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 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? 

Feb 19th, 2015

We apply the formular ;

C1 V1 = C2 V

then we substute  is given,

 0.0800 x 25  =  0.0700 x V2

V2 =   0.0800 x 25/  0.0700

  =28.571  

Vol of KOH=28.571ml


Feb 19th, 2015

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Feb 19th, 2015
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