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Let's say we have $5$ $\ce{HCl}$ molecules and we dump them into a container full of water. It's a strong acid so all $5$ $\ce{H+}$ ions would react with water and create $\ce{H3O+}$. It reacts with the water, right?

Now, we add $5$ $\ce{NaOH}$ molecules down there. They all react with some water molecules of their own as well and create $5$ $\ce{OH-}$. So now we have $5$ $\ce{H3O+}$ and $5$ $\ce{OH-}$. Those two don't react with each other, correct? They just react with the water molecules and balance each other out, giving us a neutral solution.

I am unsure of this. Is this true? Thanks in advance.

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closed as unclear what you're asking by Jan, porphyrin, bon, Todd Minehardt, Klaus-Dieter Warzecha Jan 22 '17 at 18:35

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    $\begingroup$ They certainly react with each other: $$\ce{H3O+ + OH- -> 2H2O}$$ $\endgroup$ – DHMO Jan 22 '17 at 15:30
  • $\begingroup$ Hmm ... so they react separately with water molecules and then the products react with each other? $\endgroup$ – chemistry101 Jan 22 '17 at 15:30
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    $\begingroup$ Yes. The steps would be: $$\ce{HCl + H2O -> H3O+ + Cl-}$$ $$\ce{NaOH -> Na+ + OH-}$$ $$\ce{H3O+ + OH- -> 2H2O}$$ $\endgroup$ – DHMO Jan 22 '17 at 15:31
  • $\begingroup$ Thank you. But is it then wrong to say that acids and bases balance each other out in a solution and thereby make it neutral? Provided we have an equilibrium $\endgroup$ – chemistry101 Jan 22 '17 at 15:35
  • $\begingroup$ That question doesn't make sense. $\endgroup$ – DHMO Jan 22 '17 at 15:36
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$$\ce{NaOH + HCl -> NaCl + H2O}$$

$$\ce{Na+ + Cl- + OH- + H+ -> Na+ + Cl- + H2O_{(l)}}$$ Both sides $\ce{Na+}$ ans $\ce{Cl-}$ cancel out remaining reaction

$$\ce{H+ + OH- <=> H2O_{(l)}}$$

Any strong acid and strong base reaction the remaining reaction is hydrolization of water.

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