Acid-Base Equilibria

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Acid-Base Equilibria
Chapter 17
The simplest acid–base equilibria are those in which a weak acid or
a weak base reacts with water.
We can write an acid equilibrium reaction for the generic acid,
HA.
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
Acetic acid is a weak acid. It reacts with water as follows:
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2-(aq)
Here A from the previous slide = C2H3O2-(aq)
The equilibrium constant for the reaction of a weak acid with
water is called the acid-ionization constant (or aciddissociation constant), Ka.
Ka

H

O
3

A 

HA 
Liquid water is not included in the equilibrium constant
expression.
In order of decreasing Ka:
HF
Ka
6.8 × 10-4
Highest pH
HCN
HNO2
HC2H3O2
4.5 × 10-4
HC2H3O2
1.7 × 10-5
HNO2
HCN
4.9 × 10-10
HF
Lowest pH
Calculations with Ka
Given the value of Ka and the initial concentration of HA, you can
calculate the equilibrium concentration of all species.
Given the value of Ka and the initial concentration of HA, you can
calculate the degree of ionization and the percent ionization.
Given the pH of the final solution and the initial concentration of
HA, you can find the value of Ka and the percent ionization.
We can be given pH, percent or degree ionization, initial
concentration, and Ka.
From pH, we can find [H3O+].
From percent or degree ionization, we can find Ka.
Using what is given, we can find the other quantities.
Sore-throat medications sometimes contain the weak acid
phenol, HC6H5O. A 0.10 M solution of phenol has a pH of 5.43
at 25°C.
a. What is the acid-ionization constant, Ka, for phenol at 25°C?
b. What is the degree of ionization?
HC6H5O(aq) + H2O(l)
H3O+(aq) + C6H5O-(aq)
Initial
Change
Equilibrium
We were told that pH = 5.43. That allows us to find
[H3O+] =
= x = [C6H5O-].
Now we find [HC6H5O] = 0.10 – x =
Finally, we write the expression for Ka and substitute the
concentrations we now know.
The degree of ionization is the ratio of ionized concentration
to original concentration:
Degree of ionization
x

0.100
Degree of ionization

3.7  10
6
0.10
 3.7  10
5
Percent ionization is the degree of ionization × 100%:
Percent ionization = 3.7 x 10-3% or 0.0037%
Simplifying Assumption for Acid and Base Ionizations
The equilibrium concentration of the acid is most often ([HA]0 – x).
If x is much, much less than [HA]0, we can assume that subtracting x
makes no difference to [HA]:
([HA]0 – x) = [HA]0
This is a valid assumption when the ratio of [HA]0 to Ka is > 103. If it is
not valid, you must use the quadratic equation to solve the problem.
Para-hydroxybenzoic acid is used to make certain dyes. What are
the concentrations of this acid, of hydronium ion, and of parahydroxybenzoate ion in a 0.200 M aqueous solution at 25°C?
What is the pH of the solution and the degree of ionization of the
acid? The Ka of this acid is 2.6 × 10-5.
We will use the generic formula HA for para-hydroxybenzoic
acid and the following equilibrium:
HA + H2O
H3O+ + A-
HA(aq) +
Initial
Change
Equilibrium
H2O(l)
H3O+(aq) +
A-(aq)
Polyprotic Acids
A polyprotic acid has more than one acidic proton—for example,
H2SO4, H2SO3, H2CO3, H3PO4.
These acids have successive ionization reactions with Ka1, Ka2, . . .
The next example illustrates how to do calculations for a polyprotic
acid.
Tartaric acid, H2C4H4O6, is a diprotic acid used in food products.
What is the pH of a 0.10 M solution? What is the concentration of
the C4H4O62 ion in the same solution?
Ka1 = 9.2  104; Ka2 = 4.3  105.
First, we will use the first acid-ionization equilibrium to find
[H+] and [HC4H4O6-]. In these calculations, we will use the
generic formula H2A for the acid.
Next, we will use the second acid-ionization equilibrium to
find [C4H4O62-].
H2A(aq) +
Initial
Change
Equilibrium
H2O(l)
H3O+(aq) +
HA-(aq)
We eliminate
physically
the negative
impossible
value because
to have a negative
At the end of the first acid ionization
concentrat
Now we
ionization
it is
concentrat
equilibriu m, the
ions are
use these for the second acid equilibriu m.
ion.
HA-(aq) +
Initial
Change
Equilibrium
H2O(l)
H3O+(aq) +
A2-(aq)
Base-Ionization Equilibrium
The simplest acid–base equilibria are those in which a weak acid or a
weak base reacts with water.
We can write a base equilibrium reaction for the generic base, B.
B(aq) + H2O(l)
HB+(aq) + OH-(aq)
Ammonia is a weak base. It reacts with water as follows:
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
Here the generic B from the previous slide is NH3(aq).
The equilibrium constant for the reaction of a weak base with water
is called the base-ionization constant, Kb.

HB O H 


Kb

B 
Liquid water is not included in the equilibrium constant expression.

Writing Kb Reactions
The bases in Table 16.2 (previous slide) are nitrogen bases; that is, the proton
they accept adds to a nitrogen atom. Next we’ll practice writing the Kb
reactions.
Ammonium becomes ammonium ion:
NH3+ H2O  NH4+ + OHEthylamine becomes ethyl ammonium ion:
C2H5NH2 + H2O  C2H5NH3+ + OH-
Dimethylamine becomes dimethylammonium ion:
(CH3)2NH2 + H2O  (CH3)2NH3+ + OHPyridine becomes pyridinium ion:
C5H5N+ H2O  C5H5NH+ + OHHydrazine becomes hydrazinium ion:
N2H4+ H2O  N2H5+ + OH-
We can be given pH, initial concentration, and Kb.
From the pH, we can find first [H3O+] and then [OH-]. Using what is
given, we can find the other quantities.
We can also use a simplifying assumption: When [B]0 / Kb > 103, the
expression ([B]0 – x) = [B]0.
Aniline, C6H5NH, is used in the manufacture of some perfumes.
What is the pH of a 0.035 M solution of aniline at 25°C?
Kb= 4.2 × 10-10 at 25°C.
We will construct an ICE chart and solve for x.
C6H5H(aq) + H2O(l)
C6H5NH+(aq) + OH-(aq)
Initial
Change
Equilibrium
We are told that Kb = 4.2 × 10-10. That allows us to substitute into the
Kb expression to solve for x.
The question asks for the pH:

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