A 25.00-mL sample of propanoic acid,CH₃CH₂COOH,of unknown concentration was titrated with 0.111 M KOH.The equivalence point was reached when 44.54 mL of base had been added.What is the concentration of the propanoate ion at the equivalence point?

Question

Quizplus · Accepted Answer

At the equivalence point, the amount of propanoate ion equals the amount of KOH added since they react in a 1:1 molar ratio. Using the formula $C_1V_1 = C_2V_2$, where $C_1$ and $V_1$ are the concentration and volume of KOH, and $C_2$ and $V_2$ are the concentration and volume of propanoic acid, we find $C_2 = \frac{C_1V_1}{V_2} = \frac{0.111 \, \text{M} \times 44.54 \, \text{mL}}{25.00 \, \text{mL}} = 0.199 \, \text{M}$. However, this is the concentration of propanoic acid before dilution. After dilution, the total volume is $25.00 \, \text{mL} + 44.54 \, \text{mL} = 69.54 \, \text{mL}$, and the concentration of propanoate ion is $0.199 \, \text{M} \times \frac{25.00 \, \text{mL}}{69.54 \, \text{mL}} = 0.0711 \, \text{M}$.

A 2500-ML Sample of Propanoic Acid,CH3CH2COOH,of Unknown Concentration Was Titrated with with 0.111

A 2500-ML Sample of Propanoic Acid,CH₃CH₂COOH,of Unknown Concentration Was Titrated with with 0.111