Physical Properties of Solutions
Dr. Ramy Y. Morjan
Chapter 13
Physical Properties of Solutions
Overview
Composition of Matter
All matter can be divided into two major classes which are pure substance and
mixture. The pure substance could be Element or could be compound. The mixture
can be either Homogeneous mixture or heterogeneous as in figure 1.
Matter
solid, liquid gas
Pure Substance
constant composition
Elements
Metals
Non
-metals
Mixture
Physical methods
Separating
mixture
Compounds
Preparing
mixture
variable composition
Homogeneous
Heterogeneous
Semimetals
(metalloids)
Fig. 1 Composition of Matter
Homogeneous Mixture:
The prefixes "homo"- indicate sameness A homogeneous
mixture has the same uniform appearance and composition throughout. Many
homogeneous mixtures are commonly referred to as solutions as example dissolving
sugar in H2O.
Heterogeneous Mixture: The prefixes: "hetero"- indicate difference.
the individual components of a mixture remain physically separated and can be seen
as separate components, for example mixing sand with H2O
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Physical Properties of Solutions
Dr. Ramy Y. Morjan
What is a Solution?
A solution is a homogeneous mixture of two or more substances dissolved in each
other. Solutions mainly consist of two components one is called solvent and the other
is called solute.
Solvent: the substance in which the substances (solute) are dissolves to produce a
homogeneous mixture (solution) and it is the substance that present in large amount.
Solute: the substances that dissolves in a solvent to produce a homogeneous mixture
Although we shall concentrate mainly on liquid solutions, solutions can be solids,
liquids or gases.
Classification of solution: Solutions can be classified on the basis of their state:
solids, liquids or gases. Examples of various ways of preparing a two component
solution in each of the three states are summarized in the following table.
Solution type
solvent
solute
Solid solution
Solid
Solid
Metalloids, copper dissolved in nickel
Solid
Gas
Hydrogen dissolved in palladium
Solid
Liquid
Gas
Gas
Oxygen dissolved in Nitrogen (air)
Gas
Solid
Dry ice dissolved in (sublimed) nitrogen
Gas
Liquid
Liquid
Solid
Gas solution
Liquid solution
Liquid
Liquid
Example
Mercury dissolved in gold
Chloroform dissolved in (evaporated
into) nitrogen
Sugar dissolved in water
Gas
Carbon dioxide dissolved in water (soft
drinks)
Liquid
Ethanol dissolved in water
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In order to form a liquid solution in which the solute and solvent are both liquids, the
two liquids must be miscible.
Miscible liquids mean that the two liquids can dissolve completely in each other at
any proportions (‫ )ﺑﺄي ﻧﺴﺒﮫ‬as example water and ethanol. In the other hand immiscible
means they can’t dissolve in each other as example oil and water.
Types of Solutions
Well, as we have seen in the above section that solution are formed by mixing solute
and solvent with each other. In order to such as this process to takes place the
molecules of the solute and the solvent must get in direct contact with each other
leading to solubility of the solute in the solvent and forming the solution. The ratio of
solute and its solubility in the solvent as well as the amount of the solvent used
determines the nature (type) of the produced solution.
Solubility: solubility is defined as the maximum amount of substance (solute) that
will dissolve in a solvent at that temperature. Usually solubility is measured by g/L.
Chemist usually classifying solution according to the capacity of solvent to dissolve a
solute at a specific temperature; in other words solution can be classified according to
the ratio between solute and solvent at a given temperature. This classification led to
the following three groups of solution.
1) Saturated solution
Contains the maximum amount of solute dissolved in solvent. In other words; when
the solvent has dissolved all the solute it can, and left some un-dissolved solute it’s a
saturated solution. For example; if you add 20 g of sugar to 100mL of water at room
temperature, then all of the sugar will dissolved. However if you add 300g to the same
amount of water, some of the sugar remains un-dissolved. We said this solution is
saturated. In saturated solution there is equilibrium between solution and solute that
mean if you for example keep string the solution some of the un-dissolved solute will
dissolve and in the same time some of dissolved solute will come out so they are in
equilibrium.
