Chapter 13 Physical Properties of Solutions

Remember -

Ease of solute dissolving in a particular solvent depends on
Solution - homogeneous mixture of two or more substances
Solvent => that part INXS
Solute => all other components (not exactly a rigorous definition).
1. Solute - solute interactions have to be overcome (1H)
2. Solvent - solvent => energy needed here as well (2H)
3. Solute - solvent => releases energy (3H)

Break down as solH = 1 H + 2H + 3 H
The overall magnitude of each determines magnitude (and sign) of solH.
Examples
KCl
sol H = 17.2 KJ/mol  3H is large but not large enough to compensta e the
energy needed to overcome 1H and 2H.
LiCl solH = - 37.1 KJ/mol  3H is large and “-“
If we have an exothermic solH, favourable solute - solvent interaction (i.e., the ion dipole
interactions for LiCl are strong!)
For an endothermic solH, the driving force for the solution process is related to the degree of
randomness or disorder in the system. This measure of disorder is called entropy (return to entropy
later). An increase in disorder or randomness favours the solution process
We will discuss mostly solutions of liquids in liquids (e.g. ethanol in water) or solids in liquids. We
will examine solutions of gases in liquids in the context of Henry’s Law.

Note: the “like dissolves like” rule of solubility is based on the action of intermolecular forces.
e.g.

water and oil do not mix!!
oil and CCl4 do mix!!
water and salt (NaCl) mix
NaCl + gasoline do not mix
These can be rationalised easily on the basis of intermolecular forces.

Exceptions to “like dissolves like rule”
CH3 COOH/H2O => intermolecular H - bonding is important here.
CH3 COOH/C6H6  Intramolecular H-bonds between CH3COOH molecules, i.e., allows
it to dissolve in a solvent like C6H6 (nonpolar).
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Solids in liquids
1.
Ionic e.g. NaCl in water stabilised by ion -dipole interactions
2.
Covalent Complete => generally do not dissolve (beach sand)
3.
Molecular crystals => generally weak intermolecular forces => dissolve fairly easily in
nonpolar solvents
nonpolar C10H22  does not dissolve in water, dissolves in CCl4.
polar solvents  due to H bonds, urea (NH2 CONH2) dissolves readily in polar, Hbonding solvents like H2O, CH3OH
4.
Metallic solids  generally do not dissolve.
Example
 Would LiCl be more soluble in water or benzene?
 Answer => water ion-dipole interactions are very strong, therefore LiCl would be soluble in
water, but not in benzene.
Concentration Units
 We are already familiar with MOLARITY
M = Moles/L
There are others
1.
percent by mass (% mass)
=
mass of solute
x 100%
mass of solute + mass of solvent
2.
mole fraction (X)
=
3.
molality (m)
= moles of solute
kg of solvent
4.
percent by volume (%v/v)
moles of solute
moles of solute + moles of solvent
=
volume of solute x 100%
volume of solution
Note the molarity - molality distinction!
M => moles of solute / L of sol’n
m => moles of solute/ kg of solvent
Examples
1.
A typical shot of hard liquor drink is 40% by mass of C2H5OH. Assuming the rest is
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water, how much alcohol would we consume if the drink has a mass of 30g?
NOTE ETOH is solute!
% by mass of C2H3OH = 40%, mass of drink = 30g
2.

0.40 = g of solute (Etoh)
g of Etolh + g of H2O

0.40 * 30g = 12g of C2H5OH consumed!
Calculate the molality and the mole fraction of C2H5OH in a typical shot.
mole fraction =
moles of solute
moles of solute + moles of solvent
for a typical shot
30g total (EtOH + H2O)
moles of H2O = 18g * 1 mole
12g EtOH/18g H2O in the solution.
Moles H2O = 18 g H2O/ 18.02 g/mole
= 1.0 moles
moles of EtOH = 12g * 1mol = 0.26 moles EtOH
46.1g
XEtOH =
O.26 moles EtOH
1.0 moles H2O + 0.26 moles EtOH
XEtOH = 0.21

