[MUSIC]
Lighthouse Scientific
Education presents a lecture
in the Solutions
and Water series
The topic; Concentration
and Dilution.
The lecture can be viewed in
its entirety or searched for
particular concentration
scales or dilution problems.
Material in this lectures
relies on an understanding
of the previous lectures;
Solutions and Solubility
and Molarity, Molality
and Mole Fraction.
The lecture begins with
some basic definitions
and accompanying equations.
Some of the more common
concentration scales or
ratios are discussed including
percent concentration.
That is a mass to volume ratio.
Percent concentration:
mass to mass.
Percent concentration,
a volume to volume ratio
and parts per
million or billion.
Those are extended
into concentration
as a conversion factor.
Example problems of
conversions are offered as well
as an example of preparing
a solution in a lab setting.
The second part of the lecture
is given over to dilutions.
Dilution can be summed
up with the equation
C1V1 equals C2V2
This is shown with
example dilution problems.
An all important topic in
chemistry is concentration.
We will focus the discussion
in this lecture on the
concentration of solutions
with a liquid solvent.
Some extension into gas
solutions can be made from
the discussion. Let's
start with a review of solutions.
Some compounds dissolve,
or dissociate in liquids
and form solutions.
The dissolved compound
is called the solute
and the dissolving
liquid is the solvent.
The amount of solute
per volume of solution,
or in some cases the
amount of solvent,
is called the concentration.
Concentration is a
ratio or a fraction.
The numerator is always
the amount of solute
but depending on the scale
the denominator is usually
the amount of
solution or solvent.
The concentration scales
covered in this lecture will
differ by the units for the
solute, solvent and solution.
There are a couple
qualitative descriptions most
people are aware of.
A dilute solution is where
there is a small
amount of solute.
And a concentrated solution
is where there is a large
amount of solute.
Dilute and concentrated
are usually condition
dependent and not absolutes.
Giving some values to
concentration, if 3.5 g of NaCl
is dissolved in 1.0 liters of
water then the concentration
can be described as;
the amount of solute, 3.5 grams
over the amount of
solvent, 1 liter.
An all important
note can be made here.
If the amount is solute is
small compared to the amount
of solvent then the volume
of solution is approximately
equal to the volume
of the solvent.
This only works with
very dilute solutions
but it allows us to write a
second ratio with volume of
the solution in
the denomination.
These two concentrations
are approximately equal since
this is a very dilute solution.
As for units; this concentration
scale is grams over liters.
It is a mass (or
weight) to volume scale.
Concentration is a
3 variable equation.
In the previous example
those 3 variables were
concentration, mass or
weight of solute and volume
of solution or solvent.
If we stop to think about,
we've seen a similar
set of 3 variables.
It is the density with
its mass to volume ratio.
The variable 'D' equals
the variable 'm' over the
variable 'v'.
A direct comparison of
their equations shows
just how similar they are.
The point here is that the
study of concentration can be
built upon the study of density.
Concentration problems
using this equation can be
solved in the same manner
as the 3 variable
density problems.
Later on in the lecture we
will explore this process
in more detail.
With the basic form of
concentration scale in hand we
can move into specific scales.
Notice that the definition
of concentration does not
explicitly say what the
units are for the solute,
solvent or solution.
That is where the different
concentration scales vary.
We have already covered
concentration scales
that use moles as the
unit of the solute.
There are 3 common mole
based concentration scales
and they are discussed in detail
in the 'Molarity, Molality
and Mole Fraction'
lecture in the Mole series.
We will briefly
review them here.
The first and most important
is molarity; units given
as capital M.
This scale is the number
of moles of solute dissolved
in 1 liter of solution.
Capital M means
moles per liter.
A second and less frequent
used scale is molality;
units given as small m.
It is the number of
moles of solute per
kilogram of solvent.
Little m means mole per kg.
Molality plays a prominent
role in the 'Colligative
Properties' lecture.
The final scale is mole
fraction; its unit is the
lower Greek letter chi.
Mole fraction is
the moles of solute over
the total moles of solution.
That is a much as we are
going to cover in this lecture
of the mole based scales.
What we will cover in depth
are scales in which the amount
of solute, solvent and solution
are given in mass or volume.