2) Unsaturated solution
An unsaturated solution contains less of solute than it could hold i.e. solutions can
dissolve more solute under normal condition. For example a saturated solution of
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NaCl/H2O contains 25g of NaCl in 100mL water; the unsaturated solution of
NaCl/H2O will contain less than 25g NaCl in 100mL water. In an unsaturated solution
there is no equilibrium exists between the solute and the solution because there is no
undissolved solute
3) Supersaturated solution: A solution that is temporarily holding more than it can,
a seed crystal will make it come out. A saturation solution can be prepared by adding
excess solute to a saturated solution. Unsaturated solutions are not stable solutions, for
example if just one crystal of solute is added, or the liquid is shaken, the excess solute
will crystallize immediately and saturated solution will form.
Crystallization: The process in which dissolved solute comes out of solution and
form crystals.
Crystals: Crystals are solids that form by a regular repeated pattern of atoms or
molecules connecting together. In crystal the arrangements of the building blocks
(atoms and molecules) lie in an orderly array. In crystals, however, a collection of
atoms called the Unit Cell is repeated in exactly the same arrangement over and over
throughout the entire material. Diamond the very expensive jewel which is 3D
covalently bonded network of carbon atoms. Diamond is a gigantic molecule ( ‫ﺟﺰﯾﺊ‬
‫)ﺿﺨﻢ‬with crystalline structure which gives it its unique properties.
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Dr. Ramy Y. Morjan
Aqueous solutions
An aqueous solution is a solution in which the solvent is water. It is usually shown in
chemical equations as a subscript (aq). The word aqueous means pertaining to, related
to, similar to, or dissolved in water. As water is an excellent solvent as well as
naturally abundant, it logically has become a ubiquitous solvent in chemistry.
Substances that do not dissolve well in water are called hydrophobic ('water fearing')
whereas those that do are known as hydrophilic ('water-loving'). An example of a
hydrophilic substance would be the sodium chloride (ordinary table salt).
H 2O
Cl- Na+
Na+ Clsolute
Na+
H2O Cl-
A Molecular View of the Solution Process
In the previous sections we have discussed the formation of solutions, and you learned
that solutions are formed by dissolving a solute in a solvent. Now we need to learn
how the dissolving process takes place. To understand this process we need to
remember what we have learned about the intermolecular forces and intramolecular
forces (Chapter 12), however these forces play very important rolls in interpretation
the solubility and solution formation. (Please go back to chapter 12)
To make a solution the solute and solvent are mixed together, and in order to obtain a
solution the solute molecules will dissolve in the solvent molecules that means the
solute molecules will take the positions of some solvent molecules.
What happens when you add a solid solute to a liquid solvent?
Immediately after the addition of a solid solute to a liquid solvent, the solid –state
structure begins to disintegrate (‫ )ﯾﺘﺤﻠﻞ‬little by little. The solvent molecules chip away
at the surface of the crystal lattice, prying out solute particles and surrounding them,
and finally dispersing them throughout the body of the solution. This process
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(solubility) depends on the relative strength of three attractive forces (intermolecular
forces) which are:
1) Solvent-solvent interaction (before dissolution)
2) Solute –solute interaction (before dissolving)
3) Solute- solvent interaction (during the dissolving process)
When the solution is formed i.e. for the solute to dissolve in the solvent both the
solvent –solvent intermolecular force and solute –solute intermolecular forces will be
replaced (‫ )ﺗﺴﺘﺒﺪل‬by the solute- solvent interaction (see figure 3). From this disscation,
we can conclude (‫ )ﻧﺴﺘﻨﺘﺞ‬that if the solvent –solvent intermolecular force and solute –
solute intermolecular forces are weak , and the solute- solvent interaction is strong the
formation of the solution will be easy and vies versus. (‫)اﻟﻌﻜﺲ ﺻﺤﯿﺢ‬
In solution, there is intermolecular
interaction between solve and solute.
