculate the boiling point for that ethylene glycol solution we worked with in
the previous examples
0-X = 1.86 oC@ kg/mol x mc solute
mc solute = moles solute /kg solvent
In the previous example we had 50 grams of ethylene glycol (C2H6O2) in 50
grams of water, so kg of solvent = .05 kg water
EG had a molar mass of 62 so 50/62 = .806 moles
m = .806/.05 = 16.1 mol/kg
i=1 (nonelectrolyte), so mc = 1(16.1) = 16.1
0-X = 1.86 oC@ kg/mol x 16.1 mol/kg = 29.9 oC
0-29.9=X
X = -29.9oC
Practice problem 2
Which solution will have the lower freezing point, 10 g of NaCl, 10 g of FeCl3
or 10 g of Ethylene glycol in 100 mLs of water?
All three solutes use the same equation and same Kf, so you need to find the
solution with the largest mc
10 g of EG is 10/62 = .16 m
NaCl has a molar mass of 58.44 so 10/58.44 = .171 m, but NaCl has two ions so
the I or van’t Hoff factor is 2 so .171 x 2 = mc =.342
FeCl3 has a molar mass of 55.84+3(35.45) = 162.19
moles of FeCl3 = 10/162.19 = .0617
but I = 4 so mc = 4(.0617) = .247
Looking at the three mc, NaCl will be have the lowest melting point
Clicker question
Which solution will have the highest boiling point 20 g of Ethylene Glycol, 20 g of
NaCl or 20 g of FeCl3 in 100 ml of water?
The book has an interesting discussion on antifreeze and using salt to melt ice, I
will leave that for you to read on your own
I will spend a little more time on the van’t Hoff factor. When I first introduced it a
few pages ago I said the I factor was the number of ions you get from a strong
electrolyte when it goes into solution. But what about a weak electrolyte, one
that doesn’t completely ionize?
Well here you can’t know what the I factor is without a few experiments, but the
experiments are useful because with can turn the colligative properties equations
around to experimentally determine how much a weak electrolyte ionizes
Practice question:
If a .0500m solution of acetic acid has a freezing point of -0.095oC,what is the
van’t Hoff factor (I) for Acetic Acid at this temperature
0-(-.095) = 1.86 mc
.095/1.86 = mc = .0512
mc = i m
.0512 = X(.0500)
X=i=.0512/.0500) = 1.024
If Acetic acid was did not ionize, i=1, if it was a strong acid and ionized
completely i=2. The % that it has ionized = (1.024-1)/1 x 100 = 2.4 %
16-5 Osmotic Pressure
Has anybody ever had a bag of salt that you use for throwing on the sidewalk
site in the garage for a few years? What happens to it? It picks up water. Why?
The bag of salt in an open garage is not a well defined closed system so let’s
simplify it
Figure 16.11
Why does the water level of the pure water go down and the salt water go up?
2 ways to explain.
Book way:As come to equilibrium rate of condensation of water from air over
both is the same. But rate of evaporation of the salt water is lower because
there are less molecules of water /unit surface area. So doesn’t lose water as
fast so net effect is to gain the water that the pure water is losing.
My way: VP of Pure solvent > VP of solution. Gas goes from high pressure to
low pressure, so moves into the salt water with the lower pressure
This worked because the vapor phase presented a barrier between the two
solutions that the salt could not pass through. Rather than using a vaporbarrier,
let’s use a physical barrier, a semi-permeable membrane
Key definitions:
A semipermeable membrane is a material that will let solvent molecules pass
between two solutions, but will not allow solute molecules to pass.
Osmosis is the name for the spontaneous passage of a solvent through a
semipermeable membrane from a dilute solution into a more concentrated solution
Let’s use a semipermeable membrane to separate these two solutions
Figure 16.12
Now the same thing happens water passes from the pure water side solution
side. But the way we have it set up here, notice that we begin to get a pressure
build up because there is a difference in the heights of the two sides.
Water will continue to move from the pure water side to the solution side until the
water pressure from the height difference equals the vapor pressure difference
between the solvent and solution sides
Key defintion:
Osmotic pressure is the pressure that is required to prevent the passage of
solvent from the dilute side of a semipermeable membrane to the concentrated
side of that membrane
van’t Hoff found that
Key Equation
Ð = RTMc
Where R is the gas constant in L@atm/K@mol or L@bar/K@mol
T is in K
and Mc is the colligative Molarity of the solution = iM
Practice Problem 1.
What is the osmotic pressure of .55M NaCl at 25oC in units of atm?
