Biochemistry

posted by .

Calculate the pH that would result if 0.40mL of 1.50 M HCl is added to 12.0 mL of a 0.065 M solution of the fully deprotonated form of the tripeptide glu-asn-leu.

I know I'm supposed to do an ICE table... and the equation will look like this initially:

deprotonated + HCl ---> protonated + Cl
0.00078 mols 0.06 mols 0 0

but I'm not sure where to go from there! I also know I'm supposed to use the formula pH= pKa + log (deprotonated)/(protonated).

Thanks for any help

  • Biochemistry -

    I think you have made an error in HCl. M x L = mols = 1.5 x 0.4/1000 = 0.0006

    deprotonated is the base.
    protonated is the acid.
    base + HCl ==> acid form
    base = 0.00078 mols.
    HCl = 0.0006 mols.
    volume = 12.0 + 0.4 = 12.4 mL = 0.0124 L

    So the HCl is the limiting reagent and it will form 0.0006 mols acid form. That will leave 0.00078-0.0006 = 0.00018 mols base remaining.
    So in the equation, you know pKa (although it isn't in the problem you posted), base (deprotonated) = 0.00018 mols and acid (protonated form) = 0.0006.
    Technically, the formula says that pH = pKa + log[(base)/(acid)] and
    (base) = 0.00018 moles/0.0124L = ??
    (acid) = 0.0006 moles/0.0124 L = ??
    and those go into the pH = pKa + log [(base)/(acid)] BUT the 0.0124 L (volume of the solution) cancels and mathematically you can get away with not including it;i.e., just using mols. Some profs count off it you don't do it with concn (mols/L) and some don't. You must be the judge of how to set it up.(I always counted off if concns were not used BUT I always told the students they didn't need to go through the step of actually calculating the concn. I would let them show (base) = 0.00018/v and (acid form) = 0.0006/v), the volume cancels and leaves the mols/mols.

  • Biochemistry -

    Thank you, I understand all of it but I'm not sure how I'm supposed to know what the pKa is

  • Biochemistry -

    There should be a table in your text.

  • Biochemistry -

    If not a pKa value, then a Ka or Kb value from which pKa can be calculated.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Biochemistry

    Do you know what the line angle formula for the tripeptide Ser-Ala-Asp as it would appear at pH 7?
  2. Chemistry

    1.Calculate the pH of a solution prepared by mixing 20mL of the .07222M aspirin solution with 10mL of .2M NaOH. 2. Calculate the pH for the titration of 40mL of .1M solution of C2H5NH2 with .1M HCl for a)0 mL added and b)20 mL added …
  3. chemistry

    Glutamate (Glu–) is the conjugate base form of glutamic acid (HGlu). The Ka of glutamic acid is 5.012 x 10^–5. You titrate 50 mL of 0.10 M sodium glutamate solution with 0.05 M HCl solution. Calculate the pH of the solution at …
  4. Chemistry

    I'm asked to tell which of the following peptide segments is most likely to be part of a stable Alpha Helix at physiological pH?
  5. Chemistry

    I'm given sequences and asked to classify them as "Most likely an amphipathic alpha helix", "Most likely an amphipathic beta sheet", "Most likely a turn/loop" or "Not Amphipathic" Sequence 1: Arg-Phe-Gln-Ile-His-Val-Gln-Phe-Glu Sequence …
  6. Analytical chemistry

    Below i wrote a few problems, i did them all but in the book i don't have results, so i don't know if i did them right or not. I would really appreciate if someone would go through the problems and tell me if i got the right results. …
  7. Biochemistry

    1. What is the overall charge of the tripeptide below pH 1.0?
  8. Biochemistry

    You are performing site-directed mutagenesis to test predictions about which residues are essential for a protein's function. Which of each pair of amino acid substitutions listed below would you expect to disrupt protein function …
  9. Analytical Chemistry

    What volume, in liters, of 1.09 M KOH solution should be added to a 0.102 L solution containing 9.35 g of glutamic acid hydrochloride (Glu, FW = 183.59 g/mol; pKa1 = 2.23, pKa2 = 4.42, pKa3 = 9.95) to get to pH 10.26?
  10. Biochm

    The deprotonated form of the R group of cysteine. The ratio of protonated to deprotonated form depends on the pKa of the R group and the pH of the solution. => Select all the pH values at which the protonated form of the R group …

More Similar Questions