In this module we are going to look at the stoichiometry, when dealing with solutions. By the end of this module you should be able to use concentration units as relationships in stoichiometric calculations. So we've seen this map before. We're going to go back to it again now that we're dealing with concentrations, and what we see is that the key part of this map is that we have to have moles of A. We have to have coefficients from our balanced equation to get to moles of V. That doesn't change when we're dealing with solutions. The only thing that changes is how we get to moles. Instead of starting with, say, the mass of a substance or the number of molecules, now we're going to start with liters. And we're going to use molarity, in order to get to moles of A. Now we can also substitute other units of concentration with other information about that solution to get moles as well. It just depends on what we have available to us. And since molarity is the most commonly used concentration unit, we're going to refer to that in, with examples of molarity. When I get to the other side or my unknown side. I see that I can also go back to the concentration or go back to a volume. However just because I start with one set of units, doesn't mean I have to go to that same unit on the opposite side. There's nothing wrong with starting with litres of one substance and finding the mass of another or vice versa. It just depends on what's being asked for and what type of sample we're preparing. The main type of solution stoichiometry that we look at is titration. And in titration we have a solution, which has an accurately known concentration, being added gradually to a solution of an unknown concentration. We monitor the progress of that reaction, and the way we monitor that will depend on the type of reaction it is. But we're going to monitor the progress of that reaction. And when we determine the reaction is complete, we'll be able to use the quantitative measurements to identify the concentration of our unknown solution. This is just a stoichiometry calculation, but dealing with units of molarity and volumes instead of dealing with grams and molar masses. So I usually say that titration problems are simply fancy stoichiometry problems. While there are other types of titrations, we're going to focus on acid-base titrations. This involves the reaction of an acid and a base. We have some standard solution for which the concentration is known. We run the reaction til we reach the equivalence point, and this is when the acid has completely reacted with the base. And it's the quantitative measurements at this point that allow us to calculate the concentration of our unknown solution. The other thing frequently present is an indicator, and this is a substance that has different colors in acid and base solutions, and so it becomes a marker to determine when we've reached the equivalence point. We can also use things such as pH sensors or pH meters to determine this point as well. Let's look at an example of how we can solve these problems. Remember, titration problems are simply stoichiometry problems. They just look a little bit different because we're dealing with volumes and concentrations instead of masses and molar masses. So this question asks what volume of 0.2010 molar NaOH is required to neutralize, and that just means to get to the equivalence point, 20 millilitres of 0.1030 molar HCl in an acid-base titration. Because we're looking at a stoichiometry problem, the first thing we have to do is actually get our balanced chemical equation. So I look at my two reactants, I have HCl plus NaOH, and for an acid-base reaction, the products are salt, NaCl plus water. Similar to the way that we looked at precipitation reactions where we were swapping the anions, we're basically doing the same thing here. We see that we have a balanced equation, now we can start our stoichiometry problem. Now what I want to do is I'm going to rewrite my concentrations. In terms of moles per liter. That's going to make it easier for me see half or to use them when I set up my calculation. Because now these values are seen more easily as relationships between moles and liters,as opposed to a just a single value. I see that my concentrations are relationships, the only other number I have is my 20 millilitres. So I'm going to start with the 20 millilitres of HCl. Then I need to look at the concentration of HCl because I want to use that to get to moles and I see that I have units of liters. So I first need to convert from milliliters to liters. Now that I have liters of HCL, I can use the concentration 0.1030 moles of HCL per liter of solution. Now I can use my mole to mole ration just as I did in every other stoichiometry problem. So, I have one mole of HCL to one mole of NaOH, and then I can use the molarity of NaOH for 0.2010 moles per liter of NaOH. And if I want to get this back to milliliters since the value is started out in millilters, I can do that as well. So 1 liter in 1,000 milliliters are equal. Now I can see that volume cancels with volume of HCl. Liters of HCO with liters, moles and moles of HCL, moles and moles of NaOH, liters and liters of NaOH, and now what I'll be left with are millilitres of NaOH. After I do the calculation I end up with 10.25 milliliters of NaOH. So I know that if I have .201 molar NaOH, it's going to take 10.25 milliliters to completely consume all of the 0.103 molar HCl that I have in my 20 millilitre sample. Now let's look at an example for you to try. What volume of 0.175 molar HCl is needed to titrate 30 millilitres of 0.200 molar BaOH2? Hint, write a balanced equation. Make sure you start out with a balanced equation, because this is not going to be a one to one ratio as we saw with the HCl NaOH conc, concentration problem. So we see that we get an answer of 68.6 millilieters. I'm going to first write my balanced chemical equation, which will be HCL plus BAOH2, going to BaCl2 plus H2O. And what I find is that I'll need 2 HCl and 2 waters to get a balanced chemical equation. Now I can put the information I know about HCl and my BaOH. So I have 30 millilitres of the BaOH, that's 0.200 molar. And I'm trying to find the volume of the HCl. So what I want to do is start with my 30 millilitres, which I'm going to go ahead and convert into liters of my BaOH2. I'm going to use my molarity of 0.200 moles per liter of BaOH2. Now I need to use my mole to mole ratio. And this is where the balanced chemical equation is so important. Because I no longer have a one to one ratio, I know that for every one mole of BaOH2, I'm going to react with 2 moles of HCl. Now, I know my moles of HCl, now I can use the molarity of HCl to figure out my unknown volume. And so I have 0.175 moles per liter of HCl. My units liters cancel. Moles of BaOH2 cancel. Moles of HCl cancel. And my answer will be left in liters of HCl. So I get 0.0686 liters of my HCl solution, which is equal to 68.6 milliliters of HCl.
