Chapter 13: Physical Properties of Solutions

Key topics:
Molecular Picture (interactions, enthalpy, entropy)
Concentration Units
Colligative Properties

terminology:

• Solution: a homogeneous mixture

• Solute: the component that is dissolved in solvent
-- usually in smaller amount
• Solvent: medium (often water) into which solutes are mixed
-- usually in greater amount
• concentration: describes how much solute there is
-- dilute: small amount of solute
-- concentrated: large amount of solute
-- saturated: contains as much solute as can be dissolved
-- unsaturated: can dissolve more solute
-- supersaturated: more than saturated, unstable
-- solubility: amount of solute dissolved in a given volume
of a saturated solution

a supersaturated solution can be prepared by forming a

saturated solution at a high temperature, and then cooling it
(many solutes are more soluble at higher temperature)
 Some of the solute will fall out of solution (as a solid) if the
solution is disturbed (vibration, dust, etc)

Addition of a crystal (nucleation site) causes precipitation from a

supersaturated solution.

Molecular view of a soluble

ionic compound in water.
Molecular view of the solution process:

Ultimately, we can say that any process is governed by

ΔG = ΔH - TΔS (Equation 14.10, ΔG < 0 for spontaneity)

• ΔH < 0 helps (exothermic, H = enthalpy)

• ΔS > 0 helps (increased disorder, S = entropy)
• ΔH > 0 (endothermic) is possible if ΔS compensates
First examine ΔH:
Since ΔHsoln is a state function, we can think of it as:

ΔH1 > 0; ΔH2 > 0; ΔH3 < 0

where, in terms of intermolecular forces, we have

• ΔH1 = disruption of solute-solute interactions

• ΔH2 = disruption of solvent-solvent interactions
• ΔH3 = formation of solute-solvent interactions
• ΔHsoln = ΔH1 + ΔH2 + ΔH3 = heat of solution

For a gas dissolving in a liquid, ΔHsoln is usually exothermic

because ΔH1 = 0 (gas molecules don’t interact)
The saying “like dissolves like” is useful in predicting solubility.
This expression is about the intermolecular forces present in
the solute vs. solvent.

• CCl4 dissolves in benzene (C6H6) : both nonpolar

• water (H2O) and methanol (CH3OH) dissolve in each
other : both polar

ionic compounds dissolve in polar solvents due to solvation

ionic compounds are not soluble in nonpolar solvents

example: vitamin C and vitamin D: water or fat soluble?

Two liquids that are soluble in each other in all proportions are
called miscible (opposite is immiscible)

Next examine ΔS:

ΔS > 0 helps (increased disorder, S = entropy) in solution

formation, and there is a natural tendency for entropy to
increase in any process. A solution is usually more disordered
than the separate solute and solvent molecules.
 Schematic depiction of the mixing of solvent and solute.

Concentration Units:

moles of solute
molarity = M =
liters of solution

moles of solute
molality = m =
mass of solvent (in kg)

For very dilute aqueous solutions, molarity and molality are

equal since 1 L of H2O weighs 1 kg.

moles of A
mole fraction of A = A =
sum of moles of all components
 mass of solute
percent by mass = ⇥ 100%
mass of solute + solvent
mass of solute
= ⇥ 100% = parts per hundred
mass of solution
If we multiply by 106 instead of 100 we get parts per million
(ppm); if we multiply by 109 we get parts per billion (ppb).

e.g., the drinking water standard for arsenic is 10 ppb

(EPA’s Maximum Contaminant Level)

Choice of units depends on the purpose of the experiment:

w Mole fraction: used for gases and vapor pressures of
solutions
w Molarity: commonly used since volumes of solutions are
easy to measure
w Molality: temperature independent
w Percent by Mass: temperature independent and we don’t
need to know the molar mass

Unit Conversions:

e.g., Determine the molality of an aqueous solution that is

4.5% urea by mass.

Solution:
The molar mass of urea is 60.06 g/mol.
There are different ways to approach this: we will assume we
have 1 kg of water.
Then, with the mass of urea as x,
x
4.5 = ⇥ 100 ) 100x = 4.5(x + 1 kg); ) 95.5x = 4.5 kg
x + 1 kg
Which gives x = 47.12 g. The molality is then
(47.12 g)/(60.06 g/mol)
= 0.78 m
1 kg

e.g., At 20.0°C, a 0.258 m aqueous solution of glucose

(C6H12O6) has a density of 1.0173 g/mL. Calculate the molarity
of this solution.

Solution:

In 1 kg of water, we have 0.258 mol glucose. Also, 0.258 mol

glucose weights (0.258 mol)(180.156 g/mol) = 46.48 g.
Therefore we have (1 kg) + (46.48 g) = 1046.48 g solution.
This occupies (1046.48 g) / (1.0173 g/mL) = 1028.68 mL.
Thus the molarity is (0.258 mol) / (1.0287 L) = 0.251 M.

