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Titrating weak acid buffer with strong acid after the base is used up

  • Thread starter roq2
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  • #1
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Homework Statement


I'm making a table on excel with the calculated pH of the solution vs the actual measured pH from the lab. My problem is finding the pH of a acetic acid-acetate buffer that has been titrated with HCl, once the moles of HCl are greater than the initial quantity of mols of acetate.

Homework Equations


pH = -log(H3O+) = 14-pOH = pKa + log([acetate]/[acetic acid])
Ka = [H3O][acetate]/[acetic acid] = 1.7 x 10^-5
HCl + H20 --> H3O
H3O + CH2COO- = CH2COOH
[acetate] = (initial moles acetate + moles HCl added)/total volume

The Attempt at a Solution



Up to a point I just used the Henderson-Hasselbalch equation. This worked until the moles of HCl were greater than the moles of initial acetate, since this gives a negative number for the concentration of base, which is impossible in reality and mathematically, assuming I was using it correctly. I was getting the concentration of the base by subtracting the initial moles of the base by the moles of HCl and dividing by the total volume.

At that point I turned to the Ka equation. All the HCl goes to H3O, and additional H3O reacts with acetate to form acetic acid as long as there is still acid. So H3O would = [HCl] - [initial acetate]. If that were right, I could take the -log of that for pH. But it doesn't match measured data, or the trend of calculated data up to that point. It gives a more basic result.

I guess the acetic acid is still reacting and adding more H3O, in order to get the Ka. So next I tried:

Ka = ([H3O]+x)(x)/([AA initial]+[initial acetate]-x)

I went through the quadratic with the same sample data I used last time and the x was so small the answer was nearly the same, which is what I would expect, since in this hypothetical the ionization of AA would be supressed by the H3O.

I don't know what I did wrong and am stuck here. I'd really appreciate some help.
 

Answers and Replies

  • #2
133
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first off, you cant make a buffer solution with two acids, you have acetic acid and hydrochloric acid, second, you should have been using an ICE table from the start, third, this is a titration problem, a point will come when you transition from a pH value to a pOH value; once you pass the point where the base is in excess and move to the point where the acid is in excess, you have to find the Ka value, you can do that using the Kb and Kw value, then you should use an ICE equation to determine the unknown values.
 
Last edited:
  • #3
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I already know the Ka for acetic acid, it's .000017. I solved the equation for x, which was almost nothing, and wound up with about the same pH as if I didn't add x to the H3O.

I've never heard of a ICE equation before, could you explain what that is please?
 
  • #4
133
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go here: http://www.brynmawr.edu/Acads/Chem/Chem104lc/study/buffer6.html [Broken]

its an example similar, they are titrating with HCl also.
 
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  • #5
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Thanks. I actually tried to find an example myself but only found stuff with titrations involving weak acid and bases or weak bases and acid.
 
  • #6
133
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btw, compare the values you get for starting, equivalence, and ending pH/pOH with the general case found here: http://www.ausetute.com.au/titrcurv.html that way you can tell if you're on the right track.
 
  • #7
Borek
Mentor
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titration curve calculation

Note: when you have added more HCl than initial amount of acetate, you may calculate pH with just excess HCl, as if the acetic acid was not present. Below pH 3.0 acetic acid is dissociated in less then 2% so its contribution to the final pH can be most likely neglected.

You may try BATE (pH calculator) to compare results of exact calculations with results of your calculation.

Borek
 
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