Determining the Concentration of Acetic Acid in Vinegar by Direct Titration

Key Concepts

• Vinegar is produced by the oxidation of ethanol.
• Table vinegar typically contains between 4 and 8 % v/v acetic acid (ethanoic acid).
• Acetic acid is a weak, monoprotic, organic Brønsted-Lowry acid:

  acetic acid (ethanoic acid)				acetate ion (ethanoate ion)
  H |   O ||   H- C - C -O-H   | H		H+	+	H |   O ||   H- C - C -O-   | H
  acid				conjugate base

• Acetic acid (ethanoic acid) will react with a strong base in neutralisation reaction:

  acid	+	base	→	water	+	salt
  H |   O ||   H- C - C -O-H   | H	+	NaOH(aq)		H-OH(l)	+	H |   O ||   H- C - C -O-Na+   | H
  acetic acid (ethanoic acid)	+	sodium hydroxide	→	water	+	sodium acetate ( sodium ethanoate)

• It is therefore possible to determine the concentration of acetic acid in vinegar by titrating the vinegar with a strong base such as aqueous sodium hydroxide solution.

  moles of acetic acid in vinegar sample	=	moles of sodium hydroxide used in titration
  n(CH3COOH)	=	n(NaOH)
  concentration of acetic acid	=	moles of acetic acid volume of vinegar
  c(CH3COOH)	=	n(CH3COOH) V(vinegar)

  Using the known density of acetic acid it is possible to calculate the concentration of acetic acid in vinegar as a v/v%:

  v/v % =

  volume of acetic acid volume of vinegar

  × 100

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Experiment: Determining the Acetic Acid in Vinegar by Titration

When a weak acid such as acetic acid is titrated with a strong base such as aqueous sodium hydroxide solution, the pH at the equivalence point will be greater than 7.
Suitable indicators for this experiment are phenolphthalein or thymol blue.

Procedure to titrate the acetic acid in vinegar:

Rinse a clean 250 mL conical (erlenmeyer) flask with water. Rinse a clean 25.00 mL pipette (pipet) with vinegar. Pipette 25.00 mL of vinegar into the 250 mL conical (erlenmeyer) flask. Add 2 drops of phenolphthalein indicator to the vinegar. (The solution will remain colourless) Rinse a clean 50.00 mL burette (buret) with standardised 1.00 mol L-1 aqueous sodium hydroxide solution. Fill the burette (buret) with this standardised 1.00 mol L-1 NaOH(aq). Set up the equipment as in the diagram on the right. Run NaOH(aq) from the burette (buret) into the conical (erlenmeyer) flask until the solution changes colour from colourless to pink. Repeat the titration carefully several times until concordant titres are achieved.

Sample Results

Calculating the Concentration of Acetic Acid in Vingear in mol L^-1

1. Write the balanced chemical equation for the neutralisation reaction:

   word equation	acetic acid (ethanoic acid)	+	sodium hydroxide	→	sodium acetate (sodium ethanoate)	+	water
   balanced chemical equation	CH3COOH(aq)	+	NaOH(aq)	→	CH3COO-Na+(aq)	+	H2O

2. Extract all the relevant data from the experiment.

   balanced chemical equation	CH3COOH(aq)	+	NaOH(aq)	→	CH3COO-Na+(aq)	+	H2O
   volume /mL	25.00		21.81
   concentration /mol L-1	?		1.00

3. Check the data for consistency:
   Concentrations are usually given in M or mol L^-1 but volumes are often given in mL.
   You will need to convert the mL to L for consistency.
   The easiest way to do this is to multiply the volume in mL × 10^-3 (which is the same as dividing the volume in mL by 1000)

   relevant species	acid		base
   	CH3COOH(aq)		NaOH(aq)
   volume /mL	25.00		21.81
   volume /L	25.00 ÷ 1000 = 0.02500		21.81 ÷ 1000 = 0.02181
   concentration /mol L-1	?		1.00

4. Calculate the moles of NaOH(aq), n(NaOH)

   moles = concentration in mol L^-1 x volume in L = n = c x V

   volume of NaOH(aq) = v(NaOH) = 0.02181 L
   concentration of NaOH(aq) = c(NaOH) = 1.00 mol L-1

   moles NaOH(aq) = n(NaOH) = c(NaOH) × V(NaOH)
   n(NaOH) = 0.02181 × 1.00 = 0.02181 mol

5. Use the balanced chemical equation to determine the stoichiometric (mole) ratio of acid to base:

   n(acid):n(base)
   n(CH₃COOH):n(NaOH)
   1:1

6. Use the stoichiometric (mole) ratio to calculate the moles of acetic acid

   1 mole of NaOH neutralises 1 mole of CH₃COOH
   therefore 0.02181 moles of NaOH neutralises 0.02181 moles of CH₃COOH
   moles of acetic acid = n(CH₃COOH) = 0.02181 mol

7. From the volume of vinegar (acetic acid solution) and the moles of acetic acid, calculate its concentration (c) in mol L^-1 :

   concentration (mol L^-1) = moles ÷ volume (L)
   concentration of acetic acid = moles of acetic acid ÷ volume of acetic acid in L

   moles of acetic acid = n(CH₃COOH) = 0.02181 mol
   volume of acetic acid solution (vinegar) = v(CH₃COOH) = 0.02500 L
   concentration of acetic acid solution (vinegar) = n(CH₃COOH) ÷ v(CH₃COOH)
   c(CH₃COOH) = 0.02181 ÷ 0.02500 = 0.8724 mol L^-1

Concentration of acetic acid in vinegar in mol L^-1 (molarity) is 0.8724 mol L^-1

Calculating the Concentration of Acetic Acid as a Percentage By Volume (v/v %)

1. Write an expression for calculating v/v % concentration.

   v/v % = volume of solute ÷ volume of solution × 100
   acetic acid in vinegar v/v % = volume of acetic acid ÷ volume of vinegar × 100

2. Extract the relevant data from the experiment

   volume of acetic acid = ? (can be calculated using mass and density)
   volume of vinegar = 25.00 mL

3. Calculate mass of acetic acid

   mass = moles × molar mass
   moles acetic acid = n(CH₃COOH) = 0.02181 mol (see calculations in previous section)
   molar mass acetic acid = 2 × 12.01 + 4 × 1.008 + 2 × 16.00 = 60.052 g mol^-1
   mass acetic acid = moles acetic acid x molar mass acetic acid = 0.02181 × 60.052 = 1.3097 g

4. Calculate volume of acetic acid using its mass and known density

   density (g/mL) = mass (g) ÷ volume (mL)
   volume (mL) = mass (g) ÷ density (g/mL)

   density of acetic acid = 1.049 g mL^-1 (at 25°C)
   mass acetic acid in vinegar = 1.3097 g
   volume of acetic acid = mass acetic acid ÷ density acetic acid
   volume acetic acid = 1.3097 ÷ 1.049 = 1.249 mL

5. Calculate concentration of acetic acid in vinegar as v/v %

   acetic acid in vinegar v/v % = volume of acetic acid ÷ volume of vinegar × 100
   volume of acetic acid = 1.249 mL
   volume of vinegar = 25.00 mL
   acetic acid in vinegar v/v % = 1.249 ÷ 25.00 × 100 = 4.996 %

v/v % acetic acid in vinegar is 4.996%