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A2 Chemistry Equilibrium

The answer is 4 but I'm getting something like 0.5

What have I done wrong?
(edited 8 years ago)
Not sure if it's attached but here is the question
20151031_161849_resized.jpg418.6KB
Reply 2
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myabzz
when i did it, i got 3.99,
should i try and write up my solution?
Yes please!

Original post by myabzz
when i did it, i got 3.99,
should i try and write up my solution?
Reply 4
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myabzz
1 mol ethanoic acid + 2 mol ethanol -> equilibrium
diluted to 250cm3 and 25cm3 is taken out
0.0155 mol of NaOH required to neutralise the 25cm3 which means there are 0.0155 mol of acid in the 25cm3 and therefore 0.155 mol of acid in 250cm3 in equilibrium.
As the acid dropped from 1 to 0.155 mol, it dropped by 0.845, and so did the ethanol (from 2 to 1.155).
the water and the ethyl ethanoate increased by the same amount and are now both 0.845 each.
putting it into the kc calculation, it is products over reactants which is (0.845*0.845)/(0.155*1.155) = 3.9884
Reply 5
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myabzz
Original post by michelecorrrea
Yes please!


did i get something wrong?
Original post by myabzz
did i get something wrong?


No,no. Thank you for doing that. I've been doing a couple more and I think I'm getting it now
Xx
Reply 7
Avatar for myabzz
myabzz
Original post by michelecorrrea
No,no. Thank you for doing that. I've been doing a couple more and I think I'm getting it now
Xx


Haha. Cool good luck XD

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