AP Chemistry 2026 Free Response Question 3 - SOLVED!

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[00:00:00]Hi there, my name is Jeremy Kug and this is the place for all things AP Chemistry. As I record this video, the free response questions for the 2026 AP Chemistry exam have just been released.

[00:00:12]And this is my walkthrough for free response question number three, which is a long FRQ worth 10 points. If you like what you see or you learned something from this video, remember to smash that like button and leave a comment down below. If my videos have been helpful to you this year, I'd be very honored if you'd recommend my videos to next year's AP Chem students. And just for full disclosure, this answer key is my best prediction for the point breakdown. Full scoring guidelines get released by College Board later in the summer. And as always, other answers that are chemically correct are also acceptable.

[00:00:48]Now, here's my walkthrough for question three. So question three was primarily acid base with a few other uh units tossed in there as well for good measure. We have nitrous acid HNO2 is a weak acid that ionizes according to equation one and we're given the equation along with its Ka value of 5.6 * 10 -4 at 298 Kelvin. Part A says identify a conjugate acid base pair in the equation. Be sure to clearly label which one is the acid and which one is the base. So we should remember that conjugate acid base pairs involve the conjugate acid having exactly one more H+ than its conjugate base. So we could select HNO2 as the acid and NO2 negative as the base. Or if you prefer, you could also select hydrronium as the acid and water as the base. Either of those pairs would be acceptable. I'm guessing that some common mistakes would include some students uh labeling HNO2 as an acid and maybe water as a base. You know, water is a base here, but that's not a conjugate acid base pair. If you said that or one of those, then give yourself one point for that part of the question.

[00:02:01]Moving on to part B. It says, a solution of HNO2 with an initial concentration of.125 molar has a pH of 2.09.

[00:02:10]Calculate the value of hydrronium ions in the solution and show your work.

[00:02:14]Well, we know that if you take the negative anti-log of the pH, that will give us the hydrronium ion concentration. So, 10^ the negative pH.

[00:02:23]So, if we take 10 -2.09 power, we find that the hydrronium concentration here is about 8.1 * 10 -3rd moles per liter. So, if you said that, give yourself one point for that part of the question. Part B2 says, "Calculate the concentration of HNO2 at equilibrium and show your work." Now, there are a couple different ways to do this. You can actually write out an ice box. Uh, I chose not to do this this time. What you can realize is that the concentration of HNO2 at equilibrium is just equal to the initial HNO2 that you had starting out minus the concentration of hydrronium that was produced over the course of the reaction. So in that case we know that the initial concentration of nitrous acid is 0.125. So I'll plug that in for the initial HNO2 and we just calculated that the hydrronium concentration produced was 8.1 * 10 -3rd. So that gets plugged in for the hydrronium. Once we subtract that we find that the nitrous acid at equilibrium is about.12 molar. So since this is a weak acid, we would expect the concentration not to change very much and as we can see it did not. So give yourself one point if you did something close to that and showed your work for that. So part C says in an experiment the solution of.125 molar nitrous acid is heated to 333 Kelvin. The new equilibrium concentrations are determined to be HNO2=.114 molar. NO2 negative equals 0.0109 molar and hydrronium ions equal 0.0109 molar. And part one here says to calculate the value of Ka for HNO2 at 333 Kelvin and show your work. So we need to write out the equilibrium constant expression for Ka. Once again that's products over reactants raised to the power of the coefficients omitting liquids and solids of course. So we have hydrronium time nitrite all over HNO2.

[00:04:27]We don't include the water there. So when you plug and chug those numbers in there, you take the 00109 of the nitrite and the hydrronium and then in the denominator we have.114 for the HNO2. So our calculations should be about 1.04 * 10 -3rd for Ka at this new temperature. So give yourself one point if you got that answer correct.

[00:04:55]Now the next part here, part two says, is the reaction represented by equation one endothermic or exothermic? And justify your answer by comparing the Ka values at 298 Kelvin and 333 Kelvin.

[00:05:11]Well, notice that since we raise the temperature, you know, we add heat basically, it caused the value of Ka to increase. So it was 5.6 * 10 -4 at 298 Kelvin. But when we raise the temperature to 333 Kelvin, it went up to a much larger value. So this shows that adding heat or raising the temperature causes equilibrium to favor the products more heavily. So what that tells us according to Lantier's principle is that heat must be a reactant. If we add heat, it pushes the equilibrium to favor the products more and raise the equilibrium constant. So that tells us heat is a reactant and the reaction is endothermic. So if you said something like that, give yourself one point for that answer and that reasonable explanation.

