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### Course: AP®︎/College Chemistry > Unit 3

Lesson 7: Solutions and mixtures# Dilution

A common method of making a solution of a given concentration involves taking a more concentration solution and adding water until the desired concentration is reached. This process is known as dilution. We can relate the concentrations and volumes before and after a dilution using the following equation:

*M*₁*V*₁ =*M*₂*V*₂ where*M*₁ and*V*₁ represent the molarity and volume of the initial concentrated solution and*M*₂ and*V*₂ represent the molarity and volume of the final diluted solution. Created by Sal Khan.## Want to join the conversation?

- can you just multiply the 500 mls by the .125 molars and get the same answer?(5 votes)
- Yes, that's essentially what Sal did. A more simplified way of solving this is by using the dilution formula: (M1)(V1) = (M2)(V2), where M's are molarities and V's are volumes. 1 means the initial state and 2 mean the final state.

So for this problem here where we want 500 mL of a 0.125 M solution of sodium sulfate and start with 1.00 M solution of sodium sulfate, we want to know how much volume of the 1.00 M solution we need to add. So M1 = 1.00 M, M2 = 0.125M, V2= 500 mL, V1= is unknown and what we solve for. So V1 = (0.125 M)(500 mL)/(1.00M) = 62.5 mL. So that's what Sal's doing essentially and what you suggested to do, both are just following the dilution formula.

Hope that helps.(14 votes)

- How can you know when to use the M1*V1=M2*V2 formula?(4 votes)
- Well whenever you are trying to create a more dilute solution you would use that formula. A more dilute solution meaning a solution with a lower concentration than the original. You use this quite often in chemistry when you want to work with a solution with a specific concentration.

Another use for the dilution formula is that it allows you to know how much acid/base to add to an analyte when doing acid/base titrations using strong acids and strong bases.

Hope that helps.(3 votes)

- Why is 500 mL = 0.500L, I thought that 500 has 1 significant figure and 0.500 has 3? Shouldn't it be 0.5L?(3 votes)
- Are molarity and moles the same thing? Do they have the same unit?(1 vote)
- Molarity and moles are measuring different things and hence have different units.

A mole is the unit for the amount of substance, or how much of something there is. It is defined as an Avogadro's number of particles, or 6.02214076 x 10^(23) particles. In the same way that a dozen of something is 12 particles. So a dozen eggs is 12 eggs, a dozen people is 12 people, a dozen atoms is 12 atoms. A mole of eggs is 6.02214076 x 10^(23) eggs and a mole of atoms is 6.02214076 x 10^(23) atoms. We use such a large number in chemistry because atoms are so small that having even a small amount of atoms like a gram could already be a mole of atoms. So it's more convenient to use moles of atoms instead of saying 6.02214076 x 10^(23) atoms each time we do a calculation.

Molarity is a unit of concentration, with units of moles of solute/ liters of solvent. Concentration being how much of a substance is in a given volume. A solvent being a liquid into which something is dissolved into, which is referred to as the solute. Together a solute and a solvent are called a solution. So if you have a glass of salt water, you have a solution of water where the water is the solvent and the salt is the solute. If you have a lot of salt in the water then it is a concentrated solution which we would express with a large molarity. And a small amount of salt in the water is an unconcentrated solution with a small molarity.

Hope that helps.(2 votes)

- why I don't see the equation M1V1=M2V2(1 vote)
- Well the dilution formula is in the about of the video.

But Sal used the dilution formula at4:30even though he didn’t explicitly say he did. It's the same math, just expressed differently.(2 votes)

- if we have molars of different compounds, can we use the formula MiVi=MfVf or do we have to relate the molars first?(1 vote)
- Why is the 1M also equal to 0.0625 moles Na2SO4?(1 vote)
- google says

The molarity of a solution is defined as the number of moles of solute per liter of solution. Therefore, 1M Na2SO4 solution contains 1 mole of Na2SO4 per liter of solution 1.

The molar mass of Na2SO4 is 142.0421 g/mol 1. Therefore, 1 mole of Na2SO4 weighs 142.0421 grams.

To calculate the number of moles of Na2SO4 in a 1M solution, we need to multiply the molarity by the volume of the solution in liters. Since the volume of the solution is not given, we cannot calculate the number of moles of Na2SO4 directly.

