0:00:00.180,0:00:01.140
- [Presenter] Your[br]friends are coming over.

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

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

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This is the molarity of the[br]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[br]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[br]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[br]make about, I don't know,

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five liters of this solution,[br]this drinkable Kool-Aid,

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and let's say the concentration[br]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[br]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[br]this diluted solution,

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how much of the concentrated[br]syrup should you take?

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What should be the volume that[br]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,[br]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[br]and decimals over here?

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Well that's because we[br]have precisely measured

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this to three significant figures.

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I mean, we take our Kool-Aid[br]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[br]you have doesn't change.

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Even this dilute solution[br]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,[br]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[br]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[br]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[br]moles, molarity, and volume?

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Hey, we know that!

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Molarity is the amount[br]of moles per volume.

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So from this, I can rearrange[br]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[br]molarity times volume.

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So I can plug in over here

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the molarity times[br]volume for this solution

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and over here,

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molarity times volume[br]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[br]here, the molarity is four,

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

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

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of this concentrated[br]solution, concentrated syrup.

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But that should equal the[br]molarity times volume here,

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the molarity is 0.200 and[br]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[br]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[br]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[br]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[br]significant figures over here.

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So 0.250 liters,

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that's the volume of the[br]concentrated solution

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that I should take, and the[br]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[br]molar solution that I want.

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Now, actually, we can generalize this.

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So if the concentration

0:04:09.780,0:04:12.810
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[br]had a concentration of M2,

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

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then after equating the moles,[br]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[br]can use for dilution.

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So we can write that down[br]as our dilution equation,

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which means whenever we're[br]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[br]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[br]the moles of salute.

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Here, the salute is sugar,[br]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[br]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

0:05:05.670,0:05:08.700
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[br]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[br]liter of three molar acid?

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So we have a 12 molar[br]H2SO4 solution with us,

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

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it's very concentrated.

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So that's the concentrate[br]solution that we have.

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Now what we need to do is create

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a much more dilute[br]solution, as you can see.

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So we want to create 0.5[br]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[br]dilute solution of H2SO4

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

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

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And the question is,

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how much of the concentrated[br]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

0:06:01.020,0:06:03.900
of the concentrate H2HSO4,[br]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[br]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[br]solution that's given to us

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and V2 would be the volume[br]of the dilute solution

0:06:18.390,0:06:19.710
that's given to us as well.

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So we're given M1, we are[br]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[br]free to pause the video

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and try it on your own[br]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[br]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[br]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[br]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[br]we have and there we have it.

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So this is the amount[br]of concentrated stuff

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that we need to add to water[br]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[br]0.125 liters of this

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in a pipet or usually in a[br]graduated cylinder like this.

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Then we'll take the[br]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[br]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[br]must be this minus this,

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so we do minus V1,

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so this is the exact[br]amount of water I need.

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So I'm gonna take that[br]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

0:08:08.430,0:08:09.600
because this is concentrated stuff.

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The water over here can[br]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,

0:08:14.190,0:08:15.630
you should have your safety goggles,

0:08:15.630,0:08:17.580
your lab coat and all of that stuff.

0:08:17.580,0:08:20.670
But anyways, never add water[br]to the acid to dilute it out,

0:08:20.670,0:08:22.020
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

0:08:26.765,0:08:31.765
and finally, that's how you're[br]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

0:08:38.310,0:08:42.900
if we only have 10 ml of 12 molar HCL,

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Why don't you pause the video

0:08:43.830,0:08:45.360
and see if we can solve this problem

0:08:45.360,0:08:47.223
using the dilution equation?

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Alright, let's see.

0:08:49.170,0:08:52.590
So we have 10 ml of 12 molar HCL,

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

0:08:55.950,0:08:59.010
into a much more dilute[br]three molar HCL solution,

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which means we are[br]going to add water to it

0:09:00.960,0:09:02.760
and if we're going to increase its volume,

0:09:02.760,0:09:04.890
the big question is what[br]would that volume be

0:09:04.890,0:09:07.830
in order for it to become three molar HCL?

0:09:07.830,0:09:09.090
So this is the concentrated stuff,

0:09:09.090,0:09:10.170
so let me color code that.

0:09:10.170,0:09:12.180
So this is our...

0:09:12.180,0:09:14.340
I'm gonna use dark red[br]for the concentrated one,

0:09:14.340,0:09:15.713
so this is our concentrated stuff

0:09:15.713,0:09:18.300
and we are going to convert it

0:09:18.300,0:09:20.520
into a much more dilute stuff

0:09:20.520,0:09:24.180
and the question is,[br]what's the volume for that?

0:09:24.180,0:09:27.065
So we can now write down what[br]our M1, V1 and M2, V2 are.

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

0:09:32.054,0:09:34.830
then this would be our M2,

0:09:34.830,0:09:36.808
and then we have to figure out what V2 is.

0:09:36.808,0:09:38.220
I mean, of course

0:09:38.220,0:09:40.326
you can call anything[br]one and anything two.

0:09:40.326,0:09:42.030
But now that we have this

0:09:42.030,0:09:43.830
or we can just plug into this equation

0:09:43.830,0:09:45.690
and figure out what V2 is going to be.

0:09:45.690,0:09:48.406
So if we do that, we'll get M1, V1,

0:09:48.406,0:09:51.483
12 times 10 should equal three times V2.

0:09:53.288,0:09:54.210
And now we can just simplify this,

0:09:54.210,0:09:56.670
so this molar cancels out over here

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and I'll be left with milliliters,

0:09:58.260,0:09:59.448
do I need to convert this milliliter?

0:09:59.448,0:10:00.780
I mean, in this particular case,

0:10:00.780,0:10:02.832
you don't have to because[br]this is the only unit,

0:10:02.832,0:10:05.730
our V2 answer will be[br]in terms of milliliters,

0:10:05.730,0:10:07.104
we can just keep it that way.

0:10:07.104,0:10:08.190
So what will I get?,

0:10:08.190,0:10:10.770
I get divide by three on both sides,

0:10:10.770,0:10:12.060
so twelve by three would be four,

0:10:12.060,0:10:13.800
four times ten would be 40.

0:10:13.800,0:10:18.181
So I would actually end up with 40

0:10:18.181,0:10:20.940
and I have to put three[br]significant figures,

0:10:20.940,0:10:25.940
so it's gonna be 40.0[br]ml, that will be our V2.

0:10:26.790,0:10:29.460
So this means I can make 40 ml

0:10:29.460,0:10:32.190
of the dilute three molar HCL solution

0:10:32.190,0:10:34.023
from this concentrate stuff.