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Calculating max pH of aniline

  1. Sep 26, 2013 #1
    Hi!
    I've been struggling with this for hours now, but my head doesn't seem to work properly...

    1. The problem statement, all variables and given/known data
    Aniline, C6H5NH2, is a weak base and reacts with water, forming the phenyl ammonium ion C6H5NH3+ and OH-.
    Kb at 25oC is [itex]3.2 \cdot 10^{-10}[/itex]

    a) What is the equilibrium concentration of aniline at pH=9.00?

    b) The solubility of aniline is 3.8g / 100mL of water at 25oC. Calculate the maximum PH when aniline is dissolved in water at 25oC.

    2. Relevant equations

    [itex]K_b = \frac{\left[OH^{-} \right]\left[C_6 H_5 NH_3 ^{+} \right]}{\left[C_6 H_5 NH_2 \right]} = 3.2 \cdot 10^{-10}[/itex]

    pH + pOH = 14

    pOH = -log[OH-]


    3. The attempt at a solution

    I think I did the a) one right:

    [itex]pH = 9.00 \, \Longrightarrow \, pOH = 5.00 \, \Longrightarrow \, - \log \left( \left[ OH^{-} \right] \right) = 5.00 \, \Longrightarrow \, \left[ OH^{-} \right] = 10^{-5} = \left[ C_6 H_5 NH_3 ^{+} \right][/itex]

    Then;

    [itex]\left[ C_6 H_5 NH_2 \right] = \frac{\left[ C_6 H_5 NH_3 ^{+} \right] \cdot \left[ OH^{-} \right]}{K_b} = \frac{(10^{-5})^2}{3.2 \cdot 10^{-10}} = \underline{0.3125 \textrm{ M}}[/itex]

    So the equilibrium constant of aniline at pH=9.00 is 0.3125 M.
    Is this correct?


    Then, I've tried b) for hours... Here are my thoughts:

    [itex]n_{\textrm{aniline}} = \frac{m}{M} = \frac{3.8 \textrm{ g}}{93.126 \frac{\textrm{g}}{\textrm{mol}}} = \underline{0.041 \textrm{ mol}}[/itex]

    Because it says that 3.8g = 0.041 mol of aniline is dissolved in the water, I assume that this means that there are 0.041 mol of C6H5NH3+ and, equally, 0.041 mol of OH- in the 100 mL of water.

    If so, this means that the concentration of OH- is 0.41 M.

    pOH = -log(0.41) = 0.389 ==> pH = 13.64

    My first thought here is that totally neglecting the Kb in this task feels wrong. Also, a pH of >13 for such a weak base also feels very wrong. It also feels wrong to have such high concentrations of ions, when I found in a) that we only had 10-5 of them when we had a 0.3125 M aniline solution.

    However, the task said that 3.8g of aniline was dissolved, and 3.8g of aniline is 0.041 mol of aniline, and 0.041 dissolved aniline in 100 mL water, will be 0.41 M of OH-... At least that is how I interpret the task... So I'm confused right now.

    So I guess I screwed it up badly. Where did I go wrong?

    Hope you can help me! Thank you very much!!
    Daniel
     
  2. jcsd
  3. Sep 27, 2013 #2

    Borek

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    Staff: Mentor

    There is no correct answer to a.

    Let's start with the reaction equation:

    An + H2O ↔ AnH+ + OH-

    (where An stands for aniline, I am too lazy to type the whole formula out each time). Basic dissociation constant is

    [tex]K_b = \frac {[AnH^+][OH^-]}{[An]}[/tex]

    It can be easily rearranged to

    [tex]\frac {[AnH^+]}{[An]} = \frac {K_b}{[OH^-]}[/tex]

    So, for a given pH ([OH-] is a function of pH) you can calculate ratio of concentrations of protonated and not-protonated aniline, but this ratio alone is not enough to calculate concentration. For that you need additional information about the total concentration of aniline.

    Your approach to b is wrong - that is, you correctly started trying to calculate anilline concentration in the solution, but then you assumed wrongly that it means concentration of OH- is identical. It is not, and you are right - you can't ignore Kb, you have to use it in your calculations. Do you know how to use ICE tables?

    This page on pH calculation of weak acids and bases can be helpful, but if you are struggling with the basics it can be a little bit over your head.
     
  4. Sep 27, 2013 #3
    Thank you for replying!

    Do you mean that the task can't be solved?
    Trying again, using your tips:
    I still find that pH=9.00 implies [OH-] = 10-5..

    So;

    [itex]\frac{[AnH^{+}]}{[An]} = \frac{K_b}{[OH^{-}]} = \frac{3.2 \cdot 10^{-10}}{10^{-5}} = 3.2 \cdot 10^{-5}[/itex]

    This means that there are [itex]3.2 \cdot 10^{-5}[/itex] mol of AnH+ for every mol of An. (Correct?)
    Then what? Can't we find the concentration of base dissolved when we know the Kb and the pH?


    Yes, I know about these tables. But in the task it says that 3.8g=0.041mol An is dissolved in the water. Doesn't this imply that there are 0.041 mol AnH+ and 0.041 mol OH- in the solution? I mean... 0.041 mol is dissolved... If we were putting 3.8g of aniline into the water and then looking to see how much of it got dissolved, I would understand that only a small portion of it got dissolved into the ions. But I interpret it as that 0.041 mol is already dissolved, that is, have become ions An+ and OH-.. Why is that wrong?

