9. Solubilities of Ionic and Molecular Substances
What you will accomplish in this experiment
You’ll investigate the “like dissolves like” rule for predicting the ability of a solute to dissolve in a given solvent.
You’ll confirm your predictions by assessing the solubility of a variety of ionic and molecular substances in
solvents which have varying degrees of polarity and hydrogen-bonding capability.
Concepts you need to know to be prepared
Most of the substances you encounter in everyday life are mixtures.
You should recall from one of your earliest lecture discussions that a
mixture that’s uniform throughout and doesn’t separate into phases is
called a ―homogeneous mixture.‖ If a sample is taken from any part
of a homogeneous mixture and chemically analyzed, the same
components in the same concentration would be found in each and
every sample.
The most familiar type of homogeneous mixture is a liquid solution:
a solute (which may be solid, liquid, or gaseous) is thoroughly and
uniformly dispersed throughout a solvent (a liquid). If a closed
container of a solution is allowed to remain standing, the
components will not separate out, no matter how much time passes.
There’s usually a limit as to the amount of a
particular solute that can be dispersed or
dissolved in a given amount of a particular
solvent. This limit is the solubility, and it’s
defined as the maximum mass of solute that
can be dissolved in 100 grams (or 100 mL) of
a solvent at a stated temperature.
For example, you can see in the table of
solubilities to the left that sucrose (table
sugar) has a solubility of 179.2 grams per 100
grams of water at 0 oC. This means that you
can dissolve up to 179.2 grams of sugar, but
no more, in 100 grams of water at 0 oC.
If more sugar is added, the excess will fall to
the bottom of the container and remain
undissolved. A solution in this state is said to
be saturated. A solution with less than the
maximum amount of dissolved solute is
unsaturated.
C. Graham Brittain
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Solubility varies with temperature: solids and
liquids will generally have a much higher
solubility as the temperature of the solvent is
increased.
The figure to the right shows that the solubility
of sucrose increases from 179.2 grams (per 100
grams of water) at 0 oC to approximately 300
grams at 60 oC (140 oF).
You’ve observed this variability yourself, if
you’ve closely observed sugar dissolving in a
glass of iced tea as compared to a cup of hot tea.
Gaseous solutes tend to become less soluble as
the temperature of the solvent increases.
On the figure to the right, note the decrease in
the solubility of sulfur dioxide (SO2) gas in
water as the temperature is increased.
And consider your own experience in the
observed the loss of carbonation (the solute
CO2) in a glass of soda when the temperature is
raised from ice temperature to room
temperature.
There are some liquid substances that can be
homogeneously mixed in ALL proportions; these
liquids are said to be completely miscible. Ethyl
alcohol (ethanol) and water are two such
substances. (You can see that the table on the previous page indicates the ―infinity‖ symbol as the solubility for
ethanol and water.) The solution of commercial antifreeze (ethylene glycol and water) that used in automobile
radiators is another example of a homogenous mixture of miscible liquids.
How can you predict the solubility of a particular solute in a given solvent?
The rule of thumb for assessing solubility is “like dissolves like.” This means that the more similar the polarities
of the solute and solvent, the more likely it is that the two substances will form a homogeneous mixture.

Polar solutes, with partial positive (+) and partial negative charges (-), will tend to dissolve in polar
solvents such as water.

Nonpolar solvents (such as hydrocarbons) will dissolve in other nonpolar solvents, such as oil or gasoline.
Put another way, a solute is likely to dissolve in a solvent if the two substances are capable of SIMILAR types of
intermolecular attractive forces.
As you learned from the One Page Lesson posted with Lab 8, there are three types of intermolecular attractive
forces:
1) Dipole-dipole interactions (dipolar forces),
2) Hydrogen bonds, and
3) Instantaneous dipole-dipole interactions (abbreviated as IDDI, and also called ―dispersion forces‖).
A dipole-dipole interaction is the attractive force that exists between the partial positive charge of a polar group on
one molecule and the partial negative charge of a polar group on a neighboring molecule. Another way to say this
is that a region of slight excess positive charge on a molecule is attracted to a complementary region (of slight
excess negative charge) on a neighboring molecule.
C. Graham Brittain
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Only molecules that contain polar groups of atoms will have this charge imbalance (a dipole). Polar groups in
organic molecules typically contain either an O or a N atom.