Molality
m = moles solute
18g H2O* 1 kg = 0.018kg H2O
kg of solvent
1000g
= 0.26 moles EtOH
0.018 kg H2O
= 14 molal
Converting between molality  molarity

To convert between molarity  molality, we need to know the density ( mass / volume )
of the solution!
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Example
Convert the molality of the drink to molarity, given that the density of the solution is 0.95g/cm3.
We need to know the total mass of the drink (30g)
 the total volume of the solution u
30g* 1.0
0.95g
* 1L
1000mL
= 32 mL = 0.032L
# of moles of solute = 0.26 moles of ethanol
 M moles of solute
L of solution
= 0.26 moles of solute = 8.1
0.032L
Compare with a molality of 14 molal!!
Example #2
A 2.00 M NaCl solution has a density of 1.08 g/mL. What is the molality of the solution? (solvent
is water)
0.200 M =
2.00 moles of NaCl
l L of solution
how much does l L of soution weigh?
mass of 1.000 L = 1.08g /mL * 1000ml / L = 1080g (mass of solute + solvent)
how much of the 1080g of the total solution is water?
2.00 moles of NaCl * 58.449g/mole
= 117g of NaCl.
Total mass = (g NaCl + g H2O) = 1080 g
 g of solvent = 1080g- 117 g NaCl
=
963g of H2O = 0.963 kg of H2O
molality

= moles of solute / (kg of solvent) = 2.00 moles/ 0.963 kg = 2.08 molal
Note: in general, the molaity of a soluion and the molarity of the solution are different.
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However, for dilute solutions (i.e., < 0.010 mol/L), we can safely assume that the molality
and the molarity are equal.
Effect of Temperature and Pressure on Solubility
 What do we mean by solubility?
 Solubility is defined as the maximum amount of solute that can be dissolved in a given
amount of solvent at a fixed temperature. Note that the definition implies that the solubility
depends on the temperature, pressure, and the amount of solvent.

Usual units (mass of solute g) /100g H2O

We will discuss three types of solutions.
1. Solids in Liquids

In general, most solid substances increase their solubility in a given solvent, as the
temperature increases. However, there are exceptions!!

There is no clear correlation between the dependence of solubility on T and sol H (although
one would be expected). Above the solubility limit, we have equilibrium between dissolved
solute and the solute that’s undissolved. Note that the enthalpy of solution is usually
measured for the dissolution of 1 mole of solute in a specified quantity of solvent. Hence,
the enthalpy of solution is measured at a concentration far away from the saturation
concentration. In order to get the true

The process in which dissolved substances come out of solution and form crystals is called
crystallisation.
We can also have supersaturated solutions, i.e., where we have exceeded the solubility limit but
no precipitation takes place. These solutions are thermodynamically unstable.
Due to the fact that solubilities of solids have different temperature dependencies, we can use
this for fractional recrystallisation, i.e., separate out the pure components of the mixture based
on their differing solubilities at different temperatures.
The pressure dependence of the solubility of solids in a particular solvent is very weak!



2.
Solutions of liquids in liquids.
 Compare miscible vs. immiscible liquids
 Miscible  two liquids completely soluble in all proportions;
 Immiscible  the two liquids aren’t mutually soluble in one another in all proportions.
 The solubility of liquids in liquids may either increase or decrease as a function of
temperature.
 The pressure dependence of the solubility of liquids in liquids is very weak!
3. Gas solubility
 The solubility of gases usually decreases with temperature. This is explained in terms of the
kinetic energy of the gas molecules. As T increases, the gas molecules have a higher kinetic
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
energy and can more easily escape from the surface of the solution. As more molecules
escape from the surface of the solution, the solubility decreases.
effect of increasing pressure on gas solubility this is described by Henry’s Law
C = kJ P  P = pressure
C is the gas solubility in molar units (mol/L). kJ – the Henry’s Law constant (mol L-1 atm-1);
depends on the nature of the gas.

