Solution Dilution

Alright so now I'm going to talk about solution dilution. So to dilute a solution simply means to add solvent to the solution. So remember that the solvent is our majority component. In other words, the solvent is what does the dissolving, not what becomes dissolved. That's the solute if you remember. And when we dilute a solution, the solution becomes less concentrated. So for instance, if I had a concentrated aqueous solution, aqueous meaning that the solvent is water, if I had a concentrated aqueous solution and I diluted it, that simply means adding solvent to it. And once I added a solvent to it, the solution is now less concentrated. So there's a pretty easy way to quantify solution dilution and, in general, the following equation is used: M1V1 = M2V2 So, what M and V stand for well lets just write it down. M is the molarity and remember that is in moles per liter or simply molar. So not moles per liter times molar moles per liter OR molar. And, V just stands for volumes. The V is the volume and that's typically in liters but, make no mistake, it could be given in milliliters, gigaliters just whatever prefix. It's more commonly just liters or milliliters though, maybe in microliters - anyhow -so let's go over what M1 and V1 and M2 and V2 are. M1 and V1 apply to the concentrated solution. So I'll just stick it over here - concentrated. And M2 and V2 apply to the diluted solution. So, my concentrated solution has a molarity in a volume. And my and diluted solution has a molarity in a volume. And they so happen to be related by this equation. And the reason why they are related by this equation Well, what is molarity? Remember, molarity is moles per liter. So we have the Moles 1 over the Liters1 times the volume 1 which is in liters usually and then we have the moles in 2 over the liters in 2 times the liters in 2 Well, we can cancel out liters on both sides and we're left with nothing but moles. So what this equation is really saying is that the amount of moles is conserved when we dilute a solution. Which is true because we're not messing with the solute. We're messing with the solvent. So, hopefully that gives a little bit insight as to why this equation, how it works. So, let's go through an example where we actually use this equation. And I like this coming up example because it sort of resembles a problem that you might encounter on an exam. So a lot of them happen to be worded kind of like this one. The problem says: What volume of a 12 molar HCl solution must be diluted to 10 milliliters to make 3 molar HCl solution? So I wrote the volume in blue because that's going to be our unknown. So I'm just going to call that V1, so V1 is our unknown. So that's what we're trying to figure out. And it says 'what volume' of a 12 molar solution, so I assume that the 12 molar solution from which you're taking this volume applies to solution 1 as well, the more concentrated one. So I'm going to say that the molarity of 1 is equal to 12 molar and then it says what volume of a 12 molar solution So we have the volume unknown, we have the molarity of 1, that is 12 molar, and it says that we have to dilute it to 10 milliliters to make a 3 molar HCl solution. So we're going to have a V1 and a V2 and a M2 The volume given is 10 milliliters and the molarity given is 3 molar. So, that's basically how to choose which one is M1 and with one is V1 and so forth. Typically, it doesn't matter what the subscripts 1 and 2 are as long as they're both in the same. So, in other words, I could switch this around and I could put I could put the 2 here and the 2 here and I can put the 1 here and the 1 here, it doesn't matter as long as these 2 are the same, and these 2 are the same. So, let's solve the equation. Remember that the equation is M1 V1 equals M2 V2 Well if V2 is our unknown then I can divide both side of the equation by our M2 because we're trying to get V2 all by itself M2 the M2s on the right side of the equation cancel and you end up getting V2 is equal to M1V1 over M2 So the algebra to get to the final result isn't really that bad. At this point all we have to do is just plug in our numbers and we're good to go. So M1 and V1 - M1 is 3 molar V1 is 10 milliliters And M2 that's 12 molar Molar here cancels with molar here and so we have nothing but 3 time 10 over twelve which is 30 over 12 and you can actually reduce this by dividing the top and the bottom of the fraction by 6 And you will end up getting 5 halves and this is in millimeters So that is our answer 5 halves milliliters (5 over 2) or .4 whichever one Typically you wouldn't want to express it as a fraction because you're trying to keep tabs on the number of sig figs (significant figures) and things like that. But I just did it because I'm not really trying to go over sig figs in this video I'm going over the equation. But that's how to use the equation, very straightforward all you have to do is algebraically solve for your unknown by dividing both sides by the other variable that's associated with it; with the unknown. Typically, like I said, three of them (the variables) have to be given so that you can solve for the one unknown. So there you go!