The most common of these
types of concentrations is the
percent concentration scales.
3 versions will
be presented here.
The first is percent
concentration with the ratio of
mass or weight
(grams) of solute per
volume (ml) of solution;
Since it is a percent
the ratio of weight per
volume is multiple by 100.
A second way percent
concentration is given is as
mass or weight (grams) of
solute per mass or weight
(grams or kilograms)
of solution.
It is a weight per weight scale
and also multiplied by 100.
A final way to present percent
concentration is volume (ml)
of solute per volume
(ml) of solution.
It is a volume per
volume scale and is also
multiplied by 100.
There are 3 representations
of percent
concentration because
there are 3 combinations
of weight or volume
for solute and solution.
The one new scale covered here
that is not a percent
concentration is
parts per million
or parts per billion.
Often these scales are
mass (grams) of solute per
mass (grams) of solution.
Like the percent concentration
this ratio is multiplied
by a large number.
These numbers are, however,
are very large; 10 to the 6th,
1 million, or 10 to
the 9th, 1 billion.
Let's spend some time
with each one of the 4 scales
and show some
sample calculations.
Starting with the first percent
concentration scale: mass
or weight to volume.
Often weight to volume is
designated by
'w over v':
This calculation of percent
concentration will have
the mass or weight
of the solute
divided by the
volume of the solution
and multiple by
100 to get percent.
An example of such a solution
is the 5% (w/v) NaCl.
The solvent of water
is implied here.
A couple of example
problems of calculating
this type of concentration
will be helpful.
Keeping the concentration
formula handy,
what is the percent
concentration
weight to volume) of a 100
ml solution that contains
10 g of NaCl?
Do we know if this is a
concentration problem?
Well, it is a
concentration problem
and it looks like it is the
percent concentration with
'a solid solute dissolved
in a liquid solvent'.
It is a mass or weight
to volume concentration.
The solvent type is not
mentioned here so it is
implied to be water.
Inserting the problem values
into the percent
concentration w to v equation
has a weight of 10 grams for
the solute sodium chloride
and a volume of 100
ml for the solution.
The equation is complete when
the ratio is multiplied by 100.
That is because this is
a percent concentration
and percent means per 100.
The math comes to
a 10 percent weight to
volume NaCl solution.
Is the percent sign required
in the concentration?
Yes is it. It lets a person
know that the 10 per 1 hundred.
is really a 1/10 solution.
Since we have this problem
up and solved we can
tweak the problem to cover
a variation that comes up
regarding units which will
require an additional step.
Say that instead of being
given 100 ml of solution the
problem offered 0.100 liters.
The problem can be solved the
exact same way if liters are
converted into milliliters.
The given units are
0.100 liters and using the
metric conversion into
ml oriented to cancel
liters, will produce the
desired units of ml. 100 of them.
Yes this is same value as
the original problem so the
conversion will be the same
10 percent weight to volume
NaCl solution. What
this tweak to the
original problem highlights,
aside from the need of a
metric conversion, is that if the
liter value is inserted directly
into the percent concentration
equation instead of
an equivalent milliliter value
a very different concertation
will come out from the math.
Even though the liter
and milliliter volumes
are the same. This given
definition of percent
concentration specifically
calls for milliliter and that is
what it should get.
A second practice problem
with this concentration scale
says that a solution is formed
by adding 25.0 g of ammonium
nitrate to water making a
final volume of 0.150 liters.
What is the percent
concentration
(weight to volume)?
In scouring the problem it can
be seen that the units of the
solute and solvent are weight
and volume respectively.
Just as the problem asks for
in the concentration scale.
However, the weight is given
in kg but the definition
calls for grams.
This requires a metric
conversion with the
conversion factor in
the proper orientation to
cancel out kg and to
convert 0.025 kg to 25 g.
Now, applying the definition.
25.0 g of ammonium nitrate
divided by 150 ml of
solution, times 100
is 16.7 % (weight to
volume) ammonium nitrate.
Onto the next scale and
it is percent concentration
(mass to mass).
It is usually called
weight to weight.
It is weight (in grams)
the previous scale.
