Chapter 14: Solutions and Their Behaviour
A solution is a homogeneous mixture of two or more substances
in a single phase. It consists of a _____________, the substance
in which things are dissolved, and one or more _____________,
the substance(s) dissolved.
While we tend to think of solvents as liquids, solutions can exist
for virtually any combination of states of matter:
Solvent
Solute
Type of Solution
gas
gas
homogeneous gas mixtures (e.g. air)
liquid
gas
dissolved gases
liquid
liquid
homogeneous liquid mixtures
liquid
solid
homogeneous solutions
solid
liquid
amalgams
solid
solid
solid solutions (e.g. alloys, glasses)
The amount of a given solute that can be dissolved in a given
solvent is its __________________. This property is usually
reported in grams of solute per 100 grams of solvent at a
pressure of 1 atm and specified temperature.
e.g. I2 has a solubility in water of 0.03 g/100g at 20˚C.
HIO3 has a solubility in water of 308 g/100g at 25˚C.
Why is there such a difference between these two values?
Reporting Concentration of Solutions
There are many different units commonly used to describe the
concentration of a solution (i.e. the amount of solute dissolved
in a given amount of solvent).
1.
You should be familiar with molarity from Chemistry
1000. The molarity (M) of a solution is the number of
moles of solute dissolved per liter of solution.
2.
The molality (m) of a solution is the number of moles of
solute dissolved per kilogram of solvent.
Note the differences between molarity and molality!
3.
The mole fraction (X) of a solute is the number of moles of
solute divided by the total number of moles of all solution
components. We used this unit of concentration when
discussing gas mixtures in Chemistry 1000.
4.
Weight percent (wt%) is similar to mole fraction but using
mass and converting to a percentage. It is the mass of
solute divided by the mass of solution multiplied by 100%.
5.
Parts per million (ppm) is a unit used to describe dilute
solutions. It is the mass of solute divided by the mass of
solution multiplied by 1,000,000.
Note that 1 ppm = 1 mg/kg = 1 µg / g
Similarly, use parts per billion (ppb; 1 ppb = 1 ng/g) or
even parts per trillion (ppt; 1 ppt = 1 pg/g) for extremely
dilute solutions.
e.g. 10.00 mL of ethanol (C2H6O, d = 0.789 g/mL) was
dissolved in 90.0 mL of water (d = 1.00 g/mL). After
thorough mixing, the volume of the resulting solution was
measured to be 99.0 mL.
(a) Calculate the weight percent of ethanol in this solution.
(b) Calculate the mole fraction of ethanol in this solution.
(c) Calculate the molarity of ethanol in this solution.
(d) Calculate the molality of ethanol in this solution.
(e) If 24 µg of a nickel salt were added to this solution as a
colouring agent, calculate the concentration of the nickel
salt in ppm.
Colligative Properties
Colligative properties are changes in the behaviour of liquids
induced by the presence of solutes. In the ideal case, they
depend only on the number of solute particles in solution – not
on the nature of the solute.
Which units of concentration would be useful for discussing
colligative properties?
The ideal case (colligative properties independent of solute
identity) applies fairly well for low-concentration solutions. It is
much less applicable to high-concentration solutions. Why do
you think this might be?
We will consider four closely related colligative properties:
1. vapour pressure depression,
2. boiling point elevation,
3. freezing point depression, and
4. osmotic pressure.
Vapour Pressure Depression and Boiling Point Elevation
Figure 14.14 in Kotz (at right)
shows vapour pressure curves
for pure benzene (in red) and a
2 mol/kg solution (in blue) of a
nonvolatile solute in benzene.
Note that adding the solute has
lowered the vapour pressure at
any given temperature.
By
definition, lowering the vapour
pressure also raises the boiling
point. (see chapter 13 notes)
Why does this occur?
Consider the difference
between a pure liquid and a
solution with a nonvolatile
solute. The example shown
at the right is pure water
(right) vs. salt water (left).
The surface of the pure water consists entirely of water
molecules, a fraction of which have enough energy to evaporate.
The surface of the salt water consists of some water molecules
and some salt molecules. The same fraction of water molecules
have enough energy to evaporate but, since fewer water
molecules are on the surface, fewer evaporate. Essentially, the
salt ions are “blocking the surface” of the solution.
