I'm stuck in solving this problem. I have two solution of $\pu{5 M}$ $\ce{H2SO4}$ that needs two consequential $\mathrm{pH}$ increasing: the first from the natural $\mathrm{pH}$ of $\pu{5 M}$ $\ce{H2SO4}$ (around zero) to $\mathrm{pH}$ 2 and then from $\mathrm{pH}$ 2 to $\mathrm{pH}$ 7. The second solution from the natural $\mathrm{pH}$ of $\pu{5 M}$ $\ce{H2SO4}$ to $\mathrm{pH}$ 2 and then from $\mathrm{pH}$ 2 to $\mathrm{pH}$ 7. I would like to understand how can I calculate the volume of $\ce{NaOH}$ (let's say $\pu{5 M}$) I have to add for each step. at the beginning, to calculate the Volume from $\mathrm{pH}$ 0 (as you said) to 2 I was considering this formula:
$$\ce{[H3O+]} \ \text{(after adding $\ce{NaOH}$)} =(V(\ce{H2SO4}) \times \ce{[H2SO4]} \ \text{(initial)} − V(\ce{NaOH}) \\ \text{added} \times [\ce{NaOH}])/V(\ce{H2SO4}) + V(\ce{NaOH}) \ \text{added}$$
But somebody told me this calculation is wrong because I have to consider that $\ce{H2SO4}$ is diprotic.
The value I used are: $\ce{[H3O+]}$ (after adding $\ce{NaOH}$) $= 10^{-2}$; $V(\ce{H2SO4})= \pu{1L}$; and $\ce{[H2SO4]} \ \text{(initial)} = \pu{5 M}$