If you have a graphed titration curve and have points near the equivalence point like

6.85 pH at 54.60mL titrant added and 8.01 pH at 54.73mL titrant added

is there any good way to find a more exact volume at which the equivalence point was reached other then taking a guess between those two volumes?

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    $\begingroup$ You can take the derivative of the graph to find exactly where the slope is the steepest. $\endgroup$ – Jess L Mar 17 '16 at 0:22
  • $\begingroup$ how do you do that? I know what a derivative is, but derivative of the graph? $\endgroup$ – ThIrde110 Mar 17 '16 at 0:28
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    $\begingroup$ So essentially the slope of your slope. I've only ever done it using some wacky program in my school. But you can use excel to solve for the slope at each point, then graph that vs volume. This graph will tell you which of your volume points is closer to the actual eq. point, but unfortunately won't give you any new volume points. $\endgroup$ – Jess L Mar 17 '16 at 0:42
  • $\begingroup$ Hmmm. Alright then. $\endgroup$ – ThIrde110 Mar 17 '16 at 1:28
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    $\begingroup$ With the limited information given assume the equivalence point is pH 7.00, and do a linear interpolation between the two known points. $\endgroup$ – MaxW Mar 17 '16 at 2:58

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