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AP Chemistry FRQ: How to Ace the Free Response Questions

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Posted by Samantha Lindsay | Apr 23, 2022 9:41:00 AM

Advanced Placement (AP)



Practicing free-response questions is one of the best things you can do to improve your AP Chemistry score. Not only will you excel on the free-response section, but you'll also know the material so well that the multiple-choice questions will be a piece of cake.

In this article, I'll tell you all about the free-response section of the exam, give you some tips on how to solve AP Chemistry FRQs, and go through a couple of sample problems from recent exams so you can get a feel for what they're like!


What's the Format of the AP Chemistry Free-Response Section?

The free-response section of the AP Chemistry exam seems intimidating because it's longer than free-response sections on most other AP tests, and it includes lots of calculations and experiments that you have to interpret.

The most important thing you can do is remain calm and stay focused and methodical in your approach to each question. It's not as scary or difficult as it looks if you've prepared well for the test and use your common sense!

Here's an overview of the format of the free-response section:

  • 105 minutes (1 hour 45 minutes)
  • Calculator use permitted
  • Seven questions total
  • Three long-response worth 10 points each
  • Four short-response worth 4 points each
  • The long-response questions always come first!


You'll be tested on the following skills (which may be tied to any topic in the AP Chemistry curriculum):

  • Experimental design
  • Analyzing real lab data to identify patterns and explain phenomena
  • Creating or analyzing diagrams of molecules and atoms to explain observations
  • Translating between different representations of data
  • Following logical steps to analyze and solve problems

OK, that all makes sense, but how exactly do you solve these long, complicated questions? In the next section, I'll go through a step-by-step guide for how to approach the AP Chemistry FRQs.


How to Solve AP Chemistry Free-Response Questions

It's important to have a game plan for the free-response section. My first piece of advice is not to feel obligated to do the questions in order! Take a couple of minutes (no more than 5-10) at the beginning of the section to look through everything and decide which question you want to tackle first. It's best to start with your strengths so you'll have more time at the end for challenging questions.

You should spend a maximum of 20 minutes on each long free-response question and 10 minutes on each short free-response question. Keep an eye on the time so it doesn't get away from you! Here are some basic steps you should take to solve free-response questions:


Step 1: Figure Out What You Know

First, assess the information the question gives you. It can be confusing to extract the data that's embedded in the introduction over and over again as you go through different parts of the question. Make things easier for yourself by writing down the values you're given next to the question so that they're easily accessible when you need them. You should also take time to understand (at least on a general level) the experiment being described so that you don't feel confused and overwhelmed when you start reading the question.


Step 2: Dive Into the Question

For each part of the question, read the instructions and ask yourself the following:


Do you need to do any calculations?

Decide which equations you'll need (if applicable), and write them down. Do the necessary calculations based on the numbers you extracted in the previous step and any numbers you were given in this part of the question. Make sure you show your work! Don't erase your calculations, and double-check everything to make sure you have the correct units and your answer makes sense logically.


Does the question ask you to justify or explain your answer?

If so, don't ignore these instructions. In many cases, you'll only get points for your answer if you can explain it adequately. Use concrete evidence to back up your response (we're talking hard data). Even if something seems obvious to you, spell it out as clearly as possible to ensure that you earn those points!


Do you need to draw a diagram?

Even though neither of the free-response questions I'll go over in the next section requires this, some questions will ask you to draw diagrams. For example:



If you have to answer a question like this, try to be as clear as possible. Draw out your answer on scrap paper first if you feel unsure so that the final product is neat and unambiguous. Here's the answer, by the way:



Step 3: Double Check

Look back at the question to make sure you didn't miss anything or leave out any explanations. Reread each part and connect it directly to its corollary in your response so you're scooping up all the points you possibly can!

You should also walk through how you found each answer to make sure you didn't make any weird errors you missed the first time around.

Finally, check again for appropriate units. Errors in unit conversion are common silly mistakes that are really, really frustrating if you knew how to do the problem correctly otherwise.


body_unitmeasurement.jpgDid you remember to convert milliliters to liters?


AP Chemistry FRQ Examples

I'll go through the solution process for a sample question of each type so you can get a better idea of what the test will be like. Notice that these questions look complex and overwhelming at first, but if you stay calm and break them down methodically, they don't end up being that bad!


Long Free-Response Sample Question

Here's a sample long free-response question from the 2014 exam:



There's a ton of information here, so let's start from the beginning. What we know from the short intro is that the pH of a 0.20 M 50 mL sample of propanoic acid is 2.79 at 25 degrees Celsius. There's also an equation that shows how the acid reacts with water and which products are created by this reaction.