It is weight (in grams)
of solute per weight
(in grams) of solution.
To get percent that ratio
is multiplied by 100.
Units are grams per gram.
The calculation for percent
concentration (weight per
weight) starts with the weight
of the solute over the weight
of the solution times 100.
This ratio can actually be
expressed in a manner that
is more conducive to
solution preparation.
Remember, the solution is
really the combination of
solute and solvent.
So the weight of the solution
is the sum of the weight of
the solute and the
weight of the solvent.
This is a more detailed
way of describing the scale.
Unless, of course, there is
more than 1 type of solute
in the solution.
While the addition of the
other solute makes the
equation look more difficult
all it really does is
change the denominator by
adding a second
solute since the weight
of the solution is the
weight of all components
in the solution. One
other point to make that
will help reduce the overall
confusion. Most of the
time liquids solvents
are measured by volume.
Maybe a graduated
cylinder is used.
However, with this
scale weight is needed.
There is a convenient way,
though, to still use volume
if the solvent is water and
its temperature is near room
temperature. That is
taking advantage of
the 1 to 1 relationship between
water's weight and volume.
1.00 ml of water
is 1.00 g of water.
This comes from the density
of water near room temperature.
It is 1 gram per ml. So 25
ml of water is 25 g of water.
1000 ml or 1.00 L of
water is 1000 g of water.
Ok, keeping this equation
for percent concentration,
weight to weight, let's
do an example calculation:
37.0 g of HCl is dissolved
in 63.0 g of water.
What is the percent
concentration
(weight to weight) of
HCl? An examination of the
problem shows a weight for
the solute and a weight for
the solvent. The equation
for percent concentration
(weight to weight) has the
weight of the solute over
the weight of the solution.
But we were not given the
weight or volume of the solution.
Or were we?
Isn't the solution in this
problem really just the sum of
the solute and the solvent?
Adding their weights together
gives a weight of
the solution as 100 g.
That value can be directly
added to the percent
concentration equation as
can the sum of the solute
and solvent. We will
use that description
of the weight of the solution
because it will be useful in
making a separate point
later in the problem.
Both ways work.
The final step in the set-up
of the equation is to multiply
by 100.
A bit of calculator work has
a final answer of 37.0 percent,
weight by weight, HCl.
This is our third concentration
calculation problem and
the process should start
to feel more comfortable.
One component that may
add confusion to a problem
like this is if the solvent
water is given in units of volume
instead of weight as asked
for in the concentration scale.
What is to be done?
Well fortunately we know a
direct relationship between
the volume of water and its
weight at room temperature.
At this temperature it is pretty
much a 1 to 1 relationship:
63 ml is equivalent
to 63 grams of water.
That was the gram amount
in the original problem.
It can be directly inserted
into the percent concentration
equation, as written,
and solved to the
same concentration.
Alright, one last
twist with this problem.
How would the concentration
of the HCl change if the HCl
and water contributions
are left as they are but
an additional solute was added.
5.00 grams of sodium iodide
will also be dissolved
in this solvent. We can
call the sodium iodide
solute 2. Earlier
we saw that an
additional component to
the solution predominantly
effects the denominator of the
equation because the solution
is the sum of it's
all components. Unless the
solutes have common
chemicals there will be no
change to the numerator.
Essentially this new problem
is solved the same way
as the previous one.
The equation begins with the
weight of the solute HCl.
If the question asked
for the concentration of
sodium iodide its
5.00 grams would need
to be added here instead.
The denominator
of the equation has the
weights of both solutes
and the solvent. The full
weight of the solution is
37.0+5.00+63.0
for a value 105 grams.
This weight can be substituted
into the equation
for ease of viewing.
Multiplying the ratio by
100 gives a concentration
of the HCl as 35.2%
weight to weight.
This concentration is less
than the concentration without
the additional
solute but that is to
be expected because
the volume of solution
is larger with the
addition of a second solute.
The last of the percent
concentration scales is the
volume to volume one.
That is the ml of solute
per ml of solution.
And percent says
multiple by 100.
A real world example of this
concentration scale is
with alcoholic beverages.
Wine is about 12% volume
per volume ethanol.
