WEAK ACIDS, WEAK BASES, AND IONIZATION CONSTANTS (15.5 15.7)
STUDY OBJECTIVES
- Write the acid ionization constant expression for any
weak acid and base ionization constant expression for any weak base.
- Calculate the concentrations of H+, A,
and undissociated weak acid HA, given Ka.
- For a weak base B in water, calculate the concentrations of OH,
H+, BH+, and B, given Kb.
- Determine the percent ionization of a weak acids and weak bases.
- Given Ka of a weak acid determine Kb
of its conjugate base, and vice versa.
Ionization Constants. Weak acids and weak
bases are ionized only to a small extent in aqueous solution. In both cases
the ionization reaction is reversible and equilibrium is established. For instance,
hydrocyanic acid ionizes in water as follows:
HCN(aq) + H2O(l) 
H3O+(aq) + CN-(aq), or simply
HCN(aq) H+(aq)
+ CN(aq)
The equilibrium constant for the ionization of a weak acid is called the acid
ionization constant, Ka. The expression
for this constant is:
Notice that the concentration of water does not appear in the equilibrium
expression as is the case for any pure liquid (or solid).
The value of Ka is experimentally determined, and
Table 15.3 (text),
Table 15.5 (text) and Table
15.3 below list Ka values for a number of weak acids.
Table 15.3 Values of Ka for Some Monoprotic Acids
|
| Formula |
Name |
Ka Value |
|
| HSO4- |
Hydrogen sulfate ion |
1.3 x 102 |
| HF |
Hydrofluoric acid |
7.1 x 104 |
| HNO2 |
Nitrous acid |
4.5 x 104 |
| HCOOH |
Formic acid |
1.7 x 104 |
| CH3COOH |
Acetic acid |
1.8 x 105 |
| H2CO3 |
Carbonic acid |
4.2 x 107 |
| HSO3- |
Hydrogen sulfite ion |
6.3 x 108 |
| HCN |
Hydrocyanic acid |
4.9 x 1010 |
| HCO3- |
Hydrogen carbonate ion |
4.8 x 1011 |
|
Methylamine is an example of a weak base. It ionizes in water as shown in
the following equation:
CH3NH2(aq) + H2O(l)
(aq)+ OH(aq)
The equilibrium constant for the ionization of a weak base is called the base
ionization constant, Kb.
Table 15.4 in the text and
Table 15.4 below list ionization constant values (Kb) for a
number of weak bases. Note that the concentration of water is not shown
in the Kb expression. The concentration of water is not
measurably affected by the reaction and is essentially a constant.
Table 15.4 Values of Kb for Several Weak Bases
|
| Formula |
Name |
Kb Value |
|
| C2H5NH2 |
Ethylamine |
5.6 x 104 |
| CH3NH2 |
Methylamine |
4.4 x 104 |
| CN |
Cyanide ion |
2.0 x 105 |
| NH3 |
Ammonia |
1.8 x 105 |
| C5H5N |
Pyridine |
1.7 x 109 |
|
EXAMPLE
Nitrous acid dissolves in water as follows:
HNO2(aq) + H2O(l)
H3O+(aq) + NO2-(aq)
and ethylamine dissolves in water as follows:
C2H5NH2(aq) + H2O
OH-(aq) + C2H5NH3+(aq)
Write the equilibrium expressions, Ka and Kb respectively,
for each reaction
The acid ionization constant, Ka, is the equilibrium constant for the ionization
of a weak acid. The equilibrium expression is written as the concontration of the products
over the concentration of the reactants. Water is not included in the equilibrium
expression.
The base ionization constant, Kb, is the equilibrium constant for the ionization
of a weak base. The equilixrium expression is written as the concentration of the products
over the concontration of the reactants. Water is not included in the equilibrium
expression.
EXAMPLE
Calculate the equilibrium concentrations of all species in a 0.10 M HNO2(aq)
solution.
Correct!
Click a Hint button for help.
First write the acid ionization contsont expression for nitrous acid. The value for
Ka can be found in Table 15.3.
HNO2(aq) + H2O(l)
H3O+(aq) + NO2-(aq)
Ka = 
= 4.5 x 10-4
The equilibrium concentrations of all species must be written in terms of a
single unknown. Let x equal the moles of HNO2 ionized per liter of solution. This
means that x is the molarity of H3O+ ion at equilibrium because
one H3O+ ion is formed for each HNO2 molecule that ionizes.
Also [H3O+] = [NO2-] = x because one NO2-
ion is formed for each hydrogen ion formed. It is helpful to tabulate the initial
and equilibrium concentrations of all species in an ICE box. The initial concentrations are
those that existed before any ionization of HNO2 occurred. The change in concentration
is x. When x moles of HNO2 have ionized per liter, x moles of H3O+
and x moles of NO2- are formed per liter. The equilibrium
concentrations are found by adding algebraically the change in concentration
to the initial concentration.
