This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.
5.06 Solution Concentration
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來自 University of Kentucky 的課程
Chemistry
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Week 5
This unit will delve into the quantitative relationships we can determine from a balanced chemical equation to determine the relative amounts of substances needed to react or the amount of products formed. We will also explore limiting reagents and percent yield to address practical aspects of chemical reactions because there may be leftover reactants and/or incomplete formation of products.
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Dr. Allison SoultLecturer
Chemistry
Dr. Kim WoodrumSr. Lecturer
Chemistry
In this module, we're going to talk about the concentration of solutions.
By the end of this module, you should be able to calculate the concentration of
solutions in a variety of different units.
So when we talk about the concentration,
it's a quantitative way of describing a solution.
A solution contains a solute or
more than one solute, which is present in a smaller amount.
It contains a solvent, which is a substance present in a larger amount.
And together they make up the solution.
Concentration is just a way of measuring how much solute is in
that particular solution.
We are going to look at a variety of units of concentration including molarity,
molality, mole fraction, percent, parts per million and parts per billion.
While we see all of these used in chemistry,
the most common is probably molarity.
And that's what we'll look at first.
Molarity is defined as the moles of solute over the liters of solution.
While most of our concentration units use the volume or
mass of solution in the denominator, please pay attention to the details,
because some of them do use the volume of solvent or the mass of the solvent.
So let's look at an example to see how we would use this definition of molarity to
calculate the quantitative value for a solution.
What is the molarity of a solution prepared from 25 grams of Sodium Chloride,
or NaCL, in 250 millimeters of solution?
So the first thing I need to do is find the moles of my NaCl.
I see that this substance is present in the smaller amount.
So this will be my solute.
And so I need to find the moles of solute.
So I use 25.0 grams [SOUND] and divide by
the molar mass which 58.5 grams per mole of NaCL,
and I'm able to find the moles of sodium chloride,
which is 0.427 moles of NaCL.
I can then use my definition of molarity.
0.427 moles of the NaCl, over the liters of my solution,
I see that I have 250 millimeters, so
that is going to be equal to 0.250 liters, because I divide by 1000 [SOUND].
Now I can take 0.427 and divide by 0.250, and
find that my molarity is equal to 1.71 molar.
Notice tha we use a capital M here, so
capital M represents molar or the molarity of a solution.
The definition molality is moles of solute over kilograms of solvent.
Notice that this is difference than what we saw for
molarity, which was moles of solute over liters of solution.
So we have a different unit and a difference substance in the denominator.
Let's look at an example of how this would be used.
What is the molality of a solution prepared from 25 grams
of NaCl in 175 mills of water?
So, I notice that in my definition of molality,
I have the units of moles of solute.
So the first thing I need to do is convert from 25 grams of NaCl,
[SOUND] to moles of NaCl.
And I do that using the molar mass of NaCl, which is 58.5 grams per mole.
Now I could do the calculation to find my moles of NaCl.
And I find that I have 0.427 moles of NaCl.
So I see if I have the moles of my solute.
Now I need to look for my kilograms of solvent and
I'm told that I have a 175 millilitres of water, which is my solvent, and
I remember that the density of water is 1.00 gram per millilitre, and
I need to get this into kilograms, and 1000 grams equals 1 kilogram.
So now millilitres cancel, grams cancel, and I'm left with units of kilograms.
So, I get 0.175 kilograms of water.
Now, I have both the moles of solute and the kilograms of solvent.
So, I can plug in those values, 0.427 moles,
over 0.175 kilograms, and
I get that the molality is 2.44 molal.
Notice we use a lowercase m to indicate molality.
Now, we can look at mole fraction.
And a mole fraction is calculated similar to a percentage.
The main difference, is that we don't multiply by 100.
And the amounts going in to it, must be in units of moles.
But notice we have moles of our solute.
So relative to a percent, this is our part.
Over moles of solute plus solvent, or
moles in the total solution, and that would be our whole amount.
But it has to be in terms of moles.
So what is the mole fraction of a solution prepared from 25.0 grams NaCl and
175.0 milliliters of water?
Well, we've already seen from the previous examples that 25 grams of
NaCl is 0.427 moles of NaCl.
Now we have to figure out how many moles of water we have.
We know we have 175 millilitres of water.
We can use our density of 1 gram per milliliter of water to
get into units of grams.
And then we can use the molar mass of water, 18.02 grams
per mole of water to find the moles of water.
And once we do the calculation which is 175 times 1,
divided by 18.02, we find that
we have 9.71 moles of water.
Now we can take the values we know, and plug in and find our mole fraction.
So we have the Greek symbol chi to represent mole fraction.
On the top I put our moles of solute, which is 0.427 moles.
And on the bottom I had to put both the moles of solute, the 0.427 moles,
plus the moles of the solvent, which we calculated and found to be 9.71 moles.
We don't multiply by the hundred, that's not the definition of mole fraction.
We can now find our mole fraction
of sodium chloride in this particular solution.
And so we get 0.0421 as our mole fraction.
Notice this is one of the few numbers that we write that do not
have a unit associated with them.
So our mole fraction of sodium chloride is 0.0421.
