Titrations experiment and derivatives

Question

I just got done with an experiment on titrations in which I graphed the volume of NaOH added against 3 things (pH, first deriv: ΔpH/ΔVolume, and second deriv: Δ[ΔpH/ΔVolume]/ΔVolume) I somewhat understand the first derivative, since it gives a defining point (a peak) to select when considering the equivalence point.

So, my question is, why get into the mess of a second derivative? What benefits does it offer that the first derivative does not?

Also, it seems that Yahoo! Answers does not recognize titrations to be a word (it continuously offers 't*t rations,' a comically inappropriate phrase). Is there someone I should inform to have that fixed?

Answer 1

The second derivative does not offer a lot of benefit.  The second derivative will go from near zero to a maximum just before equivalence, then swing through zero to a minimum just after equivalence.  The point at which the second derivative curve crosses 0 in between the min and max will be the equivalence point.  That may be helpful if you don't have a lot of points in the vicinity of equivalence.  In that case, the first derivative peak may be a little flat and hard to pinpoint.

Can't help with the spell checker.  Just consider it a bit of comic relief.