**Solving Equilibrium
Problems Involving Weak Acids**** **

Example: Consider the process by which we would calculate the
H _{3}O^{+}, OAc^{-}, and HOAc
concentrations at equilibrium in an 0.10 *M* solution of
acetic acid in water.

We start this calculation by building a representation of what we know about the reaction.

HOAc(aq) |
+ | H_{2}O(l) |
H_{3}O^{+}(aq) |
+ | OAc^{-}(aq) |
K_{a} = 1.8 x 10^{-5} |
|||

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

Equilibrium: | ? | ? | ? |

We then compare the initial reaction quotient (*Q*_{a})
with the equilibrium constant (*K*_{a}) for
the reaction and reach the obvious conclusion that the reaction
must shift to the right to reach equilibrium.

Recognizing that we get one H_{3}O^{+} ion and
one OAc^{-} ion each time an HOAc molecule dissociates
allows us to write equations for the equilibrium concentrations
of the three components of the reaction.

HOAc(aq) |
+ | H_{2}O(l) |
H_{3}O^{+}(aq) |
+ | OAc^{-}(aq) |
K_{a} = 1.8 x 10^{-5} |
|||

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

Equilibrium: | 0.10 - C |
C |
C |

Substituting what we know about the system at equilibrium into
the *K*_{a} expression gives the following
equation.

Although we could rearrange this equation and solve it with
the quadratic formula, it is tempting to test the assumption that
*C*
is small compared with the initial concentration of acetic acid.

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

*C* 0.0013

*C* is small enough to be ignored in this
problem because it is less than 5% of the initial concentration
of acetic acid.

We can therefore use this value of *C*
to calculate the equilibrium concentrations of H_{3}O^{+},
OAc^{-}, and HOAc.

[HOAc] = 0.10 - *C* 0.10 *M*

[H_{3}O^{+}] = [OAc^{-}]
= *C* 0.0013 *M*

We can confirm the validity of these results by substituting
these concentrations into the expression for *K*_{a}.

Our calculation must be valid because the ratio of these
concentrations agrees with the value of *K*_{a}
for acetic acid, within experimental error.

**Hidden Assumptions In
Weak-Acid Calculations **

When solving problems involving weak acids, it may appear that one assumption is madethat is small compared with the initial concentration of HOAc. In fact, two assumptions are made.

The second assumption is hidden in the way the problem is set up.

HOAc(aq) |
+ | H_{2}O(l) |
H_{3}O^{+}(aq) |
+ | OAc^{-}(aq) |
K_{a} = 1.8 x 10^{-5} |
|||

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

Equilibrium: | 0.10 - C |
C |
C |

The amount of H_{3}O^{+} ion in water is so
small that we are tempted to assume that the initial
concentration of this ion is zero, which isn't quite true.

It is important to remember that there are two sources of the
H_{3}O^{+} ion in this solution. We get H_{3}O^{+}
ions from the dissociation of acetic acid.

HOAc(*aq*) + H_{2}O(*l*)
H_{3}O^{+}(*aq*) + OAc^{-}(*aq*)

But we also get H_{3}O^{+} ions from the
dissociation of water.

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

Because the initial concentration of the H_{3}O^{+}
ion is not quite zero, it might be a better idea to write "0" beneath
the H_{3}O^{+} term when we describe the initial
conditions of the reaction, as shown below.

HOAc(aq) |
+ | H_{2}O(l) |
H_{3}O^{+}(aq) |
+ | OAc^{-}(aq) |
K_{a} = 1.8 x 10^{-5} |
|||||

Initial: | 1.0 M |
0 | 0 | ||||||||

Equilibrium: | 1.0 - C |
C |
C |

Before we can trust the results of the calculation for acetic acid in the previous section, we have to check both of the assumptions made in this calculation.

- The assumption that the amount of acid that dissociates is small compared with the initial concentration of the acid.
- The assumption that enough acid dissociates to allow us to ignore the dissociation of water

We have already confirmed the validity of the first assumption. (Only 1.3% of the acetic acid molecules dissociate in this solution.) Let's now check the second assumption.

According to the calculation in the previous section, the
concentration of the H_{3}O^{+} ion from the
dissociation of acetic acid is 0.0013 *M*. The OH^{-}
ion concentration in this solution is therefore 7.7 x 10^{-12}
*M*.

All of the OH^{-} ion in this solution comes from the
dissociation of water. Since we get one H_{3}O^{+}
ion for each OH^{-} ion when water dissociates, the
contribution to the total H_{3}O^{+} ion
concentration from the dissociation of water must be 7.7 x 10^{-12}
*M*. In other words, only about 6 parts per billion of the H_{3}O^{+}
ions in this solution come from the dissociation of water.

The second assumption is therefore valid in this calculation.
For all practical purposes, we can assume that virtually none of
the H_{3}O^{+} ion in this solution comes from
the dissociation of water. As might be expected, this assumption
only fails for dilute solutions of very weak acids.

