Molecules and ions that differ by one proton (i.e., H
+
) are referred to as a conjugate acidbase pairs. e.g., NH
3
+ H
2
O

=====

NH
4
+ + OH -
(Base) (Acid) (Con. Acid) (Con. Base)
The conjugate acid-base pairs are: NH
3
/NH
4
+
and H
2
O/OH
-
.
Hint: Look for related atoms between the reactants and the products. Both NH
3
and
NH
4
+
contain nitrogen and the other substances do not.
Q: So what is a conjugate acid (i.e., con. Acid)?
A: A substance that has one additional H
+
in the formula.
Practice: Determine the con. acids for the following:
1.
HS
-
2.
S -2
3.
CN
-
4.
PO
4
-3 the con. acid is H
2
S the con. acid is HS the con. acid is HCN the con. acid is HPO
4
-2
Q: What is a conjugate base?
A: A substance that has one less H + in the formula.
Practice: Determine the con. base for the following:
1. H
2
SO
4 the con. base is HSO
4
-
2. H
2
PO
4
the con. base is HPO
4
3. H
3
PO
4 the con. base is H
2
PO
4
-2
-
4. O
-2 there is no con. base b/c there are no H
+
ions to remove!
The importance of being able to determine conjugate acid-base pairs seems pretty trivial, but the importance of this identification exercise will become clear later in the chapter when we begin calculation style questions.

Practice: In the following equilibrium expressions, identify the acid/base reactants and the conjugate acid/base products.
1. HCN + H
2
O

==========

H
3
O
+
+ CN
-
(Acid) (Base) (Con. Acid) (Con. Base)
Make sure the correct charge is recorded on each species.
Another way to identify Con. Acids/Bases on the product side of the equation is by considering the direction of proton transfer for the reverse reaction. Notice that the H
3
O
+ must donate a proton to CN
-
for the reverse reaction to work.
Q: Identify the two conjugate acid/base pairs.
A: HCN/CN
-
and H
3
O
+
/H
2
O
2. HCO
3
-
+ H
2
PO
4

=========

H
2
CO
3
+ HPO
4
-2
(Base) (Acid) (Con. Acid) (Con. Base)
H
2
CO
3
/HCO
3
-
and H
2
PO
4
-
/HPO
4
-2
are the two con. acid/base pairs.
3. HC
2
O
4
-
+ CH
3
COO

======

C
2
O
4
-2
+ CH
3
COOH
(Acid) (Base) (Con. Base) (Con. Acid)
HC
2
O
4
-
/C
2
O
4
-2
and CH
3
COOH/CH
3
COO
-
are con. acid/base pairs.

What’s the difference between a strong acid and a weak acid?
Common misconceptions : - strong acids corrode metals faster
- the one with higher concentration
Correct Answer : Strong acids fully dissociate and weak acids form equilibria.
Strong Acid - An acid that fully dissociates in water. e.g., HCl

H
+
+ Cl
-
If 1.0M of HCl dissolves in solution, we know that 1.0M H
+
and 1.0M Cl
-
is produced because all of the HCl dissociates. e.g., HClO
4
, HBr, HNO
3
or any acid that fully dissociates (equations with a single arrow)
Note: the conjugate bases of strong acids are so weak that they cannot react with H
2
O to form OH
-
, for this reason, the top 5 on the right of the chart are not usually referred to as
“bases” in aqueous solution. They are called
Spectators .
Weak acid – forms an equilibrium in water (does not fully dissociate, usually <5%). e.g., HF

H
+
+ F
-
If 1.0M of HF is placed into solution, we do not know what the concentrations of the H
+ and F
-
will be except that the values will be less that 1.0M. Remember that we need to have some reactant in order to maintain equilibrium.
Look at the Ka values on page 6 of the data booklet. Ka is the same idea as Keq except these values all make reference to acids.
Q: Which acids have a Ka value listed?
A: The weak acids (i.e., those in equilibrium).
Q: What is similar between all the Ka values listed?
A: All values are small (i.e., less than 1.0)
Q: What do the Ka values communicate about the relative concentrations of reactants vs. products.
A: The reactants are favoured because Ka < 1.0, which means that we make very little product when we have a weak acid.
Q: The arrow that goes up page 6 (of the data booklet) on the left hand side indicates that acid strength increases as you move up the table. What happens to the value of Ka as you move up the table?
A: The Ka value increases as you move up the table. In other words, we can compare the relative strengths of weak acids by looking at the Ka values.

