0:04 so here in the second half of section
0:07 6.3 molarity we're going to focus on
0:09 being able to perform dilution
0:16 so dilution is the process whereby the
0:18 concentration of a solution is lessened
0:21 by the addition of solvent and it's very
0:24 common in chemistry to have stock
0:26 solutions that are concentrated and then
0:30 dilute them for use at a later time so
0:33 as noted here dilution is a common means
0:35 of preparing Solutions of a desired
0:37 concentration and by adding solvent to a
0:39 measured portion of a more concentrated
0:42 stock solution we can achieve a
0:45 particular concentration that we desire
0:49 so um here's an example where we have
0:51 two solutions they both contain the same
0:54 mass of copper but the solution on the
0:57 right is more dilute because it has a
0:59 greater volume right the copper
1:01 nitrogate nitrate is dissolved in more solvent
1:07 so the molar amount of solute in a
1:09 solution is equal to the product of the
1:12 solution's molarity and its volume in
1:14 milliliters so that's what's shown in
1:16 this equation here number of moles
1:19 equals molarity times volume in liters
1:22 and we can see that that's the case by
1:25 looking at the units so molarity has
1:30 units of moles per liter and liters are
1:33 and volume in liters obviously has the
1:36 units of liters so that our leaders
1:38 could cancel out and we're left with
1:41 number of moles
1:43 so expression like expressions like
1:44 these may be written for a solution
1:49 before which is uh represented by the
1:52 number one and after represented by the
1:56 number two a dilution occurs okay
2:00 and furthermore since the dilution
2:02 process doesn't change the amount of
2:05 solute in the solution uh the number of
2:07 moles at the beginning and one is equal
2:10 to the number of moles of solute at the
2:13 end okay so again just to be clear the
2:16 number of moles that you begin with uh
2:18 of solute is the same at the beginning
2:20 of the process and at the end of the
2:22 dilution process
2:26 so what that means then is that N1
2:31 equals N2 and therefore we can set the
2:34 two equations that we had earlier equal
2:36 to each other so
2:39 um we have that the molarity in the
2:40 beginning times the volume in the
2:43 beginning is equal to the molarity at
2:46 the end times the volume in liters at
2:47 the end
2:51 and we can actually write a very general
2:54 equation for any type of concentration
2:56 units that we might want to use and for
2:59 any volume units that we want to use so
3:02 typically what we'll simply say is that
3:07 C1 V1 equals C2 V2 meaning the initial
3:10 concentration times the initial volume
3:14 is equal to the final concentration
3:17 times the final volume
3:19 so let's see how we can use this to
3:22 calculate a concentration or volume
3:26 so here we have a problem it says if
3:30 0.850 liters of a five molar solution of
3:33 copper nitrate is diluted to a volume of
3:37 1.8 liters by the addition of water what
3:40 is the molarity of the diluted solution
3:43 so what is the final molarity so the
3:45 final molarity that we're interested in
3:49 is C2 right so that's the thing we don't know
3:50 know
3:53 um and so we what we can do is solve for that
3:54 that
3:58 so dividing both sides by V2 we have
4:03 that C2 equals C1 V1 divided by V2
4:06 and then we can
4:10 put in our values the initial
4:14 concentration is 5 molar so that goes here
4:15 here
4:17 the initial volume
4:20 is 0.850 liters so that goes here
4:25 and the final volume is 1.80 liters so
4:27 that's V2 that goes in the bottom here
4:30 and then we simply work the math out
4:34 notice that liters will cancel out and
4:36 we'll be left with the concentration in
4:39 molarity which turns out to be 2.36
4:41 molar now just a final word on this is
4:44 we can use this equation to um to solve
4:46 for any one of these four quantities
4:48 right so as long as we know the other
4:50 three then we can solve for the fourth
4:52 so if you wanted to know for instance
4:55 what volume you needed to achieve a
4:57 certain final concentration we could
5:01 solve for v2 or if you wanted to know
5:04 um uh how much volume of a particular
5:07 stock solution say C1 that you should
5:11 use that would be V1 we can solve for V1
