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TBR weak acid shortcut conditions

Discussion in 'MCAT Study Question Q&A' started by dudewheresmymd, 05.12.14.

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  1. dudewheresmymd

    dudewheresmymd Slowly Drifting... 5+ Year Member

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    Would the MCAT ever ask a weak pH question that didn't qualify for use with the TBR shortcut? The acid bases chapter didn't have any questions that didn't use the shortcut in the answer key. Is this type of question too time-consuming for the MCAT? What do we do if the shortcut doesn't work? 10^-7 would go in the initial for [H+] from autoionization so ... (x+10e-7)(x) =Ka?
     
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  3. Sammy1024

    Sammy1024

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    The shortcut is to make the calculations faster and yes the mcat would ask you to solve it.
     
  4. Teleologist

    Teleologist 2+ Year Member

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    Are you talking about the pH of a very dilute solution? It sounds like you are because you mention auto-ionization. In case of a very dilute solution, we must write out:

    1) A statement of charge balance. Solutions are electrically neutral. If you don't believe me plug charges into Coulomb's law and observe that even for a tiny amount of excess negative/positive charge in two beakers, there would be thousands of pounds of force between them. You may also be able to equate the statement of charge balance to charge sum.

    2) A statement of mass conservation. Your initial molarity of the solute is equal to the sum of equilibrium concentration of the solute and the equilibrium concentration of all the species derived from the solute.

    3) A Ka or Kb expression.

    4) Kw. Kw = [OH3^+][HO^-] = 1.0 * 10^-14.

    These four equations should be sufficient for you to solve for any unknown.
     
  5. Chrisz

    Chrisz 2+ Year Member

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    This shortcut is an approximation of the dissociation of weak acid, based on the assumption that the relative dissociation of water auto-ionization is negligible. Of course, It does not apply to every situation in reality. For example, when the pKa of HA is very close to pKa of water or the concentration of initial HA is very low. Notice, pKa of water is not the samething as pKw. pKa of water is around 15. However, no matter what kind of situation it is, you can always start with the shortcut. If the H+ contributed by water auto-ionization is within the golden rule, 5%, of the H+ calculated using the shortcut, then the initial concentration of H+ is negligible, which means our approximation is valid. If exceeding the 5% golden rule, we have to use the traditional and more complicated approach to solve the question, using Ka instead of pKa. So the calculation always starts with the shortcut, as a basis for determining if our approximation is valid.
     
  6. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    For the MCAT, if you are asked to calculate the pH of a weak base 99.9% of the time, they will provide a weak acid that dissociates only very little. The golden rule is, if it dissociates less than 5% of the original species then it's safe to ignore the amount that dissociates relative to the initial concentration of weak acid. You can assume this is usually the case. This simplification makes the problem significantly easier to solve because we avoid time using the quadratic. It simply boils down to this equation: [H+] = sqrt([HA] x Ka]), and then solving for the pH of it.

    We also make this assumption for many other equilibrium problems as well. In particular, scenarios were we are told there are some reactants and some products in a given reaction that is at non-equilibrium conditions, and then asked what those concentrations will be at equilibrium. These too would require you to make an ICE chart and then solve for "x" by plugging in these expressions into the equilibrium expression for that reaction. But again, because this is a timed test, generally, we'll assume that the amount of products produced is very little and this again, simplifies the question.

    From a conceptual understanding, if you're looking for some basis as to when to ignore something, look at the equilibrium expression. For instance, weak acids have a Ka value that is generally some value to the power of 10^-6 (for example). If you recall, values under 10^-3 favor reactants. Therefore, if you start with excess reactants (compared to equilibrium), we know some products must be produced to re-establish equilibrium (LeChatleir's Princ), and because it favors reactants (defavors products), very little of the reactants will actually dissociate. The same can be said about most equilibrium expressions under 10^-3 (that is, an increasingly negative exponent).

    And one other thing to note about Acid/Base chemistry in general, another thing to watch out for is this common trick. They may give you, let's say 1x10^-8 M HCl and ask you to find the pH. We know strong acids fully dissociate, which means 1x10^-8 H+ will be produced. What is the pH of solution? You would be wrong to say 8.
     
  7. dudewheresmymd

    dudewheresmymd Slowly Drifting... 5+ Year Member

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    if not 8, what would the pH be then?
     
  8. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    Water ionizes in solution to form 1x10^-7 H+ and 1x10^-7 OH- ions. The pH would be a little under, but rounded to 7.
     
  9. dudewheresmymd

    dudewheresmymd Slowly Drifting... 5+ Year Member

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    so why can't we do -log(1e-8) in that instance? I didn't see that example come up in TBR and i just finished the chapter?
     
