Solving Equilibrium Problems Involving Bases

With minor modifications, the techniques applied to equilibrium calculations for acids are valid for solutions of bases in water.

Consider the calculation of the pH of an 0.10 *M* NH_{3}
solution. We can start by writing an equation for the reaction
between ammonia and water.

NH_{3}(*aq*) + H_{2}O(*l*)
NH_{4}^{+}(*aq*) + OH^{-}(*aq*)

Strict adherence to the rules for writing equilibrium constant expressions leads to the following equation for this reaction.

But, taking a lesson from our experience with
acid-dissociation equilibria, we can build the [H_{2}O]
term into the value of the equilibrium constant. Reactions
between a base and water are therefore described in terms of a **base-ionization
equilibrium constant**, *K*_{b}.

For NH_{3}, *K*_{b} is 1.8 x 10^{-5}.

We can organize what we know about this equilibrium with the format we used for equilibria involving acids.

NH_{3}(aq) |
+ | H_{2}O(l) |
NH_{4}^{+}(aq) |
+ | OH^{-}(aq) |
K_{b} = 1.8 x 10^{-5} |
||||

Initial: | 0.10 M |
0 | 0 | |||||||

Equilibrium: | 0.10 - C |
C |
C |

Substituting this information into the equilibrium constant expression gives the following equation.

*K*_{b} for ammonia is small enough to
allow us to consider the assumption that *C*
is small compared with the initial concentration of the base.

Solving this approximate equation gives the following result.

*C*
1.3
x 10^{-3}

This value of
is small enough compared with the initial concentration of NH_{3}
to be ignored and yet large enough compared with the OH^{-}
ion concentration in water to ignore the dissociation of water.
It can therefore be used to calculate the pOH of the solution.

pOH = - log (1.3 x 10^{-3}) = 2.89

Which, in turn, can be used to calculate the pH of the solution.

pH = 14 - pOH = 11.11

Equilibrium problems involving bases are relatively easy to
solve if the value of *K*_{b} for the base is
known. The first step in many base equilibrium calculations
involves determining the value of *K*_{b} for
the reaction from the value of *K*_{a} for
the conjugate acid.

As an example, let's calculate the pH of a 0.030 *M*
solution of sodium benzoate (C_{6}H_{5}CO_{2}Na)
in water from the value of *K*_{a} for
benzoic acid (C_{6}H_{5}CO_{2}H): *K*_{a}
= 6.3 x 10^{-5}.

Benzoic acid and sodium benzoate are members of a family of
food additives whose ability to retard the rate at which food
spoils has helped produce a 10-fold *decrease* in the
incidence of stomach cancer. To save time and space, we'll
abbreviate benzoic acid as HOBz and sodium benzoate as NaOBz.
Benzoic acid, as its name implies, is an acid. Sodium benzoate is
a salt of the conjugate base, the OBz^{-} or benzoate
ion.

Whenever sodium benzoate dissolves in water, it dissociates into its ions.

H_{2}O |
|||||

NaOBz(aq) |
Na^{+}(aq) |
+ | OBz^{-}(aq) |

The benzoate ion then acts as a base toward water, picking up a proton to form the conjugate acid and a hydroxide ion.

OBz^{-}(*aq*) + H_{2}O(*l*)
HOBz(*aq*) + OH^{-}(*aq*)

The base-ionization equilibrium constant expression for this reaction is therefore written as follows.

The next step in solving the problem involves calculating the
value of *K*_{b} for the OBz^{-} ion
from the value of *K*_{a} for HOBz.

The *K*_{a} and *K*_{b}
expressions for benzoic acid and its conjugate base both contain
the ratio of the equilibrium concentrations of the acid and its
conjugate base. *K*_{a} is proportional to
[OBz^{-}] divided by [HOBz], and *K*_{b}
is proportional to [HOBz] divided by [OBz^{-}].

Two changes have to made to derive the *K*_{b}
expression from the *K*_{a} expression: We
need to remove the [H_{3}O^{+}] term and
introduce an [OH^{-}] term. We can do this by multiplying
the top and bottom of the *K*_{a} expression
by the OH^{-} ion concentration.

Rearranging this equation gives the following result.

The two terms on the right side of this equation should look
familiar. The first is the inverse of the *K*_{b}
expression, the second is the expression for *K*_{w}.

Rearranging this equation gives the following result.

*K*_{a} x *K*_{b}
= *K*_{w}

According to this equation, the value of *K*_{b}
for the reaction between the benzoate ion and water can be
calculated from *K*_{a} for benzoic acid.

Now that we know *K*_{b} for the benzoate
ion, we can calculate the pH of an 0.030 *M* NaOBz solution
with the techniques used to handle weak-acid equilibria. We
start, once again, by building a representation for the problem.

OBz^{-}(aq) |
+ | H_{2}O(l) |
HOBz(aq) |
+ | OH^{-}(aq) |
K_{b} = 1.6 x 10^{-10} |
||||

Initial: | 0.030 M |
0 | 0 | |||||||

Equilibrium: | 0.030 - C |
C |
C |

We then substitute this information into the *K*_{b}
expression.

Because *K*_{b} is relatively small, we
assume that *C*
is small compared with 0.030.

We then solve the approximate equation for the value of *C*.

*C*
2.2 x 10^{-6}

The assumption that *C*
is small is obviously valid. We can therefore use *C*
to calculate the pOH of the solution.

[OH^{-}] = *C*
2.2 x 10^{-6}

pOH = - log [2.2 x 10^{-6}] = 5.66

The problem asked for the pH of the solution, however, so we use the relationship between pH and pOH to calculate the pH.

pH = 14 - pOH = 8.34

Two assumptions were made in this calculation.

- We assumed that
*C*was small enough compared with the initial concentration of the base that it could be ignored in the [0.030 -*C*] term. - We assumed that all of the OH
^{-}ion at equilibrium came from the reaction between the benzoate ion and water. (In other words, we ignored the contribution to the OH^{-}ion concentration from the dissociation of water.)

We have already confirmed the validity of the first
assumption. What about the second? The OH^{-} ion
concentration obtained from this calculation is 2.1 x 10^{-6}
*M*, which is 21 times the OH^{-} ion concentration
in pure water. According to LeChatelier's principle, however, the
addition of a base suppresses the dissociation of water. This
means that the dissociation of water makes a contribution of
significantly less than 5% to the total OH^{-} ion
concentration in this solution. It can therefore be legitimately
ignored.

Two factors affect the OH^{-} ion
concentration in aqueous solutions of bases: *K*_{b}
and *C*_{b}. **We can ignore the
dissociation of water when ***K*_{b}*C*_{b}**
for a weak base is larger than 1.0 x 10**^{-13}**.
When ***K*_{b}*C*_{b}**
is smaller than 1.0 x 10**^{-13}**, we have to
include the dissociation of water in our calculations.**

Practice Problem 5:Calculate
the HOAc, OAc |