**Implications of the
Assumptions In Weak-Acid Calculations**** **

The two assumptions that are made in weak-acid equilibrium problems can be restated as follows.

- The dissociation of the acid is
*small enough*that the change in the concentration of the acid as the reaction comes to equilibrium can be ignored. - The dissociation of the acid is
*large enough*that the H_{3}O^{+}ion concentration from the dissociation of water can be ignored.

In other words, the acid must be *weak enough* that *C*
is small compared with the initial concentration of the acid. But
it also must be *strong enough* that the H_{3}O^{+}
ions from the acid overwhelm the dissociation of water. In order
for the approach taken to the calculation for acetic acid to
work, the acid has to be "just right." If it's too
strong, *C* won't be small enough to be ignored. If
it's too weak, the dissociation of water will have to be included
in the calculation.

Fortunately, many acids are "just right." To
illustrate this point, the next section will use both assumptions
in a series of calculations designed to identify the factors that
influence the H_{3}O^{+} ion concentration in
aqueous solutions of weak acids.

**Factors that Influence
the H**_{3}**O**^{+}**
Ion Concentration in Weak-Acid Solutions **

The following examples probe the
relationship between the H_{3}O^{+} ion
concentration at equilibrium and the acid-dissociation
equilibrium constant for the acid.

Practice Problem 1:Calculate
the pH of 0.10 |

As expected, the H_{3}O^{+}
ion concentration at equilibrium and
therefore the pH of the solutiondepends
on the value of *K*_{a} for the acid. **The
H**_{3}**O**^{+}**
ion concentration decreases and the pH of the solution increases
as the value of ***K*_{a}**
becomes smaller. **The next exercise shows that the H_{3}O^{+}
ion concentration at equilibrium also depends on the initial
concentration of the acid.

Practice Problem 2:Calculate
the H |

**The concentration of the H**_{3}**O**^{+}**
ion in an aqueous solution gradually decreases and the pH of the
solution increases as the solution becomes more dilute. **

The results of the previous two examples provide a basis for
constructing a model that allows us to predict when we can ignore
the dissociation of water in equilibrium problems involving weak
acids. Two factors must be built into this model: (1) the
strength of the acid as reflected by the value of *K*_{a},
and (2) the strength of the solution as reflected by the initial
concentration of the acid.

**Solving Equilibrium
Problems Involving Not-So-Weak Acids **

We need to develop techniques to handle problems for which one
or the other of our assumptions is not
valid. (Either the acid is not weak enough to ignore the value of
*C*,
or the acid is so weak we have to include the dissociation of
water in our calculations.)

In this section, we will consider acid solutions that aren't
weak enough to ignore the value of *C*. Let's start by
calculating the H_{3}O^{+}, HClO_{2}, and
ClO_{2}^{-} concentrations at equilibrium in an
0.10 *M* solution of chlorous acid (*K*_{a}
= 1.1 x 10^{-2}).

HClO_{2}(*aq*) + H_{2}O(*l*)
H_{3}O^{+}(*aq*) + ClO_{2}^{-}(*aq*)

The first step, as always, involves building a representation of the problem.

HClO_{2}(aq) |
+ | H_{2}O(l) |
H_{3}O^{+}(aq) |
+ | ClO_{2}^{-}(aq) |
K_{a} = 1.1 x 10^{-2} |
|||

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

Equilibrium: | 0.10 - C |
C |
C |

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

The value of *K*_{a} for this acid is
close enough to 1 to make us suspicious of the assumption that *C*
is small compared with the initial concentration of the acid.
There is nothing wrong with trying this assumption, however, even
if we suspect it isn't valid.

Solving this approximate equation gives a value for *C*
that is 33% of the initial concentration of chlorous acid.

*C* 0.033 *M*

The assumption that *C* is small therefore fails
miserably.

There are two ways out of this difficulty. We can expand the
original equation and solve it by the quadratic formula. Or we
can use successive approximations to solve the problem. Both
techniques give the following value of *C*
for this problem.

*C* = 0.028

Using this value of *C* gives the following results.

[HClO_{2}] |
= 0.10 - C |
= 0.072 M |

[H_{3}O^{+}] |
= [ClO_{2}^{-}] |
= C = 0.028 M |

Chlorous acid doesn't belong among the class of strong acids that dissociate more or less completely. Nor does it fit in the category of weak acids, which dissociate only to a negligible extent. Since the amount of dissociation in this solution is about 28%, it might be classified as a "not-so-weak acid."

**Solving Equilibrium
Problems Involving Very Weak Acids **

It is more difficult to solve equilibrium problems when the
acid is too weak to ignore the dissociation of water. Deriving an
equation that can be used to solve this class of problems is
therefore easier than solving them one at a time. To derive such
an equation, we start by assuming that we have a generic acid,
HA, that dissolves in water. We therefore have two sources of the
H_{3}O^{+} ion.

