vinegar titration was completed using a pH meter as an indicator of the changing

Question

vinegar titration was completed using a pH meter as an indicator of the changing pH. A 25.00 mL sample of diluted vinegar (diluted by a factor of 5) was placed in a beaker and subsequently titrated with 0.1098 M NaOH. A derivative curve of the titration suggested that the equivalence point occurred at 40.90 mL. Calculate the mass percent of CH3COOH. (The density of vinegar is 1.008 g/mL.) Moles of base at the equivalence point Moles of acid = Mass of acid = Volume of concentrated vinegar- Mass of vinegar solution = Mass percent of acetic acid = 141

Explanation / Answer

No of mole of NaOH = 40.9/1000*0.1098 = 0.00449 mol

No of mole of CH3COOH present in vinegar sample = 0.00449 mol

mass of ch3cooh present in vinegar sample = 0.00449*60 = 0.2694 grams

volume of vinegar solution = 25/5 = 5 ml

mass of vinegar solution = 5*1.008 = 5.04 grams

mass percentage of vinegar = 0.2694/5.04*100 = 5.345%

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