[bluecalculator]

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 7. 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 (as you said) 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 !

[AWK]

1. Read the Forum Rules carefully
2. Show your calculations and we can help you if you need help.

The pH calculations for such high concentrations are very approximate because they cannot be done accurately - the actual pH of 5 M H₂SO₄ is around -0.7.

[bluecalculator]

Yes, I'm sorry my fault. 
Anyway at the beginning, to calculate the Volume from pH 0 (as you said) 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

[AWK]

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

This "somebody" was right.
You should use K₂ of sulfuric acid.