Thus, the salt water has a lower vapour pressure than the pure
water. Because of this, the salt water will have to be heated to a
higher temperature to raise its vapour pressure to make it equal
to atmospheric pressure. Thus, the boiling point of salt water is
higher than that of pure water.
Raoult’s Law defines the relationship between the vapour
pressure of a pure liquid and the vapour pressure of a solution
with a nonvolatile solute:
Psolvent = Xsolvent P˚solvent
where P˚solvent is the vapour pressure of the solvent as a pure
liquid, Psolvent is the vapour pressure of the solvent in solution,
and Xsolvent is the mole fraction of solvent molecules. Note that
both vapour pressures must be in the same units and at the same
temperature!
Vapour pressure depression (∆Psolvent) is the difference
between P˚solvent and Psolvent:
Since we know that Xsolvent + Xsolute = 1, we can calculate vapour
pressure depression directly from Xsolute:
An ideal solution is defined to be one which obeys Raoult’s
Law. While no solution obeys the law perfectly, it is a good
approximation for most low-concentration solutions.
While vapour pressure depression is calculated using mole
fraction (X), boiling point elevation is calculated using the
molality of the solute (msolute):
∆Tbp = Kbp msolute i
where ∆Tbp is the boiling point elevation, msolute is the molality
of the solute, i is the van’t Hoff factor (number of particles
generated per molecule of solute; i = 1 for nonelectrolytes), and
Kbp is a constant (in ˚C kg/mol) specific to the solvent.
e.g. At 100.00 ˚C, the vapour pressure of water is 760 mmHg.
A cup of sugar (230 g, C12H22O11) is added to 1.000 L of
water. dH2O = 1.000 g/mL; Kbp(H2O) = +0.5121 ˚C kg/mol
(a) What is the normal boiling point of pure water?
(b) Calculate the normal boiling point of the sugar-water
solution.
(c) Calculate the vapour pressure of the sugar-water solution at
100 ˚C.
(d) If the solution had been prepared with table salt (NaCl)
instead of sugar, would the answers to parts (b) and (c) be
the same? Why or why not?
If you already know Kbp of the solvent and i for the solute,
measuring the boiling point elevation of a solution is one way to
determine the molar mass of your solute.
e.g. You determine that an unknown solid is not ionic and not
volatile. You then dissolve 0.300 g of it in 30.0 g of
chloroform (Kbp = 3.63 ˚C kg/mol) and determine that the
boiling point has increased by 0.392 ˚C. What is the molar
mass of the unknown solid?
You will see in lab that measuring freezing point depression can
also be used for this purpose.
Freezing Point Depression
Freezing point depression is calculated similarly to boiling point
elevation:
∆Tfp = Kfp msolute i
where ∆Tfp is the freezing point depression, msolute is the
molality of the solute, i is the van’t Hoff factor, and Kfp is a
constant (in ˚C kg/mol) specific to the solvent.
e.g. Icy roads are often salted to melt the ice. The normal
freezing point of water is 0.00 ˚C. If 20.0 g NaCl is added
to 1.000 L of water, what is the freezing point of the saltwater? dH2O = 1.000 g/mL; Kfp(H2O) = -1.86 ˚C kg/mol
Why does adding a solute lower a liquid’s freezing point?
Recall that a substance’s freezing point is the same as its melting
point. When a substance melts, it is because the molecules have
enough energy to overcome the intermolecular forces holding
them in place. These forces will be stronger in a uniform crystal
than in a disordered mixture. Thus, it takes more energy to melt
a pure solid than a mixture – and the melting point/freezing
point of a pure solid is higher than that of a mixture.
Osmotic Pressure
If a semipermeable membrane is used to separate two
solutions, it acts as a barrier to most/all solute molecules while
allowing solvent molecules to pass. This may be due to
discrimination by size, charge or other properties.
The movement of solvent molecules (usually water) through a
semipermeable membrane is called osmosis.
In osmosis, the average movement of solvent molecules is
driven toward the more concentrated solution. This continues
until the two solutions have the same concentration or until it
stops due to buildup of osmotic pressure.
Consider a semipermeable bag of sugar solution put in a beaker
of pure water. On average, water molecules will pass through
the membrane into the sugar solution, diluting it. This process,
results in a net increase in the number of molecules in the bag.
In a closed system, the bag would eventually burst due to the
increased pressure from the increased number of molecules.