All alcohol containers will
provide this concentration
scale of ethyl alcohol.
Solution preparation using
this scale is a common
practice in chemistry labs.
There are a 2 ways to prepare
a solution using percent
concentration
(volume to volume).
They share the first step:
measure out the
volume of the solute.
The first method and
more common method is to
add solvent, to the solute,
until the final volume of
solution is met. A second
method is to separately
measure out the calculated
amount of solvent.
The solute and
solvent are combined.
The underlying principle
here is that the volume of
solution is the same as the
volume of solute plus the
volume of solvent. This is
the same argument that we
used with weight of
solution being the sum of the
weights of the solute and
solvent. The equation for
percent concentration (volume
to volume) has the volume of
the solute over the sum
of the volume of the solute
and solvent.
Multiplication by 100 follows.
For the more dilute
solutions, the solution volume
is the volume of the solute
plus volume of the solvent.
Sometimes, and often with
concentrated solutions,
the sum of the volume of the
solute plus the volume of the
solvent does not equal
the volume of the solution.
That is seen when 30 ml
of propanol is added to
70 ml of water.
The mixture does
not equal 100 ml.
It's an intermolecular
interaction thing and such
changes in volume cannot be
predicted in advance. To be
safe, for more concentrated
solutions, it's probably
best to change the denominator
of the ratio back to
volume of solution.
When solving textbook
problems using this
scale, in either form of
the denominator, will usually
be acceptable. In fact we
will take that first form
of the equation with
the summed volumes into
the example problem. 7.5
ml of isopropyl alcohol
(C3H7OH) is added
to 142.5 ml of water.
What is the percent
concentration
(volume to volume) of
the isopropyl alcohol?
This is the concentration
problem of the
volume to volume kind
because it is a liquid
solute dissolved
in a liquid solvent.
Calling up the concentration
equation, in the numerator
the solute volume is added
and in the denominator the
sum of the volumes of the
solute and solvent are placed.
Pulling the values of
the denominator out we see
that it solves to a
tidy value of 150 ml.
Placing the volume
of the solution in the
denominator and multiplying
by 100 gives a concentration
of 5.0 percent volume
to volume C3H7OH.
That is a relatively
dilute concentration.
This exact problem
could have been reworded
with the water added to 'final
volume of 150 ml' and it
would yield the exact
same concentration.
The volume of the solution can
be given out right or summed
from the solute and the solvent.
The last type of
concentration scale is the
parts per million (ppm) or
parts per billion (ppb).
These scales are most often
used in environmental
measurements including
those of the soil
and atmosphere.
They are is a mass to
mass scale with the solute
and solvent being given in
the same unit denomination.
Both in grams for example.
The scales are similar to
the percent concentration in
that the ratio of solute
to solution is multiplied by
a large number. For the
parts per million the ratio
is multiplied by a million (10
to the 6th) and for parts per
billion the ratio is multiplied
by a billion (10 to the 9th)
The equation for ppm and
ppb will look a lot like the
percent concentration in
the weight to weight scale.
The weight of the solute over
the weight of the solution.
times 1 million or billion
depending on the scale.
There is an important shortcut
in dealing with the very small
amounts of solute found
with the ppm and ppb scales.
When the solute is a tiny
fraction of the solution:
the weight of the solvent
is approximately equal to
the weight of the solution.
The equation can be modified
to acknowledge this without
much change in the actual
concentration value.
Keeping the concentration
equation and bringing up
an example problem:
0.0080 g of oxygen,
O2, (in the gas
phase) is dissolved in
1000 ml of H2O (liquid
phase) at 22 degree Celsius.
What is the ppm and ppb of O2?
There are two issues
regarding the solvent water.
The first is that it is given
in the volume units of ml but the
concentration scales want
weight in the same unit as
the solute; grams.
We've seen this before.
It is handled by making
the equivalent statement of
1000 ml of water equals 1000
grams of water at 22 degree C.
That is based on the density
of water at room temperature.
The second issue is that,
as written, the ppm and ppb
equation asks for the weight
of the solution and not the
weight of the solvent. That
too was addressed earlier.
Since the solute makes
up such a small part of
the weight of
the solution it is
acceptable for the weight
of the solvent to be
used in place of the
weight of the solution.