Ð=RTMc
=.08206 L@atm/K@mol x 298 x 1.1 mol/L
= 26.9atm
(i=2 for NaCl)
Practice Problem 2
Biochemists use osmosis to calculate the molecular mass of an unknown.
Say we just purified a new protein, say that makes hair grow in men. We want to
know it’s molecular weight.
First we dissolve exactly 1.00 mg in 1.00 ml of solution. At this point we don’t
know the molarity of the solution because we don’t know the MM of the protein.
We can , however, put this solution on one side of a semipermeable membrane
and measure its osmotic press and experimentally fine the molarity, and hence
the MW of the protein. Let’s try it.
We make our solution up and find it has an osmotic pressure of 1.35 torr
Our osmotic pressure is then 1.35 torr x 1 atm/760 torr = 1.78x10-3 atm
when the experiment is done at rt (298 K)
Using the equation:
ð = Mc RT
1.78x10-3 = ?x .08206 Latm/K mol x 298K
? = 1.78x10-3/.08206/298 = 72.8x10-6 mol/L
We will assume our protein doesn’t ionize so i=1 and MC =M
Our solution was 1 mg/ml = 1 g/L therefore
72.8x10-6 mol = 1g/MW
MW = 1g/72.8x10-6 mol
MW = 13,700 MW Note this is a ‘small’ protein!
Reverse Osmosis
So far we have talked about the pressure that builds up on the concentrated side
of the membrane spontaneously as the water moves from the low concentration
side to the high concentration side. But what if we put a + pressure on the high
concentration side? Now we are supplying energy to do work on the system, the
result is that we can push pure water out of the other side of the membrane.
This is how reverse osmosis water purification works.
Osmotic pressure in cells
The membrane around a cell is a semipermeable membrane. The inside of a
cell has high concentrations of lots of proteins, ions and small molecules that
cannot pass through this membrane, so there is always an osmotic pressure
inside a cell.
When a biologist suspends cells in a solution he or she has to be careful of the
salt concentration of that solution (Figure 16.18) to try to match the osmotic
pressure of the cells. If the salt is too dilute, you make a hypotonic solution, and
the cells begin to swell. If you add to much salt, you make a hypertonic solution
and the cells begin to collapse. An isotonic solution is one in which the osmotic
pressure of the medium matches the osmotic pressure of the cell so the cell is
stable.
If the salt concentration is too low you can actually make cells explode. Most
animal cells will burst if Ð > 7.5 bar. Plant cells, with a rigid cell wall can sustain
higher pressures
16-6 Ideal Solutions
Solutions where both solute and solvent have a vapor pressure
So far we have dealt with solutions where the solvent was volatile and the solute
was nonvolatile. Let’s up the ante. What happens when both the solvent and
the solute have a vapor pressure (are volatile). For this discussion we will only
deal with Raoult’s law, and will not look at BP or FP
In applying Raoult’s law before I assumed that the solute was nonvolatile, so
only so solute contributed to the VP over the solution. This might apply to
something like NaCl, but definitely doesn’t apply to something like alcohol which
also has a vapor pressure. When the solute is volatile we use a modified form of
Raoult’s Law:
Key Equation:
Raoult’s Law for an ideal solution when both solute and solvent are volatile
Ptotal = PA + PB = ÷APoA + ÷BPoB
And a plot of P vs mole fraction looks like figure 16.20
Ideal and Non-ideal solutions
Just as we had ideal and non-ideal gases, we can have ideal and non-ideal
solutions. A non-ideal solution is a solution that does not obey Raoult’s law, an
ideal solution does. You usually observe ideal solutions when the two liquids in
the solution are similar in structure, like Benzene and Toluene.
Solutions can act in a non-ideal manner for many reasons, let’s look at two
1. Negative deviations from Raoult’s law
If the solute and the solvent have strong interactions, like they form
hydrogen bonds, then the solvent will hold on to the solute more tightly, and
the solute will have a lower vapor pressure than expected, so the net
pressure will be lower than expected and have a negative deviation from
Raoult’s Law (Make sketch on board + figure 16.19a )
What other property can we associate with a strong solute-solvent interaction?