Factors that affect solubility:

We can sometimes rationalize solubility using Le Chatelier's

principle.
e.g., dissolving oxygen in water releases heat:
O2(g) O2(dissolved) + heat
Therefore raising the temperature (adding heat) shifts the
equilibrium to the left, which means that less oxygen dissolves
at higher temperature.

Solubility of some compounds vs temperature.

For an endothermic process,

heat + solute(undissolved) solute(dissolved)
raising the temperature shifts the equilibrium to the right, which
means that more solute dissolves.

But in general other factors (e.g. entropy) make it hard to

predict the effect of temperature on solubility.
For gases, pressure also affects the solubility.

P1 < P2

gas + solution(little gas) solution(more gas)

Increasing the pressure will result in a shift of equilibrium to
the right to reduce the pressure (remove gas).

e.g., open a can of soda (pressure drops, shift to left)

Henry’s Law makes this quantitative: c = kP

c molar concentration (mol/L) of dissolved gas
k Henry’s Law constant
P gas pressure (atm)

e.g., Calculate the molar concentration of O2 in water at 25ºC

under atmospheric conditions (air contains 22% oxygen). k =
1.3 x 10-3 mol/L • atm.
Solution:
c = (1.3 x 10-3 mol/L • atm)(1 atm x 22/100) = 2.9 x 10-4 mol/L

e.g., Calculate the pressure of O2 necessary to generate an

aqueous solution that is 3.4 x 10-2 M in O2 at 25ºC.

Solution:

From the previous problem, we know that a pressure of 0.22

atm O2 gives a solution of 2.9 x 10-4 M. From Henry’s Law, we
know that doubling the pressure also doubles the amount of
dissolved gas. Therefore,
3.4 ⇥ 10 2
= 117.2 ⇥ 0.22 atm = 25.8 atm
2.9 ⇥ 10 4

Colligative Properties:

Properties that depend only on the number of solute particles

in solution and not the nature of the solute.

Non-volatile solute interferes with the solvent’s liquid-vapor

equilibrium (from chemwiki.ucdavis.edu).
 1. Vapor-Pressure Lowering

For a non-volatile solute, the vapor pressure of the liquid

decreases.
Another way to think of it: solute shifts equilibrium to the left
solvent (l) solvent (g)

o
Raoult’s Law makes this quantitative: P1 = 1 P1
P1 partial pressure of a solvent over a solution
1 mole fraction of the solvent in the solution
P1o vapor pressure of the pure solvent
Note: If there is only one solute, χ1 = (1-χ2) where χ2 is the
mole fraction of the solute, and thus
o o o
P1 = (1 2 )P 1 ) P 1 P1 = P = P
2 1

e.g., Calculate the mass of urea that should be dissolved in

225 g of water at 35ºC to produce a solution with a vapor
o
P
pressure of 37.1 mmHg. (At 35ºC, H2 O
= 42.2 mmHg)

Solution #1:
225 g
= 12.49 mol
How many moles of water do we have? 18.02 g/mol

P1 37.1 mmHg 12.49 mol water

1 = = = 0.879 =
P1o 42.2 mmHg x mol urea + 12.49 mol water
Thus moles of urea is 1.72. The mass of urea needed is (1.72
mol)(60.06 g/mol) = 103 g.
Solution #2:
P 42.2 37.1 x mol urea
2 = = = 0.121 =
P1o 42.2 x mol urea + 12.49 mol water
If both components are volatile, the vapor will contain both
substances, and Raoult’s Law still applies:

PA = A PAo
PB = B PBo
PT = PA + PB (Dalton, s Law of Partial Pressures)
PT = A PAo + B PBo

Ideal hexane-octane solution (from Ross Church). Note that when the
solution mole fractions are equal (0.5), the vapor above the solution
is enriched in the more volatile species. This is the principle behind
fractional distillation.
 2. Boiling-Point Elevation

§ Boiling-point is when vapor pressure = atm pressure

§ Non-volatile solute lowers the solvent vapor pressure
§ Thus need higher temperature to reach boiling point

quantitative expression: ΔTb = Kb m

Tb = Tb Tbo boiling-point elevation

Kb molal boiling-point elevation constant
m molality of the solution

3. Freezing-Point Depression

§ Solute interferes with the formation of an ordered solid

§ In order to form a solid, a lower temperature is required

quantitative expression: ΔTf = Kf m

Tf = Tfo Tf freezing-point depression

Kf molal freezing-point depression constant
m molality of the solution
 Phase diagram illustrating boiling-point elevation and freezing-point
depression. (from www.boundless.com)

e.g., Determine the boiling point and freezing point of a

solution prepared by dissolving 678 g of glucose in 2.0 kg of
water. For water, Kb = 0.52°C/m and Kf = 1.86°C/m.

