In this module we're gonna talk about the dilution of solutions or how we create less concentrated solutions from more concentrated ones. By the end of this module you should be able to calculate the concentration of a diluted solution. A dilution is the preparation of a less concentrated solution from a more concentrated one through the addition of additional solvent. When we add additional solvent to our solution, we see that the moles of our solute, represented here by the green spheres, do not change. We still have the same number on the left as the right, the only that changes is the volume of the solution. And this changes because we have additional solvent in the solution. We have a formula we use that's only used for dilutions to help us calculate one of our unknowns. We have an initial concentration and volume, and we have a final concentration and volume. It doesn't really matter what we call our initial conditions or final conditions, the math will still work out. What we have to worry about is whether or not we pair up the correct concentration and volume with each other. It doesn't matter what our units of concentrations are, as long as the units of C1 match C2 and likewise for volume. The units of volume one must match that for volume two. It doesn't matter what the units are as long as they are the same. Let's look at an example to see how we can use this equation. So we have C1V1 = C2V2. There are some words I have to look for to see if this is in fact a dilution problem. The most obvious is diluted, or dilution. We can talk about the addition of solvent. We can talk about the amount of solute staying same, but the volume of the solution changing. And these are all key words to clue us in that this is a dilution problem. The other thing we notice is that there's only one substance listed. Here, we only mention NaOH. We don't talk about anything else present which it could react with. So what I want to do is sort my information and group it together correctly. I see that a solution was 1.45 liters of .875 molar NaOH. So that tells me that these two numbers actually belong together. So I'm gonna call them C1 and V1. Then I said it was diluted to a new volume up to 2.25 L, what is the new concentration? So I'm gonna call this V2, and what we're gonna actually be solving for is C2 with that unknown concentration. So, I can plug my numbers in, 0.875 molar NaOH, for C1 times 1.45 L equals concentration 2 or C2, which I don't know, times the volume, 2.25 liters. Now I can simply do the calculation to solve 4 the concentration 2 which is the concentration of the dilute solution. What I find is that I get 0.564 molar. I know it's going to be in units of molarity because my concentration 1 was in units of molarity and they must be the same. When I look at this I also see that my answer is reasonable. When we have a dilution happening, the more dilute solution is always going to have a lower concentration than the original concentrated solution. Now let's look at another dilution problem that is slightly different from the last one. It's still a dilution. We're still going to be using C1V1 = C2V2. But we need to be careful in reading the problem to see what's actually being asked for. It says, what volume of H2O is required to dilute 205 mLof 1.15 molar HCL solution to exactly point 81 mL. So this is our C2 and then we have our V1 and our C1. It also reminds us that we can assume that volumes are additive. So what I want to look at is, I want to plug in the values I know, so I have C1 is 1.15 molar times my V1, 205 mL. And I can leave that in milliliters because I'm just gonna make sure that V2 is also in milliliters. And C2 is 0.81 molar. And, what I don't know is V2. So, in this case, I'm solving for V2 and now it's going to be in units of milliliters because that's the unit of V1, and what I find is 291 mL. Now, that is the final volume of the solution with the concentration, of 0.81 molar. However, that's not actually what the question is asking for. It's not asking us for the final volume of the solution. It's asking, what volume of water is required. In other words, how much water do I need to add to this solution in order to prepare 291 mL of solution that has a concentration of .81? And so I'm gonna have to add water, I'm adding solvent because that's what a dilution is. Moles of solute stays the same, volume of our solution changes so that the volume of solvent has to be the thing that changes. So here I know my final volume equals 291 mL, my V1, or my initial volume was 205 mL and so what I see is that I'm gonna have to add 86 mL of water, To my original solution in order to prepare the solution being asked for in the problem. Now let's take an example where we have to find the final concentration. A 20 mL solution of 2.50 M NaCl is diluted to a volume of 45.00 mL. What is the final concentration? This is just a simple dilution problem, C1V1 = C2V2. My initial concentration is 2.50 molar, my initial volume is 20 mL equal to, I don't know C2, but I do know that my final volume is 45 mL. Again, same units doesn't matter, but they're both in milileiters. Now I'm going to actually solve for C2. And what I find is that it is equal to 1.11 molar. And this is reasonable because I have little more than doubled my volume from 20 to 45, and they've done a little bit more than cut my concentration in half. In the next module, we're gonna a look at solution stoichiometry.