A 25.00-mL sample of propionic acid,HC₃H₅O₂,of unknown concentration was titrated with 0.141 M KOH.The equivalence point was reached when 37.48 mL of base had been added.What is the concentration of the propionate ion at the equivalence point? A)0.141 M B)0.147 M C)0.0846 M D)0.211 M E)0.128 M

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Quizplus · Accepted Answer

The balanced equation for the reaction is: HC₃H₅O₂ + KOH → KC3H5O2 + H2O From the equation, we can see that 1 mole of propionic acid reacts with 1 mole of KOH to produce 1 mole of propionate ion. Therefore, the number of moles of propionate ion at the equivalence point is equal to the number of moles of KOH added. Number of moles of KOH added = 0.141 mol/L x 0.03748 L = 0.00529 mol Volume of propionic acid = 25.00 mL = 0.02500 L Concentration of propionic acid = Number of moles/volume = 0.00529 mol/0.02500 L = 0.2116 M At the equivalence point, all of the propionic acid has reacted, so the concentration of the propionate ion is also 0.2116 M. However, we need to calculate the concentration of the propionate ion in terms of Molarity. Since one mole of propionate ion is produced from one mole of KOH, the concentration of the propionate ion is also 0.141 M (0.141 moles of KC3H5O2 per litre of solution). So, the concentration of the propionate ion at the equivalence point is 0.141 M (Option A).

A 2500-ML Sample of Propionic Acid,HC3H5O2,of Unknown Concentration Was Titrated with with 0.141

A 2500-ML Sample of Propionic Acid,HC₃H₅O₂,of Unknown Concentration Was Titrated with with 0.141