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Homework Help: Measurement of acetylcholine by pH change

  1. May 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Concentration of acetylcholine (ACh) determined from pH change after hydrolysis. Sample incubated with acetylcholinesterase is quatitatively converted to choline + acetic acid which then dissociates to acetate + H+. A 15mL solution has unknown amount of ACh with pH 7.65, and when incubated decreases to 6.87. Determine number of moles of ACh.

    2. Relevant equations
    [HA] = [H30][A-]/Ka

    3. The attempt at a solution
    initial molarity of H3O
    10^-7.65 = 2.2387211e-8

    molarity after incubation
    10^-6.87 = 1.3489629e-7

    difference in molarity gives increase in concentration of H3O
    1.3489629e-7-2.2387211e-8 = 1.1250908e-7

    acetic acid dissociates to equal amounts of H3O + Acetate and Ka is 1.8e-5
    1.1250908e-7^2/1.8e-5 = 7.032385e-10

    therefore the concentration of acetic acid is 7.032385e-10. If I add this to the H3O concentration to give the total concentration of the acid before it dissociated I get 1.1321232e-7 which when multiplied by the volume of the sample in litres gives 1.6981848e-9 which is the molar amount acetic acid, and since all of the Ach is converted it is also the amount of Ach.

    I had initially done this wrong, however after writing it out here I figured it out :). Since I have written it all out though, is there any better way of doing it or does that look ok?

  2. jcsd
  3. May 22, 2008 #2


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    Staff: Mentor

    I don't like the question nor the method. Why initial pH is 7.65? Obviously solution contains some substances responsible for that pH. Could they simply be ignored? Most likely these are weak bases, so one may expect some buffering effect, that'll "eat" some of the acetic acid, slowing down pH change. Then, pH is close to 7, which means that ignoring water dissociation can be dangerous for exactly the same reason.

    Somehow my intuition fails me now and I don't have enough time for detailed analysis, but I doubt your result is OK. OTOH I am not sure if there is enough information to correctly calculate amount of acetic acid liberated in hydrolysis.

    Last edited by a moderator: Aug 13, 2013
  4. May 22, 2008 #3


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    Staff: Mentor

    OK, did some math. If you ignore water autodissociation, your result is more then 4 times too low. Two assumptions made: that acetic acid is fully dissociated (justified by pH), and that initial pH was raised above 7 due to the presence of the strong base.

    Note that if the initial pH is raised by a weak base, difference can be higher by orders of magnitude.
    Last edited: May 22, 2008
  5. May 22, 2008 #4
    Hmm, well first off the question did say assuming there are was no buffer but i forgot to add that part, sorry about that. The answer i came up with was correct according to the book though. Its Lehninger priciples of biochemistry btw if that's any indication of how many errors it contains. I may be wrong here but I thought that "ACh is quantitatively converted into choline and acetic acid" meant that it was fully converted? Also I don't quite understand what you are saying here "and that initial pH was raised above 7 due to the presence of the strong base".

    Thanks for the help though.
  6. May 22, 2008 #5


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    Staff: Mentor

    Won't be short (nor fast to enter), but let's try.

    First - there must be some base present in the solution for the pH to be higher then 7. As I already wrote if pH is high because of the presence of a buffer, buffer capacity can be killing. Thus we will do the calculations assuming pH was raised above 7 with strong base, this way buffer capacity is as low as possible.

    pH after is 6.87, before it is 7.65, difference in concentrations is 10-6.87 - 10-7.65 = 1.13x10-7 (note that it is identical to the result you have found using different approach; this is no accident).

    Now let's check if we have not made mistake ignoring water autodissociation. For that we will first calculate amount of strong base present in the solution, then we will calculate amount of acid that have to be added to change pH from 7.65 to 6.87.


    We have strong base BOH in the solution, it is fully dissociated to B+ and OH-. Its analytical concentration equals [B+]. Obviously solution contains also H+. And from the question we know that

    [H+] = 10-7.65

    This solution must be electrically neutral. This is one of these conditions that have to be fulfilled no matter what the solution composition is:

    [B+] + [H+] - [OH-] = 0

    We will also use water ion product:

    [H+][OH-] = 10-14

    (note: see link to the pH calculation lectures at the end of this post, I am explaining general approach to pH calculation there - and what I am doing here is in fact no different)

    So, if we know pH we can easily calculate [B+] as

    [B+] = 10-14/10-7.65 - 10-7.65 = 4.24x10-7

    Our pH 7.65 solution is just a solution of strong base with concentration 4.24x10-7M.


    Hydrolysis means we have added some amount of acid - let's call it HA. Let's assume it is strong acid, fully dissociated - so it adds H+ and A- to solution. We can use identical approach as above to calculate amount of acid added. pH is now 6.87, so

    [H+] = 10-6.87

    Charge balance is very similar, but it accounts for the A- presence:

    [B+] + [H+] - [OH-] - [A-] = 0

    Concentration of B+ has not changed, [H+] and [OH-] are given by pH - so calculation of [A-] is straightforward:

    [A-] = 4.24x10-7 + 10-6.87 - 10-14/10-6.87 = 4.85x10-7

    and this is amount of acid that was added (per liter of solution, as these numbers were just concentrations).

    Comparing results - first approach gave 1.13x10-7 and the second 4.85x10-7. You can't ignore water autodissociation in this case. Book is wrong.

    Moral: don't believe biochemist when it comes to pH calculation :wink:

    Geez, my hands hurt from writing.
    Last edited by a moderator: Aug 13, 2013
  7. May 23, 2008 #6
    Thanks Borek that was quite interesting. I wasn't aware of that method. Thanks for the link as well, the lectures on that site are great.
  8. May 23, 2008 #7


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    Staff: Mentor

    Note: this method is just a general approach to ANY equilibrium calculation. What I have ignored here are dissociation constants for both base and acid. Adding them will make calculations harder, as we will end with 3rd degree polynomial, but the result will not change substantially - as both substances are strong and 100% dissociated.
    Last edited by a moderator: Aug 13, 2013
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