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

1. 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
2. Could you post a specific exam question?
3. (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
4. 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.
5. 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

6. (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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