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Calculating [H+] with pH known. Does the "ratio" affect? watch

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    I'm just assuming we are asked something like this for example:

    Calculate [H+] present in the H2SO4 solution, provided that the pH of the solution is 2.

    We know that pH = -log[H+].

    Based on that, will the answer simply be:

    2 = -log[H+] = 0.01 mol dm-3?

    In other words, although having the ratio between H2SO4 and H+ as 1:2, this does not affect the way of calculation when pH is known?

    Thanks.
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    I believe there is something else to take into account due to the fact that H2SO4 is a di-protic acid, which, if I remember correctly either sees the concentration either x2 or divided by 2. I can't quite remember which though or if in-fact I am alright.
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    (Original post by SWEngineer)
    I'm just assuming we are asked something like this for example:

    Calculate [H+] present in the H2SO4 solution, provided that the pH of the solution is 2.

    We know that pH = -log[H+].

    Based on that, will the answer simply be:

    2 = -log[H+] = 0.01 mol dm-3?

    In other words, although having the ratio between H2SO4 and H+ as 1:2, this does not affect the way of calculation when pH is known?

    Thanks.
    Your answer is correct. The relationship between [H+] and pH is

    pH=-log_{10}[H^+]

    the fact that the acid is dibasic is irrelevant. It would only need to be taken into account if you were trying to work out pH or [H+] from the concentration of the sulphuric acid.
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    (Original post by Plato's Trousers)
    Your answer is correct. The relationship between [H+] and pH is

    pH=-log_{10}[H^+]

    the fact that the acid is dibasic is irrelevant. It would only need to be taken into account if you were trying to work out pH or [H+] from the concentration of the sulphuric acid.
    Thanks a lot for your reply. Yes, I think that makes sense.

    And, thanks @MrCasperTom for your reply.
 
 
 

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