CHM116
Dr. Mencer

If the solute is a solid or liquid, it must first be dispersed — that is, its molecular units must be pulled apart.
This requires energy, and so this step always works against solution formation.
The solute must then be introduced into the solvent. Whether this is energetically favorable or unfavorable depends on the nature of the solute and solvent.
Exothermic if A-B attractions stronger than A-A + B-BEndothermic if attractions between like molecules are stronger than those between unlike molecules.

Henry’s Law works for ideal
Cases: note that in the real
World, there can be deviations
From ideal behavior.

Suspended SiO
2
(sand) settles very quickly.
Each colloidal particle of
SiO
2
(Ludox ® ) attains a (–) charge, which repels other colloidal particles.

Why are there no gasin-gas colloids?

• Net charge on colloidal particles
• Balanced by other particles (ions) . . . But
• Colloidal particles repel one another
– Stay in solution
– Do not sediment

When a strong electrolyte is added to colloidal iron oxide, the charge on the surface of each particle is partially neutralized …
… and the colloidal particles coalesce into a suspension that quickly settles.

Light of visible wavelengths scatter from colloidal particles due to similar size scale.

Most concentration units are expressed as:
Amount of solute
Amount of solvent or solution
• Molarity: moles of solute/ liter of solution
• Percent by mass: grams of solute/ grams of solution (then multiplied by 100%)
• Percent by volume: milliliters of solute/ milliliters of solution (then multiplied by 100%)
• Mass/volume percent: grams of solute/ milliliters of solution (then multiplied by 100%)

How would you prepare 750 g of an aqueous solution that is 2.5% NaOH by mass?
Answer: 19 g of NaOH and 731 g of water.
At 20 °C, pure ethanol has a density of 0.789 g/mL and USP ethanol has a density of 0.813 g/mL. What is the mass percent ethanol in USP ethanol (which is
95% v/v)?
Answer: 92.3% by mass EtOH in USP ethanol.

Most concentration units are expressed as:
Amount of solute
Amount of solvent or solution
• Parts per million (ppm): grams of solute/ grams of solution (then multiplied by 10 6 or 1 million)
• Parts per billion (ppb): grams of solute/ grams of solution (then multiplied by 10 9 or 1 billion)
• Parts per trillion (ppt): grams of solute/ grams of solution (then multiplied by 10 12 or 1 trillion)
• ppm, ppb, ppt ordinarily are used when expressing extremely low concentrations (a liter of H
2
O that is 1 ppm fluoride contains only 1 mg F – !)

The maximum allowable level of nitrates in drinking water in the United States is 45 mg NO
3
–
/L. What is this level expressed in parts per million
(ppm)?
Answer:
For dilute aqueous solutions (i.e. solutions that are nearly pure water) the density is so close to
1.00 gcm -3 that we can say that – ppm = mg per L so 45 mg NO
3
– /L = 45 ppm NO
3
–

Most concentration units are expressed as:
Amount of solute
Amount of solvent or solution
Molality (m): moles of solute/ kilograms of solvent.
• Molarity varies with temperature (expansion or contraction of solution).
• Molality is based on mass of solvent and is independent of temperature.
• We will use molality in describing certain properties of solutions.

What is the molality of a solution prepared by dissolving 5.05 g naphthalene [C
10
H
8
(s)] in 75.0 mL of benzene, C
6
H
6
(d = 0.879 g/mL)?
Answer: 0.598 m C
6
H
6
How many grams of benzoic acid, C
6
H
5
COOH, must be dissolved in 50.0 mL of benzene, C
= 0.879 g/mL), to produce 0.150 m C
6
H
5
6
H
6
COOH?
(d
Answer: 0.805 g C
6
H
5
COOH

Most concentration units are expressed as:
Amount of solute
Amount of solvent or solution
• Mole fraction (x i
): moles of component i per moles of all components (the solution).
• The sum of the mole fractions of all components of a solution is ____.
• Mole percent: mole fraction times 100%.