    But OK, if I try to setup such a table...

    Initially:
    0.41M an --- 0 anH+ --- 0 OH-

    Change/after reaction:
    (0.41-x) M an --- x M anH+ --- x M OH-

    Then;

    [itex]K_b = \frac{[OH^{-}][AnH^{+}]}{[An]} \, \Longrightarrow \, [OH^{-}]^2 \approx K_b \cdot [An]_0 = 3.2 \cdot 10^{-10} \cdot 0.41 = 1.312 \cdot 10^{-10}[/itex]

    Such that:

    [itex]x = [OH^{-}] = \sqrt{1.312 \cdot 10^{-10}} = 1.145 \cdot 10^{-5}[/itex]

    And then pOH = -log(1.145*10-5) = 4.94... resulting in pH = 9.06 ...

    Is that correct?
     
    Last edited: Sep 27, 2013
  5. Sep 27, 2013 #4

    Borek

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    Yes.

    OK

    There is 3.2×10-5 moles of AnH+ for a mole of An, that's OK, but that's not the best way of describing the situation. First, it is better to deal with concentrations and not with amounts of substance, second, in this case 1 mole of An means An in equilirbium with protonated AnH+ - in other words, total amount of aniline is not 1 mole, but 1+3.2×10-5. But you can't use it to calculate concentration, as you don't know in what volume it is present.

    No, that's exactly the problem here. Concentration of protonated aniline at given pH is a function of total concentration of aniline - for 0.01 M solution of aniline concentration of protonated form is around 4×10-7 M, for 0.001 M solution it is 4×10-8 M and so on.

    You are mistaking dissolved with dissociated. Aniline is a weak base, so not every molecule present in the solution produced OH-.

    Yes, that's it.
     
  6. Sep 27, 2013 #5
    Thanks for your help!

    However, I was at my university just now and asked a professor. He said I had done the a) correctly. And also, with the good help from you, the b) was correct.
     
  7. Sep 27, 2013 #6

    Borek

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    OK, I think I know what is the problem with a - you translated it in an ambiguous way.

    doesn't suggest aniline is a source of pH, if anything, it says - there is a solution that has pH of 9.0, it contains some amount of aniline, what are concentrations of the different forms? This has no solution.

    What the question really asks is - you put aniline into water till pH is pH=9.0, what will be the concentration of non-protonated aniline in this solution? And your solution of this problem was OK.
     
  8. Sep 27, 2013 #7

    epenguin

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    I worked this out and the results agreed with Norway's; whilst I was writing the answer Borek's post arrived.

    :approve: Student doing well, not every one gets this, recognising the warning of a ridiculous result. Another indication which kind of supports the conclusion is that at pH 9, the concentration is less than the solubility (you have to worry whether, but it was) but not much less than the solubility. Question (b) is about the saturation concentration of aniline; as you were not far from saturation at pH 9, reciprocally at saturation pH is not going to be very much more than 9.

    It was at first a bit disconcerting they talked of 'the equilibrium' concentrations.

    Another general issue not to say difficulty encountered in these problems is recognising what you can safely ignore. But at pH 9 you can ignore [H+] in comparison with [OH-] and so get to good approximation [OH-] = [AnH+].

    [H+] << [AnH+] << [An] . Getting to things like that can feel a bit circular reasoning at first is probably often a cause of difficulty so may take time and practice.
     
    Last edited: Sep 27, 2013
  9. Sep 27, 2013 #8

    Borek

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    I guess someone with a nick "Norway" was given a problem in Norwegian, so the wording could be not that of the original :wink:
     
  10. Sep 27, 2013 #9
    Hi.
    The task was given in Norwegian, as you say, but if this was ambiguous in English, then it was ambiguous in Norwegian aswell. I've translated it directly from Norwegian, and even though my English might not be flawless, I'm still certain that what I wrote in English means the same as what I read in Norwegian.

    The task was given:
    Hva er likevektskonsentrasjonen av anilin ved pH=9,00?
    which literally means
    What is the equilibrium concentration of aniline (at/when) pH=9.00?

    And there is nothing lost in translation. If this doesn't imply that aniline causes the pH-value in English, then it doesn't imply this in Norwegian either.

    So I'm sorry if this confused you. No wonder I got confused myself, then. :)
     
  11. Sep 27, 2013 #10

    epenguin

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    I love the "likevekt" - like weight! :smile: More transparent than our Latin term. I wonder how many people know the word 'equilibrium' without thinking or realising how it's made up?

    Will surely remember that, even if it will not get me far in travel or literature.
     
    Last edited: Sep 27, 2013
  12. Sep 27, 2013 #11

    Borek

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    I know nothing about Norwegian (well... for some strange reason I know skytebane is a shooting range :biggrin:) but what you wrote makes sense.

    Note: in what your wrote one thing clearly marked you as a non-native speaker. What you called "task" is a "problem" in English. This is a common error between Germans and Slavs - apparently also for Norwegians :wink:
     
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