The old solvent -solvent interaction and
solute-soluteinteraction disappear
there is intermolecular force
between the solvent molecules
solvent -solvent interaction
there is intermolecular force
between the solvent molecules
solute-soluteinteraction
liquid solvent
solid solute
solution
Fig. 3 Mechanism of dissolving solid solute in liquid solvent
Formation of solution & energy changes
The solubility process of solid solute in a liquid solvent is companied with changes in
the energy of both the solute, solvent and the formed solution. As we have just
learned that for a solid solute to dissolve in a liquid solvent the intermolecular force
(interaction) between the solvent –solvent, and solute – solute must be destroyed. In
the same time a new interaction between solute- solvent start to take place. In both
cases the energy cotenants of the solute and solvent and solution is changed. To make
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Dr. Ramy Y. Morjan
this clear and easy; let us assume that the formation of a solution to takes place in
three steps as the following.
a) Separation of the solvent –solvent interaction
b) Separation of the solute – solute interaction
c) Formation of new interaction between solute- solvent
In both steps a and b the process need energy in order to overcome the intermolecular
force (interaction) between the solvent molecules alone and between the solute
molecules alone. Steps a & b are endothermic steps. In any process or a chemical
reaction if energy (heat) is absorbed, the process is said to be endothermic process or
reaction. In the other hand, in step 3 the solute and solvent are mixed together, in
terms of energy; this process can be either endothermic process or can be exothermic
process. In any process or a chemical reaction if energy (heat) is released the process
is said to be exothermic process or chemical reaction. The total change of energy
during the dissolving process is the summation of each change in each step i.e. a and b
and c in our example. The total change of heat is known as the heat of solution ∆H or
enthalpy ∆H of solution (reaction).
For steps a, b and c the
∆Hsol = ∆Ha + ∆Hb + ∆Hc
If ∆H is positive (∆H > zero) that indicates that energy (heat) is absorbed i.e.
endothermic process or reaction.
If ∆H is negative (∆H < zero) that indicates that energy (heat) is released i.e.
exothermic process or reaction.
When the value of ∆Hsol is positive and when it is negative?
v
If
the interaction between solute and solvent is stronger than both the
interaction between solvent-solvent molecules and the interaction between
solute-solute molecules the value of ∆Hsol will be negative (∆Hsol < zero) and
the process is exothermic. In this the solubility process is favourable and the
solid solute will dissolve easily (easily here means that there is no great energy
deficit) in the liquid solvent to form a solution.
v
If
the interaction between solute and solvent is weaker than both the
interaction between solvent-solvent molecules and the interaction between
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Dr. Ramy Y. Morjan
solute-solute molecules the value of ∆Hsol will be positive (∆Hsol > zero) and
the process is endothermic. In this the solubility process is unfavourable and
the solid solute will not dissolve or slightly dissolved in the liquid solvent to
form a solution. Figure 5 is explaining the energy changing during the
dissolving process.
Hstep d
potential energy
positive value
endothermic
solubility is
unfavourable
step b
expanded
solvent
expanded
solute
step c
step a
solvent
expanded
solute
net ene
rgy cha
negative value
exothermic
nge
H
solute
solvent
solution
Fig. 5 The energy changes during the dissolving process
Input of energy is required to separate solute-solute (step a) and solvent-solvent (step
b) attractions. Energy is released due to solute-solvent attractions (step c). If the
amount of energy released in step c is greater than the energy absorbed in steps a and
b the process is exothermic (∆Hsol is negative) and favoured for dissolution, but if the
amount of energy in step c is less than the absorbed energy in steps a and b the
process is endothermic (∆Hsol is positive) and un-favoured as in step d in figure 5.
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In addition to the energy factor, there is another factor that effect on formation of
solution which is the tendency toward disorder. When solute and solvent molecules
are mixed together; to form a solution, there is an increase in randomness or disorder.
In the pure state, the solvent and solute have enough degree of order, this order is
destroyed when the solute dissolve in the solvent. Therefore, the solution process is
accompanied by an increase in disorder. This increasing in disorder of the system is
favouring the solubility of any substance. Figure 6 is explaining the above discussion.
Does any solute dissolve in any solvent?