Now we're ready to tackle part (a):

Where's the conjugate acid-base pair in the equation? There are two pairs that you could potentially list for this answer:

CH3CH2COOH (acid) and CH3CH2COO-(base)


H30+ (acid) and H20 (base)

We know that the first compound is propanoic acid, and the loss of the hydrogen atom through the reaction creates the basic compound on the right side of the equation. Conversely, the water in the first half of the equation is a base that becomes an acid when it picks up the hydrogen atom from the propanoic acid.

To get this point, you would need to label each compound indicating which is the acid and which is the base. Always read instructions carefully, or you may lose out on points - I can't say this enough!


Let's move onto part (b):

What's the value of Ka for propanoic acid at the temperature indicated in the question? We just need to plug some values into the equation for Ka, which is given to you on the formula sheet for the test:


Hmm, looks like we can't plug in the values yet because we don't know the molarity of the H30+, which would have to go in the [H+] spot in the equation. We can find that value using this other equation from the formula sheet:


This means:

-pH = log[H+]
10-pH = [H+]
10-2.79 = [H30+]
1.6 x 10-3 M = [H30+]

This value for the molarity of H30+ is equal to the molarity of CH3CH2COO-. The same amount of each must be created by the equation because the hydrogen atoms are removed and added in a 1:1 ratio. Armed with this new information, we can go back to the Ka equation:

body_kaequation-1.pngLet's plug in our values:

Ka = [H30+][CH3CH2COO-] / [CH3CH2COOH]
Ka = (1.6 x 10-3 M)(1.6 x 10-3 M) / 0.2 M
Ka = (1.6 x 10-3 M)2 / 0.2 M
Ka = 1.3 x 10-5

For part (b), you could earn three points total:
  • One for correctly solving for [H30+]
  • One for plugging the right values into the Ka equation
  • One for solving for Ka correctly


OK, now for part (c)!

Oh crap, this one has PARTS WITHIN THE PART. Don't panic; you got this. It's just true or false plus answer explanations! The explanations on these types of questions are very important. If you don't explain your answer adequately, you won't get any points even if the answer itself is correct. You can earn two points total on this question, one for each answer AND explanation.

Part (i):

In solution, the OH ions from the NaOH will react with the CH3CH2COOH to form water and CH3CH2COO- like so (hydrolysis reaction):


The pH of the resulting solution will be GREATER than 7 because of the formation of the new basic compound at equivalence. That means it's false!

Part (ii):

If two acid solutions have the same pH, but one is with hydrochloric acid, and the other is with propionic acid, would the first solution necessarily have a lower molar concentration of the HCl?

HCl is a strong acid that ionizes completely in solution while propionic acid only partially ionizes. Fewer moles of HCl are needed to produce the same molar concentration of H30+ and reach an equivalent pH level to the propionic acid solution. This one is true!

The next part of the question offers up a new scenario, so let's take stock of what we've learned from the added description. So, the student titrates 25 mL of a ~mystery solution~ (mysterious squiggles added for dramatic flair) of propanoic acid with 0.173 M NaOH and reaches the endpoint of the titration after 20.52 mL of the NaOH has been added.


Now onto part (d)!

Based on this information, part (d) asks us to figure out the molarity of the propanoic acid.

First, how many moles of NaOH were put into the solution? We can find this by multiplying the total volume of NaOH solution by its molarity:

(0.02052 L NaOH) x (0.173 mol NaOH / 1 L NaOH) = 3.55 x 10-3 mol

A total of 3.55 x 10-3 moles of NaOH were put into the solution. Since the titration reached the equivalence point at this time, that means that the number of moles of NaOH added would have to be the same as the number of moles of propanoic acid in the original solution. If we divide 3.55 x 10-3 mol propanoic acid by the number of liters of acid in the original solution, we will get the molarity:

3.55 x 10-3 mol propanoic acid / 0.025 L propanoic acid = 0.142 M

For this part, you get one point for correctly calculating the number of moles of acid at the equivalence point and one point for providing the correct molarity.


Part (e) is a critical thinking question about a new experiment.

Would the student have to use a different indicator to figure out the concentration of a solution of an acid with pKa of 4.83? Based on our Ka calculations in part b, we can use one of the equations on the formula sheet to figure out pKa for the original propionic acid and compare the two values.

pKa = -logKa
pKa = -log(1.3 x 10-5)
pKa = 4.89

The two pKa values of 4.83 and 4.89 are pretty close to one another, so you wouldn't need to use a different indicator in the new titration. The correct response is to disagree with the student's claim. You get one point here for disagreeing with the claim and explaining why, and you get a second point for directly comparing the two pKa values.




