in this video we're going to talk about dilution problems and how to solve them the equation that you need is this one m1 v1 is equal to m2 v2 m stands for the molarity or the concentration and v is the volume now v could be in milliliters or liters but it has to match if v1 is in milliliters v2 has to be in milliliters if v1 is in liters v2 has to be of the same unit now let's say if we have a solution with a concentration of 2m and let's say it's sodium chloride how can we decrease the concentration of the solution let's say we have a volume of 200 milliliters how can we decrease the concentration if you want to decrease the concentration you need to dilute the solution you need to add water if you add water to it the solution will be less concentrated so let's say if we add 200 milliliters of water so that the final volume is 400 if you double the volume the concentration is going to be reduced by a factor of two so if you divide this by two it's going to be one so now you have a one molar sodium chloride solution so keep that in mind anytime you need to decrease the concentration of a solution simply increase the volume by adding water or whatever the solvent is now what if we want to decrease rather what if we want to increase the concentration instead of decreasing it how can we do so one way in which you can increase the concentration is by adding more salt if you add more sodium chloride the concentration will go up another way is to remove h2o if you don't want to add solute you can remove solvent and one way to do so is by evaporation or vaporization if you leave the beaker open in a dry room water will naturally evaporate and so in time the volume will decrease so let's say if you get the volume down to 100 if you decrease the volume the concentration will increase so now you'll have a four molar sodium chloride solution so any time you want to decrease the concentration and dilute it with the addition of water if you want to increase the concentration remove water by evaporation or add more solute to it now let's work on a few problems number one a 0.8 molar hdl solution i forgot that word there was diluted to a final volume of 500 milliliters if the new concentration is 0.20 what was the original volume of the solution so what can we do to figure this out for a typical dilution problem you can use this equation m1 v1 is equal to m2 v2 now if m1 is the original concentration v1 is the original volume which makes m2 the new concentration of the solution and v2 is the final volume so the original concentration is 0.8 and that's the concentration of the solution before any water was added to it our goal is to find the original volume of the solution so we're looking for v1 we have the new concentration it's 0.20 we also have the final volume it's 500. so now let's find v1 the first thing we need to do is multiply 0.2 by 500 and that's equal to 100 next we need to divide both sides by point eight a hundred divided by point eight is one twenty five so that's the original volume of the solution now let's make some sense of these numbers the volume changed from 125 to 500. so therefore the volume increased by a factor of four in a dilution problem anytime the volume increases the concentration will decrease proportionally the original concentration was 0.8 the new concentration is 0.2 so you have to divide that by 4. so if you divide the concentration by 4 the volume will increase or be multiplied by a factor of 4. so i want you to understand that concept now let's move on to part b how much water was added to make this happen so the volume changed from 125 to 500 to find out how much water was added subtract v2 by v1 500 minus 125 is 375. so 375 milliliters of water was added to the original solution to decrease the concentration from 0.8 to 0.2 number two the concentration of a 300 milliliter aqueous kcl solution was increased from 0.25 to 0.75 molar by evaporation what is the new volume of the solution so let's use the same equation m1 v1 is equal to m2 v2 our goal is to find v2 the new volume of the solution the original concentration m1 is 0.25 the original volume is 300 and the new concentration is 0.75 now let's think about this conceptually the concentration was increased from 0.25 to 0.75 that means it was increased by a factor of 3. if you increase the concentration the volume has to decrease so if the original volume is 300 to find the new volume conceptually divided by 3. so the new volume should be 100 but let's get that answer using the equation 0.25 times 300 is 75 so let's divide both sides by 0.75 75 divided by 0.75 is 100 so you'll get the same answer so that's the new volume of the solution now how much water was removed so we went from 300 to 100. to find out how much water was removed take the difference between v2 and v1 so v2 is a hundred v1 is 300 so the change in volume is negative 200 the negative sign indicates that water was removed 200 milliliters of water was taken out of solution by evaporation number three 350 milliliters of water was added to 225 milliliters of a 4.5 molar sodium hydroxide solution what is the new volume of the solution so feel free to pause the video if you want to try this problem so i'm going to use the same equation m1v1 is equal to m2 v2 so what's m1 in this problem we're looking for the new concentration so we're trying to find m2 which means m1 is the original concentration of 4.5 v1 what's a v1 in this problem we're also looking for the new volume by the way the volume that corresponds to this solution is 225 right now we don't have the value for m2 or v2 so we can't find v2 using this equation so we got to find v2 another way what other way can we find v2 let's make a list of what we have we have v1 which is 225 milliliters now the 350 what does that represent and how can we use that to find v2 the 350 milliliters is the amount of water that's added so basically that's the change in volume since we increase the volume by 350 the change in volume is positive 350. if water was removed by evaporation the change in volume will be negative so we know that delta v let me put this on the right side is equal to v2 minus v1 so delta v is 350 we're looking for v2 and v1 is 225 so basically we got to add 225 to 350 and that will give us the new volume so 350 plus 225 that's going to be 575. so that's v2 that's the new volume of the solution if you started with 225 milliliters of solution and then you add 350 milliliters of water to it then you should end up with 575 milliliters of solution now that we have v2 we could find m2 so it's 4.5 times 225 which is 1012.5 next let's divide both sides by 5 75 so m2 is 10 12.5 divided by 575 and that comes out to be 1.761 so anytime you add water the concentration will decrease in our example it