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Titration of Dibasic Compounds

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data

    For the titration of 50.0 mL of a 0.100 M solution of a dibasic compound with 0.0500 M HCl, calculate the pH at 100mL (equivalence point) and at 145 mL.
    pka1=2.46
    pka2=9.41

    2. Relevant equations
    (1)pH=pka + log[BH+]/[BH2+]
    (2)[h+]=sqrt((K1K2F)+(K1Kw)/(K1+F))


    3. The attempt at a solution
    for the equivalence point i used the second formula and I got a pH of 6.175.
    For the second part I should use the Henderson Hasslebach equation.
    I have 7.25E-3 mol of HCL and I dont know what the concentration of the base should be. If i keep using 5E-3 mol my pH would be very low. I am stuck here! Does the concentration of the base change? if it doesn't it would be 7.25E-3 - 5E-3/0.195L
     
  2. jcsd
  3. Mar 26, 2014 #2

    Borek

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    What is F?

    Then the formula is wrong, as the first equivalence point pH is close to the average of pKa1 and pKa2.

    Concentration of the base does change - it reacts with added acid. But the formula you wrote is wrong. Perhaps you forgot 5 millimoles of the acid were already consumed earlier, as the base is diprotic?

    Sadly, even if it is done correctly, pKa1 is too high for Henderson-Hasselbalch equation, and pH is lower than the one calculated with this approach. Instead of using HH equation you should treat concentrations found as input for ICE table.
     
  4. Mar 26, 2014 #3
    F is the formal concentration, so in this case it would be 5E-3 M. the formula I used is from the book and we use it in class. I agree that ph=pka1+pka/2 but I can't figure out how to get that answer using the formula

    The pka values are for tryptophan so they can't be wrong
     
  5. Mar 26, 2014 #4

    Borek

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    Doesn't make it correct.

    Exact answer (found using pH calculator, which does full equilibrium calculation) is 5.96. Average of the pKa values is 5.94. Neither is close to 6.17

    I have problems replicating 6.17, what I got after plugging numbers into your formula was 6.05. I assume you meant

    [tex][H^+] = \sqrt{\frac{K_1K_2F+K_1K_w}{K_1+F}}[/tex]

    as in

    [tex][H^+] = \sqrt{K_1K_2F+\frac{K_1K_w}{K_1+F}}[/tex]

    units don't make sense.
     
  6. Mar 26, 2014 #5

    epenguin

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    I think these exercises were meant to test grasp, which if you have you can do them quite fast. And so to send you back to your book - or thinking :tongue: - if you find you can't do them fast.

    The first answer is just the mean of the pK's which is somehow what you could expect - I mean why should it be more towards one than the other? - (very rough argument). More soundly for a dibasic acid you easily find that

    K1K2 = [H+]2[A2-]/[H2A]

    You should be able to work out easily mass and charge balances give you, at equivalence, the ratio [A2-]/[H2A] of 1. Which gives you the result Borek reminded you of.

    The pKs are well separated so at the pH exactly between them both [A2-] and [H2A] are small. But even if pK's are close so they are no longer small, this mean is still where [AH-] is maximal and the two other forms are equal to each other.

    So I got pH 5.935 for the first question. This is the simple practical-purposes calculation you were being tested for. (Then Borek mentions that it is an approximation. I think his more exact calculation involves taking into account effects of the pesky [H+] and [OH-] in complete set of equilibria. You may be, I was, surprised that the difference between the calculations is even as much as Borek says, but the difference I think is only about 5X10-8 M in [H+]. I think it would be difficult to make up that solution to give you the pH to 0.02 units. )

    The second problem I think you were also meant to do quite fast. I think you were expected to say "by the first 100 ml HCl I have essentially protonated the first group of the base, so now this is just like adding 45 ml of my HCl to 0.1 M base whose pKa is 2.46." You can in fact use the Henderson-Hasselbalch equation. Simples.

    Though it is excellent to work out your exact equations and show they will simplify using sensible approximations, to the simple practical ones.
     
    Last edited: Mar 26, 2014
  7. Mar 26, 2014 #6

    Borek

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    This is exactly kind of thinking I would employ, and I agree that's most likely the way the question was intended to be solved.

    Sadly, it doesn't work for acids with pKa below 3. They are strong enough to dissociate further, and the assumption that neutralization was stoichiometric is no longer valid.
     
  8. Mar 26, 2014 #7
    So what you're saying is that after the first equivalence point, I only take into account the excess 45 mL of HCl? That means that 45mL x 0.05 M=2.2E-3 mol of HCl and there's a total volume of 195 mL
    [BH+]=(5E-3 - 2.25E-3)/0.195 L= 0.0141 M
    [BH2 2+]= 2.25E-3/0.195= 0.0115 M
    pH=2.46 +log (0.0141/0.0115)= 2.54
    I dont think that this is correct because the pH at the second equivalence (at 200mL) point should be 3.09 and the pH can't go back up during a titration. I understand the logic but I'm sure the stoichiometry is wrong in my calculation I just can't figure out how to fix it.
     
