Bingsuze
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‘ the teacher carried out the experiment and obtained the following results:
Mass of NaOH used to make 250cm3 of solution = 3.80g
Volume of Ethan pic acid=25cm3
Mean titre of NaOH= 11.9cm3
CH3COOH+ NaOH—> CH3NOONa + H2O
Calculate the conc of the ethanoic acid in g dm3 ‘
Also does anyone know anywhere I could find the the answers to the edexcel assessment material 2021 ( the public released thingy 😂)
Thank you so much <3
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CaptainDuckie
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What have you tried so far?
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Bingsuze
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(Original post by CaptainDuckie)
What have you tried so far?
I got 25g dm3??

n NaOH= 0.095
N in 25cm= 0.0095
Ethaboic acid data:
n: 0.095
V= 0.025
C= 0.095/0.025= 0.38
This is where I’m a bit stuck: I’ve done
Because there are 3.80g of NaOH dissolved in 250cm of solution 250/ 3.80= 65.79
65.79 X 0.38= 25g dm-3

Which just doesn’t make sense!!
Last edited by Bingsuze; 1 month ago
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scimus63
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moles NaOH = 3.8/40=0.095 mols in 250ml of solution. (Mr NaOH=40)
conc of NaOH = n/vol = 0.095/.25 = 0.38 mol/dm3
moles NaOH used in titration = 0.38 x 11.9/1000 = 4.5 x 10-3 this also same as moles of acid
conc of acid = n/v = 4.522 x10-3/0.025 = 0.18 mol dm3
to change units of conc into g/dm3 multiply by Mr of ethanoic acid
check my calcs

look here for more ideas!
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Bingsuze
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(Original post by scimus63)
moles NaOH = 3.8/40=0.095 mols in 250ml of solution. (Mr NaOH=40)
conc of NaOH = n/vol = 0.095/.25 = 0.38 mol/dm3
moles NaOH used in titration = 0.38 x 11.9/1000 = 4.5 x 10-3 this also same as moles of acid
conc of acid = n/v = 4.522 x10-3/0.025 = 0.18 mol dm3
to change units of conc into g/dm3 multiply by Mr of ethanoic acid
check my calcs

look here for more ideas!
Omg thank you so much!!! I really appreciate this
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scimus63
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meant to say look here it has this type of calculation, it may help you
https://www.science-revision.co.uk/titrations.html
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