A 25.00-mL sample of propionic acid,HC₃H₅O₂,of unknown concentration was titrated with 0.103 M KOH.The equivalence point was reached when 43.88 mL of base had been added.What was the original concentration of the propionic acid? A)0.295 M B)0.090 M C)0.059 M D)0.103 M E)0.181 M

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Quizplus · Accepted Answer

The moles of KOH used at equivalence point = 0.103 M * 0.04388 L = 0.00452 mol. Since the reaction is 1:1, moles of propionic acid = moles of KOH = 0.00452 mol. The concentration of propionic acid = moles/volume = 0.00452 mol / 0.025 L = 0.181 M.

A 2500-ML Sample of Propionic Acid,HC3H5O2,of Unknown Concentration Was Titrated with with 0.103

A 2500-ML Sample of Propionic Acid,HC₃H₅O₂,of Unknown Concentration Was Titrated with with 0.103