Lecture 8 11-04

Lecture 8
November 4, 2002
Today I am going to review fundamentals of acids and bases and acid/base equilibrium
Although acids and bases can be defined in other solvents, our focus in this program is
almost exclusively on water. Thus we will talk about acids and bases in aqueous
Here a re a number of terms dealing with acids and bases.
An electrolyte is a substance that forms ions when dissolved in water to produce solutions
that conduct electricity.
Strong electrolytes
ionize essentially completely in a solvent, producing lots of ions in solution than therefore
high conductivity.
Weak electrolytes
do not dissociate completely and thus ionize only partially.
Therefore, weak electrolytes impart less conductivity per mole to a solvent than do strong
Here is a list of strong and weak electrolytes. What is a thiocyanate? - salt of an SCNanion.
This concept that substances dissociated in solution to form ions was first proposed by a
Swedish chemist, Svante Arrhenius in the 1880s. His ideas were not accepted at first. In
fact he received the lowest possible passing grade on his Ph.D. examination for proposing
such revolutionary and obviously wrong ideas.
He received the Nobel Prize for these ideas in 1903. He was also one of the first scientists
to suggest that there was a relationship between the amount of carbon dioxide in the
atmosphere and global temperature.
In 1923, two chemists, J. N. Bronsted in Denmark and J. M. Lowry in England, proposed
independently a theory of acid/base behavior that is particularly useful in analytical
According to the Bronsted-Lowry theory, an acid is a proton donor and a base is a proton
When an acid and a base react, they produce water and a salt.
In order for a species to behave as an acid, a proton acceptor (or base) must be present.
In order for a species to behave as a base, a proton donor (or acid) must be present.
In other words, a substance acts as an acid only in the presence of a base, and vice versa.
An important feature of the Bronsted-Lowry concept is the idea that the entity produced
when an acid gives up a proton is itself a potential proton acceptor and is called a
conjugate base of the parent acid.
For example, when acetic acid gives up a proton, the acetate ion is formed, as shown by
the reaction:
HOAc < == > OAc- + proton
Acidic acid is the acid and acetate is its conjugate base and the two are known as a
conjugate acid/base pair.
I tend to use the symbol OAc- for acetate which stands for CH3CO2-, or as Harris writes
Similarly, every base produces a conjugate acid as a result of accepting a proton. For
NH3 + proton < == > NH4+
When these two processes are combined, the result is an acid/base, or neutralization,
HOAc + NH3 < == > OAc- + NH4+
The extent to which this reaction proceeds depends upon the relative tendencies of the two
bases to accept a proton (or the two acids to donate a proton).
As some of you know, Bronsted-Lowry acids and bases are a subset of Lewis acids and
bases, where a base is defined as an electron pair donor and an acid an electron pair
acceptor. This is given on page 108.
Some solutes have both acidic and basic properties and are called amphiprotic species.
An example is dihydrogen phosphate ion, H2PO4-, which behaves as a base in the
presence of a strong proton donor.
H2PO4- + H3O+ < == > H3PO4 + H2O
Here, H3PO4 is the conjugate acid of the original base. In the presence of a strong proton
acceptor, such as hydroxide ion, however, H2PO4- behaves as an acid and forms the
conjugate base HPO42-.
H2PO4- + OH- < == >
HPO42- + H2O
Notice that water is an amphiprotic solvent.
HCl + H2O < == > H3O+ + ClNH2- + H2O < == > NH3 + OHAmphiprotic solvents undergo self-ionization, or autoprotolysis, to form a pair of ionic
species. Autoprotolysis is yet another example of acid/base behavior, as illustrated by
the following equations.
H2 O
+ acid2
acid1 + base2
+ H2O
< == > H3O+ + OH+ CH3OH < == > CH3OH2+ + CH3O+ NH3
< == > NH4+ + NH2-
The extent to which water undergoes autoprotolysis is slight at room temperature. Thus,
the hydronium and hydroxide ion concentrations in pure water are only about 10-7 M.
Despite the small values of these concentrations, this dissociation reaction is of utmost
importance in understanding the behavior of aqueous solutions.
This transparency shows the dissociation reaction of a few common acids in water. The
first two are strong acids because reaction with the solvent is sufficiently complete as to
leave no undissociated solute molecules in aqueous solution.
The remainder are weak acids, which react incompletely with water to give solutions that
contain significant quantities of both the parent acid and its conjugate base.
Note that acids can be cationic, anionic, or electrically neutral.
The acids in the transparency become progressively weaker from top to bottom.
Perchloric acid and hydrochloric acid are completely dissociated in water.
In contrast, only about 1% of acetic acid is dissociated.
Ammonium ion is an even weaker acid. Only about 0.01% of this ion is dissociated into
hydronium ions and ammonia molecules.
Strong acids and strong bases are discussed in Harris in section 10-1.
Common Strong bases – NaOH, KOH, LiOH
Another generality illustrated in the transparency is that the weakest acid forms the
strongest conjugate base; that is, ammonia has a much stronger affinity for protons than
any base above it. Perchlorate and chloride ions have no affinity for protons.
Acid/base equilibrium deals primarily with weak acid and weak bases that do not
dissociate completely in aqueous solutions.