Solution:
moles of glucose: (678 g) / (180.156 g/mol) = 3.76 mol
m = (3.76 mol) / (2.0 kg) = 1.88
ΔTb = Kb m = (0.52°C/m)(1.88 m) = 0.98°C ⇒ Tb = 101°C
ΔTf = Kf m = (1.86°C/m)(1.88 m) = 3.50°C ⇒ Tf = -3.5°C
 4. Osmotic Pressure

Osmosis is the selective passage of solvent molecules

through a semipermeable membrane which blocks the
passage of solute molecules. It occurs from a more dilute
solution to a more concentrated one.

quantitative expression: π = MRT

π osmotic pressure (atm)

M molarity of the solution
R gas constant (0.08206 L⋅atm/K⋅mol)
T temperature (K)

Two solutions of equal concentration have the same osmotic

pressure and are said to be isotonic to each other.
Osmotic pressure. Beginning with equal levels of pure solvent and
solution (left), the solution level rises as a result of the net flow of
solvent from left to right. The osmotic pressure π, required to stop
osmosis, can be measured directly from the difference in the final
liquid levels.

Electrolyte Solutions

Colligative properties depend only on the number of dissolved

particles, not on the type of particle. Therefore, a 0.1 m
solution of NaCl has twice the freezing-point depression of a
0.1 m nonelectrolyte solution.
Solution Ions Theoretical ΔTf
+ -
1.00 m NaCl Na , Cl (1.86)(2.00 m) = 3.72°C
+ 2-
1.00 m (NH4)2SO4 2 NH4 , SO4 (1.86)(2.00 m) = 5.58°C

The theoretical freezing-point depression is usually an

overestimate because the ions can form pairs. To be
quantitative we introduce the van’t Hoff factor i:

actual number of particles in solution after dissociation

i=
number of formula units initially dissolved in solution

and we modify the equations for colligative properties:

ΔTb = iKb m ΔTf = iKf m π = iMRT

Free ions and ion pairs in solution. Ion pairing reduces the number of
dissolved particles in solution, affecting the colligative properties.
The more dilute a solution is, the closer i is to the expected value.

e.g., What mass of naphthalene (128.2 g/mol) must be

dissolved in 2.00 x 102 g of benzene (Kf = 5.12°C/m) to give a
solution with a freezing point 2.50°C below that of pure
benzene?

Solution:
ΔTf = Kf m ⇒ m = ΔTf / Kf = 0.49 mol/kg.

mol 1 kg
0.49 · 2.00 ⇥ 102 g · 3 = 0.098 mol
Now, kg 10 g

Then, (0.098 mol)(128.2 g/mol) = 12.6 g.

e.g., What mass of insulin must be dissolved in 50.0 mL of
water to produce a solution with an osmotic pressure of 16.8
mmHg at 25°C?
Solution:
⇡ = M R T ) M = ⇡/RT =
16.8 mmHg K mol 1 4
· · = 9.035 ⇥ 10 mol/L
760 mmHg/atm 0.08206 L atm 25 + 273.15 K

Then, (9.035 × 10-4 mol/L)(0.05 L) = 4.52 × 10-5 mol

Finally, (4.52 × 10-5 mol)(5808 g/mol) = 0.26 g.

e.g., An aqueous solution that is 0.015 M in acetic acid

(HC2H3O2) is 3.5% ionized at 25°C. Calculate the osmotic
pressure of this solution.

Solution:
π = MRT. Now, 3.5% ionized means that (100 - 3.5) = 96.5%
exists as a single species in solution while 3.5% exists as two
species in solution. Thus
1 96.5 2 3.5
i= ⇥ + ⇥ = 1.035
1 100 1 100
Then π = (1.035)(0.015 M)(0.08206 L⋅atm/K⋅mol)(298 K)
= 0.38 atm

e.g., A solution made by dissolving 14.2 g of sucrose in 100 g

of water (Kf = 1.86°C/m) exhibits a freezing-point depression of
0.77°C. Calculate the molar mass of sucrose.

Solution:
ΔTf = Kf m ⇒ m = ΔTf / Kf = 0.77 / 1.86 = 0.414 mol/kg.
Now, (0.414 mol/kg)(0.1 kg) = 0.0414 mol.
Then, (14.2 g) / (0.0414 mol) = 343 g/mol.
 Colloids

A colloid is a dispersion of particles of one substance

throughout another substance. It is intermediate between a
homogenous and heterogeneous mixture.

Paint (pigment particles in solvent) and milk (fat globules in water)

are common examples of colloids.

The particle size (1 – 103 nm) is much larger than in solution.

Categories:
o Aerosols (liquid or solid dispersed in gas)
o Foams (gas dispersed in liquid or solid)
o Emulsions (liquid dispersed in another liquid)
o Sols (solid dispersed in liquid or in another solid)
o Gels (liquid dispersed in solid)
One way to distinguish a colloid from a solution is by the
Tyndall Effect : the scattering of light by dispersed particles.

Tyndall effect.

When the dispersing medium is water, we can classify colloids

as hydrophilic or hydrophobic.

Hydrophilic colloid (a globular protein).

Hydrophobic colloid stabilized by adsorbed ions. Common example: clay.

An unstable colloid can be stabilized by the addition of an

emulsifier. Soaps (sodium stearate) and detergents work by
adsorbing to the surface of hydrophobic dirt/grease particles,
allowing them to be dispersed in water and removed.

Emulsification of grease by detergent molecules.