However, if we assume that the volume of the solution is 1 liter, then the number of moles of Na2SO4 in the solution would be equal to the molarity of the solution, which is 1 mole/L. Therefore, 1M Na2SO4 solution contains 1 mole of Na2SO4 per liter of solution 1.

If we assume that the volume of the solution is 0.0625 liters (62.5 mL), then the number of moles of Na2SO4 in the solution would be equal to the molarity of the solution multiplied by the volume of the solution in liters. Therefore, 1M Na2SO4 solution with a volume of 0.0625 liters contains 0.0625 moles of Na2SO4 1.

Therefore, 1M Na2SO4 is equal to 0.0625 moles of Na2SO4 when the volume of the solution is 0.0625 liters.

Dont mind the 1's**Votes plz**(1 vote)

- Does the molarity change when we increase or decrease the volume?(1 vote)
- of course , it does .(0 votes)

## Video transcript

- [Instructor] In this
video, we're gonna talk about a concept in chemistry
that's quite important, known as dilutions. So let's do an example. So let's say we have a large
vat, as much as we need. It's a one-molar solution
of sodium sulfate, and it's an aqueous solution. So sodium sulfate is dissolved in water. And let's say we also have
as much water as we need, and what we want to do
is create a solution, another aqueous solution
of sodium sulfate, but one that has a
different concentration, in this case, one that
has a lower concentration. So we want to create
a 0.125-molar solution of sodium sulfate, and
we want 500 milliliters of this new solution. Pause this video and think about how you would approach that. All right, now let's
think about this together. So, first let's just
go over the intuition. You have a higher concentration here. You have a lower concentration here. So our intuition would tell us is that we're going to take
less than 500 milliliters of our original solution,
pour some of that in. That's going to have a
sufficient number of moles of sodium sulfate that, if we were to then fill
this up to 500 milliliters, that we would then have
a 0.125-molar solution. So the question really
is, is how much of this do we have to put in, which
we can then dilute with water to get to our goal solution? Well, to answer that question,
we just have to figure out how many moles of sodium sulfate need to be in this final
goal solution, this one or this one, depending
on how we visualize it? And then, how much of
our original solution, of our one-molar solution,
do we need to take out to have that many moles? And to think about how many moles, we just have to remind
ourselves what molarity is. We know already that molarity
is equal to number of moles, number of moles, of solute per liters of solution, liters of solution. Or another way to think about it is, if we multiply both sides
by liters of solution, we would get liters of solution times molarity, times molarity, is equal to the number of moles of solute, number of moles of solute. So what we can do is say,
all right, how many moles of our solute do we need in our goal? Well, to do that, we just
have to say, all right, we want to eventually have
500 milliliters of solution, or we could rewrite that as 0.500 liters, and this little decimal
point right over here makes it clear that we're dealing with three significant figures, that we've rounded to the nearest one, when we got to this, when we
have this goal right over here, or we would round to the nearest, to the ones place, I guess. So, our goal is to have half a liter of solution at a molarity of 0.125 molar, and then that is gonna give us the number of moles we need. And, if we multiply this
out, this is going to be zero point, let's see, half of 12 is 6 and then half of 50 is 25, 0.0625 moles, moles of solute. And, in this case, our
solute is sodium sulfate. And let's see if I got the
significant figures right. I have three right over here, one, two, three, one, two, three. So I take the product. I'd still have one, two,
three significant figures. So this is our goal. We want to have this many moles of solute. So we just have to figure out how much of our original
solution do we need in order to have that many
moles of sodium sulfate? So, one way to think about it is, there's some mystery volume
of our original solution we need, and we know what
its concentration is. It's a one-molar concentration that, when I take this product, I am going to get 0.0625 moles of sodium sulfate. And the math here is
pretty straightforward. We can divide both sides by one molar, and what are we going to get? And the units work out because we're in moles
where you have molar here. And so this is going to give
us our answer in liters. You divide both sides by one molar. You're going to get that
question mark is equal to 0.0625 liters of solution. Or another way to think about
it is, this is equivalent to 62.5 milliliters of our original solution. I want to make sure I got all
the significant figures right. Had three over there. One,
two, three, one, two, three. And so, yes, right over here. I can still have one, two,
three significant figures or sometimes called significant digits. And so there we've answered our question. What I would do is I would take 62.5, 62.5 milliliters of my original solution, so that's this over here, and then I would take my
water and then keep filling until I get to 500
milliliters, and we're done. At that point, I'm going
to have a 0.125 molar of sodium sulfate aqueous solution.