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 !