Strong Base – Any base which fully dissociates in water to produce OH
-
(hydroxide ion),
O
-2
(oxide ion), or NH
-2
(amide ion) and contain a cation from group #1/#2 of the periodic table. e.g., NaOH  Na + + OH -
CaO

Ca
+2
+ O
-2
; O
-2
+ H
2
O

2OH
-
NaNH
2

Na
+
+ NH
-2
; NH
2-
+ H
2
O

NH
3
+ OH
-
Note: the conjugate acids of strong bases are so weak that they cannot react with H
2
O to form H
3
O + , for this reason, the bottom 2 on the left of the chart are not usually referred to as “acids” in aqueous solution.
The Levelling Effect – states that all strong acids (or strong bases) are equally strong because they all dissociate completely. e.g., compare the concentration of H
+
/ OH
-
for the equations below.
HCl -----------------

H
+
+ Cl
-
1.0M
NaOH ----------------

Na
+
+ OH
-
1.0M
CaO + H
2
O --------

Ca + 2OH
-
(CaO

Ca
+2
+ O
-2
; O
-2
+ H
2
O

2OH
-
)
1.0M
Weak Base – A base that forms an equilibrium in water.
Predict the products for the following reactions
1. HCO
3
-
+ HF

=========

? + ?
Step #1: Determine the relative strength of each acid
We can do this two ways. We can select the acid that is highest on the chart on page 6 or we can compare the Ka values. Find the Ka’s for each acid.
HF
HCO
3
-
Ka = 3.5 x 10
Ka = 5.6 x 10
-4
-11
Since the Ka for HF is larger than HCO
3
, HF is the strongest acid (i.e., HF is better at donating protons than HCO
3
-
is) and HCO
3
-
is forced to accept the proton.

Step #2: Predict the products
HCO
3
-
+ HF

=========

H
2
CO
3
+ F
-
Step #3: Label each substance as an acid, base, con acid or con base
HCO
3
+ HF  =========  H
2
(base) (Acid)
CO
3
+ F -
(Con Acid) (Con Base)
Step #4: Predict whether reactants or products are favoured. We do this by comparing the relative strengths of the acid (i.e., HF) and the con acid (i.e., H
2
CO
3
).
HF
H
2
CO
3
Ka = 3.5 x 10
-4
Ka = 4.3 x 10 -7
Since the Ka for HF is greater than H
2
CO
3
, we can conclude that the products are favoured because the stronger acid “pushes” the equilibrium to the opposite side. (i.e.,
HF is more efficient at donating it’s proton, so the forward reaction is preferred).
2. HF + H
2
O

=========

? + ?
HF Ka = 3.5 x 10
-4
H
2
O Ka = 1.0 x 10 -14
HF + H
2
O

=========

F
-
+ H
3
O
+
(Acid) (Base)
HF Ka = 3.5 x 10
-4
H
3
O
+
Ka = 1.0
(Con Base) (Con Acid)
Since the Ka for H
3
O
+
is larger than HF, we predict that the reactants are favoured.
3. HClO
4
+ H
2
PO
4

=========

? + ?
HClO
4 is a strong acid and H
2
PO
4
-
is weak, so HClO
4 is automatically the acid and forces
H
2
PO
4
-
to behave like a base.
HClO
4
+ H
2
PO
4

=========

ClO
4
-
+ H
3
PO
4
(Acid) (Base) (Con Base) (Con Acid)
Since HClO
4
is a strong acid and H
3
PO
4
is weak, the equilibrium will favour products.
As a rule, the strongest acid always reacts with the strongest base and the weakest acid always reacts with the weakest base.

Q: In #1, which base is the strongest?
HCO
3
-
+ HF

=========

H
2
CO
3
+ F
-
(base) (Acid) (Con Acid) (Con Base)
HF
H
2
CO
3
Ka = 3.5 x 10
-4
Ka = 4.3 x 10
-7
A: Since HF is identified as the strongest acid, HCO
3
-
must be the strongest base.
Q: In #2, which base is the weakest?
HF + H
2
O  =========  F + H
3
O +
(Acid) (Base) (Con Base) (Con Acid)
HF
H
3
O
+
Ka = 3.5 x 10
Ka = 1.0
-4
A: Since HF is the weakest acid in this system, H
2
O is the weakest base present.
Q: In #3, which base is the strongest?
HClO
4
+ H
2
PO
4

=========

ClO
4
-
+ H
3
PO
4
(Acid) (Base) (Con Base) (Con Acid)
A: Since HClO
4
is the strongest acid, H
2
PO
4
-
must be the strongest base.
Another method for determining the relative strength of bases is by looking at page 6 under the base column. Notice that base strength increases as one moves down the table.
Rule: The strongest Acids yield the weakest con bases
(HCl is a very strong acid but Cl
-
is a very weak conjugate base)
The weakest acids yield the strongest con bases
(HPO
4
-2
is a very weak acid, but PO
4
-3
is a relatively good base) http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons2.htm