This is also explained in terms of the kinetic molecular theory of gases
e.g. O2 (g)  O2 (aq)

Increase P(O2), more collisions of gas molecules with water surface.
probability of being ‘captured’ by the solvent.
Therefore, greater
Example.
Exactly 500 mL of Coke at 298K contains CO2 under 20 atm of pressure. Given that the
proportionality constant in the Henry’s law equation is 0.034M/atm calculate the volume of (O2(g)
in a bottle of Coke at STP.
SOLUTION C = k * P(CO2)
 where P(CO2) = the partial pressure of CO2 above the liquid in the pop bottle
C (mol/L) = 0.034 mol/L1atm * 20 atm
= 0.068 mol/L
# of L of pop = 500mL * 1L/1000 * 0.500L
 # moles of CO2 = 0.068 mol/L * 0.500L
= 0.034 mol
Ideal gas eg. PV = nRT
STP = 273K
V = 0.34 mol * 0.08206/1atm *273 1 atm pressure
= 0.760L of CO2
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Colligative Properties

Colligative properties are properties of the system that depend on the quantity and not the
nature of the solute particles. We will examine three main colligative properties
Vapour Pressure lowering => Raoult’s Law
 For a non volatile solute in a volatile solvent, Raoult found that the vapour pressure of the
solvent above the solution was lower than the vapour pressure of the pure solvent, and that it
depended timely on the amount of non-volatile solute present.

mole fraction of solvent
V.P. of pure solvent
P1 = X1 P1O
V. P. of solvent
above solution

Mole fraction of solvent
Note. We said that the colligative properties are dependent on the quantity of the solute (and
not its nature).
X1 = 1-X2 where X2 = mole fraction of solute

P1
=
(1-X2) P1o
=
P1o - X2P1o
(P1o - P1) X2 P1o =
X2 is the solute mole fraction and P1 is defined as the vapour pressure lowering of the
solvent.
Example
The vapour pressure of water is 0.308 atm at 343K. Calculate the vapour pressure of water
above a solution that contains 11.9 moles of H2O and 0.139 moles of glucose
X1
P1
=
=
mole fraction of H2O
11.9 moles
=
11.9 moles + 0.139moles
0.988
= vapour pressure of water
= 0.988 * 0.308atm = 0.304atm
P1 = vapour pressure of water = 0.988 * 0.308atm = 0.304atm.

Why is vapour pressure lowered?
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
This is related to the degree of randomness (the entropy) of the system. The solution state is
more random than the pure solvent! Vaporisation increases system disorder, but since the
solution is already more disordered (a higher entropy) than the pure solvent, there is less
tendency for the vapours to form

What about a mixture where both solute and solvent are volatile?
Simple - remember Dalton’s law of partial pressure
PT = P1 + P2 + P3...
P1 = X1 P10
P2 = X2 P2o
and
Pt = P1 + P2 = X1P1o + X2 P2o
Example
Let’s go back to the drink problem. What’s the total vapour pressure (water + EtOH) and the
V.P. of each above a typical shot.
at 298K
PoH2O =
PoEtOH
0.0312atm = 23.76mm of Hg.
=
0.079atm = 60mm of Hg
 from the previous problem
# moles of EtoH = 0.26 moles
# moles of H2O = 1.0 moles
and
X(EtOH) = 0.21
X(H2O) = 0.79
P (EtOH)
=
0.21 * 0.079atm
=
0.017 tm
PH2O
=
0.79 * 0.0312atm
=
0.025atm
total vapour pressure above liquid
=
0.017atm + 0.025atm
=
0.042atm


Note => only a few solutions obey Raoult’s Law completely; these are called ideal solutions
Note for an ideal sol’n sol H = O always
Boiling Point Elevation
 Dissolution of solids in solvents raises the boiling point of the solution relative to that of the
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
pure solvent.
The boiling point elevation
Tb = Tb – Tb = Kb m2
Molal boiling pt.
elevation constant
molality of
solute
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Example
Ethylene glycol is a common radiator additive. Calculate the boiling point elevation for a typical
antifreeze mixture (40% by mass ethylene glycol in water).
g of ethylene glycol/100g of solution = 0.40 * 100g
= 40g
 g of water = 60g
molar mass of EG = 62.01g/mol
 number of moles of E.G. = 40g E.G./ 62.01 g/mole
= 0.645 moles
molality (m) = 0.645 moles/ 0.060kg
= 10.74 molal
 Tb = Kb m2
= 0.52 C/ m* 10.7 m = 5.8 C
=> Tb = Tb - Tb  Tb = 100.0C + 5.8C = 105.8 C
100C
Freezing Point Depression

