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How to find the acetic acid content of vinegar

Equipment

50 cm3 burette

10 cm3 pipette

25 cm3 pipette

100 cm3 volumetric flask

2 x 250 cm3 conical flasks

Distilled water

White tile

Clamp and stand

Chemicals

Vinegar, Standard sodium hydroxide solution (0.1 M) and phenolphthalein indicator

What to do

1. Using the pipette add 10 cm3 of vinegar to the 100 cm3 volumetric flask. Add the Distilled water to the vinegar until the mixture reaches the base of the stem of the flask. Mix well then add more distilled to the 100 cm3 mark and mix again. Fill the burette with this solution.

2. Using the 25 cm3 pipette, add 25 cm3 of the 0.1 M sodium hydroxide solution to the 250 cm3 conical flask. Add a few drops of phenolphthalein indicator.

3. Place the white tile under the conical flask and add the vinegar solution slowly from the burette into the sodium hydroxide solution. When the mixture changes from pink to colourless, record the volume of vinegar solution added.

4. Repeat steps 2 and 3. Calculate the average volume of vinegar solution.

Calculating the acetic acid content

The sodium hydroxide reacts with the acetic acid in the following equation:

NaOH + CH3COOH = CH3COONa + H2O

The concentration of NaOH is 0.1 M, that is 1dm3 (1000 cm3) contains 0.1 moles. 25 cm3 of NaOH was used therefore the number of moles present is:

25 x 0.1 /1000

= 0.0025 moles

From the above equation one mole of NaOH reacts with one mole of CH3COOH. Therefore 0.0025 moles of NaOH reacts with 0.0025 moles of CH3COOH.

If the average volume of vinegar was 20 cm3, then this volume contains 0.0025 moles of CH3COOH. Therefore the number of moles of CH3COOH in the 100 cm3 is:

100 x 0.0025 /20 = 0.0125 moles

Since 10 cm3 of the original vinegar was diluted to 100 cm3, the number of moles of CH3COOH present in the original 10 cm3 of vinegar is 0.0125.

The relative molecular mass of CH3COOH is 60 g

Therefore the mass of 0.0125 moles of CH3COOH is 0.75 g

The density of the acetic acid solution is 1.01 gcm-3 therefore the mass of the original vinegar is 10.1 g.

Therefore the percentage acetic acid content of vinegar is:

0.75 / 10.1 x 100 = 7.4 %

Therefore acetic acid content of vinegar is 7.4 %.

Equipment

50 cm3 burette

10 cm3 pipette

25 cm3 pipette

100 cm3 volumetric flask

2 x 250 cm3 conical flasks

Distilled water

White tile

Clamp and stand

Chemicals

Vinegar, Standard sodium hydroxide solution (0.1 M) and phenolphthalein indicator

What to do

1. Using the pipette add 10 cm3 of vinegar to the 100 cm3 volumetric flask. Add the Distilled water to the vinegar until the mixture reaches the base of the stem of the flask. Mix well then add more distilled to the 100 cm3 mark and mix again. Fill the burette with this solution.

2. Using the 25 cm3 pipette, add 25 cm3 of the 0.1 M sodium hydroxide solution to the 250 cm3 conical flask. Add a few drops of phenolphthalein indicator.

3. Place the white tile under the conical flask and add the vinegar solution slowly from the burette into the sodium hydroxide solution. When the mixture changes from pink to colourless, record the volume of vinegar solution added.

4. Repeat steps 2 and 3. Calculate the average volume of vinegar solution.

Calculating the acetic acid content

The sodium hydroxide reacts with the acetic acid in the following equation:

NaOH + CH3COOH = CH3COONa + H2O

The concentration of NaOH is 0.1 M, that is 1dm3 (1000 cm3) contains 0.1 moles. 25 cm3 of NaOH was used therefore the number of moles present is:

25 x 0.1 /1000

= 0.0025 moles

From the above equation one mole of NaOH reacts with one mole of CH3COOH. Therefore 0.0025 moles of NaOH reacts with 0.0025 moles of CH3COOH.

If the average volume of vinegar was 20 cm3, then this volume contains 0.0025 moles of CH3COOH. Therefore the number of moles of CH3COOH in the 100 cm3 is:

100 x 0.0025 /20 = 0.0125 moles

Since 10 cm3 of the original vinegar was diluted to 100 cm3, the number of moles of CH3COOH present in the original 10 cm3 of vinegar is 0.0125.

The relative molecular mass of CH3COOH is 60 g

Therefore the mass of 0.0125 moles of CH3COOH is 0.75 g

The density of the acetic acid solution is 1.01 gcm-3 therefore the mass of the original vinegar is 10.1 g.

Therefore the percentage acetic acid content of vinegar is:

0.75 / 10.1 x 100 = 7.4 %

Therefore acetic acid content of vinegar is 7.4 %.