2016 AP Chemistry Exam Draft Answers and Comments

Written by Adrian
On May 04, 2016
Tags: AP
Categories: AP | College Board

My 2016 AP Exam Draft Answers and comments appear below. The questions are here. Should you find an error PLEASE let me know by commenting on this post below, rather than emailing me or using the contact form me (that way all problems can be addressed here), and I will comment/change as necessary.

PLEASE NOTE: Any of my comments below about how your individual answers might be graded, are highly speculative – you should not take them as fact!

AP Chemistry 2016

Many, many, many words! Reading comprehension rules! As a recent commenter to my blog wrote, “… I would take the kid who knows tricky algebra stoichiometry/equilibrium questions (who had no access to labs) over the one who has lab experience (but is bad on algebra)…”

Amen!

DRAFT Answer 1

The obsession with lab procedures without a lab exam continues, and is ironic. I still hate mole of reaction, and particulate diagrams. The whole, ‘interactions of ions with water’ also seems to be a new obsession with the CB/TDC. Very easy, but of course answers to (c), (d) and (e) will probably drop a lot of points.

DRAFT Answer 2

Very easy, and a ridiculously complex way of asking about a Lewis structure that, for my money, could be asked in a whole host of more elegant ways. If you want to test resonance then fine, but this question is a reading comprehension nightmare.

DRAFT Answer 3

Oh look, ANOTHER lab! I wouldn’t want to have to grade (c), and the IMF interaction obsession continues in (d).

DRAFT Answer 4

‘Deprotonated’ is another buzzword that is heavily embedded in the CED, but (b) is a good question.

DRAFT Answer 5

Seems SO easy, with a complication that could confuse (see #6 here).

DRAFT Answer 6

(b) is potentially confusing, but I wrote about it here in scenario #2.

DRAFT Answer 7

I am assuming that what I have taught for almost 30 years, that the second decimal place on a buret should either be a zero (on a line), or a 5 (between graduations), is correct, but I’ll guess that we’ll see!

178 Comments

  1. Derek Stach

    What’s your thoughts on #5c? I don’t like that they give you the rate, but don’t say what substance it is the rate for. The rate of formation of the product is different than the rate of the disappearance of the reactant. If students used the graph to solve for k (which they should be able to do), they get a different answer than if they plugged the .001 into the rate equation w/ a 2nd order.

    Reply
    • Adrian

      Full comments to come later, but for now, see my post from February 2014, item #6. I think either will score full credit.

      Reply
  2. Matt

    I think 3e asks for the voltage. Probably want to show it.

    Reply
    • Derek Stach

      It does.. When I was first filling it out this afternoon, I missed it at first… Wonder how many students did as well.

      Reply
    • Adrian

      Thanks Matt, done!

      Reply
  3. Srivikram Margam S

    What is your take on the paradoxical phrasing of the HCO3- lewis dot structure? It says one CO bond is longer than the other two; doesnt that imply that the question asks you to draw two double bonds (which is scientifically incorrect but the question was like “draw something that fits this description”?

    Reply
    • Adrian

      No paradox when one considers resonance.

      Reply
      • Srivikram Margam S

        Looks like I got it wrong, but could you explain this statement “Two of the carbon to oxygen bonds have the same length and the third carbon to oxygen bond is longer than the other two”? What exactly does this mean?

        Reply
        • Adrian

          When the double bond can appear in more than one place, we get resonance, and those positions ‘average out’ to be bond orders of 1.5. The bond lengths of 1.5 would be shorter than the other bond, which is a longer, single bond (order 1).

          Reply
          • Srivikram Margam S

            Thank you very much, that clears things up; do you have any estimate on how many points that part might have been?

          • Adrian

            My guess is max. 2.

          • Srivikram Margam S

            Thanks for the prompt reply; also thought I’d mention that the pH of the phenol should probably be 2 significant figures because the concentration is only given in 2sf.

          • Adrian

            Strictly speaking, for pH, significant figures only occur AFTER the decimal place.

          • Srivikram Margam S

            Do you think there will be a penalization if I left it as 5.0?

  4. Derek Stach

    I find question #3c interesting… good luck graders!! There’s all kinds of possible answers here (I went to dissolving/conducting electricity rather than melting)

    Reply
    • Adrian

      Any open-ended question is a disaster for graders.

      Reply
    • Robin

      I answered same as Derek.

      Reply
    • Steve

      Thoughts on freezing pt depression as a possible experiment.

      Reply
      • Adrian

        Why not? (if explained correctly). Of course, there is no reason that a kid should known anything about FPD in the new curriculum.

        Reply
  5. Derek Stach

    And it doesn’t get much more open ended than that…

    Reply
    • Adrian

      2001, 6(a), second part, was MUCH worse in that regard!

      Reply
      • Derek Stach

        Just went back to that… pick a reaction, any reaction… Back when they were given the potential table. That is BAD!!!

        Reply
      • John

        For #6, isn’t a K greater than 1*10^10 showing a reaction goes to completion?

        Reply
        • Adrian

          Yes, but not sure what you are asking.

          Reply
          • John

            Is there a specific K value that indicates the reaction goes to completion?

  6. Victoria

    Isnt the second question for 6 no change since dilution doesnt affect K value and since reaction goes to completion all of Ba2+ will be depleted?