The direct and simple answer to this question is NO. There are so many solutes that
do not dissolve in particular solvent. However; solubility of any solute in a solvent
depends mainly on the type of intermolecular force of both solutes and solvents.
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Physical Properties of Solutions
Dr. Ramy Y. Morjan
To understand why some solutes do not dissolve in some solvents let us learn a
general saying which says ‘like dissolves like’. The meaning of this says is that if any
substances in this case (solute & solvent) have the same type of intermolecular forces
they will likely dissolve in each other. In other words if two or more substances have
the same polarity they most likely will dissolve in each other.
Solvents and solutes can be broadly classified into polar and non-polar. A molecule
is polar if there is some separation of charge in the chemical bonds, so that one part of
the molecule has a slight positive charge and the other a slight negative charge. The
slight negative and positive charges are arising as a result of the difference in the
electronegativities between the elements that form the bond. Polarity can be
measured as the dielectric constant or the dipole moment of a compound.
Water is a well-known example of a polar molecule, it consist of two H atoms
covalently bonded to oxygen O atoms (covalent bond). H2O is a polar because there is
difference in electronegativities between the H atoms and the O atom (figure 7)
Fig. 7 H2O is polar molecule
The dipole moment (µ) of water is 1.85 deby. The value of µ is calculated using the
following equation
Where µ is dipole moment, q is the charge and r is the distance between the two
charges. Table 1 show some compounds and dielectric constant and dipole moment
(µ) for the values.
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Compound
Dielectric
Constant
Dipole Moment (µ )
Formamide
110
3.37
Water
79
1.85
Methanol
32
1.66
Benzene
2
0.00
Table 1
Please note that benzene has µ equal zero i.e. it is non polar compound. There are no
polar bonds in benzene as it contains covalent bond between only H and C which
have nearly similar electronegativities (2.2 for H and 2.55 for C).
The polarity of a solvent determines what type of compounds it is able to dissolve
and with what other solvents or liquid compounds it is miscible. As a rule of thumb,
polar solvents dissolve polar compounds BEST and non-polar solvents dissolve nonpolar compounds BEST: "like
dissolves like".
Strongly polar compounds (ionic compound) like inorganic salts (e.g. table salt NaCl)
dissolve only in very polar solvents like water, while strongly non-polar compounds
like oils or waxes dissolve only in very non-polar organic solvents like hexane.
Similarly, water and hexane are not miscible with each other and will quickly separate
into two layers even after being shaken well. Table 12 in the next page has a list of
common solvents both polar and non-polar solvent with the values of dielectric
constant and dipole moment. You should be able to tell a good thing about the
solubility of any solvent with another solvent or the solubility of any solute (you need
to know its structure) in any given solvent.
A compound, such as water, that is composed of polar molecules. Polar solvents
can dissolve ionic compounds or covalent compounds that ionize. Non-polar
solvents, such as benzene, will only dissolve non-polar covalent compounds.
Now you should be able to expect if a given solute will dissolve in a given solvent or
will not. Of course; no need to memorize the values of the dielectric constant and
dipole moment and the electronegativities or any other numbers. However all what
you need is to use your understanding to the concept of polarity and to understand the
classification of the periodic table as this will help you to obtain a good judgment
about the solubility of compounds in each other.
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Physical Properties of Solutions
Table 2
Dr. Ramy Y. Morjan
Common polar and non-polar Solvents
100
dipole
moment
1.85
dielectric
constant
80
CH3-OH
68
1.70
33
ethanol
CH3CH2-OH
78
1.69
24.3
1-propanol
CH3CH2CH2-OH
97
1.68
20.1
1-butanol
CH3CH2CH2CH2OH
118
1.66
17.8
formic acid
100
1.41
58
acetic acid
118
1.74
6.15
formamide
210
3.73
109
acetone
56
2.88
20.7
tetrahydrofuran (THF)
66
1.63
7.52
methyl ethyl ketone
80
2.78
18.5
ethyl acetate
78
1.78
6.02
acetonitrile
81
3.92
36.6
N,N-dimethylformamide
(DMF)
153
3.82
38.3
diemthyl sulfoxide
(DMSO)
189
3.96
47.2
69
----
2.02
80
0
2.28
Name
Structure
bp, oC
water
H-OH
methanol
hexane
CH3(CH2)4 CH3
benzene
diethyl ether
CH3CH2OCH2CH3
35
1.15
4.34
methylene chloride
CH2Cl2
40
1.60
9.08
carbon tetrachloride
CCl4
76
0
2.24
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Now, let us see how ionic compounds like NaCl dissolve in polar solvent like H2O?
