# What does an acid + a base create?

Mar 28, 2016

Acid + base = salt + water

#### Explanation:

This is called a neutralisation reaction. Acids and bases react, if they are in the right proportions, to neutralise one another.

For example:

$H C l + N a O H \to N a C l + {H}_{2} O$

in this case the salt is sodium chloride, common table salt, but that's not always what happens:

$H N {O}_{3} + K O H \to K N {O}_{3} + {H}_{20}$

In this case nitric acid reacts with potassium hydroxide to give potassium nitrate (the salt). You can see that in both cases, an acid plus a base has given a salt plus water.

Mar 28, 2016

In general, the reaction forms the conjugate base of the acid and the conjugate acid of the base.

GENERIC ACID/BASE REACTION

If we suppose we had a generic acid $\text{HA}$ and a generic base $\text{B}$, then regardless of acid strength, we would get:

stackrel("acid")overbrace("HA") + stackrel("base")overbrace("B") rightleftharpoons stackrel("conjugate acid")overbrace("BH"^(+)) + stackrel("conjugate base")overbrace("A"^(-))

where the equilibrium arrows are skewed towards the products if the $\text{pKa}$ of $\text{HA}$ is lower than that of ${\text{BH}}^{+}$, or skewed towards the reactants if the $\text{pKa}$ of $\text{HA}$ is higher than that of ${\text{BH}}^{+}$.

As you can see, we have that:

• The conjugate base of an acid is the acid with one less proton.
• The conjugate acid of a base is the base with one more proton.

And if you aren't familiar, the $\text{pKa}$ is the negative base 10 logarithm of the ${\text{K}}_{a}$, and the ${\text{K}}_{a}$ is the acid dissociation constant. The higher the ${\text{K}}_{a}$, the more easily the acid loses its proton.

HOW DO I KNOW WHICH WAY IT'S SKEWED?

Remember (or recognize) that the equilibrium lies on the side of the weaker acid, i.e. the acid with the stronger bond with the proton (${\text{H}}^{+}$). You can tell what the relative acid strength is by knowing that the acid with the higher $\text{pKa}$ is the weaker acid.

For instance, ${\text{NH}}_{4}$, ammonium, has a $\text{pKa}$ of about $9.4$. It is a weaker acid than ${\text{HC"_2"H"_3"O}}_{2}$, or acetic acid, which has a $\text{pKa}$ of about $4.74$, so if it was this type of equilibrium, then it would lie on the side of ammonium like so:

stackrel("acid")overbrace("HC"_2"H"_3"O"_2) + stackrel("base")overbrace("NH"_3) -> stackrel("conjugate acid")overbrace("NH"_4^(+)) + stackrel("conjugate base")overbrace("C"_2"H"_3"O"_2^(-))
$\text{pKa" ~~ "4.74" color(white)(--a--) "pKa" ~~ "9.4}$

What happens is that since the $\text{pKa}$ of ammonium is significantly higher than that of acetic acid, the ammonia (${\text{NH}}_{3}$) is easily capable of donating its nitrogen's two valence electrons to take a proton from acetic acid (${\text{HC"_2"H"_3"O}}_{2}$) and form ammonium, so it does.

A trick to figure out to what extent is to do the calculation

color(blue)((["BH"^(+)])/(["HA"]) = 10^("pKa,product acid" - "pKa,reactant acid")).

From that you would figure out that this reaction is favored by about
$\setminus m a t h b f \left(45000\right)$ to $\setminus m a t h b f \left(1\right)$ on the products side (the concentration of ammonium is over $45000$ times that of acetic acid), since ${10}^{9.4 - 4.74} = \approx 45708.8$.