    Reply
    • Adrian

      Dilution does affect Q.

      Reply
      • Derek Stach

        Dilution affects concentration, which in turn, ends up making Q no longer equal K.

        Reply
        • John

          Can you please explain how. If this is going to reach equilibrium again, which it says that it does, won’t basically all of the Ba be used up to give a concentration of 0.010 M to the Ba(EDTA.) So if it had the same number of moles to start and they were all used up in both scenarios, how is it different?

          Reply
          • Alex

            I had the same response when I took it. I don’t understand how the dilution affected concentration.

        • John

          And Derek, you say that Q no longer equals K but this says that it re-establishes equilibrium and isn’t the very definition of equilibrium when Q=K?

          Reply
          • Alex

            My thoughts exactly.

          • Derek Stach

            John, Correct. It had reached equilbrium so Q was = to K. Then we dilute, which changes concentrations and throws Q off from K, so to reestablish equilibrium it needs to shift towards the reactants.

            Example with some easier #’s. Let’s say that I have the reaction A + B –> C. If I have one mole of each substance in 1 L and it’s at equilibrium, then K=1. Now let’s dilute it down to 2 L instead by adding more water. Now each substance has a concentration of .5 M instead of 1 and if we find Q, Q would now equal .5, which is less than what K was. Therefore the reaction would push towards the reactants to reestablish Q=K.

          • Derek Stach

            Since there are two species on the reactant side to be affected by the decrease in concentration, compared to just one one the product side, the reactant side is more greatly affected.

          • Leah

            But if K is re established and K ONLY changes with temp. And temp. Is constant than it doesn’t change with adding water and if u look at 2006 question number one part b they asked a very similar question and they said the moles do NOT change.

          • Scott Page

            Derek, each at 0.5M would mean Q = 2 which is too many products. Would have to shift to reactants to reduce products. Is Q < K, then it would shift to products.

          • Ron Brandt

            I also believe that there is no change. All of the Ba2+ is already in solution because the reaction went to completion. If you dilute, there are no additional Ba2+ ions to add. I understand that the concentration will change, but why the moles of Ba2+ ?

          • Adrian

            Ron – There is an equilibrium set up. INSTANTANEOUSLY, the addition of water dilutes ALL of the (aq) species. Since there are two aqueous species in the denominator of Q, this means that Q INSTANTANEOUSLY becomes too big. The reaction has to shift backwards for Q to equal K once more, and therefore the moles of Ba2+ increases.

          • Scott

            Ron, there are two forms of barium ions – those that are free (right side) and those that are bound(left side). Dilution causes more of the free ions and less of the bound ions. The additional free ions come from the shift to the right. Moles change because of stoichiometry. 1 mole of bound ions must change to 1 mole of free ions.

          • Ron Brandt

            TY

            Although I am still confused. Barium Nitrate was the limiting reactant, so fully consumed. The reaction goes to completion. So where are the bound ions on the left? Sorry for being obtuse. Easy to understand if this was a low Ksp problem

          • Scott

            Oops, I meant free ions on left and bound ions on right.

          • Scott

            Ron, i made a mistake saying shift right. I meant shift left. The free ions are on left and the bound barium ions are on right. Yes, you do assume that barium ions are limiting to get concentration of the complex. However, barium ion concentration is not zero, though very small. [Ba2+] = 2.6 x 10-8 at equilibrium. You get this from 0.10/(0.05)y = K where y is the barium ion concentration. It then says, the solution is diluted to 1.0 L which is a 10x dilution. Therefore, Q = 1/0.05y which is 10x greater than K. Reaction has to shift left to reduce Q so it becomes equal to K again at equilibrium. This shift frees up barium ions and roughly increases its concentration by a factor of 10. The barium ion concentration is now approximately 2.6 x 10-7. Though still small, this was a order of magnitude chagne which is significant. In the shift, the volume did not change, only the moles of barium ions increased by a factor of 10. If the AP readers ignore this factor of 10 increase, then they will be wrong in my opinion. No change is not the correct answer.

  7. Colleen

    in question number 4; how does the pH raise above 12? When only anion is present, using the Ka converted to a Kb, wouldn’t the maximum pH only reach 12?

    Reply
    • Adrian

      Well if the base became 10000 x bigger than the acid, then log (10000) = 4 and we would be at pH = 13.95.

      Reply
      • Colleen

        that seems like a stretch…10,11,and 12 seem the most reasonable pH values

        Reply
  8. Victoria

    Also, for 2C, could you have said that since flask gets cooler as reaction proceeds, particles collide with less energy, and connect it to the collision theory?

    Reply
    • Derek Stach

      That would seem like a reasonable alternative explanation to me.

      Reply
  9. John

    What is the policy as far as exact numbers are concerned? If I answered 5.05 for the pH on #4 is that incorrect?

    Reply
    • Adrian

      You’re probably OK.

      Reply
  10. Stella

    for 7a, would 37.3-5.6=31.7 mL be acceptable…..

    Reply
    • Adrian

      IMO, no. I always teach that the second decimal place on a buret must be read to either 0 (exactly ON a line), or 5 (between lines). Maybe I’m wrong, but I guess we’ll only see when the official answers are released.

      Reply
      • Stella

        would they give me part marks at all?

