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

1. 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.
2. 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.
3. (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.

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.
4. (Original post by Plato's Trousers)

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.

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