- [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.