  10. Chrisz

    Chrisz 2+ Year Member

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    [
    In this case we, should consider the water ionization.
    Kw=(H)OH
    =(x+10^-8)x
    =x^2+(10^-8)x
    solve for x
    H+=x+10^-8

    This involves solving a quadratic, i think it would be a very rare case on MCAT
     
  11. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    The ionization of water at 25 C always produces 1x10^-7 M H+ ions and 1x10^-7 M OH-. This is why the Kw is 1x10^-14 at 25C. No quadratic or calculation is needed to solve this and it IS a realistic question you can get on the exam.

    My point in bringing this example up was so you can have a conceptual understanding of what's happening. Let's say there's 1 trillion people in the United States, and 10,000 more immigrate into United States. The total population is still effectively 1 trillion people. This scenario is essentially the same thing as the example I provided. 1x10^-8 is smaller than 1x10^-7, so effectively, the concentration in solution is still 1x10^-7. But because a very small amount of H+ was added, collectively, the concentration is very slightly higher (just a small tiny fraction higher) than 1x10^-7 and so they may take this a step further and say the pH is slightly less than 7, but it's still effectively 7 when rounded up.
     
  12. Teleologist

    Teleologist 2+ Year Member

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    Wait. Are you saying that if I were given this problem:

    What is the pH of 1.0 * 10^-10000000 M HCl solution,

    The answer isn't pH = 10000000?

    Because, duh, pH = -log[H3O^+] and HCl dissociates to give one equivalent of H3O^+?
     
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  13. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    Yup, pretty straight forward ;).
     
  14. Chrisz

    Chrisz 2+ Year Member

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    you must be kidding right. Kw stays the same all the time, but the water auto-ionization is gonna dissociate to a lesser extent. This is called common ion effect. I totally agree that the pH would be lowerd. However, it is incorrect to say water auto-ionization is always going to produce 10^-7 H+ and 10^-7 OH-1 at 25 degrees. 10^-7 H+ and 10^-7 OH is only going to happen in pure water. However, Kw always stays the same. If you add product to a reaction, the reaction always shift the to the left, according to le chatelier 's principle. The concentration of H+ contributed by water autoionization would not produce 10^-7. Based on what you say that, if you want to calculate the concentration of H+, you should add 10^-8 is smaller than 1x10^-7, which is going to be 1.8 x 10^-7.

    If you take the common ion effect into consideration when formulating an equation, it would definitely be less than 1.8 x 10^-7.

    Solving the quadratic yield, x=6.77x10^-8
    H+=(8+6.77)x10^-8=1.477x10^-7
    pH=6.83

    Double check the result with Kw
    Kw=(1.477x10^7)(6.77x10^-8)=9.99977x10^-15 (close enough to 1.0x10^-14)
     
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  15. Teleologist

    Teleologist 2+ Year Member

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    This is correct. According to Daniel C. Harris' Qualitative Chemical Analysis, water "almost never" produces 1.0 * 10^-7 M hydronium or hydroxide ion. Adding an acid or base will depress the amount of either produced due to the common ion effect.

    To properly solve any equilibrium system, we must write out (at least) four simultaneous equations. These are:

    1) A statement of solution electroneutrality. For example, if you have a solution of KBr, then you would write:

    [K^+] + [H3O^+] = [Br-] + [HO-]

    If you knew the concentration; say you had 18.2 M KBr, then you would write this out:

    18.2 M = [K^+] + [H3O^+] = [Br-] + [HO-]

    2) A statement of mass conservation. Back to 18.2 M KBr:

    18.2 M = [Br-] + [HBr] = [K+]

    3) The K expression. In this case dissociation of KBr is complete so none is necessary. But we could write one for Br- (it's a very weak base but nonetheless we could write one!)

    K = [HBr][HO-]/[Br-]

    4) Water ion product: 10^-14 = [H3O+][HO-]

    Solve these four simultaneous equations and you got it.
     
  16. Teleologist

    Teleologist 2+ Year Member

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    Good. Now I know that the pH of a 10^-19023423423123 M solution of HBr is 19023423423123 :D. And I have empirical evidence; the high pH is indicative of a corrosive effect on skin and boy, does HBr solution sting when you dip your hand into it.
     
  17. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    I was referring to pure water in my earlier statement to illustrate how many ions are in solution before more ions are introduced into solution to think about how to approach this conceptually, without resorting to calculations.

    You're right though and in a General Chemistry class, you'd approach this by realizing this is a common ion effect, using the quadratic to solve for x, then adding the x term to the ions added to solution and eventually find an accurate measure for pH. But if you avoid doing that and just stop to think about it, you can save a considerable amount of time.
     