Organic molecules that are purely ―hydrocarbon‖ (contain only C and H atoms) are nonpolar,
so they are not capable of adhering by dipole-dipole interaction (electronegativities: H = 2.1,
C = 2.5 – so close that the difference is negligible).
You should be able to recognize that the molecule on the top right, with a carbon-oxygen
double bond, has a dipole (with a slight excess of negative charge on the more electronegative
oxygen atom). Two of these polar molecules would be attracted to one another by a dipoledipole interaction.

H
O
C H
H
C

H C H
H
H
H
H C
The molecule on the bottom right, made purely of carbon and hydrogen atoms, is nonpolar
(has no dipole), so it is incapable of dipole-dipole interaction.
C
H C H
Hydrogen bonds are a very special, rather elite type of dipole-dipole interaction. Although
called hydrogen ―bonds,‖ it’s essential to remember that these are still simply a type of
attraction between partially-charged groups of atoms on neighboring molecules.
C H
H
H
However, the hydrogen bond, always indicated by a dotted line (—O—H . . . N—), is an especially strong attractive
force (about one-tenth the strength of a covalent bond). This is because it consists of two extremely
electronegative atoms (two O, or two N, or one O and one N) with a H atom between them.
—O—H
...
O—
—O—H
...
N—
—N—H
...
O—
—N—H
...
N—
Using the example (—O—H . . . N—), the dotted hydrogen bond is formed because the O atom in the polar —O—
H group is so powerfully electronegative that it draws the shared electrons in the O—H covalent bond toward
itself, leaving the positive charge of H’s nucleus almost fully exposed (electronegativities: H = 2.1, O = 3.5).
The partial-positive charge of the H proton is extremely attractive to electrons on nearby
molecules, particularly to a lone pair of electrons on an electronegative O or N atom. Thus a
hydrogen bond always occurs when a + H that is covalently attached to an O or N atom on
one molecule is interacting with a -O or N atom on a neighboring molecule.
In examining the pair of molecules on the right, you should now be able to recognize that
both molecules contain a polar group of atoms. But because the molecule on the top has a H
atom covalently attached to its O atom, its polar group is capable of hydrogen bonding. As
the molecule on the bottom has no H atom attached to its O atom, it is only capable of the
weaker dipole-dipole attractive force.
All molecules are capable of interacting with one another via dispersion forces (more
descriptively called instantaneous dipole-dipole interactions, abbreviated IDDI).
 H

H
O
C
H C
C H
H H
H
H
O
H
C

H C H
C H
H
H
To understand the origin of this attraction, you’ll need to recall the ―clouds‖ of electrons (orbitals) that surround
the nucleus of every atom. You should never think of that electron cloud as being static, or frozen in time; rather,
you should think of it as a dynamic, swirling mist. Because the electrons are constantly moving, at any given
place, the electron cloud may be thin in one instant, and thick in the next.
Wherever the cloud briefly thickens, the negative charge of the electrons overshadows the positive charge of the
nucleus. And wherever the cloud is briefly thin, the positive charge on the nucleus is able to shine through.
Thus whenever two neighboring molecules closely approach one another, the temporary charges caused by their
swirling electron clouds can interact: the partially-revealed positive charge of one nucleus will be momentarily
attracted to the partially-accumulated negative charge of a denser electron cloud surrounding a nucleus on the
neighboring molecule. So for a brief instant, there’s a very slight and very fleeting dipole, and the two molecules
are weakly attracted to one another.
C. Graham Brittain
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All organic molecules are capable of this extremely weak instantaneous dipole-dipole interaction; however, this is
the only type of intermolecular attractive force available to molecules that are purely hydrocarbon, and thus
nonpolar (have no permanent dipole).
After this review of intermolecular attractive forces, you should have a better appreciation for our rule of thumb for
predicting solubility: ―like dissolves like.‖
Again, a solute is likely to dissolve in a solvent IF the two substances are capable of SIMILAR types of
intermolecular attractive forces.
For example, molecules that can
interact with one another by forming
hydrogen bonds (such as ammonia,
NH3) are likely to be soluble in a
solvent if it can also form hydrogen
bonds (such as water, H2O).
Molecules that contain polar groups
of atoms (called “hydrophilic”
groups – meaning ―water-loving‖)
tend to be soluble in water and other
polar solvents.
