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

    What have I done wrong?
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    Not sure if it's attached but here is the question
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    when i did it, i got 3.99,
    should i try and write up my solution?
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    Yes please!

    (Original post by myabzz)
    when i did it, i got 3.99,
    should i try and write up my solution?
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    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
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    (Original post by michelecorrrea)
    Yes please!
    did i get something wrong?
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    (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
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    (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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