Mr. Bartelt Presents
Solutions and their properties
Solutions part 2
Expressions of concentration

Molarity—we already know this one.
Sadly, there are many other ways
to express concentration of
solutions.
1. Mass Percent
2. Mole Fraction
3. Molality
Mass percent
 As
the name suggests this is mass of
solute (what’s being dissolved) over
mass solution (whole thing)
 Commonly used in the place of
molarity for stuff you can buy at WalMart
– Vinegar (acetic acid and water)
– Hydrogen peroxide (H2O2 and water)
Examples

1.
2.
Find the molarity of a 5% solution
of acetic acid (CH3COOH) in water.
This is the vinegar you buy at the
store. (assume D= 1.00 g/ml)
Assume a solution mass of 100. g
Now you know that you’ve got 5.0
g of acetic acid and 100. ml of
water.
Example

M=mol/Volume
–
–
–
–
–
Volume = 100. ml or 0.100 L
mol= g/MM
We have 5 grams of acetic acid
Acetic acid’s MM is 60.05 g/mol
5.0 g / 60.05 g/mol= 0.0833 mol
0.0833
M
 0.833 M acetic acid
0.100
Next example



You can buy 3.00 % Hydrogen
peroxide at the store.
I have 1.00 M hydrogen peroxide in
the lab.
Which is more concentrated?
Mole fraction



This is similar to what we looked at in
our unit on the gas laws.
Na
Mole fraction 
Na  Nb
The above equation assumes a solution
of only two components. If there were
more then the denominator would have
additional terms
Example
What is the mole fraction if you mixed
equal masses of water and ethylene glycol
[C2H4(OH)2]?
 Assume 100. grams of each

100.g *
1mol
 5.56 mol water
18.0g
1 mol
100. g *
 1.61mol ethylene glycol
62.07 g
Mole fraction 
1.61
1.61

 0.162
1.61  5.56 7.17
Mole fraction example



Concentrated sulfuric acid is 18.0 M
H2SO4. What is the mole fraction water
in 18.0 M sulfuric acid. Assume density
is the same as sulfuric acid 1.84 g/ml.
Assume 1.00 liter
Now we have 18.0 mol H2SO4
mass H2SO4  18.0 mol * 98.08 g/mol  1770g
1840 g  1770 g  70. g water *
1 mol
 3.89 mol water
18.0 g
3.89 mol water
 0.178
18.0 mol H 2SO 4  3.89 mol water
You try
 Based
on the calculations above,
determine the mass percent of
concentrated sulfuric acid.
Molality
At this point it may seem like I’m just
making these names up.
 This is actually a very important expression
of concentration that will prove extremely
useful when we determine freezing point
depression and boiling point elevation in
lab on Wednesday.
 Molality is defined as moles solute over
kilograms of solvent.

n solute
Molality 
kg solvent
Example

What is the molality of battery acid?
–

Assume 100.g of solution Hence
–

Battery acid is sulfuric acid and water with a
mass % of 33.5% acid.
33.5 g H2SO4 and 66.5 g water
I now need
–
–
moles H2SO4 (easy)
Kilograms water (easy)
n H 2SO 4
33.5 g

 0.342 mol sulfuric acid
98.08 g/mol
Example

I now need
–
–
moles H2SO4 (easy)
Kilograms water (easy)
n H 2SO 4
33.5 g

 0.342 mol sulfuric acid
98.08 g/mol
kg H 2O
66.5g

 0.0665 kg water
1000g/kg
0.342 mol
m
 5.14 m
0.0665 kg
Note
 These
are terms that need to be
memorized
 Molarity (M) = moles/liter
 Mass % = masssample/masstotal
 Mole fraction = nsample/ntotal
 Molality (m) = nsample/kgsovent
– You need to work many problems to get
this down
Like dissolves like


We have not discussed Lewis dot structures
yet but we need to discuss the fundamental
principal of solubility.
That priciple is “like dissolves like”
–
–


This means that polar solvents are good at
dissolving polar molecules and ionic compounds
Additionally, non-polar solvents are good at
dissolving non-polar molecules.
We will investigate this in a quick activity
But first the structures
Entropy
 Let’s
look at water and hexane (C6H14)
Water has O-H bonds
which are highly polar
Hexane has C-H bonds
which are non-polar
Determining “likes”
 You
don’t really need to know how to
draw the Lewis structure to
determine if a substance is polar or
non-polar.
 A good rule of thumb is
– if your molecule is exclusively carbon
and hydrogen it’s non-polar
– if it contains carbon, hydrogen, and
nitrogen or oxygen as well, it’s polar.
Determining “likes”
 There
are countless examples
contrary to this rule, but you can’t be
expected to know them so you
should be fine.
 We will look at something you should
be able to figure out now however.
 If the molecule is extremely large
and only contains one oxygen or
nitrogen the region of polarity will
not be large enough to make it polar.
Polar “tails”

Fatty acids are long chains of hydrocarbons with
polar “tails” on the ends.
The gray and black
spheres are hydrogen
and carbon
respectively. The red
spheres are oxygen.
NOTE: The size of the
non-polar “tail”
dominates the polarity
of the molecule and
hence the molecule is
considered non-polar.
Hence oil and water
don’t mix.
Another notable exception
 CO2
has 2 carbon oxygen bonds
which is highly polar, but they
happen to be exactly opposite each
other in the molecule.
 Hence they “cancel” each other’s
polarity making the molecule nonpolar.
Polar or non-polar
 CH3OH
(methanol)
Polar or non-polar
 CH4
(methane)
Polar or non-polar
 Carbon
tetrachloride
Polar or non-polar
 C12H22O11
(sugar)
Polar or non-polar
 Any
ionic compound
Polar or non-polar
 Graphite
(carbon)
Now we’re ready
Today’s lab will focus on the principle
of “like dissolves like”.
You will note this in your lab and
answer questions that relate to the
topic of like dissolves like.
Henry’s law
We will not do any calculations with
Henry’s law in this class.
 Henry’s law states that the concentration
of dissolved gas in a solution is directly
proportional to the partial pressure of the
gas above the solution.
 This explains how carbonation works.
 When brewing beer yeast breaks sugar
down into alcohol and carbon dioxide. If
the bottle is capped and the CO2 is
trapped and builds up pressure.
 This pressure drives CO2 into the solution
and makes the beer carbonated.