Molecules that are primarily
nonpolar (―hydrophobic” – meaning
―water-fearing‖) are unable to
dissolve in strongly polar solvents,
but they are quite likely to dissolve
in nonpolar hydrocarbon solvents.
The figure to the right shows a polar
solute (indicated by the partial
charges of the permanent dipole)
interacting with the partial charges
of polar water molecules.
Keeping the “like dissolves like” rule in mind, you should now be able to study the structures of the molecules
below and begin making conclusions about the water solubility of each.
H3C
CH2
CH2
CH3
CH2
CH2
CH
H3C
CH3
H3C
CH
HO
H3C
CH2
CH2
CH3
CH2
CH2
CH
HO
The molecule to the left is purely hydrocarbon, so it is nonpolar (hydrophobic) and would not be water-soluble.
The molecule at the center has a polar –O—H group that is capable of hydrogen-bonding with water. The
nonpolar hydrocarbon region of this molecule is relatively small, so the hydrogen-bonding properties of the polar
group should be able to act as a tugboat, hauling the rest of the molecule into a polar solvent and causing it to be at
least partially water-soluble.
The molecule at the right has the same –O—H hydrogen-bonding group, but the group is attached to a nonpolar
hydrocarbon chain containing seven carbons! A carbon chain this long is too hydrophobic to allow the hydrogenbonding group to coax the molecule into solution in a solvent that is as highly polar as water. This molecule is far
more likely to be soluble in a nonpolar solvent (or in a solvent that is less polar than water).
C. Graham Brittain
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An ionic compound (with FULL positively and FULL negatively-charged ions) will generally dissolve in a polar
solvent such as water (that has partial positive and partial negative charges).
The figure to the right shows that when
ionic compounds dissolve in water, the
individual cations and anions are dislodged
from the crystal lattice and ―hydrated” by
the polar water molecules.
The cations will be surrounded by water
molecules (oriented so the partial negative
oxygen atom is interacting with the positive
charge of the solute cation).
And the anions will be surrounded by water
molecules (oriented so the partial positive
hydrogen atoms are interacting with the
negative charge of the solute anion).
Since the cations and anions are no longer
locked into their crystalline lattice, they are
mobile and can move freely from one part
of a solution to the other. For this reason,
solutions of ionic compounds can conduct
electricity.
Electrolytes are solutes which can form
ions when dissolved in water and conduct
an electric current.
Solutes that cannot ionize and conduct an electric current are called nonelectrolytes.
For some ionic compounds, the attraction of the ions for one another is greater than the attraction for the partial
charges of the water molecules; thus the ions are unable to break free from the structure of the crystal lattice.
These ionic compounds are insoluble in water.
Procedure that you will follow
1. Obtain a test tube rack and four clean 6‖ test tubes. Label each test tube as follows:
 Water
 Ethanol
 Acetone
 Hexane
2. Place approximately 0.1 gram of the first solute (listed in the observation table on the Report Sheet) into each
test tube. (Your lab instructor will weigh 0.10 grams as a demonstration of the appropriate amount of solute;
use your spatula to estimate the 0.1 gram-sized sample.) Identify the solute as either an ionic or a molecular
substance.
3. Add 4.0 mL of the appropriate solvent to the labeled test tube. (You can use your graduated cylinder to add 4.0
mL of water to the first tube, then fill the other test tubes to approximately the same level.) Mix each tube
thoroughly. Sharp tapping of the test tube with your fingers or stirring with a glass rod will agitate the
contents.
C. Graham Brittain
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4. Record your observations of the solubility of the solid in each solvent: completely soluble, slightly soluble, or
insoluble.
Note any temperature changes as the solute dissolves.
Also note any color changes, remembering that the intensity of the solution’s color is a function of the amount
of solute dissolved.
For insoluble solutes, note whether the solute is more dense (sinks to the bottom) or less dense (floats on top)
than the solvent.
5. After completing all observations, empty the tubes into a large waste beaker at your lab bench. Rinse each tube
with a small amount of the specified solvent to prepare it for use with the next solute listed in the observation
table.
6. When all solutes have been tested, rinse each tube with solvent, and then with soap and water, always emptying
the contents into the large waste beaker at your lab station. When the waste beaker fills, take it to the hood and
empty it into the appropriate liquid waste container.