Now we are ready to
tackle this problem.
Beginning with the parts
per million calculation.
The weight of the oxygen
is in the numerator
and the weight of the
solvent is in the denominator.
Multiplying this ratio by 1
million gives a final value
of 8.0 ppm of oxygen. The
scale is an appropriate
one because the
numerical value is in
a comfortable range. Contrast
that with the parts per billion.
It has the same ratio but it is
multiplied by a larger number,
one billion.
It yields a value
of 8.0 times 10 to
the 3 ppb oxygen.
Given the two which one
would you rather deal with?
The choice of concertation
scales is driven by the ratio.
The smaller it is, the bigger
the number it is multiplied
to get a value in a
comfortable range;
between 1 and 10 if possible.
That is our coverage
of concentration scales.
As said earlier, concentration
is a 3 variable equation that
when given 2 of the variables
the third can be solved for.
That is concentration can
be used to move between mass
of the solutes and volume
of the solution (or the
solvent in some cases).
There is one point that needs
to be made clear before
proceeding to the set-up
for conversions. Only the ratio
of solute to solution is used
in the conversion.
If percent concentration
is given it needs to
be multiplied by 100 to
remove the percent.
A ppm value will need to be
multiplied by 1 million and
a ppb will need to be
multiplied by 1 billion in
order to just return the solute
to solution ratio. A very
similar description of
concentration as a
conversation factor is given
with the study of molarity in
the 'Molarity, Molality and
Mole Fraction' lecture.
Converting between moles
of solute and liters of
solution is probably the
most common of these
types of conversations.
Be sure to review these
transition in that lecture.
For demonstration purposes,
we will return to our basic
setup for conversion that has
current units for the given
units of a conversion and
desired units for the wanted
or asked for units.
If we are given values in
liters of solution and
asked for mass of solute
we can use the
concertation of the solution
as the conversion factor.
That is, the concentration as
written with solute
over solution.
Liters cancel out and
units of mass remain.
Flipping the problem around,
with the mass of solute as
as the given and volume of
solution as the asked for units
would require a flip of
the concentration scale.
Volume in the numerator and
mass in the denominator so
that it cancels out. This
should be a common conversion
setup for the student at
this point in their study.
Still, an example problem
couldn't hurt.
0.22 L of a 10 g/L magnesium
chloride solution has
how many grams of
magnesium chloride?
Is this a conversion problem?
It is because it is
asking for a different
way (different units)
to express the same value
of the same solution.
Begin the problem with the
set-up for the conversion.
The current or given units
is the 0.22 liters of solution
and the desired or
asked for units is the
mass of the solute.
Is there a known conversion
that relates the
liters of solution
with the amount of
solute in that volume?
Yes the concentration as
written in weight over volume.
Is the orientation of
the concentration correct?
Would flipping the fraction
be a better fit the set up?
No, this orientation allows
for the canceling out of
given units leaving the
desired units of mass.
A bit of calculator
works has that mass as
2.2 grams of
magnesium chloride.
It is given to 2 significant
figures because the
concentration scale is
only given to 2 sig. figs.
That was not too bad.
But what if the original
problem was not given
in liters of solution but
in milliliters of solution.
It is still the same overall
problem but there is a new
issue in that the concentration
scale and the given value
are not at the same magnitude.
To use the concentration
scale as a conversion
the given units will
need to be in liters.
What can be done about this?
A separate metric conversion
is needed. The 220 ml need
to be converted into
liters of solution.
That conversion factor is of
course 1 liter over 1000 ml
with ml in the denominator
so that given units cancel.
The math has the conversion
as 0.220 liters of solution.
And yes, it's the same
value as in the original problem
which means that same
conversion can be used
and the same 2.2 grams of
solute will be obtained.
To give addition perspective
to the conversion problem
a second example is given.
5.50 g of KMnO4
is added to an empty flask.
To make a 20.0 g/L potassium
permanganate solution
what volume should
the flask be filled to?
Is this another
conversion problem?
Yes, since what we are
asking for is relating the
amount of solute to
the amount of solution.
The set-up starts
with the given grams
of potassium permanganate
and asks for the
desired units of
volume of solution.
