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- [Presenter] Your
friends are coming over.

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

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

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

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so this is four moles of sugar per liter,

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which is apparently a very sweet syrup,

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you don't wanna drink that directly.

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So what you're gonna do,

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well, of course you
are going to dilute it.

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So you're gonna take a jug,

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you're gonna add some of this over here

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and then you're gonna add a lot of water

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

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that you can serve to all your friends.

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

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

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

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needs to be about 0.2 molar,

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so that's the molarity of the sugar

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that you want in this Kool-Aid solution.

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So the big question we
wanna try and answer

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is, in order for this to happen,

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in order for you to get
this diluted solution,

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how much of the concentrated
syrup should you take?

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

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So that after you add water

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and fill it up all the way to five liters,

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you'll precisely end up with 0.2 molar

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concentration solution,
how do you figure this out?

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And by the way, if you're wondering

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

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Well that's because we
have precisely measured

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this to three significant figures.

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

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(Presenter laughing)

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But again, how do you figure this out?

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How do you figure out how much

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of the concentrated syrup do we need?

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How do you do this?

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Well, here's the key idea.

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If you look at this concentrated syrup

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that you have poured in a jug,

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it contains some moles of sugar.

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Now, when you add water to it,

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the amount of sugar,

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

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Even this dilute solution
has the same amount of sugar,

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but because now the volume of water,

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the solvent has increased,

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that's why it has become more dilute.

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So the key idea is,

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when you're diluting the amount of solute,

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which is sugar over here,
that stays the same.

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And so let's write that down.

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

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in the concentrated solution over here

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and this represent the moles of sugar

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in this dilute solution over here.

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But it has to be equal
because after adding water,

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the amount does not change.

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Well next I'm thinking,

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how do I figure out moles

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if I know the molarity and the volume?

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

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Hey, we know that!

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Molarity is the amount
of moles per volume.

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

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So I can rearrange this for moles,

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so I'll get moles equals
molarity times volume.

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So I can plug in over here

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the molarity times
volume for this solution

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and over here,

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molarity times volume
for the dilute solution,

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equate it and I can figure out what V is.

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So feel free to pause the video

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and try it out yourself first.

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Alright, here we go.

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So the amount of moles over here

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

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

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

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of this concentrated
solution, concentrated syrup.

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But that should equal the
molarity times volume here,

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the molarity is 0.200 and
the volume is five liters.

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So let's simplify this.

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The moles cancels out over here.

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

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Five times 0.2 is one

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and then if I divide
by four on both sides,

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I'll get one over four.

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So I get V equals one
liter divide by four,

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which equals 0.25.

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

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because we have three
significant figures over here.

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So 0.250 liters,

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that's the volume of the
concentrated solution

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

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to fill it up to five liters

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

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Now, actually, we can generalize this.

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So if the concentration

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of the concentrate syrup was, say, M1

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and the volume of that syrup was V1,

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

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

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

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We would've gotten M1 V1 equals M2 V2.

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And you can think of that

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as an equation that you
can use for dilution.

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So we can write that down
as our dilution equation,

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which means whenever we're
solving for such problems,

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all we have to do is equate the product

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of molarity and volume.

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The product of molarity and
volume will stay the same

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even after dilution.

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What's the logic behind it?

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Why does that product stay the same?

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Well, because look,

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the product represents
the moles of salute.

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

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any dilution case, this will work.

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The whole point is when you dilute it,

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the salute and the amount
of salute does not change,

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so the moles of salute stay the same

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and that's why the product stays the same

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and we can now use this dilution equation

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to try and solve the problem.

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So let's try and solve
another problem here.

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How much of 12 molar H2SO4 sulfuric acid,

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

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So we have a 12 molar
H2SO4 solution with us,

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

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it's very concentrated.

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So that's the concentrate
solution that we have.

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Now what we need to do is create

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a much more dilute
solution, as you can see.

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So we want to create 0.5
liters of three molar.

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So you can see it's very dilute.

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We need to create a
dilute solution of H2SO4

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

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

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And the question is,

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how much of the concentrated
stuff do we need?

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So how do we solve it?

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Well, we have our dilution equation,

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

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Well, we have the molarity

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of the concentrate H2HSO4,
we can call this M1,

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so we know M1, what about V1?

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V1 would be then the volume
of the concentered H2SO4,

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

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that's what we need to figure out.

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Then M2 would now be the volume,

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

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and V2 would be the volume
of the dilute solution

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that's given to us as well.

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

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we need to figure out V1.

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We just plug in over here and do that.

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

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

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

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times V2 that's 0.500.

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

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and so how much is our V1?

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Well, three times 0.5 is 1.5,

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and then I divide by 12 on both sides,

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

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1.500 liters divided by 12, 12.0,

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and that gives me 1.5, divided by 12.125

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

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so that's going to be 0.125.

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So three significant figures,

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

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So this is the amount
of concentrated stuff

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

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So how exactly would we carry it out?

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Well, we'll first extract
0.125 liters of this

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

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Then we'll take the
required amount of water

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in a separate flask, say a clinical flask,

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

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

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this is the final volume I need,

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in this, this is the amount of acid,

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so the remaining must be water.

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So the amount of water
must be this minus this,

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so we do minus V1,

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so this is the exact
amount of water I need.

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So I'm gonna take that
in the clinical flask

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and then never add water to the acid,

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that can be very dangerous

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because this is concentrated stuff.

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The water over here can
just boil and splash

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and in fact, that's one of the reasons

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you should always have your safety,

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you should have your safety goggles,

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your lab coat and all of that stuff.

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

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it's the always the other way around,

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you add this acid to the water, slowly,

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and you keep mixing it

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

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All right, let's try another problem.

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What volume of three molar HCL can be made

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if we only have 10 ml of 12 molar HCL,

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Why don't you pause the video

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and see if we can solve this problem

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using the dilution equation?

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Alright, let's see.

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So we have 10 ml of 12 molar HCL,

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

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into a much more dilute
three molar HCL solution,

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which means we are
going to add water to it

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and if we're going to increase its volume,

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the big question is what
would that volume be

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in order for it to become three molar HCL?

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So this is the concentrated stuff,

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so let me color code that.

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So this is our...

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

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so this is our concentrated stuff

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and we are going to convert it

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into a much more dilute stuff

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

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

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

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then this would be our M2,

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and then we have to figure out what V2 is.

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I mean, of course

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you can call anything
one and anything two.

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But now that we have this

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or we can just plug into this equation

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and figure out what V2 is going to be.

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

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12 times 10 should equal three times V2.

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And now we can just simplify this,

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so this molar cancels out over here

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and I'll be left with milliliters,

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do I need to convert this milliliter?

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I mean, in this particular case,

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

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our V2 answer will be
in terms of milliliters,

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we can just keep it that way.

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So what will I get?,

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I get divide by three on both sides,

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so twelve by three would be four,

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four times ten would be 40.

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So I would actually end up with 40

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and I have to put three
significant figures,

251
00:10:20,940 --> 00:10:25,940
so it's gonna be 40.0
ml, that will be our V2.

252
00:10:26,790 --> 00:10:29,460
So this means I can make 40 ml

253
00:10:29,460 --> 00:10:32,190
of the dilute three molar HCL solution

254
00:10:32,190 --> 00:10:34,023
from this concentrate stuff.