        Reply
      • Matt

        I’ll second Adrian. The ease of this question probably means it is going to be graded severely for the correct number of significant figures: 4 on the volumes, 3 on the molarity.

        Reply
        • Derek Stach

          I agree with Adrian & Matt. This question is designed to be the “sig fig” question. You are going to need to be out to two decimal spots here.

          Also, this likely means that significant figures won’t be as important on the other questions since there was a dedicated “sig fig” question.

          Reply
      • Scott Page

        Adrian, A student told me about his problem today and said it was easy. I said most will miss it because of sigfigs.

        Reply
  11. John

    Can you please explain 6b. If the reaction is going to reach equilibrium again, which it says that it does, won’t basically all of the Ba be used up to give a concentration of 0.010 M to the Ba(EDTA.) So if it had the same number of moles to start and they were all used up in both scenarios, how is it different? Isn’t it the same.

    Reply
    • Adrian

      It can’t be an equilibrium if they are at zero.

      Reply
  12. Frank

    Where can I find a copy of the questions? I’m not having any luck with my searches.

    Reply
  13. Brad

    On 6b, I feel that they’ll need to give credit for both trains of thought.

    1) On a pure Q vs. K argument, Q clearly increases due to the dilution of all species. The system will need to shift left increasing the number of moles of Ba2+.

    2) However, the whole point of part (a) is that this isn’t really an equilibrium system. The reaction has gone until a reactant has run out. There is a specific number of moles of Ba2+ that remain at this point. Since the reaction will still go until the EDTA has run out (K is unchanged with no temperature change), the number of moles of Ba2+ would be the same.

    Reply
    • Derek Stach

      I think if you did a full ICE chart showing the calculations, you’d be okay… But to try and explain that from a conceptual standpoint without showing the work behind it, it would probably not be acceptable. You are correct in that there is virtually nothing left, but there still must be something there (not to sig figs) for there to be an equilibrium.

      Reply
    • Brad

      One correction: From memory, I was thinking it was the EDTA that essentially ran out. Of course, it is the Ba2+. I think this even further makes both explanations valid.

      Reply
  14. John

    If we are to solve this algebraically, wouldn’t the concentrations of Ba, when rounded to appropriate sig figs, be 0.0 for both of them. So in that case, comparing the algebraic values it appears there isn’t a noticeable difference once it is solved out.

    Reply
    • Scott Page

      Concentration of barium ions in part (a) is 2.6 x 10-8 M. So no, not zero. I wish they would have asked for this concentration making the problem much more challenging. You have to set it up 0.10/(0.050y) = K where y is the concentration of barium ions. I go over Kf problems in class and how to approach them.

      Reply
  15. Alex

    Okay, I’m going to continue making a stink about 6b. I’m a student though, not a teacher, so I know I probably got it wrong, but I want to understand why :).

    The question says (please look at what’s in all caps):

    The solution is diluted with distilled water to a total volume of 1.00 L. AFTER EQUILIBRIUM HAS BEEN ESTABLISHED, is the number of moles of Ba2+(aq) present in the solution greater than, less than, or equal to the number of moles of Ba2+(aq) present in the original solution before it was diluted? Justify your answer.

    Since equilibrium has been reestablished, doesn’t that mean that q=k again, and, since K is a constant, the concentration of Ba(2+) is equal to the starting concentration? Thanks again, and sorry for being a pain :)!

    Reply
    • Derek Stach

      Just because after it reestablishes Q=K again, doesn’t mean that the Q and K expression has all the same values as before. it’s asking you to compare the moles at the start and end, not the Q and K values…

      Think of it like this.. Let’s say I have a magical reaction where Red Marbles Blue Marbles and at equilibrium it is found that I have 5 Blue and 5 Red Marbles (K=1). Now I add 10 more blue marbles. I’ve knocked it out of equilibrium. Q=5/15. To reestablish equilibrium, it’s going to push to the right making more blue until it is 10 & 10 again AFTER REESTABLISHING. K=1 again, but I don’t have the same amount of each substance as before.

      Just because K hasn’t changed, doesn’t mean the values haven’t (they’ve just readjusted).

      Reply
      • Leah

        K only would change if u added a common ion NOT water. If you have 5 marbles and add 100L of water u still have 5 marbles ????

        Reply
        • Robin

          Leah, remember, K is a RATIO of [products] to [reactants]. The equilibrium position changes because you have one product concentration, but two reactant concentrations. This means the change in volume affects the denominator much more than the numerator, thereby shifting the equilibrium position to keep your K value constant. Plug in some simple numbers to test this out.

          Reply
          • Derek Stach

            It’s hard to explain over messages.. I will post a recorded video link tomorrow morning using my SmartBoard for a video explaining that problem that should be easier to understand.

    • Scott Page

      There are an infinite number of equilibrium positions for a given K.

      Reply
  16. Jessica

    First of all, thank you so much for posting these! I have two quick questions:

    1. For 2e. (the lewis structure) if we just drew one of those structures (not recognizing that there was a resonance structure) do you predict that partial credit will be awarded?

    2. What do you think the curve will be? What percentage of correct answers for a 5? (Based on the easiness of the FRQs)

    Thanks!

    Reply
    • Adrian

      1. A single structure would not match the massive explanation given, so, who knows…
      2. If I HAD to guess, probably on the higher end of a range between 69-75, but that’s pure speculation.

