Dilutions | Chemistry | Khan Academy

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: 4 moles of sugar per liter, which is apparently a very sweet syrup. You don't want to drink that directly, so what are you going to do? Well, of course, you are going to dilute it out, right?

So, you're going to take a jug, you're going to add some of this over here, and then you're going to add a lot of water, and that will give you a much more drinkable dilute solution that you can serve to all your friends. Okay, now let's say you want to make, I don't know, 5 L of this solution, this drinkable Kool-Aid. Let's say the concentration for it to be drinkable needs to be about 2 mol. So that's the molarity of the sugar that you want in this Kool-Aid solution.

The big question you're going to try and answer is: in order for this to happen, how much of the concentrated syrup should you take? What should be the volume that you should take over here? Say that after you add water and fill it up all the way to 5 liters, you'll precisely end up with a 2 molar concentration solution. How do you figure this out?

Okay, and by the way, if you're wondering why we have so many zeros 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, but again, how do you figure this out? How do you figure out how much of the concentrated syrup we need?

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, right? Now, when you add water over it, the amount of sugar, the amount of solute that you have, doesn't change, right? 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, stays the same. Let's write that down: this represents the moles of sugar in the concentrated solution over here, and this represents the moles of sugar in this dilute solution over here. But it has to be equal because after adding water, that amount does not change.

Okay, 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.

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 equated, and I can figure out what V is. So feel free to pause the video and try it out yourself first.

All right, here we go. So the amount of moles over here would be the molarity; over here, the molarity is 4, so 4 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 2.0, and the volume is 5 L.

Okay, so let's simplify this. The moles cancel out over here. On the right-hand side, I have 5 times 2.5 (5 * 2 = 10), and then if I divide by 4 on both sides, I'll get 1 over 4. So I get V = 1 L / 4, which equals 0.25. I'm going to put one more zero over here because we have three significant figures, so 0.25 L—that's the volume of the concentrated solution that I should take.

The rest I should add water to fill it up to 5 L, and then I'll get the 2 molar solution that I want. Now, actually, we can generalize this. If the concentration of the concentrated syrup was, say, M1, and the volume of that syrup was V1, and let's say the dilute syrup had a concentration of M2 and V2 was the amount of volume we needed, then, after equating the moles, what would we have gotten?

We would have gotten M1 V1 equal to 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. This 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 solute. Here the solute is sugar, but it'll work for any solute. Any dilution case this will work, right? The whole point is when you dilute it, the solute—the amount of solute—does not change, so the moles of solute stay the same.

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 L of 3 molar acid?

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 concentrated 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 L of 3 molar; you can see it's very dilute. We need to create a dilute solution of H₂SO₄ and, just like before, we're going to 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, right? So let's see what's given to us. We have the molarity of the concentrated H₂SO₄; we can call this M1, so we know M1. What about V1? V1 would be then the volume of the concentrated H₂SO₄. Hey, that's what we don't have; that's what we need to figure out. Then M2 would now be 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're given M2, we're given V2; we need to figure out V1. We'll 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, but we can do that now.

So, M1 V1 = M2, so that's 3.0 times V2; that's 0.5. We can simplify, so the moles cancel out over here. How much is our V1? Well, 3 times 0.5 is 1.5, and then I divide by 12 on both sides, so V1 = 1.5 divided by 12, which gives me 0.125, and I have to write it down to three significant figures, so that's going to be 0.125 L.

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 this out?

Well, we'll first extract 0.125 L of this in a pipette or usually in a graduated cylinder like this. Then we'll take the required amount of water in a separate flask, say a conical flask. But how do I know how much water I need? Well, think about this: this is the final volume I need; this is the amount of acid, so the remaining must be water, right?

So, the amount of water must be this minus this. So we do V2 minus V1; this is the exact amount of water I need. I'm going to take that in the conical 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 gear. You should have your safety goggles, your lab coat, and all that stuff. But, anyways, never add water to the acid to dilute it out; it's always the other way around. You add this acid to the water slowly, and you keep mixing it. Finally, that's how you're going to prepare your solution.

Okay, all right, let's try another problem. What volume of 3 M HCl can be made if we only have 10 mL of 12 molar HCl? Why don't you pause the video and see if you can solve this problem using the dilution equation?

All right, let's see. So we have 10 mL of 12 M HCl; that's what we have right now, and we want to convert it into a much more dilute 3 molar HCl solution, which means we are going to add water to it. If we're going to increase its volume, the big question is: what would that volume be in order for it to become a 3 molar solution?

This is the concentrated stuff, so let me color code that. So this is our concentrated stuff, and we are going to convert it into a much more dilute solution. The question is: what's the volume for that?

So we can now write down what our M1, V1, M2, and V2 are. We can say this is our M1; then 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. Okay, but now that we have this, well, 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 equal to 12 times 10, which should equal 3 times V2. Now we can just simplify this. The molarity cancels out over here, and I'll be left with milliliters.

Do I need to convert this to milliliters? I mean in this particular case you don't have to because this is the only unit we have. Our V2 answer will be in terms of milliliters, so we can just keep it that way.

Okay, so what will I get? I get, um, divide by 3 on both sides, so 12 by 3 would be 4. 4 times 10 would be 40, so I would actually end up with 40, and I have to put three significant figures, so it's going to be 40.0 mL. That will be our V2.

So this means I can make 40 mL of the dilute 3 molar HCl solution from this concentrated stuff.