A 0.229-g sample of an unknown monoprotic acid is titrated with 0.112 M NaOH. The resulting titration curve is shown here. Determine the molar mass and pKa of the acid.

Video transcript

Hi, everyone. Welcome back. Our next problem says a 0.451 g sample of an unknown Monro acid is titrated with 0.150 molar sodium hydroxide. The resulting titration curve is shown here, determine the molar mass and the PPK A of the acid. And we have a titration curve where our vertical axis is Ph or horizontal axis is the volume of base added in milliliters. And we see this curve sort of goes gradually up at first, then shoots up very steeply and then levels off our answer. Choices are A PK A equals 6.5 molar mass equals 72.3 g per mole. BPK A equals 7.8 molar mass equals 85.9 g per mole. CPK A equals 4.2 molar mass equals 98.1 g per mole or DPK A equals 10.1 molar mass equals 68.8 g per mole. So let's start with the easy one. The PK A, it's pretty easy to determine PK A from an equivalence curve like this. We know that when the PH shoots up sharply like this, that's what's called the equivalence point. So keep in mind that as we're adding a strong base, our acid is releasing protons, our base is just soaking them right up to create water at the equivalent point. We have used up all of the protons produced by our acid. So the only thing left of the acid is the conjugate base. And our base has been also all used up because as we've been tight, we've been titrating slowly. Each molecule of base has been reacting with a mole with a proton. At this equivalent point. We've got to the point there's no more protons available. That's why we see this ph shooting up sharply. So we can get the equivalence point from this graph. And then when we're looking for PK A, the PK A is the Ph at half of the equivalence point. So we just need to determine our equivalence point. So we look over at our graph, look for that steep rise, draw a line down and we see that that puts us sorry. My line was too far to the right there drawing down from the point of that steep rise. We see we've got 3045 of our markings here. Our line is maybe a third of the way across there. So somewhere around 35 mL, so our equivalence point is approximately equal to 35 mL. So I'll put EQ point labeling that there. So what is the half of the equivalence point, that would be when we've added 17.5 mL, we've gotten halfway there. So now we just go to our graph, we find that point of 17.5 mL and look for the ph. So again, this graph has pretty big gaps here. We're not going to be able to get an exact amount. But note we have multiple choice options. So we'll be able to compare what we estimate from our graph to our multiple choice options. So again, our horizontal units are 15 mL and 30 mL. That's where our lines are 17.5 will be really close to 15. So if we just draw a line up, sort of just a little to the left of 15, we can see that we're just under eight in terms of the Ph. So when we look at label this as one half, when we look at our answer choices, which one gives us a PK A of just under eight. Well, choice A says PK equals 6.5. So that would be too low. So cross out choice A choice B says PK equals 7.8. That sounds pretty good just under eight. So we'll put a little dot By choice B Troy says PK A equals 4.2 definitely way too low. So cross that out Android D says PK equals 10.1. Well, that's way too high. So let's cross that out as well. So note that if I were on a test time is of the essence, I will pick choice B and go on. That's the only one that can be the possible pka. I wouldn't even horse around with calculating my molar mass. But this is a practice problem. Let's be thorough. We'll walk through how to calculate that. So we've got our PK A. So PK A is approximately 7.8. Now, let's think about our molar mass. Well, what is the molar mass? We don't know what our acid is. How do we calculate that? Well, remember the definition of molar mass is grams per mole. We have grams. We're told how many grams in our sample. We just need to find moss. Well, how do we do that? Well, what we do know is we know that when we've added 35 mL of base. So add equivalence point. So equivalence point, the moles of base added are going to equal the moles of acid that we started with. Well, we know how many grams we started with. So we have, if we have the moles, we can calculate our molar mass. So how do we know how many moles of base we've added? Well, we have the molarity of that basic solution and molarity is of course, moles per liter. So we've got molarity, we've got volume, we can calculate two moles. So if we say that molarity equals moles per volume, then our number of moles will be equal to the molarity multiplied by the volume. So are moles of acid will equal our molarity, which is 0.150. And we're going to write this as moles per liter so that our units will cancel. Our volume is 35 milliliters. Now to make the units cancel, we'd better add a conversion factor to change that to liters. So then we have 1 L divided by 1000 milliliters. So we'll look at that. We see that our leaders are going to cancel and our milliliters are going to cancel will be left with moles and we end up with five oops 5.25 times 10 to the minus three moles. So now all we have to do for our molar mass is divide the number of grams of acid we started with by the moles. So I'm just going to scroll up a little bit. So we have space and I will say our molar mass is equal to the number of grams, 0.451 g divided by 5.25 times 10 to the minus three mouse. We do that math and we get 85.9 grams promo. So there we go, we've calculated our molar mass of this unknown acid. And indeed, we see on choice B which we identify by the PK A, we have a molar mass of 85.9 g per mole. So that is definitely our answer and that is how we get to PK A pretty straightforwardly and a little more involved the molar mass of our unknown acid knowing the equivalence point. So choice BPK A is 7.8 molar mass is 85.9 g per mole. See you in the next video.