Determining of the concentration of ethanoic acid in vinegar

[take-ahike]

I was given a experiment to do to find the % weight/volume of the vinegar.
this was he procedure
1. place 25cm3 of vinegar in a 250cm3 volumetric flask and dilute with water to the calibration mark, mixing well
2.stopper the flask and invert several times to ensure a homogeneous solution. label the flask
3. place 25cm3 of diluted vine gar solution in the conicl flask using a pippette, and add 3 drops of phenolphhalein indicator
4. fill the burette to the 0 cm3 mark with NaOH
5. carry out one rough and four accurate titrations
6. calculate the concentration of ethanoic acid in the diluted vinegar solution
7. calculate the percentage(w/v) of ethanoic acid in the vinegar
my results were :17.0 cm3
17.0 cm3
17.2 cm3
17.1 cm3
(readings on the burette)

how do I do this?????? I am completely lost. can anyone please help me really really quickly.....

[deleted-2574]

Hi take-ahike,

I'm not sure what your question is since you have some results. They are close enough if you used different methods to calculate them.

If it helps, there are many resources on the web:

http://www.wesleylearning.ie/resources/ ... /index.htm gives an experiment to do the calculation

Others are:
http://en.wikipedia.org/wiki/Acetic_acid
http://answers.yahoo.com/question/index ... 349AAiBuNG
http://wwwchem.csustan.edu/consumer/vin ... alysis.htm

[take-ahike]

I still dont get it... can you do the calculation?

[deleted-71588]

This sounds like classwork / homework. This site is intended to assist students with science fair projects and not with labs and homework.

Yes I could do the calculations if I knew the concentration of your NaOH stock titration solution, but that wouldn't help you learn and might be considered cheating.

Your goal was to determine the %weight/volume of the acid in the dilution.

You started out with 25 ml (cm3) of vinegar diluted it with water to some calibration mark, then used 25 ml of that diluted vinegar in the titration. Determining what volume to use in your final calculations is a bit confusing because 25 appears twice but the key to figuring out the volume is <calculate the concentration of ethanoic acid in "the diluted vinegar solution">. The volume of diluted vinegar that you used in the titration step is 25 ml if you followed step 3 correctly.

The titration process used some (presumably known concentration) of NaOH (a base). You NEED to know this concentration so that you can determine how much base was reacted to neutralize the acid. If this were a 0.1 M (or N as there is one OH- ion) solution, then there would be 0.1 moles / ml so 17.1 ml would be 1.71 moles of OH- ions needed to neutralize the solution.

Now you need to know the formula for ethanoic acid to determine how many H+ ions and its molecular weight. Not to hard to find online. Since it disassociates to form a single H+ ion, its another example of a solution where Moarity = Normailty.

How ever many moles of OH- were required in the titration identifies how many moles of H+ ions were neutralized. Given this, you know how many moles of ethanoic acid were neutrailized. This means you know how many moles of etanoic acid were in the 25 ml dilute vinegar solution which IS the concentration. If you multiply the Molarity by the molecular weight which is usually expressed in grams/mole you get the weight of the ethanoic acid that is in the 25ml of dillute vinegar solution. This maybe the answer to step/question 7. %w/v is a bit of ambiguous terminology in itself; however, it usually means grams of solute per ml of solution and this meaning is confirmed by the "percentage(w/v)" phrase in step 7.

Step 7 is even more ambiguous. <calculate the percentage(w/v) of ethanoic acid in the vinegar> because it doesn't tell you which vinegar solution. It could be either the dilute one you titrated or the original solution that you started with. If the lab/homework wants to find this w/v for the original vinegar solution, you better know the volume the that is represented by the calibration mark or you don't have enough information to answer that interpretation of the question. If it is for the diluted vinegar solution, then you have it by multiplying the molecular weight by the number of moles and dividing by 25 ml.