• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14

Analysing the ethanoic acid concentration in different types of vinegars.

Extracts from this document...


Analysing the ethanoic acid concentration in different types of vinegars. Vinegar is a mixture of various acids, ethanoic acid being the most abundant. Most vinegars are made from a sugary liquid, this liquid is fermented to produce ethanol and then oxidised to form ethanoic acid. Different vinegars use different sugary liquids and this could lead to variation between the vinegars. My aim in this investigation is to discover whether the concentration of ethanoic acid varies between vinegars. Vinegar consists mostly of water (bp = 100 degrees C) and ethanoic acid but it also contains many flavour components and other acids in low abundances. From my research I have discovered that the main acids in Vinegar are: Citric Acid C6H8O7 mp = +153 degrees C Tartaric Acid C4H6O6 mp = +170 degrees C Malic Acid C4H6O5 mp = +101-103 degrees C Succinic Acid C4H6O4 mp = +183 bp = +235 degrees C Ethanoic Acid C2H4O2 mp = +16.6 bp = +117.9 degrees C Phosphoric Acid H3O4P mp = +158 degrees C Proponic Acid C3H6O2 bp = + 141 degrees C Due to all the other acids contained in vinegar I will not be able to obtain my results for the concentration of ethanoic acid direct from the vinegar, as the H+ ions produced by the other acids will effect the results of my titrations. Therefore I need a way of separating the ethanoic acid from the solution of vinegar. I will do this via the process of fractional distillation. This is a process whereby different substances in a solution are separated from one another. It is done by heating the solution in a round-bottomed flask and as the solution begins to warm up different substances contained within the solution begin to boil off. This gas is then condensed into a liquid using the condenser and can be collected. Using a thermometer you can measure the temperature it boils off at and as each substance has its own unique boiling point you can tell what substance you are collecting. ...read more.


Ka of ethanoic acid = [CH3CO2-]eqm x [H+]eqm [CH3CO2H]eqm From my equation I know that [CH3CO2-]eqm = [H+]eqm, and also from my research I know the Ka value for ethanoic acid = 1.7 x 10-5. Hence I can now write my equation in this form. 1.7 x 10-5 = [H+]eqm2 [CH3CO2H]eqm 1.7 x 10-5 = (3.162 x 10-3)2 [CH3CO2H]eqm [CH3CO2H]eqm = (3.162 x 10-3)2 1.7 x 10-5 [CH3CO2H]eqm = 0.588 (3 sf) Now I have obtained my values for X and X-Y however to get the concentration of ethanoic acid my distillate (Y) I will need to add the two values together. [CH3CO2H] in vinegar (Y) = 0.588 + 3.162 x 10-3 = 0.591 (3sf) Therefore the values obtained from my pH meter for all my distillates is that the concentration of ethanoic acid = 0.591 mol/dm3. Now using the measurements I took of the volume produced by the distillation of the vinegar I can calculate the concentration of ethanoic acid in Vinegar. Vol. Of Malt Vinegar distillate = 47.5 cm3 Vol. Of Cider vinegar distillate = 47.9cm3 Vol. Of White wine vinegar distillate = 48.1cm3 Concentration of ethanoic acid in vinegar = Volume of distillate/ volume of vinegar x concentration of ethanoic acid in distillate. Malt vinegar = 47.5/50 x 0.591 = 0.561 mol/dm3 Cider vinegar = 47.9/50 x 0.591 = 0.566 mol/dm3 White wine vinegar = 48.1/50 x 0.591 = 0.569 mol/dm3 Also So I will be able to compare the pHs of my distillates with that of my original acid I took 3 pH readings for all of my original acids and averaged them. Malt vinegar = pH 3.0 Cider vinegar = pH 3.0 White Wine Vinegar = pH 2.8 During my experiment I used a number of techniques to try and ensure the results I obtained were as reliable and accurate as possible. Despite this it is very likely that qualitative and quantitative errors have had an effect on the accuracy of my results. ...read more.


