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
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.
Thank you, I understand all of it but I'm not sure how I'm supposed to know what the pKa is
There should be a table in your text.
If not a pKa value, then a Ka or Kb value from which pKa can be calculated.
To solve this problem, you are on the right track by setting up an ICE table. Let's continue from where you left off:
Initial concentrations:
Deprotonated: 0.065 M x 0.012 L = 0.00078 mol
HCl: 1.50 M x 0.00040 L = 0.0006 mol
Protonated: 0 mol (since initially, no protonated form is present)
Cl-: 0 mol
After reaction:
Deprotonated: 0.00078 mol - 0 mol = 0.00078 mol
HCl: 0.0006 mol - 0.0004 mol (assuming all the HCl reacts) = 0.0002 mol
Protonated: 0 mol + 0.0004 mol = 0.0004 mol
Cl-: 0 mol + 0.0004 mol = 0.0004 mol
Now, we can use the formula pH = pKa + log (deprotonated/protonated) to calculate the pH. However, to apply this formula, we need to know the pKa value for the tripeptide glu-asn-leu. This information is crucial for determining the pH. Please provide the pKa value for glu-asn-leu tripeptide, and we can proceed with the calculation.