A 15.00 mL sample of vinegar is diluted with water to a final volume of 50.00 mL

Question

A 15.00 mL sample of vinegar is diluted with water to a final volume of 50.00 mL. A 10.00 mL portion of the diluted vinegar is then titrated with 1.000 x 10-2 M NaOH. If 12.97 mL of NaOH is required to reach the equivalence point, what is the original concentration of acetic acid in the vinegar? Molecular chemical equation: CH3CO2H(aq) + NaOH(aq) -> CH3CO2Na(aq) + H2O(l)

The concentration of the initial vinegar is______x 10^_____M.  Please Explain

A 15.00 mL sample of vinegar is diluted with water to a final volume of 50.00 mL. A 10.00 mL portion of the diluted vinegar is then titrated with 1.000 x 10-2 M NaOH. If 12.97 mL of NaOH is required to reach the equivalence point, what is the original concentration of acetic acid in the vinegar? Molecular chemical equation: CH3CO2H(aq) + NaOH(aq) -> CH3CO2Na(aq) + H2O(l) The concentration of the initial vinegar is x 10^ M. Please Explain

Explanation / Answer

CH3CO2H(aq) + NaOH(aq) -> CH3CO2Na(aq) + H2O(l)

At equivalnce point, for the diluted solution
Moles of NaOH= Moles of CH3CO2H

mmoles of NaOH= 0.01*12.97=0.1297
mmoles of CH3CO2H = concentration* volume (in ml)= conc.*10
So, concentration of the diluted vinegar solution = 0.1297/10 =0.01297 M
Hence ,
the concentration of the original vinegar solution = 0.01297*50/15 =0.04323 M

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