      Reply
  17. Jon

    I see the answers but where is everyone seeing the questions? Have the CB released them yet?

    Reply
  18. aj

    Doesn’t Li have a larger atomic radius than Cl? google says that lithium has an atomic radius of 152 pm, and chlorine and 79 pm…

    Reply
    • Adrian

      The question is about ionic radii, and in addition, not about a comparison between Li and Cl, but rather a comparison of Li+ and Na+.

      Reply
      • aj

        sorry, i was referring to part (d): lithium has an ionic radius of 220 pm, and chlorine has one of 175 pm as per google

        Reply
  19. Alex

    For 2(e), I believe the proper answer requires two double bonds. The question asked for a structure with two shorter bonds to oxygen and one longer, which would likely be one single bond and two double bonds. I’m not sure where to find the question text to verify this.

    Reply
    • Adrian

      Nope, see my earlier comment about resonance.

      Reply
    • Leah

      Its on the college board website.

      Reply
    • Jessica

      I think what its referring to with the shorter bonds are the 1.5 length bonds that come from averaging the double and single bonds in the resonance structure. Two double bonds would exceed the octet, so there must be resonance (and that is still consistent with the bond lengths it describes).

      Reply
  20. Karen

    I don’t know if anyone has commmented (I did not see any) on #1. The mass is 10.0 g not 110 g.

    Reply
    • Karen

      Sorry, hit the “m” too many times. Commented.

      Reply
    • Adrian

      The mass of the solution is 110g.

      Reply
      • Karen

        Too tired to be working on problems. Totally missed reading that.

        Reply
  21. aj

    also, for 1(c), i said that Na has a lot more electrons than Li, which would create significantly more electron-electron repulsion, thus resulting in a more “stretched out atom” than Li. would that response garner any credit?

    Reply
    • Adrian

      I depends on EXACTLY what you wrote, but I’m not keen on ‘more stretched out atom’ for more than one reason! We are dealing with ions, and the fact that there is a whole other quantum level occupied is really the point.

      Reply
  22. aj

    finally, for 3(c), i tried to find the molar mass by:
    1) finding metal M’s mass by: subtracting the (mass of the beaker+metal M) minus (the mass of the beaker)
    2) dividing that by the moles of I2 that reacted.

    I understand that this was probably wrong because M is in excess; however, would I get any credit for response?

    Reply
    • Adrian

      I wouldn’t think so.

      Reply
  23. CP

    What are the sig fig rules for the AP Chemistry Exam?

    Reply
    • Adrian

      Complicated! Search my blog for ‘Sig Fig’!

      Reply
      • Zaki

        My AP Chem teacher says they do + – 1 sig fig

        Reply
        • Adrian

          It’s more complicated than that!

          Reply
  24. john

    a few questions from a stressed student – please respond!
    a) for the last part of #1, i accidentally said that they would have dipole-dipole interactions – yes, i know i slipped up -, but I correctly described ion-dipole interactions: that Li+ would interact with the O, and the H+’s with the Cl-‘s. Would I still get partial points?

    b) the question about placing Li and Cl – i messed up and said Li was larger – how points do you think that part was worth.

    c) call me dumb, but i messed up on the #7 buret problem BADLY – not only sigfigs, but i managed to get the wrong numbers. Could that be worth more than 1/4’th of the question’s points?

    d) for the question about the experiment in #3: i gave a pretty vague experiment, saying that one should measure the electronegativity of both M and I2 by testing their ability to attract electrons from other species, and that if the electronegativity difference was great, then it’d be ionic. Would that get me any points?

    e) finally, for #2, i assumed that all combustion was exothermic, so I ended up saying that both were favored. obviously that answer is wrong, but given my incorrect assumption that delta H was negative, I correctly used the gibbs’ free energy equation – will that get me some credit back?

    Reply
    • Adrian

      (a) Difficult to know until the reading
      (b) Probably 1 point
      (c) I guess part (a) will be worth 1 point
      (d) I think not! That’s not an experiment that you could carry out!
      (e) I think not, this isn’t a combustion

      Reply
      • Aaron

        wait, is this how AP graders grade? I made a similar mistake for his (e) question. I made the wrong decision, but given my incorrect understanding of a negative enthalpy, my (ii) answer was definitely correct (based on my previous answer). In multipart problems, does making one mistake on one part (like the enthalpy sign) mean that I also get the second part wrong, even if I demonstrated that I understood how enthalpy and entropy Gibbs Free Energy equation?

        Reply
        • Adrian

          Generally, consequential credit IS given and errors are NOT carried forward in a way that you are penalized twice.

          Reply
  25. john

    Oh yeah, very last question i forgot to include:
    for #1 part a, I made two errors:
    – I used 100g instead of 110
    – for (ii), I forgot to flip the sign to be negative

    assuming that both parts combined (for a) combined to be two points, do you think I would get one point (given I hadn’t made some other calculation error on these two parts)

    Reply
    • Adrian

      Again, difficult to know.

      Reply
  26. Nora Lizarraga

    I am a teacher and wanted to get access to the free response questions. Can you tell me how please? The college board website takes me to a student access site.