Ionic compounds form ionic solutions. Sodium chloride when dissolve in
water forms an ionic solution. Such ionic solutions conduct electricity. The solutions
are called electrolytes. Table salt or sodium chloride (NaCl) is an ionic compound.
Sodium is in the first group of the periodic table and chlore is in the group 7. The
difference in their electronegativities indicates that it is an ionic compound. Ionic
compound do not form hydrogen bonding with water. Salt is a solid at room
temperature. The sodium and chloride atoms are arranged in a pattern that we called
crystal
lattice.
This
process is described on
steps as following;
1) When we put salt in
water, the sodium and
chloride
atoms
released
from
crystalline
lattice
are
their
and
form positive sodium ions
(Na+) & negative chloride ions (Cl-). The (Na+) and (Cl-) are free to move within the
water i.e. over come the Intramolecular forces.
Fig. Formation of Na+ and Cl- in water
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2) Water is a polar solvent, it consist of 2 H atoms and one O atoms. These three
atoms are covalently bonded to each other (Intramolecular forces). As a result of the
difference in electronegativities between the O and H, a partial negative charge will
arise on the O atoms and a partial positive
charge will arise on the two H atoms. The
partial charges on the O and H give the
H2O its unique property as polar solvent.
However; water molecules are joining to
each other via an intermolecular force
(hydrogen bonds), this hydrogen bonds are
stands behind the high boiling point of
water, as well as it makes the H2O a good
solvent for so many compounds.
Fig H bonds (Intermolecular forces)
In order for the solute to dissolve in the solvent, the intermolecular forces i.e. the
hydrogen bond between the water molecules must be over come, this distortion
required energy to over come the interaction between the solvent (water) molecules.
As a result of the over comes of the interaction between the solvent molecules; a room
are created between the water molecules and enabling the solute to get into the
solvent.
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3) Now we have the positive ions Na+ and the negative ions Cl- and the H2O with its
partial positive on H atoms and the partial negative on the O atom. The positive ion
Na+ will be attracted to the partially negative O atom via ion-dipole interaction, in the
same way the partial H atoms on water will be attracted by the Cl- negative ion as well
via ion-dipole interaction.
Fig. Ion-Dipole interaction between Na+ and negative O, and between H and Cl-
In the actual situation; so many partial negative O on water will be surrounding the
Na+ positive ion, and so many H atoms in water surrounding the partial negative Clion via Ion-Dipole interaction and the NaCl – H2 O solution is formed. The process in
which an ion or molecules are surrounded by solvent molecules in a specific manner
is called solvation.
Ion-Dipole interaction and the NaCl – H2O solution is formed
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Calculating Concentration
Units & Dilutions
The concentration of a chemical solution refers to the amount of solute that is
dissolved in a solvent. We normally think of a solute as a solid that is added to a
solvent (e.g., adding table salt to water), but the solute could just as easily exist in
another phase. For example, if we add a small amount of ethanol to water, then the
ethanol is the solute and the water is the solvent. If we add a smaller amount of water
to a larger amount of ethanol, then the water could be the solute!
Units of Concentration
Once you have identified the solute and solvent in a solution, you are ready to
determine its concentration. Concentration may be expressed by several different
ways, using percent by mass, mole fraction, molarity, molality, or normality.
1) Percent by Mass (%)
Percent by Mass (%) also called the percent by weight or weight percent. This is the
mass of the solute divided by the mass of the solution (mass of solute plus mass of
solvent), multiplied by 100.
moles of solute
x 100%
Percent by Mass =
Mass of solution
Percent by Mass has no units because it is a ratio of two similar quantities.
