# 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

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• 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

## 1 Answer

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
• Since nobody has pointed this out explicitly, I believe this solution is wrong because it assumes the H2SO4 will donate both protons. But H2SO4 is only a strong acid with respect to the first proton. To determine how many molecules donate the second proton, you have to use the second dissociation constant for H2SO4. The funny thing is that the OP said "The answer is 0.50 mol/L" which I assume came from an instructor or an answer key, which means the instructor got it wrong as well. – Bennett Jun 6 '18 at 19:37
• @Bennett The answer is from an answer key. Probably the exercise assumed that H2SO4 donates both H+. Otherwise, it would be less as you pointed out. – TheBarbarios Jul 13 '18 at 9:59
• @TheBarbarios thanks. Yeah I figured that -- the point is that the assumption is wrong, and gives the wrong answer :) – Bennett Jul 14 '18 at 18:34