    Reply
  27. San

    A few questions too

    1. How many marks, if any, would i lose if I put kJ instead of kJ/molrxn for 1 a)

    2. How many marks would be lost for 3 e) if I circled the right answer, but didnt provide any calculations for voltage

    Reply
    • Adrian

      1. Probably OK.
      2. Well, the question ASKS for voltage so that has to be part of the answer, but it’s probably worth 2 points in total so you’d get 1.

      Reply
  28. Jessica

    For the question about testing whether it is an ionic compound, does the following experiment seem at all feasible/point-worthy?

    Heating until it boils and then collecting the boiling point and comparing it to other boiling points. If it has a very high boiling point that is comparable to other ionic BPs it is probably ionic because ionic solids have strong bonds and therefore require a lot of temperature to change phases.

    Reply
    • Adrian

      The problem is, other things have high BP’s.

      Reply
      • Jessica

        Yeah I suspected that could be an issue… do you think that question will be 2 points?

        Reply
  29. Sheva

    For 2d, wouldn’t the answer be both enthalpy and entropy, since the beaker is cooling off, which would be considered exothermic?

    Reply
    • Adrian

      Sorry, no!

      Reply
  30. Grant

    Hello. For 1 (f) I, in my haste, included ALL of the particle interactions, but included descriptions of exothermicity. Essentially, my answer looked like the following:

    The process of breaking apart the LiCl requires breaking the ionic bonds between the Li+ and Cl- ions. This is an endothermic process.

    Also an endothermic process is the “breaking” of the h-bond IMFs between the water molecules.

    The attraction of the Li+ to the O atoms in the water and the Cl- to the H atoms in the water are ion-dipole forces, and these are exothermic.

    Would I be penalized for including the superfluous information?

    Reply
    • Adrian

      If it is written out like that, I think you’ll probably be fine.

      Reply
  31. John

    Is there a difference between ion dipole and dipole dipole? Would I be marked wrong for writing dipole dipole for 1f?

    Reply
    • Adrian

      Yes, there is a difference.

      Reply
  32. Pal

    Couple of questions:

    I have an entirely different take on 6B, which could be very wrong, but bear with me. The equation says that the Ba, EDTA, AND BaEDTA are ALL aqueous…Therefore, regardless of whether it is “bound” to EDTA, isn’t the Ba in ionic form in the water? That is, because the BaEDTA is aqueous, shouldn’t it be separated into its constituent parts, of Ba and EDTA, thus making EVERY SINGLE MOLECULE of Ba in solution?

    Secondly, I disagree about the Sig.Figs. on question 7. The buret is clearly only accurate to the 1st decimal place. Therefore, you should round to that decimal. I see no reason why .05 mL is special.

    And man that stupid phrasing on the Lewis Dot got me…

    Thanks for doing all this, and remaining attentive to comments. Really puts my mind at ease and gives me a little more confidence.

    Reply
    • Adrian

      It’s a soluble complex ion, so the Ba2+ is tied up with EDTA and not free on its own.

      As for the buret, we’ll see. I was always taught (30 years ago), and I have always taught (the last 26 years), that the second decimal place on a buret should always be recorded to with a zero (on the line), or a 5 (between the lines). I guess we’ll find out if that’s what they want!

      Reply
    • Scott

      Pal, All measurements must have 1 uncertain significant digit (rightmost digit). If you have something in a buret between 5.6 and 5.7 mark as in this problem, then you must say 5.64, or 5.65, or 5.66..whatever you are estimating it is. If you just say 5.6 or 5.7 then you are saying the 6 or 7 in these recordings are uncertain but they are not. The right most digit is always uncertain. It is clear the reading is between these two marks so you need a second decimal place as an estimate making this last digit uncertain.

      As for the other reading, it should be 37.30. It is clearly 0.3 or just slightly past it. If it is clearly 0.3, then you need that extra zero as uncertainty because you are uncertain whether or not it is just past this mark.

      I am an engineer and this is the way I was taught many years ago.

      Reply
  33. Ben Ku

    I would like to suggest:
    (1) In 1d, it would be important to mention that LiCl and NaCl have the same electrical charges, because if they didn’t, then we cannot only compare ionic radii.
    (2) For 3e, the circled compound should be Na2S2O3, even though it’s the S2O32- that is the reducing agent. Just nit-picky detail.

    Otherwise, everything agrees with my answers!

    Reply
    • Adrian

      1. You raise an interesting point. The question is, given that the charges are given on multiple occasions in the question, would it be necessary to mention them? We’ll only know when the official answers are released.
      2. Replaced, thanks.

      Reply
  34. Sathwik

    I’m not so sure about number 5. I agree that the third graph was linear, but the slope of the graph was 5. For a second order reaction, k is the slope of the graph, and 5 is so off from 2.5. I put first order because the slope of the graph was about .055 and the k when solved using .001=k[.02], k equals .05. I put first order but second order is probably what collegeboard will say is correct. However, they really messed up on making the last graph, the slope was so off.

    Reply
    • Adrian

      Search my blog for, “Writing Good Answers to Kinetics Problems”, and see item #6.

      Reply
  35. Chris

    Hey, for problem number 5, I got the rate constant k, but used the units mol-1 s-1 forgetting the liters. Will this take away points as the question did not specify to include units in the answer.

    Reply
    • Adrian

      They often insist on units for k but it’s also true that they usually say, “include units” too, so we’ll see.

