Question

What does an acid + a base create?

Answer 1

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:

`HCl + NaOH to NaCl + H_2O`

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

`HNO_3 + KOH to KNO_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.

Answer 2

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 `"HA"` and a generic base `"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 `"pKa"` of `"HA"` is lower than that of `"BH"^(+)`, or skewed towards the reactants if the `"pKa"` of `"HA"` is higher than that of `"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 `"pKa"` is the negative base 10 logarithm of the `"K"_a`, and the `"K"_a` is the acid dissociation constant. The higher the `"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 (`"H"^(+)`). You can tell what the relative acid strength is by knowing that the acid with the higher `"pKa"` is the weaker acid.

For instance, `"NH"_4`, ammonium, has a `"pKa"` of about `9.4`. It is a weaker acid than `"HC"_2"H"_3"O"_2`, or acetic acid, which has a `"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^(-))`
`"pKa" ~~ "4.74" color(white)(--a--) "pKa" ~~ "9.4"`

What happens is that since the `"pKa"` of ammonium is significantly higher than that of acetic acid, the ammonia (`"NH"_3`) is easily capable of donating its nitrogen's two valence electrons to take a proton from acetic acid (`"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
`\mathbf(45000)` to `\mathbf(1)` on the products side (the concentration of ammonium is over `45000` times that of acetic acid), since `10^(9.4-4.74) = ~~45708.8`.