HA(*aq*) + H_{2}O(*l*)
H_{3}O^{+}(*aq*)
+ A^{-}(*aq*)

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

Because we get one H_{3}O^{+} ion for each OH^{-}
ion when water dissociates, the concentration of the H_{3}O^{+}
ion from the dissociation of water is always equal to the amount
of OH^{-} ion from this reaction.

[H_{3}O^{+}]_{W} = [OH^{-}]_{W}

The total H_{3}O^{+} ion concentration in an
acid solution is equal to the sum of the H_{3}O^{+}
ion concentrations from the two sources of this ion, the acid and
water.

[H_{3}O^{+}]_{T} = [H_{3}O^{+}]_{HA}
+ [H_{3}O^{+}]_{W}

We now write three more equations that describe this system. The first equation is the equilibrium constant expression for this reaction.

The second equation summarizes the relationship between the
total H_{3}O^{+} ion concentration in the
solution and the OH^{-} ion concentration from the
dissociation of water.

[H_{3}O^{+}]_{T }[OH^{-}]_{W}
= *K*_{w}

The third equation summarizes the relationship between the positive and negative ions produced by the two reactions that occur in this solution.

[H_{3}O^{+}]_{T} = [A^{-}]
+ [OH^{-}]_{W}

(This equation simply states that the sum of the positive ions formed by the dissociation of the acid and water is equal to the sum of the negative ions produced by these reactions.)

We now substitute the second equation into the third.

We then solve this equation for the [A^{-}] term.

We then substitute this equation into the equilibrium constant expression.

Rearranging this equation by combining terms gives the following result.

*K*_{a}[HA] = [H_{3}O^{+}]_{T}^{2}
- *K*_{w}

We then solve this equation for the H_{3}O^{+}
ion concentration and take the square root of both sides.

We can generate a more useful version of this equation by
remembering that we are trying to solve equilibrium problems for
acids that are so weak we can't ignore the dissociation of water.
We can therefore assume that *C *is small compared
with the initial concentration of the acid.

By convention, the symbol used to represent the initial
concentration of the acid is *C*_{a}. If *C*
is small compared with the initial concentration of the acid,
then the concentration of HA when this reaction reaches
equilibrium will be virtually the same as the initial
concentration.

[HA] *C*_{a}

Substituting this approximation into the equation derived in this section gives an equation that can be used to calculate the pH of a solution of a very weak acid.

Practice Problem 3:Calculate
the H HCN( |

**Summarizing the
Chemistry of Weak Acids**

This section compares the way in which the H_{3}O^{+}
concentration is calculated for pure water, a weak acid, and a
very weak acid.

The product of the concentrations of the H_{3}O^{+}
and OH^{-} ions in pure water is equal to *K*_{w}.

[H_{3}O^{+}][OH^{-}] = *K*_{w}

But the H_{3}O^{+} and OH^{-} ion
concentrations in pure water are the same.

[H_{3}O^{+}] = [OH^{-}]

Substituting the second equation into the first gives the following result.

[H_{3}O^{+}]^{2} = *K*_{w}

The H_{3}O^{+} ion concentration in pure water
is therefore equal to the square root of *K*_{w}.

The generic equilibrium constant expression for a weak acid is written as follows.

If the acid is strong enough to ignore the dissociation of
water, the H_{3}O^{+} ion and A^{-} ion
concentrations in this solution are about equal.

[H_{3}O^{+}] [A^{-}]

Substituting this information into the acid-dissociation equilibrium constant expression gives the following result.

The concentration of the HA molecules at equilibrium is equal
to the initial concentration of the acid minus the amount that
dissociates: *C*.

If *C* is small compared with the initial
concentration of the acid, we get the following approximate
equation.

Rearranging this equation and taking the square root of both sides gives the following result.

When the acid is so weak that we can't ignore the dissociation
of water, we use the following equation to calculate the
concentration of the H_{3}O^{+} ion at
equilibrium.

The equations used to calculate the H_{3}O^{+}
ion concentration in these solutions are summarized below.

Pure Water: | |||

Weak Acid: | |||

Very Weak Acid: |

The first and second equations are nothing more than special cases of the third. When we can ignore the dissociation of the acidbecause there is no acid in the solutionwe get the first equation. When we can ignore the dissociation of water, we get the second equation. When we can't ignore the dissociation of either the acid or water, we have to use the last equation.

This discussion gives us a basis for deciding when we can
ignore the dissociation of water. Remember our rule of thumb: we
can ignore anything that makes a contribution of less than 5% to
the total. Now compare the most inclusive equation for the H_{3}O^{+}
ion concentration

with the equation that assumes that the dissociation of water can be ignored.

The only difference is the *K*_{w} term,
which is under the square root sign.

As a rule: **We can ignore the
dissociation of water when K**_{a}**C**_{a}**
for a weak acid is larger than 1.0 x 10**^{-13}**.
When K**_{a}**C**_{a}** is
smaller than 1.0 x 10**^{-13}**, the
dissociation of water must be included in the calculation.**

Practice Problem 4:Calculate
the pH of an 0.023 |