The next 11 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?
Please note that if I call with v the volume measured in mL of the KOH added, I can write:
K = [(200-7v)(25+v)] * 10^(-2)
where K is the Equilibrium constant of the acetic acid, and its value is:
K = 1.76*10^(-5)
so substituting the value of K into the formula above, we get:
v = 28.565 mL
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