Equilibrium :  Base-Ionization Equilibrium Constant

Base equilibrium problems can be solved much like acid equilibrium problems.  If you have not already looked at the Acid-Dissociation Equilibrium Constant page, it would probably be a good idea to do so before proceeding on this page.

Base-ionization equilibrium constant (Kb) - A measure of the relative strength of a base.  For the generic base dissociation reaction with water,

The base-ionization equilibrium constant is the mathematical product of the equilibrium concentrations of the products of this reaction divided by the equilibrium concentration of the original base,  In other words, Kb is the product of the concentrations of the products over the concentration of the reactants (not including water):

Water is not included in the base-ionization equilibrium expression because the [H2O] has no effect on the equilibrium.

As the Kb value of an base increases, so does the strength of the base.

• strong base: Kb > 1
• weak base: Kb < 1

Relationship between Ka, Kb, and Kw

Therefore, Ka can be found if Kb is known and Kb can be found if Ka is known.  Also, compounds with large Ka values will have small Kb values which makes sense because strong acids are extremely weak bases.  Likewise, compounds with large Kb values have small Ka values because strong bases are extremely weak acids.

Example Equilibrium Calculations

1) Calculate the pH of a 0.10 M NH3 solution.  (Kb = 1.8 x 10-5)

Just like in the acid problems, begin by setting up the REACTION:

Note that there is a forward/backward arrow, indicating that the base ionizes only slightly since this is a weak base.

Next set up the concentration CHART with the initial and equilibrium concentrations of the base and its ions:

 [NH3] [NH4+] [OH-] Initial: 0.10 0 0 Equilibrium: 0.10 -

Next, write the BASE-IONIZATION EQUILIBRIUM EXPRESSION (Kb):

SUBSTITUTE in the values for the Kb expression:

ASSUME that is negligible when subtracted from the initial concentration of the base:

SOLVE for :

CHECK the assumption to make sure that less than 5% of the base actually ionizes.

Since less than 5% of the base ionized in water, our assumption was valid.

So according to our chart, = [OH-].  The problem, however, asks for the pH of the solution, so we must eventually arrive at a pH value.  There are two ways to get to this answer.  One way is to convert the [OH-] to [H3O+] by using the following equation:

Another way to find the pH is to find the pOH first and then convert to pH from there:

Either way, we ultimately arrive at the same answer.

Acid Salt

2) Calculate the pH of a 0.030 M NaC6H5O2 solution.  (Ka = 6.3 x 10-5 for HC6H5O2).

First, let's begin by writing the reaction for the dissociation of the acid salt in water, noting the concentrations of the ions formed:

• Since the mole ratio of the acid salt to its respective ions is 1:1, the acid salt dissociates to give equimolar amounts of each ion equal to the concentration of the acid salt.

The anion formed in the above reaction can act as a H+ acceptor (base) in water.  We can write the REACTION for the ionization of this base in water:

Set up a CHART with the concentrations of the base and the ions it forms in the reaction with water:

Write the Kb expression:

The next step would be to substitute the given Kb value and the values from the chart into the Kb expression.  However, the value given in the problem was a Ka value for the conjugate acid of the base with which we are working.  We can use the relationship between Ka, Kb, and Kw to calculate Kb:

Now we can SUBSTITUTE the values into the Kb expression:

ASSUME that is negligible when subtracted from the initial concentration of the base:

SOLVE for :

CHECK to make sure that the assumption was valid:

Since less than 5% of the initial base concentration ionized, our assumption was valid.

According to our chart, = [OH-].  However, we were asked to find pH so we must eventually arrive at a pH value:

Next:  "Buffers"