Determining the Molar Concentration of Vinegar by Titration

Objective: Determine the concentration of acetic acid in a vinegar sample

Background

Expressing Solution Concentration

mols (solute)
molarity = ------------------ = M (mol/L)
Liter (solution)
0.493 M NaOH means 0.493 mol NaOH/L
mols
mols = ------ x L
L
mols = M x V
In a titration procedure, 40.57 mL of 0.493 M NaOH solution was used. How many mols NaOH did this volume of NaOH solution contain?
mols = M x V
0.493 mols NaOH
mols = ----------------------- x 0.04057 L
L
mols = 0.0200 mols NaOH

Volumetric glassware: buret and pipet

Reading the buretUsing the pipet
Buret reading = 0.76 mL
Determining the Volume of Titrant Delivered in a Titration
Final buret reading: 49.37 mL
Initial buret reading: 0.74 mL
volume delivered: 48.63 mL

Procedure

You may work in groups:

Sample Calculation

Calculating the concentration (M) of CH3COOH in commercial vinegar.
  1. From the balanced chemical equation:
    mols CH3COOH(vinegar) = mols NaOH(titrant)

  2. mols NaOH = MNaOH x VNaOH,L (from titration)

    0.493 mols NaOH
    mols Na OH = ------------------------ x 0.04863 L
    L

    mols NaOH = 0.240 = mols CH3COOH(vinegar)

  3. mols CH3COOH(vinegar)
    M CH3COOH(vinegar) = -----------------------------
    volume(vinegar)
    0.0240 mols
    M CH3COOH(vinegar) = ---------------- = 0.96 M
    0.0250 L

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