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

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    $\begingroup$ @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. $\endgroup$ – Mathew Mahindaratne Jun 16 at 20:36
  • $\begingroup$ @MathewMahindaratne Got it! Thanks for noting that. $\endgroup$ – Blade Jun 16 at 20:45
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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:

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

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

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