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    Please can someone explain the answer to these

    This is the June 2012 paper





    Moderator edit to clear up the unnecessary text.
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    Sorry about that

    you can just have a look yourself. Its question 3b

    Many thanks
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    (Original post by 786:))
    Sorry about that

    you can just have a look yourself. Its question 3b

    Many thanks


    Bronston-Lowry acid = proton donor
    Bronston-Lowry base = proton acceptor

    So for example in:
     H_2O + NH_3 \rightarrow H_3O^+ + Cl^-

    the acid would be the proton donor, which is clearly the  HCl as this has lost it's proton ( H^+ ) and so  H_2O is the base as it has accepted a proton and is now  H_3O^-



    Applying this to
    3. b(i): the  CH_3COOH has donated the proton to  H_2O so it is the acid, and so  H_2O is the base.
    Be careful in this question as it is a reaction in equilibrium so there will be a mix of products and reactants.


    See if you can apply this to 3. b(ii) and b(iii)
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    That's the answer I got, but it says in the markscheme BB. :/
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    (Original post by 786:))
    That's the answer I got, but it says in the markscheme BB. :/

    Which is base and base according to the text above the equations.

    Look where the boxes are placed, as I've said the  H_2O is a base which was worked out above. The other box is on the other side of the equation, as like I said as it's in equilibrium the  CH_3COO^- also forms  CH_3COOH which is gaining a proton (i.e. accepting a proton) therefore the  CH_3COO^- is a base, it can accept protons to form  CH_3COOH


    Does that help?
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    Yes that does, I can't believe I didn't get that first time round lol!
    Thank you for the help
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    hello everyone, i need help with my upcoming isa but i dont know what to revise, has anyone done their isa yet? thanks a lot
 
 
 
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