Post a New Question

chemistry

posted by .

An enzyme-catalyzed reaction is carried out in a 50-mL solution containing 0.1 M TRIS buffer. The pH of the reaction mixture at the start was 8.0. As a result of the reaction, 0.002 mol of H+ were produced. What is the ratio of TRIS base to TRIS acid at the start of the experiment? What is the final pH?

If needed: TRIS(Trizma base) mw= 121.1; pka= 8.3;
TRIS-HCl mw= 157.6; pka= 8.3

  • chemistry -

    Can you tell me what 0.1 M TRIS means? Is it
    a. 0.1 M in the base,
    b. Start with 0.1 M of the base and add HCl to a specific pH,
    c. other. If other, then what?
    I can work buffer problems but in quant we ALWAYS said, "A buffer of pH 4.2 was prepared by adding ?? g acetic acid to ?? g sodium acetate, " etc etc.I never took a course in biochemistry and this 0.1 M TRIS is a wording I don't understand. The Internet doesn't help with that either. You have posted this question numerous times and I would like to be of some help but I can't help if I don't know the terms.

  • chemistry -

    TRIS is a buffer used in biochemistry

  • chemistry -

    I KNOW that. But I don't know the biochemical terminology when you say it is 0.1 M. What does that mean? See my other post. Is it 0.1 M in TRIS ONLY, was the solution made by taking 0.1 M TRIS and adding HCl to it to the desired pH, just what do you mean when you say 0.1 M. 0.1 molar in what??

  • chemistry -

    0.1 M means 0.1 mol/liter

  • chemistry -

    Here are instructions on how to solve it. Would you help me step by step with the math and instructions?:

    Heres how to solve it:

    Use the Henderson-Hasselbalch equation.
    pH = pKa + log ([Tris] / [Tris HCl]). Here the pH is 8 and the pKa of the Tris is 8.3, so you can solve for the ratio ([Tris] / [Tris HCl]).

    After the reaction, when 0.002 moles of H+ were produced, this H+ will CONSUME the Tris base to FORM more Tris HCl From the first part of the problem you have the ratio ([Tris]/ [Tris HCl]) = ???. You also know you have 50mL of 0.1M Tris buffer, or 0.005 moles of total Tris (in acid or base forms), which can be written out as "moles Tris + moles TrisHCl = 0.005. Combine this with the equation for the ratio, and solve to find # of moles of both the base and acid form. This will tell you how many moles of acid and base form Tris you had BEFORE the reaction.

    Now subtract 0.002 moles from the amount of Tris base, and add 0.002 moles to the amount of TrisHCl you calculated (the H+ consumes base, makes TrisHCl). Now using the same pKa, and the new calculated concentrations of Tris and TrisHCl, you can again use the Henderson Hasselbalch equation, this time solving for pH, the final pH.

  • chemistry -

    Give me a few minutes to digest this and I'll get back. I don't know exactly when. Let me call you attention to the sentence that starts with, "You also know you have 50 mL ...........moles of total TRIS (in acid or base forms) which ....." That's the question I've been asking. If you START with 0.l M TRIS in the BASE form OR the ACID form, you have no buffer until some of the other is present. Let me work through this. The instructions seem to be in good detail.

  • chemistry -

    ok. I now know the answer to my question. The answer is this. When a biochemist says that s/he has prepared a 0.1 M TRIS buffer, s/he means that it is 0.1 M counting BOTH the base AND the acid forms. So it isn't 0.1 M in either although s/he may have started with that amount initially. Here is the remainder of the problem.
    The initial pH is 8.0 so we use
    pH = pKa + log B/A (Let's make this easy typing by noting B stands for mols B and A stands for mols A.
    8.0 - 8.3 = log B/A
    -.3 = log B/A
    B/A = 0.5012 (you aren't allowed that many significant figures so make your work fit your prof's rules on significant figures.).
    Now you know total mols = 50 mL x 0.1 M = 0.5 millimols = 0.005 mols.
    You also know that
    B + A = 0.005 (that is the total # mols, either acid or base, is 0.005).
    Now you add the 0.002 mols H^+.
    Here is the equation.
    base + H^+ ==> acid

    We add 0.002 mols H^+ which means we now have B-0.002 for the base and A + 0.002 for the acid.We plug that into the equation
    A + B = 0.005
    (A + 0.002) + (B - 0.002) = 0.005
    A + B = 0.005 since the 0.002 cancels.
    Now we substitute from the ratio (which is the answer to the first part of the problem where B/A = 0.5012 or B=0.5012A for B.
    Rewriting the first equation as
    A + B = 0.005, then substitute,
    A + 0.5012A = 0.005
    1.5012A = 0.005
    A = 0.00333 mols A
    B = 0.005 - A = 0.005 - 0.00333 = 0.0167. That gives us
    mols A = 0.00333
    mols B = 0.00167
    before the H^+ was added.

    So we had 0.00333 mols A, that is now 0.00333 + 0.002 = ??
    We had 0.00167 mols B, that is now 0.00167 -0.002 = ?? and here we run into trouble for we have a negative number for mols B and we can't take the log of a negative number. Check my work but I don't believe I have made an error. I worked this far a couple of days ago and didn't post my answer because I knew something was wrong. However, I have followed the instructions, I think, exactly. But you should read through to make sure I did that. If you find a made an error, please post to let me know and give details about the error.

  • chemistry -

    I see I made a typo. 50 mL x 0.1 M = 5.0 millimols = 0.005 mols. But 0.005 mols is what I used throughout the remaining part of the problem.

  • chemistry -

    I just made the neg to a pos and finished the problem. I got pH=6.3 for a final answer. hope that is what you came up with.

  • chemistry -

    I didn't get an answer since the log of a negative number is not possible. The problem would work if it was 100 mL of 0.1 M (instead of the 50 mL). Then we would have
    B = 0.00334 - 0.002
    A = 0.00667 + 0.002
    pH = 8.3 + log 0.00134/0.00867
    pH = 8.3 + log 0.155
    pH = 8.3 -0.811
    pH = 7.49

  • chemistry -

    WHAT RECTION WITH TRIC_HCL AND ATHENOL

  • chemistry -

    drbob22 I think the reason you are getting a negative number instead of the positive you need to find the log is because you put your A and B in the places. Try inverting them. I'm sure you will get a positive. I followed your instructions to a Tee and never came up with a negative log. Thank you for posting an answer to this it really help me get an understanding.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question