2. Positive deviations from Raoult’s law (Sketch on board +16.19b)
On the other hand if you mix two liquids and they give you a higher vapor
pressure than expected, this indicates that the solute-solvent interactions are
weaker than the solute-solute and solvent-solvent interaction. This means that
the solute molecules are bound less well to the solution than they are to the pure
state so they clump togethr and try to escape, and the VP of the solute and the
total VP will be higher. This is a Positive deviation form Raoult’s Law
An example is a mixture of Ethanol and Hexane. Polar ethanol is fine by itself,
non-polar hexane fine by itself. Polar and non-polar won’t interact so don’t pick
up any new interactions when they mix, and, in fact, you lose some of the Hbond interactions in the pure ethanol as it makes room for the hexane.
Practice calculations
1. Calculate the vapor pressure of a solution that has an equal number of moles
of benzene and toluene, given that the VPo of Benzene is 183 Torr and that of
Toluene is 59.2
If moles of Benz = moles of Tol, then
÷Benz = ÷Tol = .5
VP Solution = VP Bez + VP Tol
VP Benz = .5(183) = 91.5 Torr
VP Tol = .5(59.2) = 29.6 Torr
VP Solution = 91.5 + 29.6 = 121.1 Torr
2. What is the mole fraction of Benzene and Toluene in this vapor?
Going back to the chapter on gases
÷ Benz = VPbenz/(VPBenz + VPtol)
= 91.5/121.1 = .756
÷ Tol = VP Tol/(VPBenz + Vptol)
= 29.6/121.1 = .244
Distillation
Notice in the above example. The ÷ of benzene n the liquid was .5, but it the
vapor it is now .756. You have significantly increased the purity of the benzene
This is how a still works
You start with a wine solution that is say 6% alcohol. The VP of alcohol is almost
twice that of water so, when you boil the solution the vapor is enriched in alcohol.
When you condense the vapor with a still, the resulting solution has a higher
alcohol content (and has become Brandy or Cognac)
You can make the still more efficient by packing with with inert materials that will
add to the number of times the vapor is condensed and re-evaporated before the
final vapor is taken from the still head. This is a lab you will probably to with Dr.
Dixson if you take organic chem next year.
Even with the best still in the worlds, the best purity you will ver get for your
ethanol water mixture is 95.6% alcohol
This is because the water alcohol mixture is not an ideal solution, and the
solvent-solute interactions make the ÷ of vapor and liquid equal to each other
when you reach 95.6% alcohol.
Key definition:
Azeotrope: A solution that distills without a change in compostion
If you want to find out how to get your hooch up to 200 proof alcohol you will
have to read your book!
16-7 Henry’s Law
Now let’s look at the solubility of gases in liquids
Let’s do a thought experiment
draw on board similar to fig 11.5-Zumdahl
At start have a closed cylinder with dissolved gas in equilibrium with gas phase.
What happens if press down on cylinder?
Smaller volume over liquid, higher pressure, some of the gas molecules are
forces back into the liquid, so solubility must have increased!
Thus we find that the concentration of a gas in a liquid is directly proportional to
the partial pressure of the gas over the liquid. This is summarized in Henry’s
Law
Key Equation:
Henry’s Law Pgas = kh Mgas
Where Pgas is the partial pressure of the gas
Mgas is the molar concentration of the gas in the solution
and kh is the Henry’s Law constant for that gas
The Henry’s law constant is different for each gas, each solvent and every
temperature
Practice Problems:
Part 1. A soft drink is canned so that the bottle at 25o C contain CO2 gas at a
pressure of 5 atm over the liquid. When you open the can the pressure naturally
drops to 1 atm, but further, because the air is not made of CO2the partial
pressure of CO2 is only 4.0 x 104 atm (.0004 atm). Calculate the concentration of
CO2 in the liquid before and after the bottle is opened.
Needed fact, the k for Henry’s law for CO2 is 29 L atm/mol at 25oC (table 16.3)
Henry’s law
Pgas = kh Mgas
In sealed can, P=5, k=29L atm/mol so C =
5=29(X)
X=5 atm /29Latm/mol
= .172 mol/L = .172M
In the open can we have
P = .0004= 29(X)
X=.0004/29Latm/mol
= 1.38x10-5 M
Part 2. How big is your belch from a can of soda?
a can of soda is 355 mL = .355 L; so in full can you have
.156 mol/L x .355 L = .055 moles CO2
After you open the cab you have:
.0000138 x .335 = 4.6x10-6 moles of CO2 Remaining
That means you have released .055 - .0000046 ~ .055 moles of CO2
Using Good old PV=nRT; V = nRT/P
=.055 (.08206)298/1 = a 1.3 L belch, or serval smaller belches
Agin the chapter ends with some interesting discussion on the Bends and how
He/ O2 gas mixtures are used in deep diving to avoid the bends.