As with the previous example
the concentration serves
as the conversion factor.
Is this the correct orientation
of the concentration?
Do given units of
grams cancel out?
No they do not.
A flip of the
concertation is needed.
Grams of potassium
permanganate have to be in
the denominator to cancel out
with the given units of grams.
The division of 5.50 by 20.0
gives a volume of 0.275 liters.
Another way of
saying that is 275 ml.
Being able to make this
kind of transition between ml
and liters is a useful skill.
These types of conversion are
not only necessary for
homework and testing
problems they also come
up in a lab setting.
Students can be asked to
prepare a solution for a lab.
That solution will
have a set concentration
and a final volume.
For example the student might
be asked to prepare 0.20 L
of a 25g KCl per liter solution.
It's an aqueous solution.
How should they go about this?
Step 1 is to calculate the
appropriate mass of solute
from the concentration
and solution volume.
It is the only piece of
information not given and
will be solved for in the
same manner as the previous
conversion problems.
Given units of 0.20 L;
concentration oriented to
cancel liters yielding a value
of 5.0 grams of KCl. With
that piece of information
weigh out the calculated
mass of solute. (2.5 grams)
The second step is to add the
solute to an empty graduated
cylinder or volumetric flask.
Be sure that as much of the
weighed out solute as possible
finds its way into the vessel.
A stir bar can be added here if
the solution is on a stir plate.
Finally add solvent (water
in this case) until the
final volume is reached.
Stirring is advised.
This is a straight forward
process but focus is needed to
prevent errors that may or
may not be easily detected.
That completes our
concentration portion
of the lecture. The next
section deals with dilutions.
Solutions are not only
common in the lab they
are common in everyday life.
Dilution is the act of
reducing the concentration of
a solute in solution. In
practice that generally
means to add more solvent
without adding more solute.
In every instance the
concentration of the solute is
lower in the dilution.
Examples that we are familiar
with include making
orange juice by adding
4 cans of water to frozen
orange juice concentrate.
Dilutions can come up in
art when a darker color
is made lighter.
Each case the dilution results
in a weaker product.
Looking at the practice of
making a dilution from a
more mathematical perspective,
the amount of solute is
constant but the amount
of solvent is increased.
Say we have a concentration
where x grams of some
solute is dissolved in
y ml of some solvent.
Dilution occurs with the addition
of more solvent. We'll
call that additional amount z ml.
Looking at our
concentration ratio after
the dilution shows that
the denominator is larger
due to the addition of solvent.
From a fraction view
a larger denominator produces a
smaller fraction or a smaller
concentration in our case.
While dilutions can initially
seem confusing dilution
problems are almost
always solved the same way.
Fortunately, there is an
equation that relates the
concentration of a solute
before and after a dilution.
It works with the
concentrations that has
volume of solution
in the denominator
and the solute, in
the numerator, having
units of mass or count.
That includes concentration
with weight for the
solute; like mg/ml or g/L
Also, percent concentration,
(the weight to volume kind).
and the molarity scale,
moles per liter with
moles being a unit of count.
The key is that the
concentration be given
with volume in the solution.
So what is the magical
equation that relates the
concentration of a solute
before and after a dilution?
Simply C1V1 equals C2 V2.
The 1 subscript designates
values from before the dilution
and the subscript 2 designates
values after the dilution.
Initial and final work as well.
C stands for concentration.
C1 is the initial
concentration of the solute
before the dilution.
V stands for volume.
V1 is the initial volume of the
solution before the dilution.
V1 is the initial volume of the
solution before the dilution.
C2 is the final
concentration of solute
after the dilution and V2 is
final volume of solution
after the dilution.
As a heads up sometimes
the C1V1 = C2 V2 equation
is written with M's
in place of the C's.
It just means that the
concentration scale being
used is molarity.
Nothing else is changed
including the process of
solving dilution problems.
Dilution problems are tackled
step by step and do well
with a problem
solving strategy.
On first encountering a
problem, take the time to
identifying the known
and unknown variables.
For dilution problems those
terms will generally be the
concentration and volume
values of before and
after the dilution.
