Blueprint MCAT Full-Length 1: Passage 5 — Titration

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MP 187: Blueprint MCAT Full-Length 1: Passage 5 — Titration

Session 187

Phil is back for some more MCAT passage. Today, we cover Passage 5 of Blueprint’s full-length one which is all about titration. If you haven’t yet, go sign up for their full-length one at Blueprint MCAT.

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[03:55] Passage 5 (Questions 23-26)

Potentiometric titration is a useful means of characterizing an acid. Instead, the cell potential is measured across the analyte solution. When cell potential is plotted against titrant volume added, the equivalence point is the cell potential at the inflection point, the midpoint of the steep segment of the titration curve.


It’s similar to regular hydration where you add a base, and we have an equivalence point. But we’re also adding in this ability to measure the voltage with the electric potential. And so that’s why it’s a potentiometric or potentio-measuring thing. And so that’s telling us when we reach this equivalence point, rather than looking at the actual indicator.

Continuation of Passage…

For polyprotic acids, acidic hydrogen will produce an inflection point only if it is not very weakly acidic and if its ionization constant differs from that of any other acidic hydrogen of the acid by at least a factor of 104.

Captopril (molecular weight: 217.29 g/mol), shown in Figure 1, is a competitive inhibitor of angiotensin-converting enzyme (ACE).

Figure 1 Captopril

Students studying captopril were provided the following in vivo IC50 values (the minimum plasma concentration need to inhibit 50% of target enzyme activity in vivo) for captopril inhibition of ACE under different pH conditions.


IC50 is the minimum concentration of whatever it is that we’re talking about needed to inhibit 50%. So it’s the inhibitory concentration at 50%.

Table 1 Captopril IC50 values

pH rangeIC50 (mean 士 standard deviation)
1.5-3.70.058 士 0.013 pM
3.8-9.50.012 士 0.004 pM
9.6-12.50.069 士 0.017 pM


The pKa for the salt hydro group is 9.8. And the pKa of the carboxylic acid is 3.7, which are those middle points in those pH ranges. They have the pH up to 3.7, from there to 9.6, and then above 9.6. So they’re telling us that what’s happening at these different points is we’re dealing with the molecule.

IC50 is the amount of the drug we need to shut down 50% of the ACE enzymes. And so a smaller number is a more powerful drug because I don’t need very much of it to shut down 50% of angiotensin-converting enzyme.

While bigger IC50 is actually a less powerful drug. And so that’s something that seems counterintuitive, where the smaller the number, you need less of it because it’s telling you how much you need to shut down the enzymes.

[09:33] Continuation of Passage 5

Students then performed a potentiometric titration of captopril in order to determine the captopril content contained in a tablet formulation. Two tablets were ground and homogenized, producing 104.4 grams of fine powder. The powder was then dissolved in 100 mL of water and titrated with a solution of 2x 10-2 M NaOH. The potentiometric titration curve obtained, along with a plot of the rate of change of potential during the titration, is shown in Figure 2.

Figure 2 (a) Potentiometric titration curve of captopril in NaOH solution; (b) Rate of change of cell potential


You could see the traditional titration curve appearance with that s-shaped curve. And then we have this sharp peak of the cell potential, right around 7.5 or something like that. It’s telling us that that’s probably where the equivalence point is happening.

Remember from the first paragraph that they told us that we get this indicator through the potential when we hit the equivalence point rather than using an indicator. And so they use that and Figure 2.

[11:31] Question 23

How many noles of captopril were present in the original analyte solution tested?

(A) 7.5 x 10-5 moles

(B) 1.5 x 10-4 moles

(C) 7.5 x 10-3 moles

(D) 1.5 x 10-2 moles

Thought Process Behind the Correct Answer:

They’re trying to figure out just how much captopril is in this container. How many moles of it is in a container and they did this by titrations curves. The titration is to figure out how much is there. The key here is that anytime you hit an equivalence point, the moles of the acid are equal to the moles of the base. And that’s the point of the equivalence point which is equivalent acid-base.

7.5 mL is not how many actual moles of the NaOH is in there. It’s just how many drops we put in there. But they do give us the concentration of it.

You can just multiply those two because you got mol/L x L, if you convert the mills to liters. So then you’re just left with moles, and so it’s going to be 2 x 10-2 molarity times 7.5 x 2-3 once you convert that to liters, so 2 x 7.5 is 15.

And then you end up with 1.5 times 10-4 moles per liter when you convert that back to scientific notation, so the answer is B.

Correct Answer: B

[16:21] Question 24

If the students perform an enzyme inhibition assay using captopril, which of the following changes in the kinetic parameters of ACE should be expected?

(A) Vmax decreased; Km unchanged

(B) Vmax decreased; Km increased

(C) Vmax unchanged; Km decreased

(D) Vmax unchanged; Km increased

Thought Process Behind the Correct Answer:

You have to know how these different inhibitors work in terms of competitive, noncompetitive. Then it says a competitive inhibitor right above Figure 1.

So the Vmax is how fast this angiotensin-converting enzyme can work at maximum speed. So generally, you add an inhibitor and it slows down a reaction.

