- [Presenter] Your
friends are coming over.

So you decide to make
some Kool-Aid for them.

You happen to have a very
concentrated Kool-Aid solution.

This is the molarity of the
amount of sugar that you have,

so this is four moles of sugar per liter,

which is apparently a very sweet syrup,

you don't wanna drink that directly.

So what you're gonna do,

well, of course you
are going to dilute it.

So you're gonna take a jug,

you're gonna add some of this over here

and then you're gonna add a lot of water

and that'll give you a much
more drinkable, dilute solution

that you can serve to all your friends.

Now, let's say you wanna
make about, I don't know,

five liters of this solution,
this drinkable Kool-Aid,

and let's say the concentration
for it to be drinkable

needs to be about 0.2 molar,

so that's the molarity of the sugar

that you want in this Kool-Aid solution.

So the big question we
wanna try and answer

is, in order for this to happen,

in order for you to get
this diluted solution,

how much of the concentrated
syrup should you take?

What should be the volume that
you should take over here?

So that after you add water

and fill it up all the way to five liters,

you'll precisely end up with 0.2 molar

concentration solution,
how do you figure this out?

And by the way, if you're wondering

why do we have so many zeroes
and decimals over here?

Well that's because we
have precisely measured

this to three significant figures.

I mean, we take our Kool-Aid
very seriously, okay?

(Presenter laughing)

But again, how do you figure this out?

How do you figure out how much

of the concentrated syrup do we need?

How do you do this?

Well, here's the key idea.

If you look at this concentrated syrup

that you have poured in a jug,

it contains some moles of sugar.

Now, when you add water to it,

the amount of sugar,

the amount of solute that
you have doesn't change.

Even this dilute solution
has the same amount of sugar,

but because now the volume of water,

the solvent has increased,

that's why it has become more dilute.

So the key idea is,

when you're diluting the amount of solute,

which is sugar over here,
that stays the same.

And so let's write that down.

I write this way, so this
represents the moles of sugar

in the concentrated solution over here

and this represent the moles of sugar

in this dilute solution over here.

But it has to be equal
because after adding water,

the amount does not change.

Well next I'm thinking,

how do I figure out moles

if I know the molarity and the volume?

What's the connection between
moles, molarity, and volume?

Hey, we know that!

Molarity is the amount
of moles per volume.

So from this, I can rearrange
and find out what moles is.

So I can rearrange this for moles,

so I'll get moles equals
molarity times volume.

So I can plug in over here

the molarity times
volume for this solution

and over here,

molarity times volume
for the dilute solution,

equate it and I can figure out what V is.

So feel free to pause the video

and try it out yourself first.

Alright, here we go.

So the amount of moles over here

would be the molarity over
here, the molarity is four,

so 4.00 molar times the
volume, which I don't know,

that's what I need to
figure out, the volume

of this concentrated
solution, concentrated syrup.

But that should equal the
molarity times volume here,

the molarity is 0.200 and
the volume is five liters.

So let's simplify this.

The moles cancels out over here.

On the right hand side, I
have five multiplied by 0.2.

Five times 0.2 is one

and then if I divide
by four on both sides,

I'll get one over four.

So I get V equals one
liter divide by four,

which equals 0.25.

And I'm gonna put one more zero over here,

because we have three
significant figures over here.

So 0.250 liters,

that's the volume of the
concentrated solution

that I should take, and the
rest, I should add water

to fill it up to five liters

and then I'll get 0.2
molar solution that I want.

Now, actually, we can generalize this.

So if the concentration

of the concentrate syrup was, say, M1

and the volume of that syrup was V1,

and let's say the dilute syrup
had a concentration of M2,

molarity was M2 and V2 was the
amount of volume we needed,

then after equating the moles,
what would we have gotten?

We would've gotten M1 V1 equals M2 V2.

And you can think of that

as an equation that you
can use for dilution.

So we can write that down
as our dilution equation,

which means whenever we're
solving for such problems,

all we have to do is equate the product

of molarity and volume.

The product of molarity and
volume will stay the same

even after dilution.

What's the logic behind it?

Why does that product stay the same?

Well, because look,

the product represents
the moles of salute.

Here, the salute is sugar,
but it'll work for any salute,

any dilution case, this will work.

The whole point is when you dilute it,

the salute and the amount
of salute does not change,

so the moles of salute stay the same

and that's why the product stays the same

and we can now use this dilution equation

to try and solve the problem.

So let's try and solve
another problem here.