In an open system, the sugar solution
would be pushed upward as more water
entered the bag – until the pull of
gravity on the elevated solution is equal
to the pull of water into the sugar
solution. Equilibrium is reached, and
we can determine the osmotic pressure
from the height of the sugar solution.
More generally, we can calculate osmotic pressure using a
formula similar to the ideal gas law:
Π=MRT
where Π is osmotic pressure (in atm), M is the molarity of the
solution, R is the ideal gas constant (in L atm mol-1 K-1) and T is
temperature (in K).
e.g. An aqueous solution contains 30.0 g of a protein in 1.00 L.
The osmotic pressure of the solution is 0.167 atm at 25 ˚C.
What is the approximate molecular mass of the protein?
When comparing two solutions, we say that they are isotonic if
they have the same osmotic pressure. If not, the one with the
higher osmotic pressure is hypertonic and the one with the
lower osmotic pressure is hypotonic.
e.g. A cell membrane is an excellent example of a
semipermeable membrane. As such, cells must be stored in
an isotonic solution. What would happen if they were
stored in a hypotonic solution? …a hypertonic solution?
If some pressure can be used to stop osmosis, more pressure can
be used to drive it in the opposite direction. Such reverse
osmosis is used to purify drinking water. Salt water is subjected
to pressure (~50 atm), forcing water molecules to pass through
a semipermeable membrane into an area of reduced pressure
(~1 atm). The salt ions cannot pass through the membrane so
fresh water is obtained. This is literally “osmosis backwards”.
Colloids
Colloids are heterogeneous mixtures which can’t settle out into
two phases. This makes colloids different from ‘normal’
heterogeneous mixtures (e.g. sand/water) and is due to the small
size of the suspended molecules. Small molecules have high
surface-area-to-volume ratios. This means that there is more
opportunity for intermolecular interactions between small
suspended molecules and solvent molecules. These interactions
maintain a “permanent suspension” of one substance (the
dispersed phase) in another (the dispersing medium).
Type of
Colloid
aerosol
aerosol
foam
foam
emulsion
gel
sol
solid sol
Dispersing
Medium
gas
gas
liquid
solid
liquid
solid
liquid
solid
Dispersed
Phase
liquid
solid
gas
gas
liquid
liquid
solid
solid
Example
clouds
car exhaust
whipped cream
Styrofoam
milk
Jello
mud
alloys
Note that all of these examples are opaque substances. Because
the molecules of the dispersed phase are generally large, they
scatter visible light. This is the Tyndall effect.
Colloids can also be classified according to the polarity of the
dispersed phase and dispersing medium. A hydrophilic colloid
contains a polar substance dispersed in water. A hydrophobic
colloid contains a nonpolar substance dispersed in water.
Which would you expect to form more easily and why?
An emulsifying agent (aka surfactant) is often used to help in
the formation of a ________________________ colloid.
Surfactants
Surfactants are substances that affect the properties of surfaces –
including the barrier between two phases. A familiar example
of a surfactant is soap:
H3C
H2
C
C
H2
H2
C
C
H2
H2
C
C
H2
H2
C
C
H2
H2
C
C
H2
H2
C
C
H2
H2
C
O
C
H2
C
O Na
+
Soap molecules contain a polar “head” (CO2Na) and a nonpolar
“tail” (CH3(CH2)n where n is ~12-20). The hydrophobic “tail” is
attracted to other hydrophobic molecules (like grease/dirt) while
the hydrophilic “head” is attracted to water. Thus, the soap
molecule ‘pulls’ grease into the water, cleaning dirty dishes.
The difference between “soap” and “detergent” is the nature of
the hydrophilic end of the surfactant molecule. In soap, this can
be a good ligand for metal ions like calcium and magnesium. In
detergent, different polar groups are used that are poor ligands.
Which would be more effective to use in hard water, soap or
detergent?
Important Concepts from Chapter 14
• units for concentration
o molarity
o molality
o mole fraction
o weight percent (102), ppm (106), ppb (109) and ppt (1012)
• colligative properties
o vapour pressure depression (Raoult’s Law)
o boiling point elevation (∆Tbp = Kbp m i)
o freezing point depression (∆Tfp = Kfp m i)
o osmotic pressure (Π = M R T)
o when is a solution “ideal”
o van’t Hoff factor
• hypertonic, hypotonic and isotonic solutions
• volatility
• colloids
• surfactants