# Find the concentration of proton after adding a acid to NaOH [closed]

I have this simple problem that I cannot figure out though, only the first part.

100 ml of 1.0 $\frac{mol}{L}$ $NaOH_(aq)$ contains ( a ) g of NaOH. After mixing 100 ml of 1.0 $\frac{mol}{L}$ $H_2SO_4(aq)$ with the first solution, the concentration of proton becomes ( b ) $\frac{mol}{L}$

I got that there are 4 g of NaOH in the solution, but I don't have clear ideas about what should I do to find the concentration of proton. Could you advise me?

The answer is 0.50 $\frac{mol}{L}$

## closed as off-topic by aventurin, Jon Custer, Mithoron, airhuff, pH13 - Yet another PhilippMay 27 '18 at 22:25

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

• Remember that both protons of the H2SO4 are available to react with the NaOH – Waylander May 26 '18 at 16:42
• When you say "The answer is 0.50 mol/L", how sure are you about that? Because I think it's wrong. The 0.50 mol/L answer seems to assume that the H2SO4 donates both protons, but that's not the case. – Bennett Jun 6 '18 at 19:34

The reaction taken place in the mixture is: $$\ce{2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + 2H2O (l)}$$
(a) You already have done this but I'll show you how to do the unit conversion: $$\text{g of}~\ce {NaOH} = \pu{1.0 \frac{mol~of~\ce {NaOH}}{L}} \times \pu{0.100 L} \times \pu{40 \frac{g}{mol~of~\ce {NaOH}}} = \pu{4.0 g }$$
(b) Now similar to above calculation, find out number of moles of $\ce {[OH-]}$ in the original base solution ($\pu {0.10 mol}$) and number of moles of $\ce {[H3O+]}$ in the original acid solution ($\pu {0.20 mol}$), because each $\pu {\ce {[H2SO4]} mol}$ gives $\pu {2\ce {[H3O+]} mol}$s. Note that, $\ce {H3O+}$ ions are in excess. When solutions are mixed, each $\ce {OH-}$ ion react with one $\ce {H3O+}$ ion to give one $\ce {H2O}$ molecule, according to the equation. At the end, $\pu{0.10 mols}$ of $\ce {H3O+}$ ions remain unreacted, because they are in excess. However, total volume now is $\pu{200 mL}$. Now, I assume you may able to calculate the final $\ce {H3O+}$ concentration.
• So, if I understood, when solution are mixed, 0.10 mol of $OH^-$ react with 0.10 mol of $H_3O^+$ so that only 0.10 mol of protons is in excess. Therefore, since the solution now is 200 mL I have to do $\frac{0.10 mol}{0.200 L} = 0.50 \frac{mol}{L}$. Thanks for your help! – TheBarbarios May 27 '18 at 10:33