Having obtained the titration curve (Potential vs Volume added), one can attempt to identify the inflection point, which corresponds to the equivalence point. This might prove to be difficult in some cases, hence, another method is to caluclate the first derivative of the curve. In this case the volume at which the extremum of the derivative occurs denotes the equivalence point.
The matter is laid out quite well in the following book "Potentiometry and Potentiometric Titrations (Chemical Analysis: A Series of Monographs on Analytical Chemistry and Its Applications)", E.P. Serjeant, Wiley-Interscience, 1984, ISBN: 978-0471077459. For a "quick guide", please find enclosed a web-available laboratory manual on the subject.
Dear Mr Jarosz, there is many possibilities. One is 1st derivative, but is possible to use second derivative. Potentiometric curves with short inflection are caused by low concentration of titrant and titrand between strong base or acid, but it also depends of the pKa, in case of titration acid-base (weak vs strong). Another manner to find the equivalence point in potentiometric titration is using Gran functions. The Gran functions depend of the kind of titration ( acid-base, redox, precipitation, complexometric, etc). The Gran function converts from logarithm function to exponential function. Plot the Gran function versus titrant volume is linear. There are two function before and after equivalence point. Where the functions find in the X axis is the equivalence point. See the classic article: http://www.wfu.edu/chemistry/courses/jonesbt/280L/Experiment%203/Gran1952.pdf or http://en.wikipedia.org/wiki/Gran_plot
There is nothing else to add to answers given before. The common guide is: Derivate your "pH (or E) vs titration Vol " curves. See Harrys and others authors.
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