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  18. Chrisz

    Chrisz 2+ Year Member

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    Teleologist, do you realize HBr is a strong acid, and a concentration of 10^-19023423423123 M is an extremely low concentration. Acid lowers pH, not raises it. Another thing is that such a low concentration of HBr does not affect the pH at all. You can make an analogy. You take a piss in an ocean, does the ocean water taste or smell anywhere close to your piss at all. The HBr is extremely diluted. Or you are making a joke?
     
  19. Teleologist

    Teleologist 2+ Year Member

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    I suggest that you look at the post immediately before that one you quoted to assess how much chemistry I know. Hint: it's probably more than you.
     
  20. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    Okay, before this escalates, I think you guys should just relax and realize this forum is intended to help people, not to engage in silly fights over the internet.
     
  21. Chrisz

    Chrisz 2+ Year Member

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    So I would assume that you are saying an HBr solution with concentration of 10^-19023423423123 M is extremely basic then. Lol
    pKa for HBr is -8.

    Here is the website I randomly found on the internet.
    http://cactus.dixie.edu/smblack/chem2310/summary_pages/pKa_chart.pdf

    I think I should complain to my school and ask for a refund. The general chemistry professor taught me the concept wrong.
     
  22. Chrisz

    Chrisz 2+ Year Member

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    Dont worry, bro. We do not mean to insult each other. It is sort of fun to have a nice conversation after a lengthy study
     
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  23. Teleologist

    Teleologist 2+ Year Member

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    Try solving it with charge balance and mass balance as I suggested.

    Also, what do you think the pH is?
     
  24. Chrisz

    Chrisz 2+ Year Member

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    Teleologist, Probably, you was offended by my earlier ocean joke. No insult was intended. I just thought it was a funny joke. I sincerely apologize. But i do not think dissolving such a low concentration would raise the pH that high. One thing is that HBr is an acid, so it does not raise pH and another thing is that the concentration is extremely low. I would say the pH would stay at 7, almost no effect at all.

    If you want to see to what extent the common effect of HBr on the water ionization and calculate the pH
    Here is what we generally do it.

    HBr, as its negative pKa indicates, is a very strong acid, so it fully dissociate in water, so initial H+ concentration 10^-19023423423123 M
    Now, let s see how the common effect of the initial concentration of H+ on autoionization

    H2O -->OH- + H+

    As you see H+ is in the water dissociation equation, this is why H+ is the common ion in this case. Br- has no effect in this situation.

    Kw=(H)(OH)
    Kw=(10^-19023423423123+x)x
    Since this number is extremely small
    Kw=x^2
    which means x=10^-7
    H+=10^-19023423423123+10^-7=10^-7
    10^-19023423423123 is within 5% of 10^-7, actually extremely smaller than 10^-7, the calculated x
    So our approximation is valid.
    so pH=7

    if you want to solve it with 100% accuracy
    You multiply out
    Kw=10^-19023423423123x+x^2
    0=x^2+(10^-19023423423123)x-Kw
    Kw equals to 10^-14

    Then you use the quadratic formula to solve it, which would be really nasty, but we still can do some modification to the quadratic to make the calculation easy.
    x=[-b+sqroot(b^2-4ac)]/2a

    since the b^2 is extremely small compared to 4ac, it becomes x= [-b+sqroot(-4ac)]/2a
    b=10^-19023423423123, a=1, c=Kw=10^-14
    After calculation, x=(-10^-19023423423123 +2x10^-7)/2
    again, the first term is so small compared to 2x10^-7, so it becomes 2x10^-7/2=10^-7
    x=10^-7
    H+=10^-19023423423123+x=10^-7+10^-19023423423123=10^-7
    pH=7

    So we have arrived the pH=7 using both approaches. Both calculations fit my earlier prediction.
     
    Last edited: 05.13.14
  25. SuperSneaky24

    SuperSneaky24

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    [​IMG]
    But all jokes aside, actually doe I still learned from reading this thread
     
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  26. Chrisz

    Chrisz 2+ Year Member

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    Nice poster. Actually, I have a question for you. How you post pic?
     
  27. SuperSneaky24

    SuperSneaky24

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    Lol thanks. You just click the image icon (looks like a scenery painting).
     
  28. Chrisz

    Chrisz 2+ Year Member

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    But this thing seems like you can only post pics that are already existing on the internet. How do I post my own images
     
  29. SuperSneaky24

    SuperSneaky24

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    I'd just upload your image to a free image hosting site and get a link from there. This site doesn't do uploading your own photos as it'll take up their web space if everyone does it.
     

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