C. Graham Brittain
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Report Sheet 9: Solubilities of Ionic and Molecular Substances
Student ______________________________ Lab Partner__________________________ Date Lab Performed__________
Section #_________ Lab Instructor__________________________________________ Date Report Received ___________
Lab Notebook: Data and Observations
Solute
Name &
Chemical
Formula
Ionic
or
Covalent?
Solvent
Water
Ethanol
Acetone
Hexane
H2 O
CH3CH2OH
CH3COCH3
CH3(CH2)4CH3
Copper (II)
Sulfate
CuSO4
Sucrose
C12H22O11
Aspirin
C9H8O4
Naphthalene
C10H8
C. Graham Brittain
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Formal Report: Results and Conclusions
The Lewis structures for the solvents ethanol, acetone, and hexane are shown below.
1. For each structure, draw circles around the polar and nonpolar regions of the molecule, and label EACH circled
region with the type of intermolecular attractive force that it is capable of using.
Ethanol
H
H C
H
H
C O H
H
Acetone
Hexane
O
H
H C
H
H
H
C
C H
H C
H
H
H H
C C
H H
H H
C C
H H
H
C H
H
2. Draw the Lewis structure of the fourth solvent you used (a water molecule), including all covalent bonds and all
lone pairs. Label the + and - regions of the molecule. Which type of intermolecular attractive force is the water
molecule capable of using?
3. Rank the four solvents (ethanol, acetone, hexane, and water) according to their degree of polarity, with #1 being
most polar, and #4 being least polar).
#1: ______________________________
#2: ______________________________
#3: ______________________________
#4: ______________________________
4. Does the ―like dissolves like‖ rule account for your observations of Copper (II) Sulfate solubility in each
solvent? Compare and EXPLAIN the solubility of Copper (II) Sulfate in EACH solvent.
Water:
Ethanol:
Acetone:
Hexane:
C. Graham Brittain
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5. The structure of the solute Sucrose is
shown to the right. In this ―condensed‖ ring
structure, it’s understood that there’s a carbon
atom at each corner of the ring (except where
an oxygen atom is shown).
C H2O H
O
H
H
OH
H
O
H
OH
H
H
O
HO
Draw circles around the various polar and
nonpolar regions of the molecule, and LABEL
each region with the TYPE of attractive force
it can use to interact with other molecules.
C H2O H
H
C H2O H
HO
OH
H
Sucrose
Does the ―like dissolves like‖ rule account for your observations of Sucrose solubility in each solvent? Compare
and EXPLAIN the solubility of Sucrose in EACH solvent.
Ethanol:
Acetone:
Hexane:
6. The structure of the solute Aspirin is shown to the right. Draw
circles around the various polar and nonpolar regions of the
molecule, and LABEL each region with the TYPE of attractive
force it can use to interact with other molecules.
H
H
C
C
Does the ―like dissolves like‖ rule account for your observations
of Aspirin solubility in each solvent? Compare and EXPLAIN the
solubility of Aspirin in EACH solvent.
Water:
Ethanol:
Acetone:
Hexane:
C. Graham Brittain
Page 9 of 10
O
Water:
H
C
C
H
C
C
C
O
O
H
C
C
O
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H
H
H
7. The structure of the solute Napthalene is shown to the right. Draw
circles around the various polar and nonpolar regions of the molecule,
and LABEL each region with the TYPE of attractive force it can use to
interact with other molecules.
H
C
C
H
H
H
C
C
C
Does the ―like dissolves like‖ rule account for your observations of
Napthalene solubility in each solvent? Compare and EXPLAIN the
solubility of Napthalene in EACH solvent.
Water:
Ethanol:
Acetone:
Hexane:
8. The concentration of a solution (amount of solute/amount of solution) is
often expressed with “molarity” units:
C
C
a) Illustrate how ascorbic acid molecules dissolve in water by drawing the
structures of at least FOUR water molecules, labeling the + and - regions
of each, and sketching the way in which they interact with the ascorbic acid
molecule to the right.
C
C
C
H
H
H2C
H
O
HC
O
HC
C
C
H
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O
C
O
O
b) Use the definition of molarity, the molar mass of ascorbic acid, and the
solubility information above to calculate the concentration of this solution
in molarity units (moles ascorbic acid/ liter solution).
C. Graham Brittain
H
H
O
1.00 gram of Vitamin C (ascorbic acid, C6H8O6, structure shown to the
right) can dissolve in water to make 3.00 mL of solution.
H
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H