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Errors on Kaplan General Chem discrete test 2?

Discussion in 'MCAT Discussions' started by Swiperfox, Apr 18, 2007.

  1. Swiperfox

    Swiperfox 2+ Year Member

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    Mar 8, 2007
    As I'm sure most of you have access to kaplan materials, I thought this would be a great place to ask. When taking this test, I got two questions wrong which I completely disagree with. I will post the question, my reasoning, and kaplan's answer. I would appreciate someone either pointing out my flawed reasoning or validating that this is in fact an error.

    Which of the following will occur if ammonia were added to a saturated solution of Na2SO4?

    A)The Ksp will decrease and additional Na2SO4 will dissolve.
    B)The Ksp will be unchanged and additional Na2SO4 will dissolve.
    C)The Ksp will increase and additional Na2SO4 will precipitate.
    D)The Ksp will remain unchanged and additional Na2SO4 will precipitate.

    Kaplan:
    A saturated solution of sodium sulfate will be acidic, since sulfuric acid will form when the sulfate ions are hydrated. If ammonia is added to the solution, additional sodium sulfate will dissolve because the acid protons will react with the base. The MCAT expects you to know that Ksp is a function of the structure of the salt and the temperature of the solution. Adding a stress to the system will not affect the Ksp if the temperature remains the same, so the answer is (B) .

    My explanation:

    Ksp doesn't change dependent upon reaction conditions, so A,C can be eliminated. However, since when is a solution to which you add a base acidic? SO42- is the conjugate base of sulfuric acid, not sulfuric acid itself. Therefore, when you add a BASE to water, you should get

    H20 + SO42- -> OH- +HSO4-

    I don't think it is any leap of faith to say this should be a basic solution, not an acidic one. Therefore, when you add NH3 two things should happen. 1) you get increased formation of OH-, and 2) NH3 will deprotonate HSO4- back to SO42- as ammonia is a stronger base. Now, we have a situation in which the number of sulfate ions in solution has increased above previously saturated levels. Thus it should precipitate out and the answer should be D.



    Next problem:

    For a given exothermic reaction, how are ΔH, Ea (forward) and Ea (reverse) related?

    A) ΔH + Ea (forward) = Ea (reverse)
    B) ΔH - Ea (forward) = Ea (reverse)
    C) ΔH = Ea (forward) + Ea (reverse)
    D) ΔH = Ea (forward) - Ea (reverse)

    Kaplan:
    The activation energy for the forward reaction is the amount of energy it takes to overcome the barrier between the products and the reactants. Hence, the activation energy for the reverse reaction is the amount of energy it takes to overcome the barrier between the reactants and the products. The difference between the two is the enthalpy change for the reaction. This can be written as Ea (reverse) - Ea (forward) = ΔH, or ΔH + Ea (forward) = E a (reverse), choice (A).


    My explanation:
    The problem with kaplan's answer is this: If this reaction is spontaneous, enthalpy change is negative, and Ea reverse >Ea forward. Thus, if you plug it into their equation you get something that looks like:

    negative number + small positive number = larger positive number :eek: wtf?

    They flipped the sign of dH, and thus D is my answer.
     
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  3. estairella

    estairella Senior Member 10+ Year Member

    403
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    May 7, 2006
    You are completely right about the first question. Kaplan is on crack when they say "A saturated solution of sodium sulfate will be acidic, since sulfuric acid will form when the sulfate ions are hydrated. ". And by crack, I mean the type of crack you find from dirty dealers on alleyways, laced with PCBs and urine.

    Kaplan is maybe-a-little right for the second question, but you're still completely right. They're basically equating deltaH with heat. In the context of exothermic/endothermic reactions, I don't think you can do that. However, enthalpy IS a measure of heat content, so I bet it works in other contexts.

    Anyways, lesson here - don't trust Kaplan. lol
     
  4. prettymonkey

    prettymonkey 2+ Year Member

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    Apr 10, 2007
    ok what is really really really scary is that i took this exact test yesterday too!!! are you watching me? for the first one i am confused... you have a salt Na2SO4. this is the salt of a strong base (NaOH) and a strong acid (H2SO4). kaplan says that the solution will have the characteristic of the stronger of the acid and base that it hypothetically came from. but the problem is that NaOH and H2SO4 are BOTH strong. so i have no idea how we would determine what the pH of the solution would be. maybe they said it would be acidic because you have 2 Na for every one SO4 and so more NaOH would form and since H2SO4 is diprotic you would still have some H+ in solution.... even if you get HSO4. i dont know. my head hurts.
     
