??? A vinegar titration was completed using a pH meter as an indicator of the ch

Question

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A vinegar titration was completed using a pH meter as an indicator of the changing pH. A 10.00 mL sample of concentrated vinegar was pipetted into a 100 mL volumetric flask and diluted to volume. A 25.00 sample of diluted vinegar was placed in a beaker and subsequently titrated with 0.1050 M KOH. The initial volume reading from the buret was 2.32 mL. The final volume reading from the buret was 45.72 mL. The total volume delivered was 43.40 mL. curve of the titration suggested that the equivalence point occurred at 18.29 mL. the mass percent o

Explanation / Answer

Titration of vinegar solution with KOH

a. moles of KOH = molarity x volume

                            = 0.1050 M x 18.29 ml

                            = 1.92045 mmol

b. moles of acetic acid at equivalence point = moles of KOH at equivalence point

                                                                      = 1.92045 mmol

c. mass of acetic acid = moles x molar mass

                                   = 1.92045 mmol x 60.05 g/mol/1000

                                   = 0.115 g

mass of acetic acid in original 10 ml vinegar solution = 0.115 g x 4

                                                                                    = 0.460 g

d. Volume of concentrated vinegar = 10 ml

e. mass of vinegar solution = volume x density

                                            = 10 ml x 1.008 g/ml

                                            = 10.08 g

f. Percent of acetic acid by mass = 0.460 g x 100/10.08 g

                                                     = 4.563%

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