# Solution Dilution

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Alright so now I'm going to talk about solution dilution.
So to dilute a solution simply means to add solvent
to the solution. So remember that the solvent is our majority component.
In other words, the solvent is what does the dissolving,
not what becomes dissolved. That's the solute if you remember.
And when we dilute a solution, the solution becomes less concentrated.
So for instance, if I had a concentrated aqueous solution,
aqueous meaning that the solvent is water,
if I had a concentrated aqueous solution
and I diluted it, that simply means adding solvent to it.
And once I added a solvent to it, the solution is now less
concentrated. So there's a pretty easy way to quantify
solution dilution and,
in general, the following equation is used:
M1V1 = M2V2
So, what M and V stand for
well lets just write it down. M is the molarity
and remember that is in moles per liter
or simply molar. So not moles per liter times molar
moles per liter OR molar.
And, V just stands for volumes.
The V is the volume
and that's typically in liters but, make no mistake, it could be given in milliliters,
gigaliters just whatever prefix.
It's more commonly just liters or milliliters though,
maybe in microliters - anyhow -so let's go over
what M1 and V1 and M2 and V2 are.
M1 and V1 apply to the concentrated solution.
So I'll just stick it over here - concentrated.
And M2 and V2 apply to the diluted solution.
So, my concentrated solution has a molarity in a volume.
And my and diluted solution has a molarity in a volume.
And they so happen to be related by
this equation. And the reason why they are related by this equation
Well, what is molarity? Remember, molarity is
moles per liter. So we have
the Moles 1
over the Liters1 times the volume 1 which is in liters usually
and then we have the moles
in 2 over the liters in 2
times the liters in 2
Well, we can cancel out liters on both sides
and we're left with nothing but moles.
So what this equation is really saying
is that the amount of moles is conserved when we dilute a solution.
Which is true because we're not messing with the solute.
We're messing with the solvent.
So, hopefully that gives a little bit insight
as to why this equation, how it works.
So, let's go through an example where we actually use this equation.
And I like this coming up example because
it sort of resembles a problem that you might encounter
on an exam. So a lot of them happen to be worded kind of like this one.
The problem says: What volume of a 12 molar HCl solution
must be diluted to 10 milliliters to make 3 molar HCl solution?
So I wrote the volume in blue because that's going to be our unknown.
So I'm just going to call that V1, so V1 is our unknown.
So that's what we're trying to figure out.
And it says 'what volume' of a 12 molar solution,
so I assume that the 12 molar solution from which you're taking this volume
applies to solution 1 as well, the more concentrated one.
So I'm going to say that the molarity of 1 is equal to 12 molar
and then it says what volume of a 12 molar solution
So we have the volume unknown, we have the molarity of 1,
that is 12 molar, and it says that we have to dilute it to 10 milliliters
to make a 3 molar HCl solution.
So we're going to have a V1 and a V2 and a M2
The volume given is 10 milliliters
and the molarity given is 3 molar.
So, that's basically how to choose which one is M1 and with one is V1
and so forth.
Typically, it doesn't matter what the subscripts 1 and 2 are as long as they're both in the same.
So, in other words, I could switch this around and I could put
I could put the 2 here and the 2 here and I can put the 1 here
and the 1 here, it doesn't matter as long as
these 2 are the same, and these 2 are the same.
So, let's solve the equation.
Remember that the equation is M1
V1 equals M2 V2
Well if V2 is our unknown
then I can divide both side of the equation by our M2
because we're trying to get V2 all by itself
M2
the M2s on the right side of the equation cancel
and you end up getting V2 is equal to M1V1 over M2
So the algebra to get to the final result isn't really that bad.
At this point all we have to do is just plug in our numbers and we're good to go.
So M1 and V1 - M1 is 3 molar
V1 is 10 milliliters
And M2 that's 12 molar
Molar here cancels with molar here
and so we have nothing but
3 time 10 over twelve which is 30 over 12
and you can actually reduce this by dividing the top and the bottom of the fraction by 6
And you will end up getting 5 halves and this is in millimeters
So that is our answer 5 halves milliliters (5 over 2) or .4 whichever one
Typically you wouldn't want to express it as a fraction because you're trying to keep tabs
on the number of sig figs (significant figures) and things like that.
But I just did it because I'm not really trying to go over sig figs in this video I'm going over the equation.
But that's how to use the equation, very straightforward
all you have to do is algebraically solve for your unknown by dividing both sides
by the other variable that's associated with it;
with the unknown. Typically, like I said, three of them (the variables) have to be given
so that you can solve for the one unknown.
So there you go!