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Now let's talk about dilutions. The governing equation for dilutions is equal to molarity one times volume one, molarity two times volume two. Now, this works out because a molarity is a mol per liter, a volume in liters. The next set, we've got moles per liter, times liters.

So essentially we're saying that if we don't add any new moles, but just add a dilution, or some liquid that doesn't contribute to the amount of the substance in concern, we won't change the moles. All we change is the volume, which affects the molarity, or the concentration. So increase the volume of the inert material or the solvent, and all we do is dilute the solute.

We don't affect the moles of the solute present, just the concentration. So how much water needs to be added to 300 milliliters of a 0.8 Molar solution of Sodium Hydroxide to dilute it to a 0.6 Molar concentration? So if we're going to do this we set up an equation, M1V1 = M2V2. Now we can plug in our values. Molarity one is 0.8 molar, volume one is 300 milliliters.

This equals sum molarity two which is 0.6 molar times an unknown volume. We can then solve for that by rearranging volume 2 = 0.8 times 300 is 240 molar milliliters over 0.6 molar. Molars cancel, leaving us with a volume in milliliters.

240 divided by 0.6 might look like a complicated division, but we can simply multiply both sides by 10. 2400 divided by six, six goes into 2400 400 times, so there are 400 milliliters in V2. Now, you can't stop here. This is not the answer.

But it's probably the answer that most people will choose. What we wanna know is how much water is added. This is the final volume, which means that to find out how much water's added, we need to take our final minus Initial to get the volume added. So we have 400 milliliters minus the 300 initial we see that we only need to add 100 milliliters.

So again, on problems, you can't stop before you actually solve them. This dilution only calculated the final volume. The added volume is a totally different number. So we need to take the final volume, subtract the initial volume to get the added volume. Don't stop short on problems.

Make sure you solve what's asked for. Now, let's do a little bit more complicated dilution problem. What is the final concentration of a solution if 250 mL of 0.8M NaOH is added to 150 mL of 0.4M NaOH, and then diluted with another 100 mL of pure water? Now when we set this up, we're gonna write our dilution equation, M1V1 + M2V2 = some M3V3.

Now on this side, M1 is 0.8 molar plus a V1 of 250 Millilitres plus an M two of 0.4 molar times 150 millilitres equals sum unknown M three times a total volume. What's the total volume? 250 milliliters plus 150 milliliters + 100 milliliters of pure water. And now we can solve these.

0.8 times 250 is 200 Molar milliliters + 0.4 times 150 is 60 Molar milliliters equals M three times 250 plus 150 plus 100 equals 500 milliliters. You're gonna divide this out, 260 Molar milliliters over 500 milliliters. The milliliters cancel and we get 260 / 500 M = M3.

If we multiply these both by 2, we can say that equals 520 / 1000. We then do this division and say that the molarity is equal to 0.52. Now, why does this work? Remember in dilutions, essentially we're just trying to keep track of the mols of NaOH and we know that M1 V1 gives us a set number of mols of NaOH.

We could've added in another M1 V1 for the water but the molarity of the water is 0, so it doesn't even matter. We just plug it in to the total volume because it adds no moles of sodium hydroxide. All the sodium hydroxide comes from our 0.4 molar solution and from our 0.8 molar solution.

Moles of sodium hydroxide plus moles of sodium hydroxide, divided by a total volume. And that's essentially what a dilution problem is, calculating the number of moles in the system and dividing it by the total volume.

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