How much NaOH I have to add to increase pH?

[bluecalculator]

Homework Statement

How much NaOH I have to add?

Relevant Equations

No relevant equations

Hi everyone, I'm stuck in solving this problem. I have two solution of H2SO4 5M that needs two consequential pH increasing: the first from the natural pH of H2SO4 5M (around zero) to pH 2 and then from pH 2 to pH 7. The second solution from the natural pH of H2SO4 5M to pH 2 and then from pH 2 to pH 11. I would like to understand how can I calculate the volume of NaOH (let's say 5M) I have to add for each step.
at the beginning, to calculate the Volume from pH 0 to 2 I was considering this formula:

[H3O+](after adding NaOH)=(V(H2SO4)×[H2SO4](initial)−V(NaOH)added×[NaOH])/V(H2SO4)+V(NaOH)added

But somebody told me this calculation is wrong because I have to consider that H2SO4 is diprotic.

The value I used are:
[H3O+](after adding NaOH)= 10^-2
V(H2SO4)= 1L
[H2SO4](initial)= 5 M

Thank you !

 

[Borek]

You should take into account not only fact the acid is diprotic, but also - especially around pH 2 - that pKa2 is close to 2.

I don't see any difference between the first and the second solution.

 

[bluecalculator]

Yes sorry..a typo. Now it's correct. Thank you, I'm going to try!

 

[epenguin]

You want to get your solution to pH7? The pH of pure water is 7, you wouldn't start with sulphuric acid. That is the pH of freshly deionised water is 7, if you leave it standing around without preventing it, it will absorb CO₂ from the air and decrease to about 4.5. Also 0.001 M NaOH should be pH 11 - what do you need the sulphuric acid for?

When we get this sort of question from someone sounding rather unfamiliar we can really only help (maybe) if they tell us the practical thing they are trying to achieve or overall problem to solve.