Example
Determine the percent composition by mass of a 100 g salt solution which contains 20
g salt.
Solution:
20 g NaCl / 100 g solution x 100 = 20% NaCl solution
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2) Mole Fraction (X)
This is the number of moles of a compound divided by the total number of moles of
all chemical species in the solution. Keep in mind, the sum of all mole fractions in a
solution always equals 1. For example if we have a solution contains 2 mol of solute
A dissolved in 10 mol of solvent B; the mole fraction of the solute A is calculated by
using the following formula
moles of solute A
Mole Fraction A
moles of solute A + moles of solute B
2
10
Mole Fraction A
Mole Fraction B
2 + 10
2 + 10
2/12 + 10/12 = 1
Example:
What are the mole fractions of the components of the solution formed when 92 g
glycerol is mixed with 90 g water? (molecular weight water = 18; molecular weight of
glycerol = 92)
Solution:
90 g water = 90 g x 1 mol / 18 g = 5 mol water
92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glycerol
total mol = 5 + 1 = 6 mol
A water = 5 mol / 6 mol = 0.833
B glycerol = 1 mol / 6 mol = 0.167
It's a good idea to check your math by making sure the mole fractions add up to 1:
A water + B glycerol = .833 + 0.167 = 1.000
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3) Molarity
Molarity is probably the most commonly used unit of concentration. It is the number
of moles of solute per liter of solution (not necessarily the same as the volume of
solvent!).
moles of solute
Molarity =
litter of solution
Example:
What is the molarity of a solution made when water is added to 11 g CaCl2 to make
100 mL of solution?
Solution:
11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2
100 mL x 1 L / 1000 mL = 0.10 L
molarity = 0.10 mol / 0.10 L i.e. molarity = 1.0 M
Example 2:
.0678 g of NaCl is placed in a 25.0 ml flask full of water. When the NaCl dissolves,
what is the molarity of the solution?
.0464 M NaCl
4) Molality
Molality is the number of moles of solute per kilogram of solvent. Because the
density of water at 25°C is about 1 kilogram per liter, molality is approximately equal
to molarity for dilute aqueous solutions at this temperature. This is a useful
approximation, but remember that it is only an approximation and doesn't apply when
the solution is at a different temperature, isn't dilute, or uses a solvent other than
water.
moles of solute
molality =
kilogram of solvent
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Example 1:
.20 mol of ethylene glycol / 2.0 Kg of solvent = .10 m ethylene glycol
Example 2:
What is the molality of a solution of 10 g NaOH in 500 g water?
Solution:
10 g NaOH / (4 g NaOH / 1 mol NaOH) = 0.25 mol NaOH
500 g water x 1 kg / 1000 g = 0.50 kg water
molality = 0.25 mol / 0.50 kg
molality = 0.05 M / kg i.e. molality = 0.50 m
5) Normality (N)
Normality is equal to the gram equivalent weight of a solute per liter of solution. A
gram equivalent weight or equivalent is a measure of the reactive capcity of a given
molecule. Normality is the only concentration unit that is reaction dependent.
Example:
1 M sulfuric acid (H2SO4) is 2 N for acid-base reactions because each mole of surfuric
acid provides 2 moles of H+ ions. On the other hand, 1 M sulfuric acid is 1 N for
sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.
Dilutions
You dilute a solution whenever you add solvent to a solution. Adding solvent results
in a solution of lower concentration. You can calculate the concentration of a solution
following a dilution by applying this equation:
M1V1 = M2V2
where M is molarity, V is volume, and the subscripts 1 and 2 refer to the initial and
final values.
Example:
How many millilitres of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?
Solution:
5.5 M x V1 = 1.2 M x 0.3 L
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V1 = 1.2 M x 0.3 L / 5.5 M
V1 = 0.065 L
V1 = 65 mL
So, to prepare the 1.2 M NaOH solution, you pour 65 mL of 5.5 M NaOH into your
container and add water to get 300 mL final volume.
