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 (K_b) - 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, K_b 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 K_b value of an base increases, so does the strength of the base.  

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

Relationship between K_a, K_b, and K_w

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

Example Equilibrium Calculations

1) Calculate the pH of a 0.10 M NH₃ solution.  (K_b = 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 (K_b):

SUBSTITUTE in the values for the K_b 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.  (K_a = 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 K_b expression:

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

Now we can SUBSTITUTE the values into the K_b 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"