In the same manner as the b.p. elevation, addition of solutes to a solvent can depress the freezing
point.
Tf = Tf – Tf = Kf m2
molality of
solute
Molal freezing pt.
depression constant
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Example
Calculate the lowest temperature at which you could (theoretically) start out your automobile
with the 40/60 mixture of EG + H2O!
note. from above problem
MEG
=
10.7 molal
Tf
=
=
=
Kf *m
1.86 C /m * 10.7 molal
19.90 C = 19.9C
 The car would be protected down to
 Tf = Tf- Tf
19.9 = 0.0C – Tf
Tf = -19.9C (pretty cold, but attainable)
Osmotic Pressure
 We have two compartments with pure solvent inside A and a cone solution in Side B, separated
by a semi-permeable membrane

Beginning  equal water levels

Semi-permeable membrane allows solvent, but not solute to pass  we have a net flow of
solvent to the RHS and the solvent level in the tube on the right begins to increase.

The flow of solvent molecules from a dilute sol (or pure solvent) to a more concentrated
solution is called OSMOSIS!
osmotic pressure is the pressure required to stop osmosis

 = osmotic pressure = M R T
molarity

Temperature
0.08206 L.atm/K.mol
uses of osmosis  one of the best is for the measurement of the molar mass of polymers, since
only a small amount of polymer can give rise to fairly large osmotic pressures.
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1. isotonic  equal conc., therefore equal II
2. hypertonic  greater osmotic pressure
3. hypotonic  smaller osmotic pressure.
Example
A solution of 4.6 g of haemoglobin is found to have an osmotic pressure of 0.0135 atm at 300 K.
Given that the volume of the solution is 0.125 L, what is the molar mass of haemoglobin?
 = MRT
M=
0.0135 atm
=>
300 K x 0.08206 L atm/ K mol =>
M = 5.48 * 10-4 mol / L
# moles of haemoglobin = 5.48 x 10-4mol / L * 0.125 L
= 6.85 * 10-5 moles
 molar mass =
4.68 g
6.85 x 10-5moles
= 6.83 * 104
mol
g
Colligative properties in electrolyte solutions

NOTE:
Electrolytes dissociate
NaCl (aq)

Na+(aq) + Cl- (aq)
Colligative properties are based on the # of particles per unit amount of solvent



For NaCl (aq), there are 2 x as many particles in solution

We would expect, e.g., the effect on the physical properties to be greater then if the solute
were non-dissociating. How do we incorporate this into our discussion of the colligative
properties?
We introduce the van Hoff factor i

i




NaCl
HCl
Na2SO4
MgSO4

# of particles in sol. after dissolution
# of formula units dissolved in solution




i expected to be 2
i=2
i = 3 ( 2 Na+(aq) + SO42- (aq))
i=2
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
We will modify our colligative properties as follows
Tf = i Kf m2
=iMRT
Tb = i Kb m2
Example
 A 0.0103M solution of K2SO4 is found to have an osmotic pressure of 0.680 atm at 299K.
Calculate the value of i.
We expect K2SO4(ag)  2KT(ag) + SO42-(ag)
i = 3 (theoretically)
 = iact R T
0.680atm = i act x 0.08206 L atm/ (K mol) 8 299 K.
i act = 2.69
which is quite a bit lower than 3.

Why => ion pairing occurs in solution. In some solutions, the hydration of ions isn’t
complete, and the water is less effective at keeping changes separated and screened.
Electrostatic attractions level to dominate at higher concentrations and we have ion - pairing
in the solvent.
compare f.p. depression of E.G./H2O mixture with NaCl (i = 2)
 NOTE NaCl => f.p. depression is 39.8 C, tf = -39.8 C
 With E.G./H2O, the freezing point depression is 19.9 C, tf = -19.9 C.