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    Calculate the pH of the buffer solution formed when 10.00 cm3 of
    0.100 mol dm–3 potassium hydroxide are added to 25.00 cm3 of 0.410 mol dm–3 ethanoic acid

    Jan 2010.

    I got a ph of 4.80, but the answer is 3.79. I calculated it correct up to the point of the equation, and I am still confused as to why you do no need to calculate the new CONCENTRATIONS by dividing by the new total volume of 35cm^3

    Why do you find new concentrations by dividing by the total volume and when do you not?
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    What is the Ka value


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    I got the right answer by your method?


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    (Original post by cutelady)
    I got the right answer by your method?


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    Did you find new concentrations by dividing by volumes?

    When are you supposed to divide by volumes? THats what is really confusing me!
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    1) find the moles of KOH
    2 find the moles of ethanoic acid
    3) find what's in excess, in this case ethanoic acid was in excess.
    4) find the new concentrations, by doing MOLES multiply by 1000 and divide by 35.
    5) rearrange the equation to find [H+]
    6) find the pH
    7) would give you the right answer



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    (Original post by cutelady)
    1) find the moles of KOH
    2 find the moles of ethanoic acid
    3) find what's in excess, in this case ethanoic acid was in excess.
    4) find the new concentrations, by doing MOLES multiply by 1000 and divide by 35.
    5) rearrange the equation to find [H+]
    6) find the pH
    7) would give you the right answer



    Posted from TSR Mobile
    Thats perfect, thank you so much
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    what'd the ka value? also why divide by 35??
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    (Original post by soLit)
    Thats perfect, thank you so much
    Welcome xx


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