0:18

A dilution is the preparation of a less concentrated solution from

Â a more concentrated one through the addition of additional solvent.

Â When we add additional solvent to our solution, we see that the moles of

Â our solute, represented here by the green spheres, do not change.

Â We still have the same number on the left as the right,

Â the only that changes is the volume of the solution.

Â And this changes because we have additional solvent in the solution.

Â 0:54

We have an initial concentration and volume, and

Â we have a final concentration and volume.

Â It doesn't really matter what we call our initial conditions or final conditions,

Â the math will still work out.

Â What we have to worry about is whether or

Â not we pair up the correct concentration and volume with each other.

Â It doesn't matter what our units of concentrations are,

Â as long as the units of C1 match C2 and likewise for volume.

Â The units of volume one must match that for volume two.

Â It doesn't matter what the units are as long as they are the same.

Â 1:28

Let's look at an example to see how we can use this equation.

Â So we have C1V1 = C2V2.

Â There are some words I have to look for to see if this is in fact a dilution problem.

Â The most obvious is diluted, or dilution.

Â We can talk about the addition of solvent.

Â We can talk about the amount of solute staying same, but

Â the volume of the solution changing.

Â And these are all key words to clue us in that this is a dilution problem.

Â The other thing we notice is that there's only one substance listed.

Â Here, we only mention NaOH.

Â We don't talk about anything else present which it could react with.

Â So what I want to do is sort my information and

Â group it together correctly.

Â I see that a solution was 1.45 liters of .875 molar NaOH.

Â So that tells me that these two numbers actually belong together.

Â So I'm gonna call them C1 and V1.

Â Then I said it was diluted to a new volume up to 2.25 L,

Â what is the new concentration?

Â So I'm gonna call this V2, and what we're gonna actually be solving for

Â is C2 with that unknown concentration.

Â So, I can plug my numbers in, 0.875 molar NaOH,

Â for C1 times 1.45 L equals concentration 2 or

Â C2, which I don't know,

Â times the volume, 2.25 liters.

Â Now I can simply do the calculation to solve 4 the concentration

Â 2 which is the concentration of the dilute solution.

Â What I find is that I get 0.564 molar.

Â I know it's going to be in units of molarity because my concentration 1 was in

Â units of molarity and they must be the same.

Â When I look at this I also see that my answer is reasonable.

Â When we have a dilution happening, the more dilute solution is always going

Â to have a lower concentration than the original concentrated solution.

Â 3:42

Now let's look at another dilution problem that is slightly different from

Â the last one.

Â It's still a dilution.

Â We're still going to be using C1V1 = C2V2.

Â But we need to be careful in reading the problem to see what's

Â actually being asked for.

Â It says, what volume of H2O is required to dilute 205 mLof 1.15 molar

Â HCL solution to exactly point 81 mL.

Â So this is our C2 and then we have our V1 and our C1.

Â It also reminds us that we can assume that volumes are additive.

Â So what I want to look at is, I want to plug in the values I know, so

Â I have C1 is 1.15 molar times my V1, 205 mL.

Â 4:30

And I can leave that in milliliters

Â because I'm just gonna make sure that V2 is also in milliliters.

Â And C2 is 0.81 molar.

Â And, what I don't know is V2.

Â So, in this case, I'm solving for V2 and

Â now it's going to be in units of milliliters because that's the unit of V1,

Â 5:05

with the concentration, of 0.81 molar.

Â However, that's not actually what the question is asking for.

Â It's not asking us for the final volume of the solution.

Â It's asking, what volume of water is required.

Â In other words, how much water do I need to add to this solution in order to

Â prepare 291 mL of solution that has a concentration of .81?

Â And so I'm gonna have to add water,

Â I'm adding solvent because that's what a dilution is.

Â Moles of solute stays the same, volume of our solution changes so

Â that the volume of solvent has to be the thing that changes.

Â So here I know my final volume equals 291 mL,

Â my V1, or my initial volume was 205 mL and

Â so what I see is that I'm gonna have to add 86 mL of water,

Â 6:28

This is just a simple dilution problem, C1V1 = C2V2.

Â My initial concentration is 2.50 molar,

Â my initial volume is 20 mL equal to, I don't know C2,

Â but I do know that my final volume is 45 mL.

Â Again, same units doesn't matter, but they're both in milileiters.

Â 6:55

Now I'm going to actually solve for C2.

Â And what I find is that it is equal to 1.11 molar.

Â And this is reasonable because I have little more than doubled my volume from 20

Â to 45, and they've done a little bit more than cut my concentration in half.

Â