[joybaker]

Now I want to measure the pKa of a weak acid (oleic acid). I understand the method of titration, using NaOH and find the point of complete reaction and its midpoint as the pKa value. The problem is now I only have its conjugated base (sodium oleate).

Previously I thought the titration curve is symmetrical, as whether a strong base titrates the weak acid or a stong acid titrates the conjugated base of the weak acid, the result is the same pKa (as calculated from the Henderson-Hasselbach equation), but I tried to do the experiment, the results were different. I think it is not simply one result of pKa and one result of pKb (as I can't get this from the equation).

So I want to know what is wrong in this process? If someone can explain the derivation process.   
What I really got from the conjugated base titration?
If I want to know the pKa of oleic acid and only have sodium oleate, what should I do?

thank you very much

[Borek]

Can you elaborate on the results that you got?

These titration curves are not symmetrical, but you should get the same pH at midpoint (so the same pKa value).

[joybaker]

Can you elaborate on the results that you got?

These titration curves are not symmetrical, but you should get the same pH at midpoint (so the same pKa value).

So what I did is titrate 10 mM sodium oleate with 0.1 M HCl and when it pass the completion point, I titrated it back with 0.1 M NaOH. I know the excessive volume of HCl should be substrated from the total volume to find the midpoint, but even we consider this, the pKa of the two graphs are different because the graphs are just chiral, which is not what I expected

[Babcock_Hall]

Borek is the expert, but I would just like to point out one thing.  Oleic acid is not very soluble in water, and I don't believe that sodium oleate is much better.  My recollection is that it can form a monolayer on top of water and can form micelles, but I don't have a citation handy.  I am not certain how this would affect a titration curve.

[joybaker]

Borek is the expert, but I would just like to point out one thing.  Oleic acid is not very soluble in water, and I don't believe that sodium oleate is much better.  My recollection is that it can form a monolayer on top of water and can form micelles, but I don't have a citation handy.  I am not certain how this would affect a titration curve.

it is not soluble at room temperature, but soluble at higher temperature, not hot but warm. So I am sure at this concentration and condition it is homogenious and dissolved

Wait for an expert:)

[Babcock_Hall]

What is the temperature?

[joybaker]

The temperature is not measured, but the beaker was placed on a magnetic stirer with heat on. But the heat is not high as I don't want temperature to be an issue of pH.

[Borek]

OK, they way you did the experiment you should get a symmetrical curve - note that while the whole curves are not symmetrical, they do have a symmetrical part visible on the combined curve attached to this post (not exactly symmetrical, as there are some dilution effects). Additional part - titration of the excess strong acid/base - should yield a symmetrical response as well.

And as far as I can tell your curve is symmetrical as expected. There are several factors that can make them not exactly symmetrical - dilution, changes in the ionic strength, precipitation of the neutralized acid out of the solution (or adsorption on the glass, or hiding in micelles) effecting in changes of the concentration and so on.

What values have you got for pKa on both titration legs?

[joybaker]

OK, so if a symmetrical graph is expected (as my graph shows), my question is how I can get the same pKa value from both method? The completion points of both graphs are the sample, corresponding to pH~5, and if I find the half volume of the first graph, a pKa around 7. And if I find the half volume of the second graph, a pKa around 3-4 is expected. So the pKa is different from the 2 methods.

So I am confused what these 2 pKa refer to. if I just want to know the pKa of oleic acid (weak acid), and only have sodium oleate (its conjugated base), what should I do?

[Borek]

You are measuring your midpoints incorrectly. Mid point is when the concentrations of HA and A^- are equal, that means slightly above 2000 μL and somewhere around 9000 μL - problem is, you don't know what was the excess of HCl used initially, so it is difficult to follow exact stoichiometry.

Look at it this way: your first endpoint is around 4500 μL of HCl added, so the midpoint is half of that - say 2250 μL. Then you add some excess of HCl. During titration with NaOH you have to neutralize this excess HCl first (unknown volume) and the endpoint you observe (around 7000 μL) is actually where the HCl ends. To get to the midpoint you need to add another 2250 μL, so the second midpoint is at about 7000+2250 = 9250 μL. And pH at both 2250 μL and 9250 μL are around 7 - which means pKa is around 7 as well.

[Babcock_Hall]

Why would the pK_a of a carboxylic acid be near 7?  There is nothing in R group that should perturb it away from normal values. 

[Borek]

No idea - I am reading the result from the titration curve as shown. I can only guess it is shifted up to high pH values (after titration without 0.1M HCl pH stays somewhere between 3-4, I would expect it to go down to 1-2). It can be badly calibrated pH electrode, or something else.

[Babcock_Hall]

@OP What is the concentration of oleic acid in the titration?

[Borek]

He already wrote it - 10 mM.

[joybaker]

You are measuring your midpoints incorrectly. Mid point is when the concentrations of HA and A^- are equal, that means slightly above 2000 μL and somewhere around 9000 μL - problem is, you don't know what was the excess of HCl used initially, so it is difficult to follow exact stoichiometry.

Look at it this way: your first endpoint is around 4500 μL of HCl added, so the midpoint is half of that - say 2250 μL. Then you add some excess of HCl. During titration with NaOH you have to neutralize this excess HCl first (unknown volume) and the endpoint you observe (around 7000 μL) is actually where the HCl ends. To get to the midpoint you need to add another 2250 μL, so the second midpoint is at about 7000+2250 = 9250 μL. And pH at both 2250 μL and 9250 μL are around 7 - which means pKa is around 7 as well.

Thanks Borek, this makes sense about the inflection point of the second graph. The first mid point (2150 μL) refers to pKa~8.3. And my total HCl is 5600 μL. So I can calculate my second mid point at 5600+(5600-2150*2)+2150=9050 μL, refers to pKa~7.7 (which is similar result). And end point at 11200 μL(red point). So I am wonder why this time the end point is not distinguishable as a inflection point like the first one? and what may cause the pKa slightly different (volume change?)