Pure substances
Chapter 13
We’ve seen that the physical
properties of a pure
substance are dependent on
the strength of the
intermolecular forces
between the molecules
Solutions and Their Physical Properties
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1050.php
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Solutions
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Solutions
Solutions are “homogenous
mixtures” of two or more pure
substances. The substance
found in greater amounts is
usually referred to as the
SOLVENT, while all other
substances in the solution are
considered to be the SOLUTE(S).
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also known as alloys
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Concentrated vs. dilute
Solution concentrations
There are many different ways to measure
the concentration of a solute in a solution.
What you have likely seen defined as
“concentration” is actually only one of
these different ways, and the correct term
for this “concentration” is the molarity,
which is the number of moles of solute
divided by the solution volume (mol L-1)
molarity (M) = nsolute / Vsolution
Concentrated solutions have
relatively large quantities of
the solute(s) dissolved in the
solvent.
Dilute solutions have relatively
small concentrations of the
solute(s) in the solvent.
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Parts per million, billion, etc.
Mass, volume and mass/volume percent
Mass percent, volume percent and
mass/volume percent are fairly straightforward measures of concentration with
“units” of percent (%):
mass percent = msolute / msolution x 100%
volume percent = Vsolute / Vsolution x 100%
mass/volume percent =
msolute / Vsolution x 100%
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If we have very dilute solutions,
the mass or volume percent is very
low, so it sometimes makes sense
to express these concentrations in
terms of how many “parts” of the
solution are “solute” out of a given
number of total “parts” of the
solution.
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Parts per million, billion, etc.
Parts per million, billion, etc.
In very dilute aqueous solutions, the
density of the solution is essentially the
same as pure water, so a liter of solution
has a mass of 1000 grams.
So, a one parts per million (ppm)
aqueous solution is the equivalent of
1 g of solute per 1,000,000 g of solution
or a mass percent of 0.0001 %
or a mass/volume percent of 0.001%
Therefore a mass percentage
or a volume percentage tells
us how many “parts per
hundred” of the solution are
solute, measured in terms of
either mass or volume.
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Parts per million, billion, etc.
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Parts per million, billion, etc.
Also, a one parts per billion (ppb) aqueous
solution is the equivalent of
1 g of solute per 1,000,000,000 g of solution
or a mass percent of 0.0000001 %
A one parts per trillion (ppt) aqueous solution
is the equivalent of
1 g of solute per 1,000,000,000,000 g of
solution
or a mass percent of 0.0000000001 %
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However, it’s likely easiest to remember
these units relative to the mass of solute
per liter of aqueous solution
1 ppm = 1 mg L-1
1 ppb = 1 µg L-1
1ppt = 1 ng L-1
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Mole fraction and mole percent
Mole fraction and mole percent
The mole fraction xi of solute i is the
number of moles of solute divided by the
total number of moles of ALL substances
in the solution.
The mole percent of solute i is the mole
fraction expressed as a percentage
mole percent = xi x 100%
xi = ni / ntotal
Mole fractions have no units, and the sum
of the mole fractions for every substance in
the solution MUST add up to ONE.
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Molality
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Problem
An 11.3 mL sample of methanol (CH3OH –
molar mass of 32.042 g mol-1 and density of
0.793 g mL-1) is dissolved in enough water
(molar mass of 18.015 g mol-1) to produce
75.0 mL of a solution with density of 0.980 g
mL-1.
Express the concentration of this solution as
a) the mole fraction of water b) the molarity of
methanol and c) the molality of methanol.
The molalilty m of a solution is the number
of moles of the solute divided by the mass
of the SOLVENT (not solution!) in
kilograms
m = nsolute / masssolvent
Units will be mol kg-1
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Problem answer
Problem
mole fraction of water is 0.928
[CH3OH] (molarity) is 3.73 mol L-1
molality of CH3OH is 4.33 mol kg-1
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A 10.00% (by mass) aqueous solution of
sucrose C12H22O11 (molar mass 342.2956
g mol-1) has a density of 1.040 g mL-1.
