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    (Original post by charco)
    0.1M ethanoic acid does not have a pH of 1

    ka of ethanoic acid is 1.78 x 10-5
    Ahh i was wondering why there was no Ka given - am i supposed to know it?
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    (Original post by kiiten)
    Ahh i was wondering why there was no Ka given - am i supposed to know it?
    you ARE supposed to know that ethanoic acid is a weak acid ...
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    (Original post by charco)
    you ARE supposed to know that ethanoic acid is a weak acid ...
    Yeah i know that, just didnt take any notice for this question

    I dont need to know what Ka for ethanoic acid is though?
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    (Original post by kiiten)
    Yeah i know that, just didnt take any notice for this question

    I dont need to know what Ka for ethanoic acid is though?
    no
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    (Original post by kiiten)
    Yeah i know that, just didnt take any notice for this question

    I dont need to know what Ka for ethanoic acid is though?
    A GCSE student would not know about Kc, but would know that EtOOH is a weak acid and therefore only partially ionises, i.e. [H+] =/= [EtOOH], (like wot i sed b4).
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    (Original post by Pigster)
    A GCSE student would not know about Kc, but would know that EtOOH is a weak acid and therefore only partially ionises, i.e. [H+] =/= [EtOOH], (like wot i sed b4).
    This is for something else but when you write ionic equations. E.g. barium hydroxide and sodium carbonate - why isnt the product separated into its ions.

    Ba2+ + 2OH- +2Na+ + CO32- ------> BaCO3 + 2NaOH
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    (Original post by kiiten)
    This is for something else but when you write ionic equations. E.g. barium hydroxide and sodium carbonate - why isnt the product separated into its ions.

    Ba2+ + 2OH- +2Na+ + CO32- ------> BaCO3 + 2NaOH
    BaCO3 is really rather insoluble, whereas NaOH is very soluble. You'd therefore show BaCO3 as a product, but Na+ and OH- remain in solution.

    The ionic equation is: Ba2+(aq) + CO32-(aq) -> BaCO3(s)
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    (Original post by Pigster)
    BaCO3 is really rather insoluble, whereas NaOH is very soluble. You'd therefore show BaCO3 as a product, but Na+ and OH- remain in solution.

    The ionic equation is: Ba2+(aq) + CO32-(aq) -> BaCO3(s)
    Ohh so if its insoluble you write it as a product and not ions
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    Agreed.
 
 
 
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