A 25.00 ml of vinegar sample was diluted to 250 ml using a volumetric flask. A 5

Question

A 25.00 ml of vinegar sample was diluted to 250 ml using a volumetric flask. A 50.0 ml aliquot portion of the dilute vinegar solution needed 35.45 ml of 0.1200 M NaOH to reach the phenolpthalein end point. (a) Calculate the no. of millimoles of NaOH required to reach the endpoint (b) Calculate the no. of millimoles of acetic acid present in the 50 ml aliquot portion that was titrated (c) Concentration of acetic acid in the titrated aliquot portion (d) The concentration of acetic acid in the volumetric flask (e) Calculate the total mass of acetic acid dissolved in the solution contained in the volumetric flask (f) The mass(g) of acetic acid dissolved in the solution contained in the volumetric flask came drom the 25.00 ml vinegar sample. Calculate the mass of acetic acid present in the 25 ml vinegar sample. (g) Calculate the % CH3COOH (w/v) present in the 25.00ml vinegar sample

Explanation / Answer

Sol: CH3COOH + NaOH NaCH3COO + H20 Convert: 18.62 mL = .01862 L moles acetic acid: .095M NaOH * .01862L = .00177 moles because it is 1:1 ration. .00177 moles * (60.05 g acetic acid/1 mol acetic aci) = 0.11 g acetic acid present. The mass percent of acetic acid present, assuming that no other acids are present is 0.11 g. Is this correct? I have doubts because I imagine I must put grams in percent by multiplying by 100; also, I did nothing whatsoever wiht the given information of 2.578 grams required. PLEASE RATE ME AND AWARD ME KARMA POINTS IF IT IS HELPFUL FOR YOU During the spoiling process, fruit juices often oxidize to create vinegar. A sample of a beverage with a mass of 2.578 grams reguires 18.62 ml of .095M NaOH for neutralization. What is the mass percent of acetic acid present, assuming that no other acids are present? Show the balanced equation in your calculations. Sol: CH3COOH + NaOH NaCH3COO + H20

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