What is the a) molarity b) molality and c)
the mole fraction of sucrose?
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Problem answer
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Enthalpy of solution
The energy change that occurs
upon the mixing of two
substances is often referred to as
the enthalpy of solution ∆Hsoln.
This enthalpy of solution can be
broken down into individual
contributions, each with their own
enthalpy change.
molarity is 0.3039 mol L-1
molality is 0.3246 mol kg-1
xsucrose is 0.005814
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Enthalpy of solution
Lastly, the mixed solvent
and solute molecules
will have “new”
averaged IMFs between
them that will bring the
molecules together,
releasing energy ∆Hc
First, we would have to take the pure
solvent molecules and “move them apart”
against their attractive IMFs to make room
for the solute molecules. This will take an
energy input of ∆Ha.
Next we have to move the solute
molecules apart against their IMFs to
make room for the solvent. This will take
an energy input of ∆Hb.
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∆Hsoln =
∆Ha + ∆Hb + ∆Hc
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Solutions
|∆Hc| > ∆Ha + ∆Hb then
enthalpy of solution is
exothermic
|∆Hc| < ∆Ha + ∆Hb then
enthalpy of solution is
endothermic
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Solutions and IMFs
Solutes and solvents only tend to
mix (are MISCIBLE) when the
intermolecular forces of the
solvent and solute(s) are of
similar strengths.
Oil (London forces) and
water (hydrogen bonds)
DON’T MIX, and are said
to be IMMISCIBLE.
“Like dissolves like”
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If molecules A and B have
large differences in
intermolecular force
strengths, then a molecule
of A will be “sucked
back” into pure A instead
of mixing with pure B.
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Solutions and IMFs
Solutions and IMFs
1)
If molecules A and B
have similar
intermolecular force
strengths, then a
molecule of A can mix
with pure B.
Alcohols (hydrogen
bonds) and water
(hydrogen bonds)
DO MIX!
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Solutions and IMFs
2)
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Solutions and IMFs
If the solvent and solute molecules have
stronger IMFs with each other than the pure
substances, they WILL mix with an
exothermic enthalpy of solution, and we
would have a nonideal solution.
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If the solute and
solvent had exactly
the same strength
and type of IMFs
then the enthalpy of
solution would be
zero since the IMFs
in in the mixture
would be the same.
We often call this
an ideal solution.
If the solvent and solute molecules have
weaker IMFs with each other than the
pure substances, they MAY mix with an
endothermic enthalpy of solution. This
would also be a nonideal solution.
4) If the solvent and solute molecules have
MUCH weaker IMFs with each other
than in the pure substance, mixing WILL
NOT occur.
3)
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Problem
Forming ionic solutions
Refer to Practice Example 13-3 A in the
text.
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Ionic solids can sometimes dissociate into
aqueous ions in solution. This would
happen due to the ion-dipole forces
between the ions and the water molecules.
However, for this to happen, the energy
that is released by taking the gas phase
ions an putting them into water (the
hydration energy) MUST be similar to or
greater than the negative lattice energy.
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Forming ionic solutions
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Forming ionic solutions
More specifically, if the sum of the
hydration energies is greater than the
negative of the lattice energy, the ionic
solid WILL dissociate into ions in solution
and the dissolution will be exothermic.
If we have an endothermic enthalpy of
solution, then there is another factor called
entropy (which you will see in Chem 1051)
that plays a role in spontaneous mixing.
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Solutions and equilibrium
Solutions and equilibrium
At some point, the rate of these two opposing
processes becomes the same and no more solid
APPEARS to dissolve. This is a dynamic
equilibrium state. The concentration of the ions
or molecules in solution no longer change and
we call this solution saturated.
Molecules or ions leave a solid to enter the
solution. However, nothing says the opposite
process where molecules or ions leave the
solution to become the solid can’t also happen.
In fact, both processes WILL happen!
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Solubility
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Solubility
This graph shows the
solubility (in grams solute
per 100 grams of solution)
of several solids as a
function of temperature.
