State the half-equivalence point in a titration curve

Reference no: EM13182844

Really confused on calculating/finding the half-equivalence point in a titration curve. This was a lab experiment done in my chem lab. We added 0.0750 M NaOH to acetic acid (pipet 10.00mL of ~0.1M HCH3CO2 into the beaker and add 40mL of DI water).

From my data, the points that jumped were inbetween 13mL and 13.5 mL.

13 mL pH-6.58

13.5 mL pH-10.28

From this jump I calculated my equivalence point at 13.25.

Now what is the half-equivalence point in mL AND pH ?

Really confused on calculating it. I don't think it's just exactly 1/2 (13.25). Because I've seen another student's work who had the same data as me and they calculated their half quiv point in (mL) to be 7.4 and their ph 4.7

Reference no: EM13182844

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