ICE box
|
| Concentration |
HNO2 |
+ |
H2O |
|
H3O+ |
+ |
NO2- |
|
| Initial (M) |
0.1 |
|
-- |
|
0 |
|
0 |
| Change (M) |
-x |
|
-- |
|
+x |
|
+x |
| Equilibrium (M) |
0.1 - x |
|
|
|
x |
|
x |
|
The value of x can be determined by substituting the equilibrium
concentration into the Ka expression.
The small magnitude of Ka indicates thot very little nitrous acid
actually ionizes. Therefore, assume the value of x is much less than 0.1,
and 0.1 - x @ 0.1. This approximation simplifies
the equation, and allows us to avoid solving a quadratic equation.
Solving for x:
The assumptionmust be thecked for validity. As a rule, if x is equal
to or less than 5% of the initial concentration of
[HNO2]0, then neglecting x in the calculation
does not introduce an unacceptable amount of error in the answer.
[HNO2] = [HNO2]0 - x and [HNO2]
@ [HNO2]0
To test the validity of the approximation, divide x by
[HNO2]0;
Since x is greater than 5 percent of the initial amount of HNO2,
the approximate solution is not valid. The equilibrium concentrations must be
determined exactly using the quadratic equation.
,
x2 = 4.5 x 10 -5 - 4.5 x 10-4x
0 = x2 + 4.5 x 10-4x - 4.5 x 10-5
The quadratic equation takes the form
,
Thus, a = 1, b = 4.5 x 10-4, and c = -4.5 x 10-5
Solving for x:
x = 6.5 x 10-3 and x = -6.9 x 10-3
Clearly, the negative root is not possible.
Substituting 6.5 x 10-3 M for x in the ICE box gives the
following
ICE box
|
| Concentration |
HNO2 |
+ |
H2O |
|
H3O+ |
+ |
NO2- |
|
| Initial (M) |
0.1 |
|
-- |
|
0 |
|
0 |
| Change (M) |
-6.5 x 10-3 |
|
-- |
|
+6.5 x 10-3 |
|
+6.5 x 10-3 |
| Equilibrium (M) |
0.0935 |
|
-- |
|
6.5 x 10-3 |
|
6.5 x 10-3 |
|
The equilibrium concentrations are tabulated in the bottom row of the ICE box.
EXAMPLE
Calculate the equilibrium concentrations of all species in a 0.25 M
C2H5NH2(aq) solution.
Correct!
Click a Hint button for help.
Write the base ionization constant expressian for ethylamine. The value of
Kb can be found in Table 15.4.
C2H5NH2(aq) + H2O(l)
OH-(aq) + C2H5NH3+(aq)
Set up an ICE box as described in Example 15.11.
ICE box
|
| |
C2H5NH2 |
+ |
H2O |
|
OH- |
+ |
C2H5NH3+ |
|
| Initial (M) |
0.25 |
|
-- |
|
0 |
|
0 |
| Change (M) |
-x |
|
-- |
|
+x |
|
+x |
| Equilibrium (M) |
0.25 - x |
|
|
|
x |
|
x |
|
With practice, you will soon realize that an approximate solution can be found when
- X is small (< 1 x 10-5) or
- [Initial]0 is large (> 0.2).
However, always check any assumption made.
Substituting the equilibrium concentrations into the Kb expressian and
making the assumption thot x << 0.25 gives:
Solving for x:
Check assumption:
,
therefore, assumption is valid.
Substituting 1.18 x 10-2 M gor x in the ICE box gives.
| [C2H5NH2] |
= 0.238 M |
| [OH-] |
= 0.012 M |
| [C2H5NH3+] |
= 0.012 M |
Finally, the [H3O+] is determined from Kw =
[H3O+][OH-].
Acid and Base Strength. The Ka
value is the quantitative measure of acid strength. The larger the Ka,
the greater the extent of the ionization reaction and the stronger the acid.
Therefore, acetic acid with Ka = 1.8 x 105 is
a stronger acid than hydrocyanic acid with Ka = 4.9 x
1010. The acids listed in Table 15.3 above are arranged in order
of decreasing acid strength as you read down in the table. The Ka
values for strong acids are never tabulated.
Strong acids are essentially 100
percent ionized. It will help you to be able to recognize 5 or 6 strong acids
by their names and formulas, so that you won't confuse them with weak acids.
EXAMPLE pH of a Weak Base Solution
What is the pH of a 0.010 M C5H5N (pyridine) solution?
pH =
Correct!
Click a Hint button for help.
Pyridine is a weak base. Write the equation for the reaction of pyridine in
water and look up its Kb value in Table 15.4.
C5H5N + H2O 
C5H5NH+ + OH
Kb = 1.7 x 109
Let x equal the moles of C5H5N ionized per L to reach
equilibrium. Note, when x mol/L of C5H5N are ionized,
x mol/L of OH and x mol/L of C5H5N+
are formed. These changes are listed in the ICE box:
|
| |
C5H5N |
+ |
H2O |
|
C5H5NH+ |
+ |
OH |
|
| Initial (M) |
0.010 |
|
-- |
|
0 |
|
0 |
| Change (M) |
-x |
|
-- |
|
+x |
|
+x |
| Equilibrium (M) |
(0.010 - x) |
|
-- |
|
x |
|
x |
|
Calculation
Substitute the equilibrium concentrations into the base constant expression.
