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25.0 mL of vinegar(acetic acid) was titrated with 1.5 M standardized NaOH.What is the concentration of the acetic acid in the vinegar if the equivalence point of titration required 8.2 ml of the 1.5 M NaOH titrant?
PLEASE HELP ASAP!

Respuesta :

mickymike92

Based on the volume of NaOH required to reach the equivalence point, the concentration of the acetic acid in the vinegar at is 0.492 M.

What is the equivalence point of a titration?

The equivalence point in a titration experiment is the point at which equal amounts of reactants have reacted.

At equivalence point, moles of vinegar = moles of NaOH

Moles of NaOH = concentration × volume

Moles of NaOH = 1.5 × 8.2 = 12.3 mmoles

Moles of vinegar = 12.3 mmoles

Concentration of vinegar = moles/volume

Concentration of vinegar = 12.3/25.0 = 0.492 M

Therefore, the concentration of the acetic acid in the vinegar is 0.492 M.

Learn more about equivalence point at: https://brainly.com/question/27136841

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