What i also know about mole fraction is that the mole fraction of sodium chloride
in this problem, plus the mole fraction of water, will equal to one.
So if I want to find the mole fraction of water,
I can do that as well by simply changing the value I put on top.
And what I should see is that the number I get for
the mole fraction of water, plus the mole fraction of sodium chloride,
should equal to one because that's the entire amount of sample.
The only two components are the solute and the solvent.
There are other units of concentration other than molarity, molarity or
mole fraction.
Percent is also a common unit of concentration.
With percent we actually to have to m,
identify what type of percent we're talking about.
If we're looking at a mass percent,
then we're looking at mass of solute over mass of solution.
For a volume percent, we're looking at the volume of solute for
the volume of solution.
But for mass volume percent,
we have the mass of solute, then the volume of solution.
Notice because the percentages they're all multiplied by 100, and
we can't just say the percent of the solution,
we always have to indicate the type of percent it is.
Two other concentration units that are much less common simply because they
measure very small amounts are parts per million and parts per billion.
We can think of these similar to a percentage, which is parts per 100.
What we usually see is that if we write something in terms of ppm,
it usually means milligrams per liter or milligrams per kilogram of solution.
So, where we see these most frequently, in our everyday lives,
is if you have a municipal water supply.
You may get a yearly statement from the company letting you
know how much of each contaminate is present in your water.
In general, these amounts are very, very small, so they're generally reported in
units of ppm, parts per million or ppb, parts per billion.
If I look at parts per billion instead of milligrams per liter, what I will see is
micrograms per liter or micrograms per kilogram of solution.
When I look at a solution,
I need to remember what happens to ionic compounds in solution.
Now if I look at the concentration of BaCl2 and
I find that the concentration is equal to 1.25 molar, that
tells me how many moles of barium chloride I put in the solution to prepare it.
1.25 moles per liter of solution.
However, because barium chloride is an ionic compound and
therefore a strong electrolyte and it dissolves in water,
what I'll actually see in solution are my chloride ions and my barium ions.
I won't, I will no longer see barium chloride units.
As a result, we have to look at the concentration of
ions a little bit differently than we look at concentrations of our solute.
So this is our solute, and these are our ions.
And this is what's actually present in solution, you know, the ions.
So if I have 1.25 moles per liter of solution, what I find is
that also I have 1.25 moles of barium ion, per liter of solution.
So that's the same as saying 1.25 molar.
If I look at the chlorine, however, what I see is,
I have twice as many chlorines as I had originally in my original molecule.
So here I would have 2.50 molar of my chloride ions.
Because, for every unit that broke apart.
Every unit of BaCL2 that broke apart.
I got one barium ion and two chloride ions.
Because our concentrations actually show a relationship between two units.
For example, molarity shows us the moles per liter of solution,
ppm shows us the milligrams per liter of solution.
We can use these as a relationship to figure out some calculations.
For example, we can figure out how many moles of HCl are in
500 millilitres of 0.3 molar HCl.
I usually recommend that as soon as you see units of molarity or that capital M,
that you rewrite it to represent it as moles over liters.
So, for 0.30 molar, I would actually rewrite that as
0.30 moles of HCL per liter of solution.
This allows me to look at it with both of it's units, it helps me to
figure out how to set up the problem so that my units cancel out correctly.
Now when I look at molarity, I see that it's a relationship there,
there are two units combined into one, so
a compound unit, whereas if I had millilitres, I only have a single unit.
So I'm going to start with my 500 millilitres.
And then I'm going to convert that to liters.
Because I see I have liters in my concentration.
So 1000 milliliters in one liter.
And then I can use my concentration.
0.30 moles of HCl per liter of solution.
Now, millilitres are cancelled, liters are cancelled and I'm left with moles of HCl.
And once I do my calculation, I find that I get 0.15 moles of HCl.
If I had chosen, I could have taken this problem one step further and
actually found the grams of HCl, but I didn't need to
go that additional step because it wasn't being asked for in the problem.
The point is, is that we can continue on past this, if we need to,
depending on what's being asked for in the problem.
Let's look at an example that does just that.
What mass of magnesium bromide is in 250 millilitres of a 3.5 molar solution?
The answer is 161 grams.
We have to look at how we find that information.
Remember, we have 3.50 of molar.
So I can rewrite that as 3.50 moles per liter of solution.
Then I want to use my 250 millilitres,
because that's the information I was given to start with.
I'm going to convert it into liters [SOUND].
Then I can use my molarity, or
3.50 Moles of magnesium bromide, per liter of solution.
Now I'm going to use my molar mass of
magnesium bromide, which is 184.1 grams per mole of MgBr2.
Now millilitres cancel,
liters cancel, moles cancel, and I'm left with a unit of grams.
And when I do the calculation, what I find
is 250 divided by 1000 times 3.50 times 184.1, and
I get 161 grams of magnesium bromide.
Now we can add another tool to our toolbox, or
values we can use as relationships to get between one unit and the other.
Here, I've shown moles to liters as an example and used the unit of molarity.
However, we can use different concentration units, including mole
fraction, percents, parts per million, parts per billion and molality.
To us relationships as well, to get from one set of units to another.
In the next module we'll talk about the dilution of solutions.