      Reply
  36. Daniel

    For #5e, I clumsily used 0.20 instead of the written 0.020 as my value for n when solving for P in PV=nRT. So I ended up with an answer with the right numbers, but with the decimal point in the wrong place. In this case, would I somehow be able to scrape up some partial credit? Thank you.

    Reply
    • Adrian

      My guess is no, but it might depend on how many points they assign to that part.

      Reply
  37. Matt

    Thoughts:

    Easy test IF you ignore all the reading and writing, which experienced teachers can do. Students are likely to be overwhelmed with the amount of extraneous information presented in some of these questions. Related: Was there a single challenging calculation on this entire test at all?

    Question 1: Many students will miss the 110g (a) and (f) is going to be a mess to grade because some students will talk about ALL interactions and others won’t use precisely “ion-dipole” terminology.

    Question 2: I like the way the thermodynamics question was done here in (d) but agree that (e) is a mess of reading comprehension. I believe many students will get (e) wrong.

    Question 3: So much information before the question starts and in reality only the data table is needed. In (b) I bet some students find the molar mass of the compound and not just M. I like (e), but (f) is a silly question.

    Question 4: I really like 4. One easy, straight forward part in (a) and one part in the (b) that requires thinking.

    Question 5: I found (b) interesting only in that a 2nd order rate law from the graph hasn’t been used as much as 0 or 1st order (c) Messed with some kids that tried to find k both ways and it didn’t match up.

    Question 6: Based on discussion from the CB discussion group, this question was a mess of approaches and thinking, many of them wrong… FROM TEACHERS. I expect this to be he worst question for students.

    Question 7: Easy, except I bet far more students than you would think don’t get the SF correct in (a) and (b). I also like that it seems to be more transparent that this is obviously the question that SF will be scrutinized on.

    Reply
    • Adrian

      Thanks Matt.

      Reply
    • Adrian

      Calculations are EVIL!

      Reply
    • Scott

      Question 1: Agree. We actually have an instructor that teaches the kids to use only mass of water and I tried to correct him but he won’t listen. In general, students hate dealing with intermolecular forces.

      Question 2: Agree. (e) is a poorly constructed question. Trying to test multiple concepts in one questions is not a good idea.

      Question 3: Agree. Is this a reading test? (b) I understand Adrian’s approach but I prefer using conservation of mass to find mass of M. Just say moles of M = moles of MI2 and then show the calculation mass over moles. Wish they would have made them balance the half-reactions in (e) before answering and gave the point from (f) to (e). Don’t need part (f) – not testing on anything important here in my mind.

      Question 4: Agree. But how do they want them to solve it since you are not suppose to use pH = pKa + Log([A-]/[HA])? I do anyhow and teach it.

      Question 5: Love that they used a second order. (c) Need to say include units. It does not make it easier.

      Question 6: I like this question. Just wish they would have asked for the barium ion concentration but I understand they wanted an easier part. Makes the studetns use the Q versus K concept which they hate.

      Question 7: I expect the mean to be low on this question though it was preceived to be easier. SIG FIGS – another concepts the kids hate. In general students don’t know how to read analogue equipment properly in this digital age.

      Reply
  38. J Richey

    For Q 6B. I get that Ba+2 would increase… but for the justification do you think referring to LeChatelier’s Principal would be enough or would you need to discuss Q and or/actually show an ice table to get the point(s)?

    Reply
    • Adrian

      ICE table wouldn’t be necessary, but an explanation clearly is required. They are very much in favor of Q versus K statements these days.

      Reply
  39. A Eggert

    No problem with the response….but just noticed that the label for question 6 said 2015.6 instead of 2016.6…typo?

    Reply
    • Adrian

      Changed, thanks!

      Reply
  40. John

    If, for question 6, I accidentally labeled part b as part a and part c as part b, but very clearly was answering part b where part a was and part c where part b was, will they still deduct full points since I mislabeled the questions or will they just deduct the point for a since I basically forgot to answer it.

    Reply
    • John

      Sorry, question 7

      Reply
    • Adrian

      If it is as clear and obvious as you say then you’ll likely get full credit for the parts you did.

      Reply
  41. Srivikram Margam S

    Can we use Le Chateliers principle for 6b? The question didn’t ask for Q vs K. I said that the overall concentration of ions decreases so the backward reaction (more moles) reaction is favored, hence increasing the number of moles of Ba2+.

    Reply
    • Adrian

      Well let’s just say that there is very often more than one way to get full credit! Obviously ONLY saying, ‘Because of Le Chatelier’s Principle’ would not be good enough.

      Reply
  42. Nicolle Arteaga

    Hi Adrian,
    For question #1, I did not add the mass of the salt to the mass of the water (100 g vs 110 g), and I doubt many of my kids did either. So my delta H came out to be -36.5 instead of -40.2 kJ. The IB test banks always indicated that graders will accept both, so I didn’t even think of it until now. Do you have any idea how strictly AP grades that one? I suppose I should have taken the soln hint a little more seriously.

    Again, many thanks for your coaching this year. It was a fun ride.
    Nicki

    Reply
    • Matt

      The ease of the question suggests to me that a point will be deducted for not using the combined mass. It has been done that way in the past, but that decision likely won’t be known until the official reading.

