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    Like the title says, how do you calculate the PH of 10^-9 M H2SO4??
    The final answer is around 7, but I don't know how to get that answer.
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    (Original post by AntiClock)
    Like the title says, how do you calculate the PH of 10^-9 M H2SO4??
    The final answer is around 7, but I don't know how to get that answer.
    Full dissociation so: pH = -log10 [H+]

    So for every mol of H2SO4 you'll have twice the H+.
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    pH = -log10 [H+]

    Assume it fully dissociates, so the [H+] is 2 x 10^-9. Plug that into the formula above!
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    (Original post by Revd. Mike)
    pH = -log10 [H+]

    Assume it fully dissociates, so the [H+] is 2 x 10^-9. Plug that into the formula above!
    -log(2x10^-9)=8.6, but the answer say it's 7. So, I guess it's a mistake on the answer key then.

    A more complex question if I may, what's the PH of:
    1.0 liter of 0.1M glycine hydrochloride (pka are 2.3 and 9.7) to which as been added 100ml of 1M NaOH.
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    You need to take into account the H+ ions in water in this case (you're never going to have an acidic solution that has a pH > 7 !).

    Kw = [OH-][H+] = 10-14 so [H+] = 10-7 (from the water) + 2 x 10-9
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    H2SO4 definitely has a ph lower than 7.

    The name sulphuric acid kind of gives that away, you'd think.
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    (Original post by EierVonSatan)
    You need to take into account the H+ ions in water in this case (you're never going to have an acidic solution that has a pH > 7 !).

    Kw = [OH-][H+] = 10-14 so [H+] = 10-7 (from the water) + 2 x 10-9
    (Original post by ste_mc_efc)
    H2SO4 definitely has a ph lower than 7.

    The name sulphuric acid kind of gives that away, you'd think.
    I understand that one now, thanks. What about this one?

    1.0 liter of 0.1M glycine hydrochloride (pka are 2.3 and 9.7) to which as been added 100ml of 1M NaOH.

    I hate these stupid PH questions in biochem :eek:
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    Work out how many moles of glycine, HCl and NaOH you have to start with then work out how much of each you'll have once they're added. Go from there
 
 
 
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