How much of 12 molar H2SO4 sulfuric acid,

do we need to create a 0.5
liter of three molar acid?

So we have a 12 molar
H2SO4 solution with us,

that's the stock solution
that we usually find in labs,

it's very concentrated.

So that's the concentrate
solution that we have.

Now what we need to do is create

a much more dilute
solution, as you can see.

So we want to create 0.5
liters of three molar.

So you can see it's very dilute.

We need to create a
dilute solution of H2SO4

and just like before, we're
gonna take a little bit of this

and add it to a lot of water
to create our dilute solution.

And the question is,

how much of the concentrated
stuff do we need?

So how do we solve it?

Well, we have our dilution equation,

so let's see what's given to us.

Well, we have the molarity

of the concentrate H2HSO4,
we can call this M1,

so we know M1, what about V1?

V1 would be then the volume
of the concentered H2SO4,

hey, that's what we don't have,

that's what we need to figure out.

Then M2 would now be the volume,

the molarity of the dilute
solution that's given to us

and V2 would be the volume
of the dilute solution

that's given to us as well.

So we're given M1, we are
given M2, we are given V2,

we need to figure out V1.

We just plug in over here and do that.

So let's do that, feel
free to pause the video

and try it on your own
first, we can do that now.

So M1, V1 equals M2, so that's 3.00

times V2 that's 0.500.

And we can simplify, so the
molar cancels out over here

and so how much is our V1?

Well, three times 0.5 is 1.5,

and then I divide by 12 on both sides,

so V1 equals 1.5, let
me use the same color,

1.500 liters divided by 12, 12.0,

and that gives me 1.5, divided by 12.125

and I have to write it down
to three significant figures,

so that's going to be 0.125.

So three significant figures,

liters, that's the unit that
we have and there we have it.

So this is the amount
of concentrated stuff

that we need to add to water
to get our desired solution.

So how exactly would we carry it out?

Well, we'll first extract
0.125 liters of this

in a pipet or usually in a
graduated cylinder like this.

Then we'll take the
required amount of water

in a separate flask, say a clinical flask,

but how do I know how
much water do I need?

We'll, think about, this is how much,

this is the final volume I need,

in this, this is the amount of acid,

so the remaining must be water.

So the amount of water
must be this minus this,

so we do minus V1,

so this is the exact
amount of water I need.

So I'm gonna take that
in the clinical flask

and then never add water to the acid,

that can be very dangerous

because this is concentrated stuff.

The water over here can
just boil and splash

and in fact, that's one of the reasons

you should always have your safety,

you should have your safety goggles,

your lab coat and all of that stuff.

But anyways, never add water
to the acid to dilute it out,

it's the always the other way around,

you add this acid to the water, slowly,

and you keep mixing it

and finally, that's how you're
gonna prepare your solution.

All right, let's try another problem.

What volume of three molar HCL can be made

if we only have 10 ml of 12 molar HCL,

Why don't you pause the video

and see if we can solve this problem

using the dilution equation?

Alright, let's see.

So we have 10 ml of 12 molar HCL,

that's what we have right
now and we to convert it

into a much more dilute
three molar HCL solution,

which means we are
going to add water to it

and if we're going to increase its volume,

the big question is what
would that volume be

in order for it to become three molar HCL?

So this is the concentrated stuff,

so let me color code that.

So this is our...

I'm gonna use dark red
for the concentrated one,

so this is our concentrated stuff

and we are going to convert it

into a much more dilute stuff

and the question is,
what's the volume for that?

So we can now write down what
our M1, V1 and M2, V2 are.

So we can say this is our
M1 and this would be our V1,

then this would be our M2,

and then we have to figure out what V2 is.

I mean, of course

you can call anything
one and anything two.

But now that we have this

or we can just plug into this equation

and figure out what V2 is going to be.

So if we do that, we'll get M1, V1,

12 times 10 should equal three times V2.

And now we can just simplify this,

so this molar cancels out over here

and I'll be left with milliliters,

do I need to convert this milliliter?

I mean, in this particular case,

you don't have to because
this is the only unit,

our V2 answer will be
in terms of milliliters,

we can just keep it that way.

So what will I get?,

I get divide by three on both sides,

so twelve by three would be four,

four times ten would be 40.

So I would actually end up with 40

and I have to put three
significant figures,

so it's gonna be 40.0
ml, that will be our V2.

So this means I can make 40 ml

of the dilute three molar HCL solution

from this concentrate stuff.