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# AQA GCSE Unit 3 Titration mass question Watch

1. Hi. I have a question on titrations in a past paper that I don't understand. It says that a student did a titration of white wine vinegar and ethanoic acid. The volume of NaOH is 20cm^3 and the concentration is 1.0 mol/dm^3. One mole of ethanoic acid has a mass of 60g, and the white wine vinegar has a volume of 25.0cm^3. It asks "Is the student able to conclude that white wine vinegar contains 5g of ethanoic acid in 100cm^3). I'm really not sure where to start. Thanks in advance
The answer is apparently no, because it is 4.8g per 100cm^3 according to the mark scheme
2. As it's a titration, start by finding the concentration of your unknown solution, the vinegar.

The equation for the neutralisation is:
NaOH + CH3COOH --> H2O + CH3COONa
So the ratio is going to be 1:1.

You'll need to find the moles of NaOH present in the 20cm3 used - then this will be the moles of acid present in 25cm3 of the vinegar.
Use your titration equation triangle (don't forget to convert cm3 to dm3) to find the Molarity of the vinegar.

Then once you have the molarity (moles per litre), you can calculate the mass of ethanoic acid in one litre, then div by 10 to get mass per 100ml.
(for example if you worked out the vinegar was 0.4 moles per litre, you'd do 0.4 * 60g then div that by 10)

Bob's yer uncle. Hope that helped you work through the steps.
3. Thank you, that helped a lot.
4. (Original post by Qaisha)
As it's a titration, start by finding the concentration of your unknown solution, the vinegar.

The equation for the neutralisation is:
NaOH + CH3COOH --> H2O + CH3COONa
So the ratio is going to be 1:1.

You'll need to find the moles of NaOH present in the 20cm3 used - then this will be the moles of acid present in 25cm3 of the vinegar.
Use your titration equation triangle (don't forget to convert cm3 to dm3) to find the Molarity of the vinegar.

Then once you have the molarity (moles per litre), you can calculate the mass of ethanoic acid in one litre, then div by 10 to get mass per 100ml.
(for example if you worked out the vinegar was 0.4 moles per litre, you'd do 0.4 * 60g then div that by 10)

Bob's yer uncle. Hope that helped you work through the steps.
I just have one last question - I know that 1dm^3 is 1 litre, but how do I calculate the mass in one litre from the molarity? Thanks
5. They tell you in the question that one mole of ethanoic acid is 60g.

When it comes to molarity, 1mol per dm^3 (1 molar, or 1M) is literally one mole of the stuff dissolved per litre of water. So if you had a 1M solution of ethanoic, it would be 1 x 60g per litre. A 0.5M solution of ethanoic would be 0.5 x 60g per litre. Et cetera...
6. Thanks again

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Updated: April 14, 2013
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