Factors affecting solubility
1) Effect of temperature
a) Solid solute
b) Gas solute
2) Effect of pressure
1) Effect of Temperature on Solubility (solid solute)
Solubility refers to the ability for a given substance, the solute, to dissolve in a solvent
at a specific temperature. Generally, an increase in the temperature of the solution
increases the solubility of a solid solute. A few solid solutes, however, are less
soluble in warmer solutions. The chart shows solubility curves for some typical
inorganic salts (all solids). Many salts behave like barium nitrate and disodium
hydrogen arsenate, and show a large increase in solubility with temperature.
The solubility increases with temperature
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Some solutes (e.g. NaCl in water) are fairly independent of temperature. A few, such
as cerium(III) sulphate, become less soluble in hot water.
There is no direct relationship between the value of ∆Hsol (if it is positive or negative)
and the variation of solubility with temperature. For example, the solution process of
CaCl2 is exothermic and that for NH3NO3 is endothermic, but the solubility of both of
them has increased by increasing the temperature. However, experimental work is the
best way to determine the effect of temperature on solubility that because the
solubility is depends on the natural of the solute and solvent, for example; only 1
gram of lead (II) chloride can be dissolved in 100 grams of water at room
temperature, while 200 grams of zinc chloride can be dissolved at the same
temperature.
b) Effect of Temperature on Solubility (Gas solute)
As the temperature increases, the solubility of a gas decreases as shown by the
downward trend in the graph.
More gas is present in a solution
with a lower temperature compared
to
a
solution
with
a
higher
temperature. The reason for this gas
solubility
relationship
with
temperature is because the increase
in temperature causes an increase in
kinetic energy. The higher kinetic
energy causes
more
motion in
molecules
which
break
intermolecular bonds and escape from solution.
This gas solubility relationship can be remembered if you think about what happens to
a bottle of Pepsi as it stands around for awhile at room temperature. The taste of the
Pepsi becomes bad since more of the "tangy" carbon dioxide bubbles have escaped.
Boiled water also tastes "flat" because all of the oxygen gas has been removed by
heating.
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Physical Properties of Solutions
Dr. Ramy Y. Morjan
2) Effect of Pressure on Solubility
Liquids and solids solutes exhibit practically no change of solubility with changes in
pressure. Gases as might be expected, increase in solubility with an increase in
pressure. Henry's Law states that: The solubility of a gas in a liquid is directly
proportional to the pressure of that gas above the surface of the solution.
CαP
C = ĸP
C = molar concentration of a gas in a liquid
P = pressure of the gas over the solution in atmospheres
K = Constant that only depends on temperature. Its unit is mol/L.atm
If the pressure is increased, the gas molecules are "forced" into the solution since this
will best relieve the pressure that has been applied. The number of gas molecules is
decreased. The number of gas molecules dissolved in solution has increased as shown
in the graphic on the left. Carbonated beverages provide the best example of this
phenomena. All carbonated beverages are bottled under pressure to increase the
carbon dioxide dissolved in solution. When the bottle is opened, the pressure above
the solution decreases. As a result, the solution effervesces and some of the carbon
dioxide bubbles off.
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Physical Properties of Solutions
Dr. Ramy Y. Morjan
What is Colligative Properties of Solutions?
Solutions have different properties than either the solutes or the solvent used to make
the solution. Those properties can be divided into two main groups’ colligative and
non-colligative properties. Colligative properties depend only on the number of
dissolved particles in solution and not on their identity. Non-colligative properties
depend on the identity of the dissolved species and the solvent.
A property that depends only on the amount of solute in a solution and not the identity
of the solute is called Colligative properties
Colligative properties are
1. Boiling point elevation,
2. Freezing point lowering, and
3. Osmotic pressure.
4. Vapour pressure lowering,
Kf and Kb
Kf and Kb are the freezing point depression constant and boiling point elevation
constant respectively. When a solute is added to a solvent, the boiling point of the
solution is always greater than the boiling point of the pure solvent. Adding a solute
also lowers the freezing point.
