A 0.229-g sample of an unknown monoprotic acid is titrated
with ...

Question

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.

Accepted Answer

hi everyone for this problem. It reads 0.127 molar sodium hydroxide was used to titrate 0.219 g of an unknown mono protic acid sample Using the tight rations curve based on the gathered data. What are the peak A. And molar mass of the acid? So the two pieces of information that we want to answer for this problem are the value for P. K. A. As well as the value of the molar mass of the acid. And were given a titillation curve. So by looking at this thai tradition curve, we can gather some information from this to be able to answer our peak A. Okay, so the equivalence point is something that's going to be important for us to answer this question of peak A. And the equivalence point is where there's a sharp ph rise. Okay, so taking a look at our titrate in curve, we have a sharp P. H. Rise at about here. Okay. And so if we take a look at our Y access and our X axis, we see that that ph is about nine and that volume Is about 18 ml. Okay, so that means our equivalence point is equal to 18 male leaders. So something that's important to know is that ph is equal to P. K. A. So we know the ph. And so our ph is equal to R P K. A. At one half the equivalence point. We know what our equivalence point is. So that means we'll be able to identify the value for R P. K. A. Okay, so our peak A. Is going to equal one half are it's going to equal our ph at one half of our equivalence point. Okay, so R P H R. The equivalence point we said is 18 mL. So we have one half times 18 mL. So this gives us P. H. At nine mL. So going back to our tight rations curve, let's look at what our ph is at nine mL. Okay. At nine mL which is about here, Our P. K. A. We can say is going to be about 5.8. Okay, so we can say P K A. Is equal to 5.8. So that's the first part. The second part of the question asks us for the molar mass of the acid. So let's just recall that molar mass Is equal to g over. Okay, so we're given some information in this problem. That's going to be useful for us to use to solve and that is our volume. We know we have 18 ml of sodium hydroxide. So let's start there. Okay, so we have 18 mL of sodium hydroxide and we have a male arat. E. Was also given. Okay, so let's go ahead and actually we don't need that. We need the we were given the grams. So let's plug that in. We're given the grams of The sample which is equal to 0.219g. So now we need to solve for the moles. Okay. And so we're told in the problem we have .127 moller of sodium hydroxide. So what moller stands for is or it's the same thing as concentration. So that 1.27 Moller is 0.127 moles over leader. And we're going to have a volume. We're going to multiply this by our volume of .025 L. Okay. Okay. So when we do that, our leaders cancels and we'll be left with our unit of moles. So this gives us 3.175 times 10 to the negative third moles. So now we can plug this in here 3.175 times 10 to the negative third. Okay, So once we solve for that, we're going to get a value. Okay? And that value is going to be 95.8 g per mole. So our molar mass is equal to let's go ahead and write this as our final answer. Our Molar mass is equal to 95.8 g per mole. And that is the second answer to our question. That's the end of this problem. I hope this was helpful

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.

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