      Reply
    • Adrian

      I suppose I should have taken the soln hint a little more seriously

      I think that’s the point. If the heat capacity of the solution is given, then we should be using the mass of the solution. Also, see this blog post.

      Reply
  43. Derek Stach

    Adrian…

    For #7, every book that I’ve seen used for AP, along with my old college textbook, just tells students to “estimate” the unknown digit. Given that, I believe that pretty much anything that goes to 2 decimal spots will be accepted. i think the key is that students go out that 2nd decimal spot, and since that spot is an “estimation” any reasonable answer that goes out that far is acceptable.

    I know I saw a multiple choice question at some point in time (either on AP, ACS, or Chemistry Olympiad) that was essentially the same, but only asked what you should record the starting value as (instead of finding the difference).

    Reply
    • Adrian

      Honestly, I’ve never really question the thing that I was taught (0 on the line, 5 for anything else’, since it made sense AND I DON’T EVEN CONSIDER THIS TO BE ‘CHEMISTRY’! I don’t want to study instrumentation, and I believe you CAN separate the two!

      Reply
      • Derek

        I wholeheartedly agree with you… I hate that I have to spend time in my classes teaching significant figures, especially since many of my AP Chem students are taking either AP Calc or AP Stats at the same time and they can see the issues with significant figures.

        Reply
  44. Ron Brandt

    Adrian, in Q1 -f, I assume they want a discussion of all 3 steps in the solution process. Your draft answer only deals with the 3rd step. What do you think?

    Reply
    • Matt

      Question specifically says: “Identify all particle-particle interactions that contribute significantly to the dissolution process being exothermic.” So no need to discuss all 3 steps. In fact discussing all 3 steps might lose a point for not answering the question, because some of those steps are endothermic.

      Reply
    • Adrian

      I have no problem with the inclusion of the other steps (with correct ENDO and EXO labels), BUT the way that the question is worded means that they CANNOT insist on the endothermic processes being included.

      Reply
  45. Anonymous

    I drew three resonance structures. Can someone explain to me why this would be wrong?

    Reply
    • Adrian

      Sure. If you draw three resonance structures, presumably you are saying that the double bond can be placed between ANY of the O atoms and the C atom. That is not true. Why not? Well, the O atom that is connected to the H atom (as the questions flags), cannot also have a double bond between it and the central C atom because that particular structure has an unfavorable formal charge, so it is not a valid resonance structure. As a result, the double bond can only exist in two places, eachh with favorable formal charge, hence two resonance structures.

      Reply
      • Matt

        While I agree with your explanation it terms of chemistry. On what grounds can you exclude that bad FC drawing, since the question doesn’t mention formal charge?

        Reply
        • Derek

          That it says that two of the bonds are shorter, and the 3rd longer… If you put double bonds on each of the three O’s, then they’d all be the same bond order and therefore the same length.

          Reply
    • Derek

      Easiest explanation: A double bond on the O that contains the H as well would cause the O to have 3 bonds (and therefore have a +1 formal charge), and the other two O’s would both have a -1 formal charge… Want to limit formal charge to as few atoms as possible (and O doesn’t really like a + formal charge)

      Reply
  46. Greg

    For 6B

    K = 7.7 x 107 = [Ba(EDTA)2-] / [Ba2+][EDTA4-] = [0 + X] / [0.1 – X] [0.15 -X]
    X = 0.09999997402
    [Ba2+] = 0.1-0.09999997402 = 2.598 x 10-8 M

    Now, dilute to 1L and see what the equilibrium position is:
    K = 7.7 x 107 = [Ba(EDTA)2-] /[Ba2+][EDTA4-] = [0 + X] / [0.01 – X] [0.015 -X]
    X = 0.00999997402
    [Ba2+] = 0.01-0.00999997402 = 2.598 x 10-8 M

    By solving the equilibrium expression, [Ba2+] remains constant. But, volume increases, so MOLES of Ba2+ increases.

    Just my take on this.

    Reply
    • Scott Page

      Greg, this is an odd approach because you are not mixing 500 mL of one solution with 500 mL of the other to give you a liter. Best approach is Q to K, reestablishing equilibrium after the dilution. If you take this approach, you get barium ion concentration to be 4.9 x 10-8M or 88% higher after dilution. In reestablishing equilibrium after dilution, the volume did not change, only a shift in moles.

      (0.01-x)/[(0.005+x)(y+x)] = K where y = 2.6 x 10-9

      Q = 0.01/0.005y

      Reply
      • Scott Page

        Greg, I think another way of saying it is the position at equilibrium will depend on the initial conditions, though the equilibrium constant will be the same. There are infinite positions for a given constant. Your initial conditions are different that what the question is saying.

        Reply
  47. Brian Boroski

    I was wondering anyone’s take on question 4b. I got the idea that pH=pKa and they the “perfect buffer” was at 9.95. I took the statement about the reaction has to take place in a buffered solution as you needed the values from the midpoint buffer to equivalence point. above the equivalence point it is no longer a buffer. So, i had answered this as a range from 9.95 to 11.91, or about 10-12. I think the statement about the reaction needed to happen in a buffered solution is confusing. Do they mean the disassociation or the “certain reaction” that is thrown in.