Temperature effects on solutions
 Typically
an increase in temperature
will result in greater solubility of
solids, but not always.
 The opposite is true of gasses. The
higher the temperature, the less gas
can be maintained in solution.
 Click here if you must know more.
Why do things dissolve?
 Disorder!!!
If something is dissolved
in something else 
 The dissolution process makes a
more disordered system 
 Now we need to look at Le
Chatelier’s Principle
Le Chatelier’s Principle
 Le
Chatelier’s Principle applies to systems
that are at equilibrium (ΔG=0)
 We’ll look at boiler scale in great detail to
learn this new principle.
Equilibrium systems
 CaCO3
is “insoluble”
– This doesn’t mean that NO CaCO3 will
be in solution, it just means that very
little will be in solution.
– But still, a little is dissolved
Ca2+ + CO32-  CaCO3
 If water contains Ca2+ ions (and
nearly all tap water does) it will bond
with any CO32- present and form
limestone in a pipe. Not good.
More systems
If water contains carbonate ions this reaction
will also take place
 CO32-(aq) + CO2(aq) + H2O(l)  2HCO3 What happens to the solubility of CO2 at
higher temperatures?
 If CO2 is removed from our system, the
equation will shift to the left to compensate
for the loss of CO2.
 Now we have extra CO32- and this reaction
proceeds
Ca2+ + CO32-  CaCO3

Le Chatelier's Principle

If a chemical system at equilibrium experiences a
change in concentration, temperature, volume, or
total pressure, then the equilibrium shifts to
partially counter-act the imposed change.
CO32-(aq) + CO2(aq) + H2O(l)  2HCO3Ca2+ + CO32-  CaCO3
If T goes up  CO2 goes down  CO32and H2O increase to compensate. The
addition of CO32- then drives the
production of more CaCO3 and your pipes
get clogged up.
Colligative properties
Colligative properties are properties of
solutions that depend on the number of
particles in a given volume of solvent and
not on the mass of the particles.
 We can use colligative properties to
determine the molar mass of substances.
 This is very useful.
 We’re going to look specifically at:

– Boiling point elevation
– Freezing point depression
You can change the freezing point?
 Of
course you can. Ever added salt
to snow to make it melt?
 Probably not because you only see
snow twice a decade, but back in my
homeland of Chicago it’s done every
winter.
 The salt is added to increase the
freezing point of the water and hence
melts the ice.
Freezing point depression
 ΔT
= Kfmsolution
m=molality
 ΔT
stands for the change in freezing
point.
 Kf is a substance specific constant
 We’re only doing this for water so
just memorize that
– Kf = 1.86 (ºC•kg/mol) for freezing point
– Kb = 0.51 (ºC•kg/mol) for boiling point
Examples
 What
freezing point would you
expect from a solution of 120.0 ml of
water and 25.0 g of sugar
(C12H22O11)
 ΔT = Kfmsolution
 First find molality of solution
– Need moles sugar
– Need kg water
Examples

ΔT = Kfmsolution
– Need moles sugar
– Need kg water 120. ml 0.120 kg water
25.0g
mol sugar 
 0.0731 mol
342 g/mol

ΔT = Kfmsolution
– 0.0731 mol sugar
– 0.120 kg water
0.0731 mol
mol
m
 0.609
0.120 kg
kg
ΔT = (1.86 ºC•kg/mol)(0.609mol/kg)
 ΔT = 1.13 ºC
 The freezing point will drop by 1.13 ºC

Boiling point
These problems are the exact same except
the Kf is now a Kb. The b stands for boiling.
 ΔT = Kbmsolution Kb = 0.51 (ºC•kg/mol)
 What boiling point elevation would you
expect from the addition of 20.0 grams of
NaCl to 200.0 ml of water?

Answer you’ll get:
ΔT = 2.62 ºC
But this answer is wrong
Why’s it wrong
When you dissolve sugar, the molecule
doesn’t break up. The moles of sugar are
the moles dissolved.
 When you dissolve salt, you produce both
Na+ and Cl- ions.
 The BPE and FPD are affected by the # of
particles, and thus when one mole of
NaCl is dissolved an mole of Na+ and Clions are produced for a total of 2 moles of
particles in solution.

New equation
ΔT = Kb•i•msolution
 i in this case would be 2
– For Na2SO4 it would be 3
– For Al(NO3)3 it would be 4
 What
boiling point elevation would you
expect from the addition of 20.0 grams
of NaCl to 200.0 ml of water?
We’ll confirm this in lab
 We’re
going to dissolve 20.0 g of salt
in 200.0 ml of water in lab tomorrow
and determine the freezing point.
 Now you know what the freezing
point should be.
 But this is only the “calculated”
freezing point, we’ll see what the
experimental freezing point really is.
But why does the FP depress?
 Think
back to Shiva
 What has more disorder, salt water
or regular water?