# How are acids in aqueous solution able to react again after reacting with water?

I hope to clear up some confusion I have about acid-base reactions in aqueous solution.

Based on the Bronsted-Lowry definition of acids, I know acids donate protons. In water, a strong acid should react completely with water to form hydronium and some anion. For example,

$$\ce{HCl + H2O -> H3O+ + Cl-}$$

However, my confusion arises when this acidic solution is reacted with another solution. For purposes of demonstration, I’ll use aqueous $$\ce{NaOH}$$. Usually, the acid-base reaction is

$$\ce{HCl + NaOH -> H2O + NaCl}$$

How is this possible if the $$\ce{HCl}$$ already reacted with water to form hydronium and chloride ions? Is it perhaps better to think of it in the Arrhenius sense where the original reaction is just $$\ce{HCl -> H+ + Cl-}$$ that way $$\ce{HCl}$$ can still be a reactant in the acid-base reaction without already being “used up”?

• In a way, you may think that HCl doesn't exist in the solution and hence doesn't react. H3O+ does. Jun 12 at 8:47
• What you have drawn as a single arrow reaction for HCl + H2O is an equlibrium Jun 12 at 9:07
• Jun 12 at 9:40

The proton transfers may happen in cascade.

For example, pure $$\ce{HCl}$$ is a gas that dissolves and reacts with water (acting as a base) according to a proton transfer from HCl to $$\ce{H2O}$$ : $$\ce{HCl + H2O -> H3O+ + Cl- ... \ (1)}$$ But $$\ce{H2O}$$ does not hold its proton very strongly. Here is why : if another base like ammonia is added to this solution, a new proton transfer occurs from $$\ce{H3O+}$$ to $$\ce{NH3}$$ which is a stronger base than $$\ce{H2O}$$ $$\ce{H3O+ + NH3 -> NH4+ + H2O} ... (2).$$ A last proton transfer happens from $$\ce{NH4+}$$ to $$\ce{OH-}$$ if some solution of $$\ce{NaOH}$$ (containing $$\ce{OH-}$$ ions) is added, according to : $$\ce{NH4+ + OH- -> NH3 + H2O} ...\ (3)$$

This shows that

• $$\ce{H2O}$$ is a base just strong enough to pick a proton from $$\ce{HCl}$$, according to ($$1$$).
• $$\ce{NH3}$$ is a stronger base as it is able to pick up a proton form $$\ce{H3O+}$$ according to $$(2)$$. Of course $$\ce{NH3}$$ is also able to pick a proton for from $$\ce{HCl}$$, although it was not shown in the preceding cascade of events. But if pure $$\ce{HCl}$$ and pure $$\ce{NH3}$$ are mixed, the following proton transfer happens from $$\ce{HCl}$$ to $$\ce{NH3}$$ direct : $$\ce{NH3 + HCl -> NH4+ + Cl-}$$
• $$\ce{OH-}$$ is the strongest base of all, as it is able to pick up a proton from $$\ce{NH4+}$$ as seen in ($${3}$$).

Of course, $$\ce{OH-}$$ is also able to pick up a proton from $$\ce{H3O+}$$ and $$\ce{HCl}$$

Just before the end, I would like to state that some teachers are reluctant to accept the reaction $$({1})$$, because nobody is sure that $$\ce{H3O+}$$ does exist. It could be admitted that the reaction is more complicated and does happen according to $$\ce{HCl + 2 H2O -> H5O+ + Cl-}$$ or even $$\ce{HCl + n H2O -> H_{2n+1}O+ + Cl-}$$. So, in order to simplify their teaching, they would recommend to simply write down the equation of $$\ce{HCl}$$ with water as : $$\ce{HCl -> H_{aq}^+ + Cl-}$$ And indeed the reaction of $$1$$ mole $$\ce{HCl}$$ with $$2$$ moles $$\ce{H2O}$$ is more exothermic that the reaction with $$1$$ mole $$\ce{H2O}$$.