Why are acids characterized by giving hydrogen and bases characterized by receiving it when acid and base reactions are double displacement reactions?

Question

One example of this is $\ce{HCl}$ and $\ce{KOH}$. Why would the acid not just give the hydrogen ion in $\ce{HCl}$ to form $\ce{Cl-}$ and $\ce{KOH2+}$?

Answer 1

This is a good question. First, we must be cognizant of what actually exists. If we don't know what actually exists then any chemistry we attempt to do is just a shot in the dark.

If we are talking about solutions, then $\ce{HCl}$ doesn't exist; there isn't much to any unionized hydrochloric acid in the system. Thus it is misleading to write $\ce{HCl_{(aq)}}$, at least for introductory chemistry students, as one might take this to mean there is unionized hydrochloric acid that is solvated by water.

What actually exists in solutions of strong acids (apart from the counterion) is at the bare minimum the hydronium ion, $\ce{H_3^+O}$. The bare hydrogen proton, $\ce{H^+}$ does not exist by itself; it's too reactive. Recent studies have also suggested that the $\ce{H_5^+O_2}$ molecule exists in acidic solution in equal quantity as $\ce{H_3^+O}$. And of course other waters of hydration may exist due to the sheer positive charge density of a bare proton.

Speaking of $\ce{KOH}$, this is a very soluble salt, and unless we have a saturated solution of $\ce{KOH}$, we generally won't have any unionized $\ce{KOH}$. What we do have however in solution is $\ce{K^+}$ and $\ce{HO^-}$.

Therefore, I suggest that before trying to predict the products of any chemical reactions, write out a complete list of what you have in your system - a principle system inventory. Ignore the very minor species such as the higher waters of hydration, such as the $\ce{H_9^+O_4}$ molecule - this likely won't be a big deal as far as introductory-level calculations go, and scientists have yet to even fully characterize the solvation of the hydrogen proton in water themselves. Only upon realizing what's really in the system can we do accurate chemistry.

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Now, if we were dealing with pure, liquid $\ce{HCl}$ and solid $\ce{KOH}$, then what we have is simply $\ce{KOH}$ and $\ce{HCl}$. There is nothing to dissolve the potassium hydroxide and no medium for the pure hydrochloric acid to completely ionize in.

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Speaking specifically of why $\ce{KOH2+}$ doesn't exist or isn't stable: Note that its components are the $\ce{K^+}$ and $\ce{HO^-}$ ions. Note the opposite charges on these two ions. Remember Coulomb's law about the attraction between electrically charged particles. The opposite charges give $\ce{KOH}$ a substantial lattice energy, allowing it to exist as an ionic solid. If you were to protonate the $\ce{HO^-}$ part of the molecule, however, you'd rob the $\ce{HO^-}$ part of its negative charge. This substantially diminishes the lattice energy of the solid, so the solid wouldn't be stable anymore. Not to mention that this protonation also creates water, and $\ce{KOH}$ is soluble in water.

Answer 2

$\ce{KOH}$ does not exist as a molecule. In water it is completely dissociated as $\ce{K+}$ and $\ce{OH-}$. It does not make sense to write $\ce{KOH_2^+}$ for this reason.

Answer 3

You have the right idea - many simple acid/base reactions (using the Arrhenius definition) follow the same pattern as a double-displacement reaction in aqueous solution.

The generalized pattern for a double displacement reaction is:

$\ce{AB + CD -> AD + CB}$

Typically, ionic double displacement reactions are driven by the formation of a precipitate:

$\ce{AB_{(aq)} + CD_{(aq)} -> AD_{(aq)} + CB_{(s)}v}$

In this example, the compund $\ce{CB}$ is insoluble in water, and so it precipitates out of solution (indicated by the arrow pointing downward). The reaction goes forward because the $\ce{C}$ ions and $\ce{B}$ ions leave the solution as soon as they meet each other.

In an Arrhenius acid/base reaction, we see a similar pattern:

$\ce{HA_{(aq)} + BOH_{(aq)} -> BA_{(aq)} + H_2O_{(l)}}$

Here, the formation of water drives the reaction - as protons meet hydroxide ions, they form water.

You can see that the pattern is the same, which is good! However, pay careful attention to the details of the pattern:

$\ce{HA}$ corresponds to $\ce{HCl}$, and $\ce{BOH}$ corresponds to $\ce{KOH}$.

In other words, $\ce{A=Cl}$ and $\ce{B=K}$.

Putting those into the pattern above, we get:

$\ce{HCl_{(aq)} + KOH_{(aq)} -> KCl_{(aq)} + H_2O_{(l)}}$

It helps to always think about these as ions when in aqueous solution (as @DavePhD mentioned). For this reason, these reaction patterns are typically taught along with solubility and dissociation of ionic compounds in aqueous solution. If you get in the habit of mentally (or explicitly on paper) dissociating all strong electrolytes, the patterns are much easier to understand.

For example:

This is the ionic equation for the general precipitation reaction pattern:

$\ce{A^+_{(aq)} +B^{-}_{(aq)} + C^+_{(aq)} +D^{-}_{(aq)} -> A^+_{(aq)} +D^{-}_{(aq)} + CB_{(s)}}$

And this is the net ionic equation:

$\ce{B^{-}_{(aq)} + C^+_{(aq)} -> CB_{(s)}}$

For acid/base reactions (only when both are strong):

$\ce{H^+_{(aq)} +A^{-}_{(aq)} + B^+_{(aq)} +OH^{-}_{(aq)} -> B^+_{(aq)} +A^{-}_{(aq)} + H_2O_{(l)}}$

$\ce{H^+_{(aq)} + OH^{-}_{(aq)} -> H_2O_{(l)}}$