The Vmax is like how fast can this enzyme work? Now, if you had a competitive inhibitor, that’s going to compete for the active site with the angiotensin because they’re both going to go into the same thing. If I keep adding more and more angiotensin, my inhibitor isn’t going to get a chance to get in there.

It’s like if you’re trying to get in the Apple Store, and there are 10 million people trying to get in. I’m being outcompeted because there are too many people trying to go for that spot. In this scenario of an infinite substrate, it tells that this enzyme is still going to work as fast as it normally does.

Hence, its Vmax is going to be unchanged because the inhibitor is not going to get a chance to go in there and shut it down.

At this point, we know it’s going to be either C or D because we know that Vmax can’t be altered.

Km is how much substrate needed to add to get to that 1/2 Vmax. If I added an inhibitor, that’s going to make it harder for the substrate in there. So I’m going to have to add more substrate than normal to get that to the half of its maximum speed.

And so our Km which is a measurement of how much substrate we need to reach our half maximum velocity, that’s gonna go up and so the answer has got to be D.

Correct Answer: D

[20:22] Differentiating Competitive Inhibitors vs Noncompetitive

So the Vmax goes or the velocity will decrease when you add an inhibitor, but if you add enough substrate, then the inhibitor can’t do anything.

Imagine that your job is to go in and shut down the Apple Store, you’re trying to inhibit the Apple Store because you work for Samsung.

A competitive inhibitor is going to try to get into the same spot that everything else is trying to get into.

A noncompetitive inhibitor is going to go slick in the back door and shut it down. It doesn’t matter how many people are trying to get in the front door, but you’re going to be able to shut down the enzyme all the time.

But in the case of a competitive inhibitor, if you keep adding substrate eventually, it’s like there are 10 million people trying to get to that front door. Then you’re trying to get it shut down but you can’t. Because there’s just you’re being outcompeted. It’s like it’s all that extra.

[21:59] Question 25

Which of the following protonation states of the captopril and carboxyl groups is required to maximize captopril’s inhibition of angiotensin-converting enzyme (ACE)?

(A) Deprotonated thiol and protonated carboxyl

(B) Protonated thiol and deprotonated carboxyl

(C) Deprotonated thiol and deprotonated carboxyl

(D) Protonated thiol and protonated carboxyl

Thought Process Behind the Correct Answer:

This protonation stuff is one of the things that causes people some confusion. The nucleus has just a proton and then it’s got one electron. So if you pull that electron, it’s just a proton. And so when we talk about protonation states, it’s just adding or removing hydrogen.

We’re going to have to use Figure 1 in determining the answer here. The pH range where this is most powerful was between 3.8-9.5, which is between the things they give us in Figure 1. So that’s going to be higher than one of these. The pH is going to be higher. Then the pKa for carboxylic acid is going to be lower than the pKa for the salt hydro group, the SH group. So just off of that, it’s probably going to be ones protonated and ones deprotonated.

If it was below both of them, they’re both going to be the same. If the pH was above both of them, they would both be the same. And so just from that, the answer is probably A or B at that point.

Now, a lot of times students struggle a little bit with this, but I think the easiest way to think about this is to figure out whether it’s going to have hydrogens on it or not at these different pH’s. Think about what’s going to happen at the lowest pH if there’s an infinite number of hydrogens. Everything’s going to have hydrogen on it. At a really low pH, the waters just full of hydrogens.

And so at a pH of one, all of these things are going to have hydrogens on them. And then as we remove hydrogens from there, aka, we raise the pH by removing the hydrogens. Hydrogens are going to start to fall off this molecule. After we pass a pH of 3.7, the hydrogens are going to start to fall off the carboxylic acid. So that’ll be deprotonated. But the salt, hydro will still be protonated. And then as you go above 9.8, then the hydrogens will fall off of that also. And so then everything will be deprotonated. So the answer is C.

Think of a low ph as being bathed in hydrogens. And as you increase the pH, those hydrogens are falling off and then use the pKa to determine where they’re falling off sooner rather than later. Those are the turning points.

Correct Answer: C

[26:54] Question 26

According to the data in Table 1, what mass of captopril must be dissolved in 3 L of plasma at pH of 7.4 to inhibit 50% of ACE enzyme activity in vivo? Assume an equal volume of distribution.

(A) 7.9 pg

(B) 7.9 ng

(C) 7.9 μg

(D) 7.9 g

Thought Process Behind the Correct Answer:

Looking in terms of the units, we have 3 L and we’re going to need the 0.012 micromolar, which is mol/L. But the answer is here is in grams, and we’re given captopril’s weight of 217.29 grams per mole.

If we multiply 3 L whatever mol/L, the L cancels out and we’re left with moles. So if we take that x grams per mole, the mole will cancel out more of the grams. And so the answer to this one is to just multiply all three of those numbers together.

3 L x 0.012 micromolar x 217 grams per mole which is going to round to 220. But we’re not even really that worried about that anyway, because the answer is going to come out to be 7.9.

0.012 x 10-6 mol/L when you convert it to micromolar then x 3 L and x 220 grams, so you get 7.9 x 10-6. So the answer is micrograms.

Correct Answer: C


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