    Reply
    • Scott Page

      Good point but I think you are reading too much into the problem. It just says in deprotonated form. Isn’t this form present even at pH > 11.91 (100%) but just not the major species? Isn’t 100% > 50%? This is why I don’t like AP and prefer ACS. AP does test for concepts; instead they test for reading comprehension + concepts and sometimes two concepts within the same question. Good question if asked properly.

      Reply
  48. bob

    For #2, is this justification right?

    Acid-base.

    Justification:
    NaHCO3, a base, reacts with HC2H3O2, an acid, to produce NaC2H3O2, a salt, and H20, water.

    Reply
    • Adrian

      We’ll see, but I think H+ transfer is much, much better!

      Reply
  49. bob

    For #3 c, I wrote that dissolve the MI2 in water and then do the conductivity as you did. Does dissolving also work or does it have to be molten?

    Reply
    • bob

      Also, for #6b, if I put equals (not greater than) but provided a good explanation that explained why I put equals. is it worth partial credit?

      Reply
      • Matt

        For 3C there are probably dozens of acceptable answers. We would know until the scoring standards are decided. For 6B, you got it wrong and will not receive credit.

        Reply
      • Adrian

        I hope not, since I think ‘the same’ is plain wrong!

        Reply
    • Adrian

      My guess is that will be OK, although there is the pesky situation where HCl, a covalently bonded molecule becomes ions when dissolved in water.

      Reply
      • Srivikram Margam S

        Are there any solid examples of this? It may not be so problematic if all of the exceptions are gases, given that MI2 was a solid

        Reply
        • Adrian

          I don’t think you have to worry, I think it will be fine.

          Reply
      • Srivikram Margam S

        That’s good. Upon further search I find that only compounds with hydrogen tend to have this problem. Also, the line kinda gets blurred, considering that HCl has an ionic character of 18%, which is essentially the same percentage as Zinc sulfide (which, interestingly enough, is considered ionic). On this spectrum, the covalent HF is at a startling 64% ionic character. Looks like the binary definition of ionic and covalent aren’t particularly reliable.

        Reply
  50. Sarah Shea

    What would happen if I drew three resonance structure, changing where the H is bounded to and making one of the left over O doubled bonded with C? Would I not gain credit since I drew 3 instead of 2?

    Reply
    • Adrian

      I’m not sure that I quite understand your description of what you drew!

      Reply
  51. surfer

    I tried not to say anything, because the kids are so sweet and earnest. I don’t want to dump on their hard work and their dreams. But.

    Either you are way more concise than I was answering chemistry questions (very possible) or these problems seem to have very little multi step calculations. [Or you are not showing all the work, needed.] Where is the Initial-Used Up-Equilibrium problem (like with an x squared and a quadratic solution)? Maybe pH change of a buffer? A stoichiometry problem with 2 equations, 2 unknowns? I’m not saying they should all be like that, but I don’t see one. I remember most of the course/book being that! Dimensional analysis out the hooha, too: atmospheres and liters and torrs, oh boy!

    http://pictures.abebooks.com/isbn/9780030578045-us-300.jpg

    I’m reaching into dim memory, but I seem to remember always having to show an equation with variables. And units. Add conversion factors. Arrange to solve for X. Insert amounts. Solve with scientific calculator, recording the several digit answer. Then box an answer with the correct Sig Figs. [Doing last two helps in the odd case, that you make a mistake and overtruncate an aswer, have still demonstrated mastery of the exact amount.]

    Reply
    • Adrian

      Well you always have to show work and be careful with units and significant figures, but boxing answers and dimensional analysis have never been ‘required’. The reason that the work looks so scant on the answers these days, is that the AP exam simply no longer needs ANY extensive calculation work. The exam is a shadow of its former self.

      Reply
  52. daveeckstrom

    I actually think the answer to 5c has to be 5, not 2.5.

    The reason I think this is that the initial rate they give is a positive number, which I take to mean it’s the rate of formation of C8H12. That would mean the initial rate of decomposition of C4H6 is double the given value (0.0020 mol/Ls), which not only yields a value for k that agrees with the slope of the graph of 1/[C4H6] vs. t, but also matches the initial slope of the [C4H6] vs. t graph.

    Reply
    • daveeckstrom

      Which I think makes 5c a lousy question to give a kid who may find himself or herself trying to deal with two different answers that both seem right. It took me 3 or 4 minutes to figure this out myself. A stressed-out 17 year old might never figure out what’s going on, even with a solid grasp of kinetics.

      Reply
    • Adrian

      Given that there is a precedent for this (albeit on the old exam) from 2008, where BOTH answers were accepted, I would be very upset if that isn’t the case this year. Also, the use of positive and negative numbers for rates of products and reactants is far from universal, and has its own issues. Aside from that there is no precedent for using negative number for rates of disappearance (see 2009B, 2).

      Reply
  53. student

    For 1C, would this be a valid explanation?:

    Na+ is larger than Li+ because the extra electron shell in Na+ causes electron electron repulsion, pushing out against the empty valence shell?

    Reply
    • student

      or is the second part about repulsion incorrect or invalid that I should have left out?

      Reply
      • Adrian

        Well, there’s simply an extra, occupied quantum shell so that’s all that’s needed!

        Reply
        • student

          thanks for the reply; is the second part of my response factually incorrect or invalid?

          Reply
          • Adrian

            I don’t think it makes any sense.

  54. Christian

    You sir, are a life saver.

    Reply

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