You are provided with a sample of vinegar which contains approximately 5g of ethanoic acid (CH3COOH) per 100cm. The aim of this experiment is to determine the exact concentration of this chemical in vinegar.

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Georgina Garfield

EXPERIMENT 4: TO DETERMINE THE CONCENTRATION OF ETHANOIC                  ACID IN A SAMPLE OF VINEGAR

You are provided with a sample of vinegar which contains approximately 5g of ethanoic acid (CH3COOH) per 100cm³. The aim of this experiment is to determine the exact concentration of this chemical in vinegar.

To do this you must carry out an acid-base titration because it is a very accurate procedure if done correctly. The solution of vinegar is the acid, and the base you will titrate it against is sodium hydroxide (NaOH). This solution has a concentration of exactly 0.100 mol dmˉ³.

The first thing you must do is convert the approximate concentration of Ethanoic Acid in vinegar from grams per 100cm³, to moles per dm³. To do this you must carry out the following calculations:

*Number of moles of CH3COOH in 5g. You know that the number of moles can be calculated by dividing the mass by the M.R.  (Relative Molecular Mass). You can find the M.R by adding the relative atomic mass of each of the elements in the molecule in the proportions given in the molecular formula.

There are 2 Carbons, 4 Hydrogens and 2 Oxygens.

Atomic Masses: Carbon → 12.0

   Hydrogen → 1.0

   Oxygen → 16.0

                Therefore: M.R CH3COOH = 2 × 12 + 4 × 1 + 2 × 16

                                                   = 24 + 4 + 32

                                                   = 60

        The mass is given so now you can proceed to calculate the number of moles:

                                  N° of moles = mass

                                                 M.R

                                                   = 5

                                                      60

This is the number of moles 1000cm³. You need the number of moles in 1000cm³ or 1 dm³. Therefore you must divide this number by 100 and multiply it by 1000.

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                        0.0833˙ × 1000

                          100

                            0.000833˙ × 1000

                        0.833˙ mol dmˉ³

Now you have an approximate concentration of Ethanoic Acid in vinegar (acid) and the exact concentration of Sodium Hydroxide (base).

Ethanoic Acid reacts with Sodium Hydroxide and forms Sodium Ethanoate

(CH3COONa) and water (H2O).

                CH3COOH + Na OH → CH3COONa + H2O

This equation is balanced. From a balanced equation you can extract a mole ratio. This equation tells us that 1 mole of CHзCOOH reacts with 1 mole of NaOH to form 1 mole ...

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