How would you know when an acid or base is paired with $\ce{H2O}$ that it will form an $\ce{OH-}$ ion or a $\ce{H3O+}$ ion? I just started the acid and base equilibrium unit, and I'm just confused on what conditions $\ce{OH-}$ or $\ce{H3O+}$ ions form.
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$\begingroup$ Why, that's simple: acids give H3O+, bases give OH-. $\endgroup$– Ivan NeretinCommented Mar 21, 2021 at 19:56
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2 Answers
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Acids are hydrogen ion donors. When the react with water, they can give a hydrogen ion to form $ \ce{H3O+}$.
For example:
$ \ce{HCl(aq) + H2O(l) \rightarrow Cl-(aq) + H3O+(aq)}$
Simple acids, such as $ \ce{HCl}$ or $\ce{H2SO4}$, can be recognized as acids by the H at the start of the formula. Other more complex acids may be written with $ \ce{COOH}$ at the end of the formula, which denotes a carboxylic acid. An example of this is acetic acid: $\ce{CH3COOH}$. Otherwise, the name of the compound should give it away.
Bases are hydrogen ion acceptors, and generate $\ce{OH-}$ when they react with water. Strong bases are generally soluble metal hydroxides, such as $\ce{NaOH}$. You should be able to recognize them from their formulas.
Simple weak bases often contain nitrogen, like ammonia: $\ce{NH3}$.
$\ce{NH3(aq) + H2O(l) \rightleftharpoons NH4+(aq) + OH-(aq) }$
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H3O+ is just the combination of a a $\ce{H+}$ ion, which we know to be released from the dissociation of an acid in an aqueous solution, with a water molecule. It happens because the $\ce{H+}$ is so, so positive (and therefore so reactive) that the water molecules, with its lone pairs (which are locally negatively charged), are willing to form a dative covalent bond with the $\ce{H+}$ ion.
So like Ivan said, whenever you see an acid in an aqueous solution you can consider $\ce{H+ \equiv H3O+}$. The key there though, is that it must be in aqueous solution so that it can dissociate (and the $\ce{H3O+}$ form in the first place). That is the condition that is needed to form it. This is the same for $\ce{OH-}$, which requires an aqueous solution to form and thus release the ions in say NaOH, forming $\ce{Na+}$ and $\ce{OH-}$.