In the vast majority of
dilution problems the student
is given 3 of the 4 variables
in the dilution equation and
asked to solve for the 4th.
The second step is to find
the relationship between the
known and unknown
variables and to solve that
relationship mathematical.
That is solved the equation
for the unknown variable.
In this example C2 is the
unknown variable but it is
not isolated by itself on
one side of the equation.
This type of equation,
however, solves with
a single algebra step.
When an equation only
having multiplication
the unknown variable
can be isolated
by dividing both
sides of the equation
by the variable that is
the neighbor of the unknown.
In this example the unknown
is C2 and its neighbor is V2.
Divide both sides of
the equation by V2.
The V2's on the right
hand side of the equation
cancel out leaving the
unknown variable isolated.
The equation is
algebraically solved.
The last step is to
take the equation and
input the numerical
values from the setup and
solve the math.
It is always a good idea to
keep an eye on the units to
make sure the final solution
has the same units as
the unknown variable.
Let's put the
process to the test.
Calculate how many of ml of
1.0 M HCl is required to make
100 ml of a 0.10
M solution of HCl.
This is one of the problems
that use molarity as
a concentration scale.
Begin with the set up.
What is known and not
known in this problem?
Well there appears to
be 2 concentrations,
A 1.0 molar and a 0.10 molar.
Which one is the before or
initial concentration and
which one is the after
or final concentration?
Remember, dilution always
reduces the concentration so
the 0.10 molar, which is less,
is the after dilution
concentration.
This brings up to the
first helpful hint in
solving dilution problems:
C1 is always bigger than C2
and V1 is always
smaller than V2.
We can add the smaller
concentration to C2
and the larger to C1.
What about volume?
There is 100 ml.
Which concentration is
it attached to? C1 or C2?
Yes it is C2.
So we add the value to V2.
All that is left is V1.
Is its value given?
No, it is what the
problem asks for.
A question mark designates
it as an unknown.
Onto step 2, finding
the relationship between
known and unknown variables.
It is of course the topic
equation C1V1 equals C2 V2.
The similar equation M1V1
equals M2 V2 could also be
used here since
concentration is in molarity.
Also part of step 2 is to
solve the relationship in terms
of the unknown.
Is this equation solved for V1?
It is not. Some algebra
is necessary and it is the
same algebra for
every dilution problem.
To isolate the unknown
variable divide both sides of
the equation by the variable
that is the neighbor
of the unknown.
The variable next to V1 is C1.
Divide both sides of
the equation by C1.
On the left hand side
of the equation the
the C1's cancel out leaving
just the unknown variable V1.
The equation is considered
algebraically solved.
The student might ask why not
just input the variables into
the C1V1 =C2 V2 equation.
The answer is that adding
numbers and units complicates
the original equation and
increases the likelihood
of math errors.
Solving the equation with
the variables in place
is a best practice.
Step 3 is to solve the
equation mathematically.
Add numerical values from
the setup into the equation.
C2 is 0.10 molar, V2 is
100 ml, C1 is 1.0 molar.
The units of molar HCl cancels
out leaving ml units which
is a good thing since we
are solving for a volume.
Some calculator work gives
us a volume for V1 of 10 ml.
10 ml of a 1.0 molar solution
diluted to a final value
of 1.0 liters will have a
concentration of 0.10 molar.
Onto a second problem
and it will be solved
in a nearly identical
manner, well almost.
100 ml of water was
added to 150 ml of
a 70% ethanol solution.
What is the new concentration?
The solution to the problem
begins with the set-up.
Looking at the problem
we see a volume attached
to a concentration. Hint
number 2 says to connect
a volume to its concentration.
If that is done for one pair
(the before pair or
the after pair) then the
other values have to
belong to the other pair.
The 150 ml belongs to
the 70% concentration.
The question to ask is
whether this pair represents
before dilution or after.
A careful reading of the
problems shows that water
is added to this solution.
These are the before
dilution values.
Add them to the V1
and C1 in the set up.
Note that ethanol
is shorted to EtOH.
The OH is the alcohol part.
There are still 2
variables to deal with.
The question asks for the
new concentration so clearly
C2 is an unknown.
And the volume,
V2, is this unknown?