Example 6
An aqueous solution of ethylene glycol HOCH
2
CH used as an automobile engine coolant is 40.0%
2
OH
HOCH
2
CH
2
OH by mass and has a density of 1.05 g/mL. What are the (a) molarity, (b) molality, and (c) mole fraction of HOCH
2
CH
2
OH in this solution?
Example 7 An Estimation Example
Without doing detailed calculations, determine which aqueous solution has the greatest mole fraction of
CH
3
OH: (a) 1.0 m CH
3
OH, (b)10.0% CH
3
OH by mass, or
(c) x
CH3OH
= 0.10.

How solutes affect the properties of solutions.

Solution: Solute dispersed in a solvent.

• An ideal solution exists when all intermolecular forces are of comparable strength,
D
H soln
= 0 and
D
V soln
= 0.
• When solute–solvent intermolecular forces are somewhat
stronger than other intermolecular forces,
D
H soln
< 0 and
D
V soln
< 0 .
• When solute–solvent intermolecular forces are somewhat
weaker than other intermolecular forces,
D
H soln
> 0 and
D
V soln
> 0.
• When solute–solvent intermolecular forces are much
weaker than other intermolecular forces, the solute does not dissolve in the solvent (no mixing = immiscible).
– Energy released by solute–solvent interactions is
insufficient to separate solute particles or solvent particles.

• Mixing is physical process; chemical properties don ’ t change
• Properties of solutions are similar to those of the pure solvent
• Addition of a foreign substance to water alters the properties slightly

• Reduces the vapor pressure of the solvent in the solution.
• Lowers the freezing point of the solution.
• Raises the boiling point of the solution.
• Generates Osmotic Pressure

• Colligative comes from colligate – to tie together
• Colligative properties depend on amount of solute but do not depend on its chemical identity (as a first order approximation)
• Solute particles exert their effect merely by being rather than doing

• Vapour pressure is always lower
• Boiling point is always higher
• Freezing point is always lower
• Osmotic pressure drives solvent from lower concentration to higher concentration

Why does vapor pressure of the solvent decrease?
The surface area of the water exposed to the vapor phase is reduced
By the presence of the solute particles at that boundary.

• Vapor pressure of solvent in solution containing solute is always lower than vapor pressure of pure solvent at same T
– At equilibrium rate of vaporization = rate of condensation
– Solute particles occupy volume reducing rate of evaporation the number of solvent molecules at the surface
– The rate of evaporation decreases and so the vapor pressure above the solution must decrease to recover the equilibrium

• Vapor pressure above solution is vapor pressure of solvent times mole fraction of solvent in solution
P solvent
=
P
0 solvent
X solv
• Vapour pressure lowering follows:
D
P solvent
=
P o solvent
X solute
We will return to Raoult’s law later for mixtures that contains two volatile components.

• Like ideal gas law, Raoult ’ s Law works for an
ideal solution
• Real solutions deviate from the ideal
– Concentration gets larger
– Solute – solvent interactions are unequal
• Solvent – solvent interactions are stronger than the solute – solvent: P vap is higher
• Solvent – solute interactions are stronger than solvent – solvent interactions: P vap is lower

• Blue curves are phase boundaries for pure solvent
• Red curves are phase boundaries for solvent in solution
• Freezing point depression
– Pure solid separates out at freezing – negative ΔT f
• Boiling point elevation
– Vapour pressure in solution is lower, so higher temperature is required to reach atmospheric – positive ΔT b

• The influence of the solute depends only on the number of particles
• Molecular and ionic compounds will produce
different numbers of particles per mole of substance (ideal behavior shown here)
– 1 mole of a molecular solid → 1 mole of particles
– 1 mole of NaCl → 2 moles of particles
– 1 mole of CaCl
2
→ 3 moles of particles

• Not all ionic substances dissociate completely
• Others form ion pairs
• Still others have non-ideal solvent interactions
• Van ’ t Hoff factor accounts for this
Van ‘ t Hoff factor:
i = moles of particles in sol’n/moles of solute dissolved
(can be non-ideal)

At higher concentrations, the values of i are significantly lower than the theoretical values; ion
pairs form in solution.
At very low concentrations, the
“ theoretical ” values of i are reached.

