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working out the hydrogen ion concentration of a strong diprotic acid? Watch

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    I know that [H+]= 10^-pH for a strong monoprotic acid, however I am confused about working out the hydrogen ion concentration of a strong diprotic acid.

    I don't understand why you would divide (10^-pH) by 2 and not times it by 2...

    I mean if [H+]= 10^-ph, then wouldn't [2H+]= (10^-ph) x2?

    I hope this all made sense
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    Could you post a specific exam question?
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    (Original post by XavierMyshkin)
    Could you post a specific exam question?
    well here is a question from a revision sheet, page 3 last question

    http://www.chemsheets.co.uk/Chemshee...&%20bases).pdf
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    Makes sense. If the pH of an acid is 1, multiplying it by 2 would be 2 which is a weaker acid, when we know it should be 0.5.
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    The question is asking for the [H2SO4] necessary to give a pH = 1.30:

    pH = 1.30 :. [H+] = 10^-1.30
    But since [H+] = 2 x [H2SO4] since per 1 mole of H2SO4 2 moles of H+ is produced assuming complete dissociation
    [H2SO4] = (10^-1.30) / 2

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    (Original post by XavierMyshkin)
    The question is asking for the [H2SO4] necessary to give a pH = 1.30:

    pH = 1.30 :. [H+] = 10^-1.30
    But since [H+] = 2 x [H2SO4] since per 1 mole of H2SO4 2 moles of H+ is produced assuming complete dissociation
    [H2SO4] = (10^-1.30) / 2

    ahhhhh, i get it now, thanks so much!
 
 
 
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