We see that the solubility
often changes with
temperature. For some
solids, solubility increases
with T, but for some solids
it decreases!
The concentration (which
can be measured many
ways!) of the molecules
or ions in the saturated
solution depend on the
amount of solid that
dissolved in a given
amount of solution. We
call such measures of
concentrations the
solubility of the solid.
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Solubility
Solubility
If we create a solution with a
concentration that falls
above the solubility curve
we have an supersaturated
solution which has “too
many” molecules or ions in
it for that given temperature.
We could then begin to see
solid form until the
concentration is reduced to
the solubility curve value at
that T.
The solubility curves
represent the saturated
solution at a given
temperature, and divides
the graph into two
areas. If we create a
solution with a
concentration that falls
under the solubility curve
we have an unsaturated
solution into which we
could dissolve more solid.
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Solubility
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Gases in solvents
It turns out whether
solubility increases or
decreases with temperature
depends on whether the
enthalpy of solution is
endothermic or exothermic.
If its endothermic, then
raising the T increases
solubility. If its exothermic,
then raising T decreases
solubility. You’ll see why in
Chem 1051!
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Gases can often dissolve in liquids.
Whether they do is governed by the
changes in IMFs, just like in liquid-liquid
solution.
Usually gases are only slightly soluble in
water, because the IMFs in many gas
molecules tend to be much weaker than
the hydrogen bonds of water. They tend to
be much more soluble in organic solvents.
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Gases in solvents
Gases in solvents
Gas solubility also depends on
temperature.
As we increase the external
pressure, more gas molecules
are “forced” into the solution.
When we lower the pressure
(like opening a pop bottle) these
molecules are now free to
“escape the solution” and we
see bubbles of gas form in the
solution.
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Gas solubility also depends
on the external pressure.
Most of the time gas solubility
IN WATER tends to
DECREASE with
temperature. This is why
when we put a pot of water on
to boil, we often see small
bubbles of dissolved gas
(NOT water vapor!) appear
before boiling occurs.
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Gases in solvents
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Gases in solvents
C = k x Pgas
William Henry showed that the increase of gas
solubility with pressure follows a simples
relationship, which we now call Henry’s Law. The
solubility C is directly proportional to the pressure
of the gas. The amount the solubility is increased
with pressure depends on Henry’s constant k.
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Henry’s constant k is a function of both the
nature of the gas AND the nature of the
solvent, since solubility is ultimately
determined by the IMFs of both!
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Problem
Problem answer
A handbook lists the solubility of carbon
monoxide in water at 0 °C and 1 atm of
pressure as 0.0354 mL CO per mL of
water. What pressure of CO must be
maintained over water to obtain a solution
molarity of 0.0100 mol L-1?
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The solubility of CO in water at 1 atm and
273 K will lead to a concentration of
0.00158 mol L-1, so Henry’s constant for
CO in water at 273 K can be expressed as
k = 0.00158 mol L-1 atm-1
To get a concentration of 0.0100 mol L-1
then requires the pressure of CO to be
6.33 atm.
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Failures of Henry’s Law
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Physical properties of solutions
Even though “like dissolves like” the
presence of solute molecules in the
solvent disrupts the bulk solvent
intermolecular forces to some extent.
Henry’s Law fails at very high pressures
because the gas can no longer be treated
as ideal. The IMFs of the gas phase will
make a difference!
Also Henry’s Law fails if the gas reacts
with water or ionizes in solution. Henry’s
Law specifically applies to the gas
molecules being dissolved in the solvent.
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This change in forces means that the
physical properties of solutions (like
vapor pressure, freezing and boiling
points) are slightly different than those
of the pure solvent.
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Physical properties of solutions
The disruption of IMFs generally
depends more on the amount of solute
(in terms of concentration) rather than
the chemical identity of the solute.
Such colligative (“tied together”)
properties of solutions therefore depend
on the concentration of the solution
while ignoring the identity of the solute.
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