• Analagous to boiling point, the freezing point depression is proportional to the molal concentration of solute particles
D
T f

K f m
• To account for dissociation, the van ’ t Hoff factor is applied to modify m:
D
T f

K f m
 i

• Depends on the solute only being in the liquid phase
– Fewer water molecules at surface: rate of freezing drops
– Ice turns into liquid
– Lower temperature to regain balance
– Depression of freezing point

• Depends on the number of particles present
• Concentration is measured in molality
(independent of T)
• K b is the molal boiling point elevation constant
• Note: it is the number of particles (use i as needed)
D
T b
=
K b mi

D
T f
= –K f
× m
D
T b
= K b
× m

• A semi-permeable membrane discriminates on the basis of molecular type
– Solvent molecules pass through
– Large molecules or ions are blocked
• Solvent molecules will pass from a place of lower solute concentration to higher concentration to achieve equilibrium

• Solvent passes into more conc solution increasing its volume
• The passage of the solvent can be prevented by application of a pressure
• The pressure to prevent transport is the osmotic pressure

Ordinarily a patient must be given intravenous fluids that are isotonic —have the same osmotic pressure as blood.
External solution is
hypertonic; produces osmotic pressure > π int
.
Net flow of water out
of the cell.
Red blood cell in isotonic solution remains the same size.
External solution is
hypotonic; produces osmotic pressure <
π int
. Net flow of water
into the cell.

• The ideal gas law states PV
 nRT
• But n/V = M and so P = iMRT
• i is this the Van’t Hoff factor
• Where M is the molar concentration of particles and
 is the osmotic pressure
• Note: molarity is used not molality

• Molar mass can be computed from any of the colligative properties
• Osmotic pressure provides the most accurate determination because of the magnitude of Π
– 0.0200 M solution of glucose exerts an osmotic pressure of 374.2 mm Hg but a freezing point depression of only 0.02ºC

• A solution contains 20.0 mg insulin in 5.00 mL develops an osmotic pressure of 12.5 mm Hg at 300 K
M
= P iRT
M
=
12.5
mmHg
1
0.0821
760 mmHg
L
· atm
300 K mol
·
K
=
6.68
´
10
-
4
M
Assuming the species does not dissociate (i = 1)

• Moles insulin = MxV = 3.34x10
-3 mmol
• Molar mass = mass of insulin/moles of insulin
= 20.0 mg/3.34x10
-3 mmol
= 5,990 g/mol

• Volatile solute in volatile solvent
• Total pressure is the sum of the pressures of the two components
P total

P
A

P
B
P total

P
A

X
A

P
B

X
B

• Total pressure in a mixture of toluene (b.p. =
110.6ºC) and benzene (b.p. = 80.1ºC) equals sum of vapor pressures of components
P total
 
P ben
X ben
 
P tol
X tol

• Real solutions can deviate from the ideal:
– Positive (P vap
> ideal) solute-solvent interactions weaker
– Negative (P vap
< ideal) solute-solvent interactions stronger

• The vapor above a liquid is richer in the more volatile component
• Boiling the mixture will give a distillate more concentrated in the volatile component
• The residue will be richer in the less volatile component

The vapor here …
… is richer in the more volatile component than the original liquid here …
… so the liquid that condenses here will also be richer in the more volatile component.

• In practice, it is not necessary to do the distillation in individual steps
• The vapor rising up the column condenses and re-evaporates continuously, progressively becoming enriched in the volatile component higher up the tube
• If the column is high enough, pure liquid will be collected in the receiver