decreased from 4.5 to 1.761 so that's the answer for part b it's 1.761 the answer for part a is 575. number four 60 milliliters of a 0.8 molar sodium chloride solution is mixed with 80 milliliters of a 0.5 molar kbr solution part a what is the new concentration of sodium chloride in the first three problems we were adding or removing water and that's it but in this problem we're mixing two solutions two different solutions with different substances in them we have two different substances nacl and kbr so how can we find a new concentration after mixing these two solutions well first we need to calculate the final volume of the combined mixture and that final volume is going to be 60 milliliters from the nacl solution plus 80 from the kbr solution so v2 the final volume for both solutions which is not one solution is 140 milliliters so let's use pictures to represent what's going on here let's say the first solution is nacl and the second solution is kbr the volume for the first one is 60 milliliters and for the second it's 80. and the concentration of sodium chloride is 0.8 and the concentration of kbr is 0.5 so once we mix the two solutions the volume of the new solution is going to be 140. it's going to be 60 plus 80. what's going to happen to the concentrations of nacl and kbr will they increase or will they decrease for nacl the volume went up from 60 to 140 so it didn't more than double so therefore the concentration will decrease any time the volume goes up the concentration will go down so since the volume increased more by more than a factor of two the concentration should decrease more than a factor of two so the final answer for nacl should be less than 0.4 now for kbr the volume increased from 80 to 140 it didn't double it increased less than a factor of two so the concentration should be reduced by less than a factor of 2. so the final concentration should be greater than 0.25 but less than 0.5 so let's start with the concentration of nacl so we're going to use the same formula m1 v1 equals m2 v2 but with reference to nacl the original concentration is 0.8 the original volume is 60. we're looking for the new concentration of nacl and the new volume is 140. so m2 is going to be 0.8 times 60 divided by 140. so the new concentration of nacl is 0.343 as you can see is less than 0.4 now let's do the same thing for kbr let's use the same formula the original concentration of kbr is 0.5 the original volume is 80. the new volume is the combined solution which is uh it has a volume of 140. so m2 is going to be 0.5 times 80 divided by 140. so the new concentration of kbr is 0.286 m so it's greater than 0.25 but it's less than 0.5 so now you know how to find a new concentration of two solutions after they're mixed with each other now what about this one 150 milliliters of a 0.6 molar calcium chloride solution was mixed with 300 milliliters of another calcium chloride solution what is the new concentration of the mixture and the last example we mix two solutions that contain different substances and so the concentrations of both solutions decreased in this example we're mixing two solutions with the same substance so the concentration of this solution the second solution will decrease but the concentration of the first solution will increase so what's going to happen is you're going to get a weighted average do you think the new concentration will be closer to 0.6 or closer to 0.9 what would you say the midpoint of 0.6 and 0.9 if you add them up and divide by 2 is 0.75 notice that we have a greater quantity of the more concentrated solution we have twice as much more therefore the new concentration is going to be somewhere between 0.75 and 0.9 because we have more of the concentrated solution so basically this problem is a weighted average problem now we need to modify the equation a bit it's going to be m1 v1 plus m2 v2 and that's equal to m3 v3 so if you're looking for m3 the new concentration it's going to be m1 v1 plus m2 v2 divided by v3 i want you to understand how this equation works and what it means whenever you multiply molarity by volume it's going to give you moles n is equal to cv or mv c and m are the same concentration and you could use the units to help you figure this out m the molarity is defined as the moles per liter and if you multiply by the volume with units of liters this will give you the moles so m1v1 is basically the moles of the first solution m2v2 is the moles of the second solution and v3 is the volume of basically the combined mixture so n1 plus n2 is basically the total number of moles of calcium chloride vt is the total volume if you take the total moles and divided by the total liters you're going to get the new concentration of the solution m3 so i want you to understand how this problem works so if you found the moles of the first calcium chloride solution and then add it to the moles of the second calcium chloride solution and divide it by the total volume you would get the same answer m3 or you could just use this equation just add a an mv component to it so let's say if we wanted to mix three solutions with the same substance i would use the formula m1 v1 plus m2 v2 plus m3 v3 is equal to m4 v4 where v4 is the combined volume of everything so you can keep adding mvs to these equations if you're adding a solution that contains the same substance if not then it doesn't work now let's finish this problem m1 is 0.6 v1 is 150 now for the second solution m2 is 0.9 and v2 is 300 v3 is the sum of 150 and 300 which is 450. 0.6 times 150 that's 90. so right now this is not 90 moles but 90 millimoles because we're multiplying molarity by milliliters if we multiplied molarity by liters this would represent moles so the first solution has 90 millimoles the second solution 0.9 times 300 contains 270 millimoles and so this is equal to m3 times 450 milliliters 90 plus 270 that's 360. so to find m3 it's going to be 360 divided by 450 which is 0.8 so that's the concentration of the combined mixture as we said before it's between 0.6 and 0.9 but it's closer to 0.9 because the midpoint between 0.6 and 0.9 we said it's 0.75 and we mentioned that the answer should be somewhere between 0.75 and 0.9 because we have a greater quantity of the more concentrated solution and so the average concentration is going to be closer to 0.9 since we have a greater volume of the concentrated solution so hopefully that makes sense just remember if you mix in two solutions with the same substance your final answer the concentration is the weighted average of these two values if the volumes were equal and then the concentration would be exactly in the middle would be 0.75 but if the volumes are unequal the final concentration will be closer to the concentration that has a greater volume you