  9. Mar 26, 2014 #8

    epenguin

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    Firstly I don't know where you get this pH at the second equivalence point from, but it cannot be 3.09. The second equivalence point would be where you have just a solution of H2B, well to the acid side of the pK. (You should calculate what it is.)

    In fact your calculation appears to be correct - I get to 2 dec. pts. 2.55 but we needn't argue about 0.01 between friends.

    Note that you divided one number by 0.195 and another number by 0.195, and then you calculated the ratio of these two results - so although that does not make the result wrong you never needed to divide by the 0.195 in the first place. As in the equation defining the equilibrium constant, or the equivalent Henderson-Hasselbalch equations you have a ratio of concentrations usually, but as these refer to one and the same solution whatever its volume, molar ratios equal mole ratios and it is more often than not convenient to reason with moles and forget volumes.

    I am not sure what Borek is saying, possibly that at such acid solutions [H+] starts to be a concentration that starts to be comparable to the other ionic components - here by the time you have diluted the total base is about 0.025 M while [H+] is about 0.0028 M so maybe a more refined calculation is in order, but it would give a pH very much inside the same ball park.
     
  10. Mar 26, 2014 #9
    Ok then in that case, how would I calculate the ph at the second equivalence point at a volume of 200 mL of HCl?
     
  11. Mar 26, 2014 #10

    epenguin

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    It is too late at night here for me to think but you can work it out or if not look back in you book and notes for such questions and exercises as how to calculate the pH of 0.01 M acetic acid.
     
  12. Mar 26, 2014 #11
    The problem is my book and my notes don't go into too much detail for polyprotic systems, which is why I am having trouble with this problem. Ill try to work on it myself but can you please give me a quick guideline on how to do it whenever you have the time? It is truly appreciated! you have helped me a lot and I can't thank you enough
     
  13. Mar 27, 2014 #12

    Borek

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    At the second equivalence point you have just a solution of diprotic acid with given pKa1,2 values. They are so far from each other you can safely ignore the pKa2. Compare http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-simplified

    pH that you have both calculated for 145 mL of titrant added is off by about 0.14 unit, exact pH of the solution is 2.69. That's because you calculated ratio of concentrations of acid and conjugate base assuming protonation was stoichiometric. It wasn't.

    If I will find time I will try to explain why this approach is not yielding a correct answer. For now just remember it doesn't work for for acids with pKa below 3.
     
    Last edited: Mar 27, 2014
  14. Mar 27, 2014 #13

    epenguin

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    The problem of the pH at the second equivalence point. Just rounding the volume to 200 ml the problem is simply: 'what is the pH of a 0.025 M solution of BH2?'

    This is, I remind you, not a polyprotic problem. Because at the pH's around 2 we are dealing with now, the second (amine) group of tryptophan is as good as all protonated (about 99.9999%). So there is only one protonation equilibrium
    BH2 ⇔ BH- + H+ to be considered.

    Calculating the pH of a weak acid solution has probably been covered in your book, notes, or earlier exercises. Anyway it would help you more to work it out yourself. You can find a very simple formula for the case where AH2 is reasonably concentrated and the pH not too extreme. I found this gives a pH of 2.05.

    And then yes, here the acid is rather dilute and not so weak and the pH pretty low so this is not quite accurate so you can do without the approximations and have to solve a quadratic equation, in which case I get pH of 2.15.
     
    Last edited: Mar 27, 2014
  15. Mar 27, 2014 #14
    I solved the pH at the second equivalence point and that's the answer I got thank you. Now, past the equivalence point, there's only HCl left, right? so we would have HCl <> H+ + Cl-. The concentration of H+ would no longer be x but would be x + the concentration of H+ stemming from the excess titrant added and we would have a new quadratic formula.
    HCl<-> H+ + Cl-
    F-x x+ [H+] x
    So for 220 mL, [H+] would be 0.02 L x 0.05 M= 1e-3 mol/(0.220+0.05 L)=3.7e-3 M
    and the formal concentration, F, of [HCl] would be 0.05 X (0.02/0.27)= 3.7e-3 M
    using ka of 3.467e-3 (pKa of 2.46) I got a pH of 2.7..which is obviously wrong... Where is my mistake?
     
  16. Mar 27, 2014 #15

    Borek

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    You have a mixture of a strong acid AND a weak acid now. HCl is fully dissociated, the other acid is dissociated only partially - but it still lowers the pH. How much of the weak acid is dissociated can be calculated using ICE table.
     
  17. Mar 27, 2014 #16
    ok so the equation would still be BH22+ <-> BH+ H+?
     
  18. Mar 27, 2014 #17

    Borek

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    Yes.
     
  19. Mar 27, 2014 #18
    Then we would have BH2 2+ <> BH+ H+
    F-x x x+ [H+]
    F= 0.1 X (0.05/0.27)= 0.0185 M
    [H+] = 0.05 x (0.02/0.27)=3.703e-3 M
    using ka= 3.467e-3, I got a ph of 2.28, which is wrong again
    what am i doing wrong?
     
    Last edited: Mar 27, 2014
  20. Mar 28, 2014 #19

    Borek

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    You forgot about H+ from the HCl.
     
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