[00:06:03]Now moving on to part D. Here we have a titration curve and this titration curve applies to parts D, E, and F. And it says in a second experiment, a student is given a bottle containing HNO2 with an unknown marity. To determine the concentration of the HNO2 in the bottle, the student titrates 35.0 milliliters of the HNO2 with.16 molar sodium hydroxide at 298 Kelvin, which reacts according to equation two. And we have that equation right here. And we have the titration curve. So part D says, identify the pH at the equivalence point. Well, it's important to realize that the equivalence point is basically that uh inflection point right there. And as we see that inflection point, it seems to be very very close to pH8. And so you should answer that the pH at the equivalence point is about 8.0. So if you said that or something close to it, give yourself a point for that. Now part E says using this same titration curve figure one calculate the marity of the HNO2 I think there's a little typo here of HNO2 in the bottle show the work that leads to your answer so I would use the titration equation for this marity of the acid times the volume of the acid equals marity of the base times the volume of the base we can do that because we're using just regular sodium hydroxide there there are no multiple hydroxy oxides here. So this equation works just fine. The marity of the acid is what we're trying to solve for. So our ma is the unknown. The volume of the acid is given to us in the problem is 35 milliliters. So that's our VA. The marity of the base is is given to us actually twice in the problem both up in the header as.16 molar and on the graph itself. So we can plug that in for M subb. And the volume of the base we have to extract that from the titration curve. Notice that we hit that equivalence point at 50 milliliters. So we use 50 milliliters as the volume of the base. So when you do the math on here, we find that M sub A is.23 molar. So if you said that, give yourself a point for part E. Now, part F says, draw an X on figure one to represent a point in the titration where the concentration of nitrous acid is greater than the concentration of nitrite ions in the reaction mixture.

[00:08:30]Well, we need to remember that at the halfway point or the half equivalence point, the concentration of both the acid and the conjugate base are going to be exactly equal. Well, what points or what part of the titration will there be more acid? Well, at the parts of the titration that are more acidic than the half equivalence point. So, basically, I would put an X or you could put an X anywhere on the titration where you know you are greater than 0 milliliters but less than 25 milliliters. So basically anywhere in this approximately in this range here would be an acceptable place to put your X. So anywhere that has a pH less than the half equivalence point of 25 ms. So if you did that, give yourself one point for part F. So that was a good titration uh curve question or series of those I should say. Now part G we have equilibrium constants at 298 Kelvin are shown in table one for the acid ionization reaction represented by equation one that we saw earlier and then we have the autoionization of water KW is given to us and this neutralization reaction that we saw earlier which is K2 and the question says calculate the value of K2 for equation two at 298 Kelvin and show your work and so that means that somehow We have to manipulate the first two equations so that they add up to give us the third equation. Well, in order to do that, it looks like we're going to have to leave equation one as is. And so I just rewrote equation one, and its K stays the same as 5.6 * 10 -4. Now, in order to make these work out, I'm going to have to flip equation number two. So, I'm going to flip that around. So we have H30+ plus hydroxide ions yield two water molecules. In order to uh manipulate my K value, I have to take the reciprocal of the original KW. So the reciprocal of 1 * 10us 14th is 1 * 10 pos 14. So we have that as our second equation. And now hopefully these two equations add up to give us the third equation. And indeed they do because the hydrronium cancels and one of the water molecules cancel. So we do get the third equation. Now the way this works is you have to multiply those two equilibrium constants together to get the final equilibrium constant. So 5.6 * 10 - 4 * 1 * 10us4th equals 5.6* 10. So if you did all that, good job.

[00:11:10]You can give yourself one point for that. Now part H says the student decides to repeat the experiment this time using an indicator and has access to the indicators in table 2. So we have methyl orange, thymol blue, clayton yellow as well. And the student conducts a second titration by adding two drops of methyl orange indicator to 35 ml of the HNO2 solution and titrating the solution with.16 molar sodium hydroxide until a color change from red to yellow occurs. The student claims that methyl orange was the best choice for the indicator. Do you agree or disagree and justify your answer? I would straight up disagree here. Now, we just said in that previous titration curve that the equivalence point is going to hit somewhere around pH8.

[00:11:59]And we can see that, you know, we still have that weak acid HNO2 and the strong base NaOH. So the equivalence point is going to be somewhere on the slightly basic side. So you know, yeah, somewhere around pH8. And notice that the appropriate indicator should always have a color change range very close to the pH of the equivalence point of the titration. And since we're looking at an equivalence point of about eight, really thymol blue would be your best choice of the three. In fact, methyl orange would be a terrible choice. And so if you did methyl orange, you're you're not going to get the right answer. So uh that is why you should disagree. So one point if you said that. So this was primarily an acid base question although we had some uh unit 7 equilibrium uh expressions in there and a few other things as well. So if you uh did uh unit 8 and uh got all 10 points for this question, that would be very fantastic. So, good job if you did that and hopefully you learned something from this uh this walkthrough.

[00:13:04]So, that's it. I hope this was useful for you as you reflect on your exam performance or get ready for a future exam. And if you are getting ready for a future AP Chem exam, remember that my Ultimate Review Packet and Ultimate Exam Slayer have dozens of practice free response questions and nearly a thousand practice multiplechoice questions with full explanations to help you slay your AP exam. The link is in the description down below. Thanks so much for watching and I hope to see you soon.