Is it 100 ml? Well it
might initially seem that way
but the volume of the
dilution cannot be smaller
than the starting volume
since solvent is added.
Hint 1 says V1 is
smaller than V2.
Are we stuck? Perhaps a revisit
to the problem would help.
It says that that 100 ml is
added to the 150 ml of the
70% concentration so the
final volume, V2, will
include the 150 ml of V1
plus the addition of 100 ml.
V2 is 250 ml.
V2 is larger than V1.
This leads to hint 3: V2 is
equal to V1 plus added solvent.
Sometimes the problem will
give the whole volume of V2
and sometimes its
value is implied
by giving V1 and the amount
of additional solvent.
Now we are ready for the
relationship expression that
connects the known
and unknown values
and is of course C1V1 = C2 V2.
Is the equation solved
for the unknown C2?
No it isn't.
We can employ the same
strategy in isolating C2
as we did in insolating
V1 in the previous problem.
That is, divide both sides
of the equation by the
variable next to the unknown.
The variable next to C2 is V2.
Divide both sides of the
equation by V2 and cancel
that variable out
leaving C2 isolated on
one side of the equation.
It doesn't matter which side.
The equation is algebraically
solved and we are ready to
insert the numerical
values from the set-up
into the equation and solve.
The initial concentration
times the initial volume,
divided by the final volume.
the unit ml cancel out
and the math gives a
diluted concentration
of 42% ethanol.
The last example asks
for 'What is final volume
when 250 ml of a
1.5 M NaCl solution
is diluted down to a
concentration of 0.25 M?'.
This is a typical type
of question one asks when
making a dilution
from a concentrate.
Starting with the set-up
and analyzing the problem.
One thing that stands out
in our reading of the problem
is a paired or connected
concentration and volume
(hint number 2). Is this
the initial or final pair?
Well, noting that the other
concentration, 0.25 molar,
is smaller than the 1.5 molar
in this pair it is safe to
say that the 1.5 M is
the initial concentration.
Hint number 1 says
C1 is bigger than C2.
This pair can be added to
the set-up as C1 and V1.
The remaining values in
the problem include the
final concentration, C2 is
0.25 molar, and an unknown,
final volume.
V2 is our unknown.
What relates the known
and unknown values?
The C1V1 = C2V2 equation and
like the previous problems it
is not solved for the unknown.
And like the previous
problems the unknown
is isolated by dividing
both sides by its neighbor
in this problem; C2.
When both sides are
divided by C2, the C2's
on the right hand side of the
equation cancel out.
The unknown is
algebraically isolated.
Step 3 is the
plug and chug part.
Insert the values from
the set-up: C1, V1, V2.
Cancel units and do the math.
1.50 times 250 divided by 0.25
is 1500 ml.
Some metric converting and
that comes to 1.500 liters.
Dilute 250 ml to 1500 ml and
the concentration will drop
from 1.5 molar to 0.25 molar.
And that concludes the
material for this lecture.
Our Recap; some compounds
dissolve in liquids
and form solutions.
The dissolved compound
is the solute and the
dissolving compound
is the solvent.
The amount of solute
per volume of solution
(or amount of solute
per volume of solvent)
is called the concentration.
We covered 4 concentration
scales in this lecture.
Percent concentration (weight
to volume) has the solute in
units of mass or weight and
the solution in volume units.
The ratio is multiplied by 100.
Percent concentration (weight
to weight): has the solute
and the solution
in units of weight.
The ratio is multiplied by 100.
Percent concentration (volume
to volume): has the solute
and the solution
in units of volume.
The ratio is multiplied by 100.
Parts per million or
billion has the solute
in units of mass and the
solution in units of mass.
The ratio is multiplied
by 1 million for ppm
and 1 billion for ppb.
Concentration is a
ratio and can be used
to move between
units of the solutes
and units of the
volume of the solution.
Dilution is the act of reducing
the concentration of a solute
(add more solvent without
adding more solute).
The equation C1V1=C2V2
relates the concentration of a
solute before and
after a dilution
The 3 step problem solving
technique fits well with
solving dilution problems.
And that concludes our lecture.
Concentration and dilution
are fundamental components
of chemistry!
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