What does an acid + a base create?

2 Answers
Mar 28, 2016

Answer:

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.

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 #"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#.