Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: rob on April 02, 2006, 09:59:33 AM

I have done a potentiometric titration in the lab for determining the dissociation constant of a weak acid which is benzoic acid, using NaOH as a titrate. I took several measurements of the pH of the solution after adding certain quantitiies of NaOH. I used the results obtained for calculating pKa by two methods: 1 plotting pH against V and determining pH midpoint which is equal to pKa and 2 ploting pH against log V/ V0V (V0 = volume of NaOH needed to reach the endpoint) and find pKa from the intercept on the pH axis. [ I'm not sure though if pH axis should be the horizontal or the vertical one??] From both these methods I got a value of approx 4.1 while the literature value is 4.21.
Now I have to outline possible sources of error that lead to a different value from the literature one and also to comment on the realtive accurency of the two methods for determining the pKa.
I would like some help with these as I don't know how to comment on the sources of error as I have to mention factors that have a major contribution on the error and haven't got a clear idea of them. Also don't know how to comment the relative accurancy of the two methods because from both of them I got a similar value about 4.1 while the correct one should be 4.21. Please I would greately appreciate if you could help me with these. Thanks in advance.

Both methods are equivalent, nothing strange that they gave identical result.
pHpKa=log(V/(V_{0}V))
is just another version of HH equation  note that right side equals zero for half titration. Try to understand how to rearrange HH to obtain this version.
It will not work in case of a very weak acid, but for benzoic is OK.
So, your error in both cases has the same theoretical source (neglectable) and identical experimental problems. Plus, you probably ignored ionic strength (http://www.chembuddy.com/?left=pHcalculation&right=ionicstrengthactivitycoefficients) of the solution.