questions are related to the titration of 25.00 mL of a 0.0800 M acetic acid solution with 0.0700 M KOH. Assume that the temperature is 25 oC.
What volume of KOH has been added at the half-equivalence point?
We apply the formular ;
C1 V1 = C2 V2
then we substute is given,
0.0800 x 25 = 0.0700 x V2
V2 = 0.0800 x 25/ 0.0700
Vol of KOH=28.571ml
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