When a weak acid or a weak base is dissolved in water, partial dissociation occurs. Thus,
for nitrous acid, we can write
HNO2 + H2O < == > H3O+ + NO2Ka =
Or as our book would write it:
< == > H+ + NO2-
Ka =
where Ka, is the acid dissociation constant for nitrous acid. In an analogous way, the
base dissociation constant for ammonia is
NH3 + H2O < == > NH4+ + OHKb =
As we have noted before, the concentration of water does not appear in the denominator in
either dissociation constant expression.
Thus in common with the ion-product constant for water, [H2O] is incorporated in the
equilibrium constants Ka and Kb.
Consider the base dissociation-constant expression for ammonia and the acid
dissociation-constant expression for its conjugate acid, ammonium ion:
NH3 + H2O < == > NH4+ + OHNH4+
< == > NH3 + H+
Kb =
Ka =
Multiplication of one equilibrium-constant expression by the other gives
Ka Kb
= [H+][OH-] = Kw
K w = K a Kb
This relationship is general for all conjugate acid/base pairs. Many compilations of
equilibrium-constant data list only acid dissociation constants because it is so easy to
calculate basic dissociation constants with this relationship.
We can use this equation to confirm the observation that, as the acid of a conjugate
acid/base pair becomes weaker, its conjugate base becomes stronger and vice versa.
Thus the conjugate base of an acid with a dissociation constant of 10-2 will have a basic
dissociation constant of 10-12, whereas an acid with a dissociation constant of 10-9 has a
conjugate base with a dissociation constant of 10-5.
Which is the stronger acid? Which is the stronger base?
Let's look at an example. What is Kb for the equilibrium
CN- + H2O < == > HCN + OHFirst, how do we tell that this is a base reaction?
Appendix G lists a Ka value of 6.2 x 10-10 for HCN. Thus,
Kb = Kw =
[HCN][OH-] =
1.00 x 10-14 = 1.62 x 10-5
6.2 x 10-10
If your like Dr. Stroh and think in log terms then pKa + pKb = pKw = 14.00
pKb = 14.00 – 9.21 = 4.79  Kb = 10-4.79 = 1.62 x 10-5
Suppose we want to calculate the hydrogen ion concentration in 0.120 M solution of the
weak acid, nitrous acid. The principal equilibrium is
HNO2 < == > H+ + NO2From Appendix G
Ka = 7 .1 x 10-4 = [H+][NO2-]
The chemical equation says that for each nitrite ion formed (the conjugate base) there will
be one hydrogen ion formed.
As long as we are dealing with concentrations of the acid that are large compared to the
dissociation of water at 10-7 molar, we can write
Charge balance
Mass balance
[NO2-] = [H+]
F = 0.120 M = [HNO2] + [NO2-] = [HNO2] + [H+]
Note that the text uses Formal concentrations in Chapter 10. Remember, this is the total
number of moles (in all forms) of the compound that is dissolved in a liter.
[HNO2] = 0.120 M - [NO2-] = 0.120 M - [H+]
When these relationships are introduced into the expression for Ka, we obtain
Ka =
[H+]2 _ =
0.120 - [H+]
7.1 x 10-4
If we now assume [H+] << 0.120, we find
[H+]2 = 7.1 x 10-4
[H ] = (0.120 x 7.1 x 10-4)1/2 = 9.2 x 10-3 M
We now examine the assumption that 0.120 - 0.0092  0.120 and see that the error is about
We have two significant figure data, so this is a pretty large error. We can improve the
value by using the quadratic equation or by doing another approximation.
0.120 - 0.0092 = 0.1108 and if we put that back into the square root we get
8.9 x 10-3 M for the hydrogen ion concentration which is the same answer I got using the
quadratic formula.
pH = 2.05
Let's use this example to bring in another concept from Chapter 10, that of fraction of
dissociation which is given the symbol , and is defined as the fraction of the acid in the Aform.
 =
[HA] + [A-]
= .
x + (F – x)
8.9 x 10-3 = 0.074
Thus nitrous acid is 7.4 % dissociated at a formal concentration of 0.120 M.
Now let's look at a weak base example.
Our generalized equation from chapter 6 is
B + H2O < == > BH+ + OHKb = [BH+][OH-]
Formal concentration here is F = [B] + [BH+]
Let's calculate the hydroxide ion concentration and pH of a 0.0750 M NH3 solution. The
predominant equilibrium is
NH3 + H2O < == > NH4+ + OHKb = [NH4+][OH-]
In Appendix G we find that the conjugate acid, the ammonium ion, has a Ka of 5.70 x 10-10.
from chapter 6 KaKb = Kw we get Kb = 1.00 x 10-14
5.70 x 10-10
= 1.75 x 10-5.
The chemical equation shows that
[NH4+] = [OH-]
and from the formal concentration we get
F = [NH4+] + [NH3] = 0.0750 M  [NH3] = 0.0750 M - [NH4+] = 0.0750 M - [OH-]
Substituting these quantities into the expression for Kb yields
0.0750 – [OH-]
= 1.75 x 10-5
which is analogous to the equation we got for the weak acid problem.
Assume that[OH-] << 7.50 x 10-2, then this equation simplifies to
[OH-]2 = 0.0750 x 1.75 x 10-5
[OH-] = 1.15 x 10-3 M
pOH = 2.939 
pH = 14.000 - 2.939 = 11.061
This value is less than 2% of F, 0.0750 so should be fine for most purposes. The quadratic
formula gives 1.14 x 10-3 M.
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