  5. oxeye

    oxeye Moderator Emeritus 7+ Year Member

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    Feb 12, 2006
    Kaplan has lots of errors. I'd say you're doing well if you can spot them. :)
     
  6. Swiperfox

    Swiperfox 2+ Year Member

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    Mar 8, 2007
    Gut instinct says you get formation of HSO4 and not NaOH. NaOH will completely dissociate, and it is an ionic compound anyway (thus in solution it exists as Na and OH ions, not as NaOH). H2SO4, while a strong acid, has a significantly less acidic second proton, and thus doesn't tend to want to completely exist as SO42-.
     
  7. prettymonkey

    prettymonkey 2+ Year Member

    567
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    Apr 10, 2007
    can we assume that the aamc won't ask such a question? damn kaplan. sometimes i feel like they ask questions that have no solution just to mess with my head. i really love when they ask you to calculate something that they didnt give you the information (or diagram) for. take kaplan full length 11 especially PS. you will never want to use anything kaplan ever again.
     
  8. The Force

    The Force 7+ Year Member

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    Apr 14, 2007
    I agree that Kaplan is wrong on the first question.
    For the second one, if you draw the diagram with the Ea and deltaH for an exothermic reaction, you will arrive at their answer
     
  9. MeCord3

    MeCord3 2+ Year Member

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    Nov 11, 2006
    H2O Na2SO4 <--------> HSO4- OH- 2Na+

    The added NH3 will deprotonate the HSO4-, causing the [HSO4-] to go down. Thus causing more Na2SO4 to dissolve.
     
  10. Swiperfox

    Swiperfox 2+ Year Member

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    Mar 8, 2007
    Think about that one for a second. Sure, it will deprotonate the HSO4-, but to what? In the solution, you have saturated levels of SO42-, Na, and some HSO4- floating around. If you deprotonate HSO4-, you have just increased SO42- beyond saturated levels.


    As for the reaction diagram, I did draw it out. Ea reverse - Ea forward for an exothermic reaction would give dH for the reaction operating in reverse (since it would be a positive number, and dH should be negative if the reaction is exothermic)
     
  11. Kraazy

    Kraazy 7+ Year Member

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    Feb 17, 2007
    MDApps:
    Wouldn't you have mostly HSO4-? The second proton is not nearly as acidic. So additional NH3 would take HSO4- to SO42-, pushing equilibrium of Na2SO4 dissolution to the product side. More Na2SO4 dissolves.

    Though I do agree that the original solution would be basic. There is simply no way it could be acidic. Kaplan has quite a few errors in their tests, and some of their explanations really lack depth and they don't help you to figure out why you got something wrong.
     
  12. Swiperfox

    Swiperfox 2+ Year Member

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    Mar 8, 2007
    Perhaps you see something that I don't, but that doesn't make any sense to me. If you deprotonate HSO4- to SO42-, why would you expect more SO42- to dissolve into the solution? You might at first think you would have to compensate by making more HSO4- by dissolving SO42-, but the common ion effect should crowd out any more Na2SO4 from dissolving, and in fact force precipitation. If you added an acid, however, formation of H2S04 would reduce HSO4- levels without increasing SO42-, and thus you would see more dissolution.
     
  13. Lshapley

    Lshapley Old Man Med Student 2+ Year Member

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    Jul 2, 2006
    Portland
    MDApps:
    Initial Species are: Na2SO4, Na+, SO4-2, NH3, HSO4-

    If you do a Hess's law path on this you will get a reaction something along the lines of: NH3 + HSO4-1 --> NH4+ + SO4-2 with HSO4 acting almost as a strong acid (it's Ka is almost 1, which is pretty good for a Ka2). It's basically a neutralization reaction.

    The next best reaction will be the Ksp reaction which will have an unchanged value (it can't change, it is a property of the salt and water). That reaction is Na2SO4 --> 2Na+ + SO4-2. However, because the "neutralization" is greater (Ksp's are not that large), and we have a new value for the concentration of SO4- to plug in according to the MRA. This larger value of SO4-2 necessitates that more Na2SO4 dissolves.

    This question requires way too much though to be on a real MCAT in my opinion. Typical Kaplan.

     
  14. estairella

    estairella Senior Member 10+ Year Member

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    May 7, 2006
    Read your last statement again. If more Na2SO4 dissolves, your value for SO4-2 increases again. According to your logic, more Na2SO4 will dissolve again. You've just started an infinite loop... I think you're forgetting that solids (Na2SO4) don't appear in Ksp's.

    Ksp = [Na]^2[SO4-2]

    You add more SO4-2. Ksp remains the same. Therefore [Na]^2 must go down.

    The only reaction which allows for [Na]^2 to go down is

    2Na+ + SO4-2 --> Na2SO4 (precipitation)

    I know that [SO4-2] will also go down, but that's fine since [SO4-2] is too big as well. It will reach a new equilibrium where a little Na2SO4 has precipitated.
     

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