Make the approximation [C5H5N] = 0.010 x 
0.010 M. Substituting
x = 4.1 x 106 M
Testing the assumption that the approximation is reasonable
Since x is less than 5% of the initial concentration of pyridine, then the
assumption is valid.
At equilibrium:
[OH] = x = 4.1 x 106 M
To find the pH, first calculate the pOH, and then use the relationship pH +
pOH = 14.
pOH = log( 4.1 x 106) = 5.38
Since, pH = 14.00 - pOH
Then, pH = 14.00 5.38
pH = 8.62
Percent Ionization. The strength of an acid
is also reflected in its percent ionization. The percent ionization is defined
as follows:
For a weak acid HA, the concentration of the acid that is ionized is equal
to the concentration of H3O+ ions (or A
ions) at equilibrium that comes from HA. HA(aq) + H2O
H3O+(aq) + A(aq)
Be sure to work through Examples 15. and 15. because they illustrate the
problem-solving techniques used to calculate [H+], [OH],
pH, and percent ionization of weak acids and bases.
EXAMPLE Percent Ionization of a Weak Acid
- Calculate the concentrations of H3O+, F, and
HF in a 0.31 M HF (hydrofluoric acid) solution.
Correct!
Click a Hint button for help.
- What is the percent ionization?
% ionization =
Correct!
Click a Hint button for help.
First write the ionization reaction and the acid ionization constant expression
for hydrofluoric acid. The value for Ka can be found in
Table 15.3.
HF(aq) + H2O(l) H+(aq)
+ F(aq)
Set up an ICE bok as described in new Example 15.11.
ICE box
|
| |
HF |
+ |
H2O |
|
H+ |
+ |
F |
|
| Initial (M) |
0.31 |
|
-- |
|
0 |
|
0 |
| Change (M) |
-x |
|
-- |
|
+x |
|
+x |
| Equilibrium (M) |
(0.31 - x) |
|
-- |
|
x |
|
x |
|
Calculation
The value of x can be determined by substituting the equilibrium concentrations
into the Ka expression.
The small magnitude of Ka indicates that very little hydrofluoric
acid actually ionizes. After all, it is a weak acid. Therefore, the value of
x is much less than 0.31, and 0.31 x @ 0.31 M. This approximation simplifies
the equation, and allows us to avoid solving a quadratic equation.
Solving for x:
Check assumption:
, therefore,
assumption is valid.
Substituting 1.5 x 10-2 M for x in the ICE box gives
[H+] = [F] = 1.5 x 102 M
[HF] = 0.31 M 0.015 M = 0.30 M
The percent ionization was already calculated above.
In Examples 15.11-13 we have followed three basic steps common to all problems
of this type.
| 1. |
Write the acid dissociation equilibrium and express the equilibrium concentration
of all species in terms of an unknown (x), where x equals the change in the
concentration of HA that is required to reach equilibrium. |
| 2. |
Substitute these concentrations into the Ka expression
and look up the value of Ka in a table. |
| 3. |
Solve for x, and calculate all equilibrium concentrations, the pH, and percent
ionization. |
The Kb
of a Base is Related to the Ka of Its
Conjugate Acid. The Ka and Kb
for any conjugate acid-base pair are always related by the equation:
Ka Kb = Kw
The Kb for a conjugate base of a weak acid is found by
rearranging:
From this equation you can see that Ka decreases, Kb
increases, as the weaker the acid, the stronger its conjugate base
a relationship we have noted before.
Fluoride ion is a base. What is its Kb value?
F(aq) + H2O(l)
HF(aq) + OH(aq) Kb = ?
The base ionization constant Kb for F
is related to the Ka for the conjugate acid HF by:
Substituting Ka for HF gives
OBJECTIVE CHECK
Complete the following questions to check your understanding of the material.
Select the check button to see if you answered correctly.
- What is the percent ionization of a weak acid in a 0.020 M HA solution
given that its Ka is 2.3 x
105?
- What is the pH of a 0.050 M C6H5COOH (benzoic acid)
solution?
- HA is a weak acid. If a 0.020 M HA solution is 2.5% dissociated, what is
the Ka value of the weak acid?
- A 0.100 M solution of chloroacetic acid, ClCH2COOH, has a pH
of 1.95. Calculate the Ka for chloroacetic acid.
- A 0.012 M solution of an unknown base has a pH of 10.10.
What are the hydroxide ion and hydronium ion concentrations in the solution?
Is the base a weak base or a strong base?
- Which of the following is the strongest acid?
HF
CH3COOH
C6H5COOH
- Which if the following solutions has the lowest pH?
0.10 M HNO2
0.10 M CH3COOH
0.10 M HCN
- Determine accurately the percent ionization of formic acid, HCOOH, in a
0.0050 M solution.
