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# Titration calculation example

## Video transcript

- [Voiceover] Let's do another titration problem, and once again, our goal is to find the concentration of an acidic solution. So we have 20.0 milliliters of HCl, and this time, instead of using sodium hydroxide, we're going to use barium hydroxide, and it takes 27.4 milliliters of a 0.0154 molar solution of barium hydroxide to completely neutralize the acid that's present. All right, so let's start with what we know. We know the concentration of barium hydroxide. It's 0.0154 molar, and we also know that molarity is equal to moles over liters. All right, so we have 0.0154 as equal to, let's make moles X, over liters. 27.4 milliliters is 0.0274 liters, right? So that's 0.0274 liters. We solve for X, and X of course represents the moles of barium hydroxide. So let's get out the calculator here, and let's do that. So let's get some room over here. So we take 0.0154 and we multiply that by 0.0274, and that gives us, this will be 4.22 times 10 to the negative fourth, right? So that's equal to 0.000422 moles of barium hydroxide. All right, next, let's write the neutralization reaction. So we have barium hydroxide reacts with HCl. So barium hydroxide plus HCl gives us, for our products, we have H plus and OH minus, so that's H20. And then our other product, this is barium two plus, right? This is BA2 plus, and over here we have Cl minus 1. So we have BA2 plus and CL minus 1, so you could cross those over. So BACl2, right? So BACl2, barium chloride, as our other product here. All right, next we need to balance our equation, right? We need to balance the neutralization reaction here. So let's start by looking at the chlorines. So over here on the left, we have one chlorine. On the right, we have two. So we need to put a two right here, and now we have two chlorines on both sides. Next, let's look at hydrogens. So on the left side, we have two hydrogens here, and then we have two over here. So we have four hydrogens on the left. On the right, we have only two hydrogens. So we need to put a two here for this coefficient to give us four hydrogens on the right. So now we have four, and we should be balanced, right? Everything else should be balanced. Let's look at the mole ratio for barium hydroxide to HCl. For every, right, there's a one here. So for every one mole of barium hydroxide, we have two moles of HCl. So we already calculated how many moles of barium hydroxide that we used in our titration, right? That's 0.000422. So therefore, we had twice as many of HCl. So we can multiply this number by two, and we'd figure out how many moles of HCL we have. Or you could set up a proportion. Right, so if we're talking about a ratio of barium hydroxide to HCl, our mole ratio is one to two. Right, and our moles of barium hydroxide, let me go ahead and use a different color here. That's up here, that's 0.000422 moles of barium hydroxide. Our goal is to find how many moles of HCl were present. And so obviously you just need to multiply 0.000422 by two. And so we get X is equal to 0.000844, right? That's how many moles of HCl we have at our equivalence points. All right, so finally, we just have to calculate the concentration of our acid solution, right? Let's go back up here so we can see what we started with. Right, so we started with 20 milliliters of HCl. Right, and 20 milliliters would be, move our decimal place, 0.0200 liters. So now we have moles, right? We have moles, and we have liters. So we can calculate the concentration. All right, so the concentration of HCl in our original solution would be, we had 0.000844 moles. All right, divide that by liters. That was 0.0200 liters, right? 20 milliliters is equal to 0.0200 liters. And so we can do our calculation here. So we can take 0.000844, and we can divide that by 0.0200, and we get for our answer here 0.0422 molar. All right, so the concentration of HCl is equal to 0.0422 molar, and we are finally done, right? That's our concentration of our acid solution. Let's see what happens if you try to use MV is equal to MV, that shortcut that we learned about in the last video. So this would be MV is equal to MV, and let's do the molarity of the base times the volume of the base is equal to the molarity of the acid times the volume of the acid. So for our base, the concentration was 0.0154 molar, and the volume of base that we used was 27.4 milliliters in our titration. For the acid, we don't know what the molarity is. That's what we're trying to find in the problem, and the volume was 20.0 milliliters, right? So let's do that calculation. So trying to use the shortcut way, 0.0154 times 27.4 gives us that number, divide by 20, right? So we get 0.0211, right? So for our answer, for X, we get 0.0211 molar. And so you can see that's not the correct answer, right? Here we've got a concentration that's half of the concentration that we got when we did it the longer way. And so if you want to use the shortcut way for this problem, you would have to multiply by two, right? So if you multiply your answer by two, then you get the correct answer, 0.0422 molar. And a lot of the time, students have a hard time figuring out what you do, right? So where do you mulitply by two? How do you do that? Of course, you can figure it out by looking at your balanced equation up here, right? But it's tricky for a lot of students. And so that's why the shortcut way isn't always the best way. You can still use it if you understand how to use it, right? But it's a little bit better for these problems, when your mole ratio is not one to one, to go through the longer way and do it, if you want to be absolutely sure that you're getting it correct.