Question

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

Answer 1

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.

Answer 2

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

Answer 3

There should be a table in your text.

Answer 4

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