titration of 25.00 mL of a 0.0800 M acetic acid solution with 0.0700 M KOH. assume temperature is 25 oC.
What is the analytical concentration of acetate ions at the equivalence point?
What volume of KOH has been added at the half-equivalence point?
CH3COOH + KOH ==== CH3COOK + H2O
25.00 mL of a 0.0800 M CH3COOH = 25.00x10^(-3)x0.0800=0.0020 moles of CH3COOH
So at equivalence point 0.0020 moles of KOH will be added.
Then the volume of KOH added = 0.0020/0.0700=0.02857 L=28.57 ml
So the analytical concentration of CH3COOH = 0.0020 / (25.00+28.57)x10^(-3)=0.0373 M
At half-equivalence point, 0.0010 moles of CH3COOH is neutralized.
So 0.0010 moles of KOH is used at the point.
The volume of KOH = 0.0010/0.0700=0.01429 L=14.29 ml
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