Short Free-Response Sample Question

Here's a sample short free-response question, also from the 2014 exam:



As you can see, the "short" questions aren't really that short, but they're not as involved as the long ones. There isn't as much information to digest, and each of the parts of the question is more direct. Each part of this question is worth one point (4 points total). Your response must include the correct answer and the correct justification/methodology to earn points!


Starting with part (a):

This is a PV = nRT question! Since we want the number of moles of CO2, we're solving for n. P is 1.04 atm, V is 1.00 L, R is the gas constant (0.08206 L atm mol−1 K−1), and T is 1100 K.

n = PV/RT
n = (1.04 atm)(1.00 L)/(0.08206 L atm mol−1 K−1)(1100 K)
n = 0.0115 mol CO2


Moving onto part (b):

In experiment 1, the original number of moles of CaCO3 would be equivalent to 50.0 g / (100.09 g/mol). The 100.09 g/mol number was calculated by adding up the atomic weights of the elements in the compound. This calculation gives us 0.500 mol CaCO3 total.

If all of it had decomposed, the figure we calculated in part a for the number of mols of CO2 produced would also be 0.500 mol, but it was only 0.0115 mol. This discrepancy means that the student's claim has to be false!


Now let's tackle part (c):

What would happen if more gas was added to the container and the pressure went up to 1.5 atm? Would it go back down to 1.04 atm afterward?

Equilibrium was reached in both experiments, and it resulted in a final pressure of 1.04 atm. The reaction would just adjust to the added gas by shifting towards the reactant. The pressure would go back down to the equilibrium figure of 1.04 atm as the excess CO2 was consumed. The final pressure would still be equal to 1.04 atm.


Finally, we'll answer part (d):

Can we find Kp with the information we're given? Yes! The pressure of the CO2 in this experiment determines the equilibrium constant as well because it's the pressure of the gas at equilibrium. There's only one gas involved in the reaction, and we already know its equilibrium pressure, which means we also know the value of the constant.

Kp = 1.04


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How to Practice AP Chemistry Free-Response Questions

You can find AP Chemistry FRQs from previous years (and their solutions) on the College Board site. The exam was last significantly changed in 2014 (seven free-response questions instead of six, and no questions asking you just to balance equations), so keep in mind that the 2014 and onward exams will be the most accurate representations of what you can expect to see on your test. Here's a link to the most recent questions and answer explanations:

2014 to 2019 AP Chemistry Free-Response Questions

You can also view the 2021 FRQ and their scoring guidelines.

There is no login required for access to these questions. You can also check out my article that lists all the AP Chemistry practice tests and quizzes that are available online. There are a few more unofficial practice tests that include free-response questions modeled after the questions on the real exam.


body_calculatorandpen.jpgAnd use a calculator when you practice (you get one for the free-response section on the real test)! Definitely don't use a quill, though. That part of this image is irrelevant and mystifying.


The free-response section is the most challenging part of the AP Chemistry exam for most students. To do well, you need to have a strong understanding of all the major concepts covered in the course and be able to apply them to a variety of experimental scenarios. Most of these questions look intimidating, but they're manageable if you take them one step at a time and break them down into smaller chunks.

Here's a review of how to solve free-response questions:

  • Step 1: Figure out what you know
    • Write down any data that's included in the question
  • Step 2: Dive into the question
    • Figure out which formulas you need
    • Do the necessary calculations
    • Justify your responses
    • Draw diagrams if applicable
  • Step 3: Double check
    • Make sure your units of measurement are correct
    • Verify that your answers make sense logically

Practice your skills on free-response questions from past AP tests until you feel comfortable. Remember, questions from tests before 2014 will be slightly different from the current free-response questions; base your expectations for the real test off the material from 2014 onward!


What's Next?

If you're looking for more AP Chemistry practice, check out my list of the best review books for this year's test. Almost all of them include free-response questions modeled after the current format of the exam.

For a holistic look at the most effective way to study for in-class assessments and the final exam, read my ultimate study guide for AP Chemistry.

Are you debating whether to start studying now or put it off a little longer? Learn more about when you need to get serious about your study plans for AP tests.


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Samantha Lindsay
About the Author

Samantha is a blog content writer for PrepScholar. Her goal is to help students adopt a less stressful view of standardized testing and other academic challenges through her articles. Samantha is also passionate about art and graduated with honors from Dartmouth College as a Studio Art major in 2014. In high school, she earned a 2400 on the SAT, 5's on all seven of her AP tests, and was named a National Merit Scholar.

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