Two of my vinegars have identical pHs and these vinegars have similar concentrations of ethanoic acid (both 0.7 to 1 dp) whereas the values for the concentrations of other organic acids vary relitivly significantly. The pHs of all my distillates were the same. This is surprising as my titration results showed that the concentrations of ethanoic acid in the distillates varied significantly. This could be due to the limited accuracy of the equipment as the pH meter can only measure to 0.1 and so may not be able to pick up subtle differences in the pHs of the distillates. Or it could be due to the fact that this method contains a lot of quantitative error that could of affected my results. As you can see from the analysis I have been able to make trends in my results and most of them fit these trends. However as you can see by the differences between the two values for the concentration of ethanoic acid I obtained, there is clearly error in my results. However the results from both the methods seem to fit in with my scientific knowledge and as I haven't obtained any anomalous results I believe that my results are fairly accurate. However there are improvements I could make to my method to improve accuracy. Instead of measuring out the vinegar in a measuring cylinder next time I would measure it out using a pipette as it is far more accurate. Also during titrations I would invert the burette to make sure that the NaOH does not collect towards the bottom. To ensure the accuracy of my titrations I would test the concentrations of the standard solution I have made by either titrating it with an acid of known concentration or using short-range indicator paper. This would ensure my titration results were accurate. This should improve the accuracy of my results if I were to perform the experiment again. Josh Rowlands March 2002 2 ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Aqueous Chemistry essays

  1. Marked by a teacher

    Thermometric Titration Investigation

    4 star(s)

    as many hydrogen ions, you will need half as many molecules as hydrochloric acid needs to neutralise the Sodium Hydroxide. This is because one sulphuric acid molocule can neutralise sodium hydroxide at a ratio of 2 sodium hydroxide molecules to one sulphuric acid.

  2. Peer reviewed

    An Investigation of Titration

    3 star(s)

    negative pressure and absorbing the water and releasing it into the measuring cylinder. Neutralization of an Alkali with an Acid HCl + NaOH => NaCl + H2O From this equation we know that we need the same molarity and the same volume of sodium hydroxide and hydrochloric acid, and this should reach neutralization.

  1. Determine the concentration or molarity of Ethanoic acid (CH3COOH) in two types of commercial ...

    There will be 20 cm3 of the diluted vinegar as this will be an easy value to deal with as it is a tenth of the amount of the original amount of diluted vinegar and it is also a convenient amount in relation to how much of sodium hydroxide will be added to it.

  2. Investigating the effects of varying pH levels on the germination of cress seeds

    I then made the 1% concentrated solution; I first measured 360cm� of H2O using the 100cm� measuring cylinder labelled 'Water', and put this into the corresponding bottle labelled '1%'. Using one of the remaining two 100cm� measuring cylinders I measured 40cm� of the 10% concentrated solution which I added to the bottle, and shook gently to mix the solution.

  1. Back Titration to find the Concentration of Vinegar (Ethanoic Acid)

    Again it will be green at the equivalence point at which the acid will be neutralised. * The first 2cm3 micropipette used for the base will be used to titrate the sodium hydroxide into each of the wells with ethanoic acid.

  2. How much Iron (II) in 100 grams of Spinach Oleracea?

    present in 100 grams of Spinach Oleracea the volume of moles in the 100cm3 spinach extract solution will need to be multiplied by 5. 0.00605832 mol dm-3 X 5 = 0.0302916 mol dm-3 Now that I know the moles present in 100 grams of spinach I can use the equation below to work out the mass of Iron (II)

  1. Planning of Titration

    Also watch fairly constantly the solution as the colour changes very suddenly which is most important in this experiment. Do not put dangerous chemicals in containers for food use (i.e.: cups, glasses, bottles) because they could be mistaken for beverages or foods.

  2. Ethanoic Acid Titration

    ethanoic acid NaOH + CH3COOH ï CH3COONa + H2O mol = 0.0020 mol Mole ratios mol n(CH3COOH)= 0.0020 mol Convert n(CH3COOH) to c(CH3COOH) Convert diluted c(CH3COOH) to undiluted c(CH3COOH) Molar Concentration *To one significant figure: Concentration Percentage mL m/v *To one significant figure: 5% m/v Discussion: Many different kinds of

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work