# How much NaOH I have to add to increase pH?

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 $$\ce{[H2SO4]} \ \text{(initial)} = \pu{5 M}$$

• @Blade: I appreciate your hard work on editing. However, keep in mind that we do not approve MathJax formatting on the title as a policy. – Mathew Mahindaratne Jun 16 '20 at 20:36
• @MathewMahindaratne Got it! Thanks for noting that. – Blade Jun 16 '20 at 20:45

Can't quite follow your equations but your friend is right. It takes twice as much $$\ce{NaOH}$$ as $$\ce{H2SO4}$$. In other words for every $$\pu{1L}$$ of $$\pu{5M}$$ $$\ce{H2SO4}$$, you need $$\pu{2L}$$ of $$\pu{5M}$$ $$\ce{NaOH}$$.

The reactions are:

$$\ce{2H+ + 2OH- -> 2H2O}$$ $$\ce{2Na+ + SO4^2- -> Na2SO4}$$

OVERALL:

$$\ce{2NaOH + H2SO4 -> 2H2O + Na2SO4}$$

• Numbers don't match in your first equation. 6 H on LHS and 4 H on RHS. Also, what are .=. and =. ? – Blade Jun 16 '20 at 20:56
• @Blade: Use $\pu{}$ for units, e.g.., $\pu{1 M}$ conveted to $\pu{1 M}$. – Mathew Mahindaratne Jun 16 '20 at 23:34
• @MathewMahindaratne Thanks! What about constants? same or \mathrm? – Blade Jun 17 '20 at 0:06
• @Blade: Check This and this. Good luck! :-) – Mathew Mahindaratne Jun 17 '20 at 0:20