Question

2A student, performing the same experiment as you, recorded the following informat orded the following information. Complete the data table below. The original solution was diluted by a factor of 5. Show all enlculations needed TRIAL 2 25.00 0.0900 14.50 38.00 Volume of diluted vinegar sample (mL) Molarity of NaOH Initial buret reading (mL) Final buret reading (mL) Volume of NaOH Moles of acetic acid (diluted) Molarity of acetic acid (diluted) Molarity of acetic acid (original)

Answer 1

The question is based on acid-base titration Volume of diluted vinegar 25 mL Molarity of Naok = 0.09M Initial Buret = 14.50 mL final " 1= 38.0mL Volume NaOH = M154438 14.5 = 23.5mU Molanty Moto of Acetic and = 0.0846M deluted Moles of Moda Acetic and = 2,115x03 I diluted) " " " (original) = 0.423M Calculation Volume NaoH= final Burette -Initial burette hlading = 38 -14-5 mL = 23.5mL CH₃COOH + Naol CH₃COo Na + H₂O MiV2 = M₂ V2 Aceticaçid Naoh (diluted) MX 25 = 0.098 23.5 mL M | Mail = 0.0846M ! Moles = molarity x volume = 0.0846 mol x 25 ml Moles = 2.115 mmol :looo mmol=4 mol moles of Acid = 2.115xlo 3

Dilution factor of 5 Vi final = 5 Vinitial or (Acid) undiluted = (5) Acid dilated = 5x (0.0846M) (Acid) undiluted = 0.423M, Thus, the dilution factor was Calculated