You are provided with a sample of vinegar which contains approximately 5g of ethanoic acid (CH3COOH) per 100cmï¿½. 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 dmï¿½ï¿½.
The first thing you must do is convert the approximate concentration of Ethanoic Acid in vinegar from grams per 100cmï¿½, to moles per dmï¿½. 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 ï¿½ 12.0
Hydrogen ï¿½ 1.0
Oxygen ï¿½ 16.0
Therefore: M.R CH3COOH = 2 ï¿½ 12 + 4 ï¿½ 1 + 2 ï¿½ 16
= 24 + 4 + 32
The mass is given so now you can proceed to calculate the number of moles:
Nï¿½ of moles = mass
This is the number of moles 1000cmï¿½. You need the number of moles in 1000cmï¿½ or 1 dmï¿½. Therefore you must divide this number by 100 and multiply it by 1000.
0.0833? ï¿½ 1000
0.000833? ï¿½ 1000
0.833? mol dmï¿½ï¿½
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 ï¿½ CH3COONa + H2O
This equation is balanced. From a balanced equation you can extract a mole ratio. This equation tells us that 1 mole of CH?COOH reacts with 1 mole of NaOH to form 1 mole of CH?COONa and 1 mole of water.
According to this mole ratio, the sample of vinegar is too acidic/too concentrated to titrate against the Sodium Hydroxide solution. Also if you carry out the titration without taking this into consideration, then you will not only be increasing experimental error, but it is also a waste of chemicals.
Therefore you must dilute the vinegar down to a suitable concentration. The ideal dilution factor would be ï¿½8, but it would be a very complicated process and you do not have the necessary apparatus. Therefore you must dilute it by a factor of ï¿½10. To do so, you will need the following apparatus and chemicals:
* A 25cmï¿½ pipette
* A 250cmï¿½ volumetric flask
* Distilled water
* Vinegar is a dilute aqueous solution of Acetic Acid. The pure acid is hazardous, yet when diluted it is not very dangerous. It has a very strong smell so use it in a ventilated environment.
*Wear a lab coat and safety goggles at all times.
* Rinse the 250cmï¿½ pipette with vinegar
* Pipette 25cmï¿½ of vinegar into the volumetric flask. (note: always make sure the bottom of the meniscus is on the graduation mark)
* Carefully fill the volumetric flask with distilled water to the graduation mark.
* Stopper the flask and shake until the solution is homogeneous.
The vinegar has now been diluted by a factor of ï¿½10 and will be used in the titration.
* Conical flask
* 25 cmï¿½ pipette
* Pipette filler
* Filter funnel
* Diluted vinegar solution
* Aqueous NaOH solution
* Distilled water
(*The end point in this weak acid – strong base titration is approximately 9.3. Methyl orange has a transition pH range of 3.1-4.4 and is therefore completely useless for this experiment.
Phenolphthalein on the other hand has a pH range of 8.3-10.0, so it will change colour right on the end point. For this reason, the indicator which will be used in this titration is Phenolphthalein.)
HEALTH AND SAFETY CONSIDERATIONS:
* Sodium Hydroxide in its pure form is very corrosive and can cause severe burns. In its diluted form it is less harmful, but still an irritant. It can also cause severe permanent eye damage.
*Wear a lab coat and safety goggles at all times.
* Rinse the pipette with the diluted solution of vinegar.
* Rinse the conical flask with distilled water
* Pipette 25cmï¿½ of diluted vinegar solution to the conical flask and add 4-5 drops of phenolphthalein indicator.
* Rinse the burette with the solution of Sodium Hydroxide and be sure to get rid of any air bubbles in the tip of the burette.
* Using the filter funnel, carefully fill the burette with the solution of Sodium Hydroxide. Record the initial burette reading in a results table.
* Set up the burette and the conical flask in the corresponding places for the titration. Make sure there are no bubbles in the burette before using it.
* Slowly and carefully run Sodium Hydroxide solution from the burette into the conical flask with swirling. Do this until the solution changes from colourless to light pink. This is the end-point. Record your final burette reading also to the nearest 0.05 cmï¿½.
* Repeat this procedure until 2 concordant results are obtained making sure you rinse the conical flask between every titration.
* All burette readings must be recorded to the nearest 0.05 cmï¿½ (1 drop).
* Consecutive titrations should be concordant.
This means that they are either identical or within 0.10 cmï¿½ of each other.
* The titration must be repeated until this is achieved.
* Burette readings must be recorded to 2 decimal places.
Results table for a titration:
Titre: final burette reading – initial burette reading.
Treatment of results:
Once you have your results table with your 2 concordant titres, you have to calculate you average titre.
To do so, add your 2 concordant titres and then divide the total by two.
Average titre = 1st accurate + 2nd accurate
= x cmï¿½
( This is the volume you will use in the further calculations.
The next step is to write down the balanced equation again with the mole ratio and the corresponding volume of each one.
CH3COOH + NaOH ï¿½ CH3COONa + H2O
v v v v
1 mole 1 mole 1 mole 1 mole
25 cmï¿½ x cmï¿½
Now, you have to calculate the number of moles of NaOH that reacted in this experiment. This is done as follows:
Number of moles (n) = Volume (v) ï¿½ concentration (c)
The exact concentration of NaOH has been given: 0.100mol/dmï¿½;
and the volume you will use is your average titre.
.?. n = x cmï¿½ ï¿½ 0.100 mol dmï¿½ï¿½
= y moles
According to the mole ratio, the number of moles of CH3COOH is the same as that of NaOH. Therefore we can rearrange the former equation for c, to work out the concentration of Ethanoic Acid in diluted water.
If n = v ï¿½ c
.? . c = n ï¿½1000
Now you can substitute into the equation.
C = y moles ï¿½ 1000
C = Z mol dmï¿½
Since the vinegar was diluted by a factor of ï¿½10, if you multiply this concentration by 10 you will have the concentration of CH?COOH in undiluted vinegar.
C = Z ï¿½ 10
C = 10z mol dmï¿½ï¿½
This now has to be converted to g/dmï¿½.
To do this you multiply the number of moles by the M.R, which we already know is 60.
Mass = n ï¿½ M.R
= 10z ï¿½ 60