Solvent Formula Melting Point (oC) Boiling Point(oC) Kf(oC/m) Kb(oC/m)
Benzene C6H6
80.2
5.065
2.61
Ethanol C2H5OH --
78.3
--
1.07
Water
100.000
1.858
0.521
H20
5.455
0.000
To determine the amount of change in boiling point you will need this equation:
Tb = Kb * m
•
Tb change in boiling point
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Dr. Ramy Y. Morjan
•
Kb boiling point elevation constant
•
m molality of solution
Then Tb is added to the normal boiling point of the pure solvent. Note that the
identity of the solute is not important, just its concentration (expressed in molality).
Therefore, boiling point elevation is a colligative property.
To determine the freezing point of a solution, you need to calculate the decrease in
freezing point caused by the addition of a solute to the solvent. Use the equation:
Tf = Kf * m
•
Tf change in freezing point
•
Kf freezing point depression constant
•
m molality of solution
Then the
Tf is subtracted from the normal freezing point of the pure solvent.
Freezing point depression is also a colligative property.
Example:
Calculate the boiling point and freezing point of a solution of .30 g of glycerol
(C3H8O3) in 20.0 g of water.
moles glycerol = (.30 g) (1 mole / 92 g) = .0033 moles
molality of solution = .0033 moles / .020 kg = .16 m
Tb = (.521 oC/m)(.16 m) = .083 oC
Boiling point = 100.00 + .083 = 100.083 oC
Tf = (1.858 oC/m)(.16m) = .30 oC
Freezing point = 0.00 oC - .30 oC = -.30 oC
In an ionic solution, the total concentration of ions is important. Therefore, another
factor (i) is included in the equations.
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Tb = Kb * m * i
Tf = Kf * m * i
"i" is the number of ions from each formula unit. In the previous example, if NaCl had
been the solute your change in boiling point and freezing point would have needed to
be multiplied by 2. (i=2 because NaCl consists of a Na+ ion and a Cl- ion
To explain the difference between the two sets of solution properties, we will compare
the properties of a 1.0 M aqueous sugar solution to a 0.5 M solution of table salt
(NaCl) in water. Despite the concentration of sodium chloride being half of the
sucrose concentration, both solutions have precisely the same number of dissolved
particles because each sodium chloride unit creates two particles upon dissolution a
sodium ion, Na+, and a chloride ion, Cl-. Therefore, any difference in the properties of
those two solutions is due to a non-colligative property. Both solutions have the same
freezing point, boiling point, vapour pressure, and osmotic pressure because those
colligative properties of a solution only depend on the number of dissolved particles.
The taste of the two solutions, however, is markedly different. The sugar solution is
sweet and the salt solution tastes salty. Therefore, the taste of the solution is not a
colligative property. Another non-colligative property is the colour of a solution. A
0.5 M solution of CuSO4 is bright blue in contrast to the colourless salt and sugar
solutions. Other non-colligative properties include viscosity, surface tension, and
solubility.
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Making solutions
1) Pour in a small amount of solvent
2) Then add the solute and dissolve it
3) Then fill to final volume.
M x L = moles
Example1: How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl
solution?
Example 2: How many grams of CaCl2 are needed to make 625 mL of a 2.0 M
solution?
Example 3: 10.3 g of NaCl are dissolved in a small amount of water then diluted to
250 mL. What is the concentration?
Example 4: How many grams of sugar are needed to make 125 mL of a 0.50 M
C6H12O6 solution?
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Dr. Ramy Y. Morjan
Dilution
Dilution is the addition of water to a solution or (solvents) that depends on the type of
solution you are making.
The number of moles of solute doesn’t change if you add more solvent.
The moles before = the moles after
M1 x V1 = M2 x V2
M1 and V1 are the starting concentration and volume.
M2 and V2 are the starting concentration and volume.
Stock solutions are pre-made to known M
Practice
2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity?
You have 150 mL of 6.0 M HCl. What volume of 1.3 M HCl can you make?
Need 450 mL of 0.15 M NaOH. All you have available is a 2.0 M stock solution of
NaOH. How do you make the required solution?
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