[Impulse29]

From a titration curve, how is it possible to determine the pKa? I titrated 1-molar hydrochloric acid with different concentrations of base (sodium carbonate). I think it'd be best if someone explained what the pKa is used for, and what I should be looking for. I know that there are supposed to be two values per curve, but I don't know whether they are the plateau, the slope part, or what?

Any help is appreciated.

[Borek]

You have to know how to calculate titration curve, then pKas will be obvious. To calculate titration curve you have to know how to calculate pH of weak acid/base.

Try these pH calculation lectures.

[mike]

Ok, so on your graph you should be able to label the half equivalence points and the full equivalence points.

Half equivalence is where only half the acid/base has been neutralised, and full equivalence points are for 1:1 or full neutralisation.

At half equivalence the change in pH is small on addition of acid/base so the graph is quite flat (ie the pH doesn't change much).

At full equivalence there is a sudden change in pH (signified by a very steep plot).

If you look at the equilibtium expression:

K = [A-][H3O+]/[HA]

at equilibrium (ie flat part of the graph or half equivalence) [A-] = [HA] so therefore:

K = [H3O+]

therefore:

pKa = pH

basically this means that the pKa is equal to the pH at half equivalence, and half equivalence occurs at the shallowest part of the graph.

[Impulse29]

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THAT IS WHAT I NEEDED! THANK YOU!
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I did the titration myself, so the curve is due to a ton (60+) plots on a graph. However, I didn't know what the pKa was, as the only thing that was given in the lab was the molarity of the acid, and the bases. It is weird, but my biology textbook covers acids and bases, but there is no mention of pKa anywhere in the book. I checked my chemistry book, but it wasn't much help either.

[Borek]

at equilibrium (ie flat part of the graph or half equivalence) [A-] = [HA] so therefore:

K = [H3O+]

therefore:

pKa = pH

basically this means that the pKa is equal to the pH at half equivalence, and half equivalence occurs at the shallowest part of the graph.

That's not whole truth. If the acid is weaker then about pKa = 9 hydrolysis starts to play important role and half-equivalence pH is no longer the same as pKa. For carbonic acid the difference is 0.07 - not much, but can't be neglected.

[mike]

That's not whole truth. If the acid is weaker then about pKa = 9 hydrolysis starts to play important role and half-equivalence pH is no longer the same as pKa. For carbonic acid the difference is 0.07 - not much, but can't be neglected.

OK, so how would you answer the original question then, with this new information? I think for the question at hand the method I suggested is quite valid. What are your thoughts?

[BSc Kin]

Hey Impulse 29

Do you by chance have the Biology textbook by Campbell and Reece.  I just had a lab in with nearly the exact problems you came against and my textbook was no help at all.  These postings helped me out a bunch. Thanks a lot guys.

[Borek]

OK, so how would you answer the original question then, with this new information? I think for the question at hand the method I suggested is quite valid. What are your thoughts?

Method is OK - should be just accompanied by a comment "doesn't work for relatively strong and very weak acids".

For example it can't be used for determination of pKa for dichloroacetic acid (1.48) and pKa₃ for H₃PO₄ (12.35). But it will work good enough for every acid with pKa between - say - 3.7 and 10.3 (error below 0.05 unit).

You know it is the same situation as with volume/volume concetrations and every other approximate approach to many tasks. Approximate approach is very good and very usefull, you just have to know where its suitability range ends (no idea if it is in English    ).