Lesson #2: Conjugate Acid-Base Pairs
Notes
Part 1. Identifying conjugate acid/base pairs
Molecules and ions that differ by one proton (i.e., H+) are referred to as a conjugate acidbase pairs.
e.g.,
NH3 + H2O
=====
(Base) (Acid)
NH4+ + OH(Con. Acid) (Con. Base)
The conjugate acid-base pairs are: NH3/NH4+ and H2O/OH-. Hint: Look for related
atoms between the reactants and the products. Both NH3 and NH4+ 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:
HSS-2
CNPO4-3
1.
2.
3.
4.
the con. acid is H2S
the con. acid is HSthe con. acid is HCN
the con. acid is HPO4-2
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.
2.
3.
4.
H2SO4
H2PO4H3PO4
O-2
the con. base is HSO4the con. base is HPO4-2
the con. base is H2PO4there is no con. base b/c there are no H+ ions to remove!
Orally: 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 +
(Acid)
H2O
==========
(Base)
H3O+ +
(Con. Acid)
CN(Con. Base)
Orally: (Comment on the importance of recording the correct charge on each species. A
0.5 deduction will be made if charges are incorrect)
Orally: 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
H3O+ must donate a proton to CN- for the reverse reaction to work.
Q: Identify the two conjugate acid/base pairs.
A: HCN/CN- and H3O+/H2O
2.
HCO3- + H2PO4(Base)
(Acid)
H2CO3/HCO3- and
3.
HC2O4- +
(Acid)
=========
H2CO3 + HPO4-2
(Con. Acid) (Con. Base)
H2PO4-/HPO4-2 are the two con. acid/base pairs.
CH3COO(Base)
======
C2O4-2
+
(Con. Base)
CH3COOH
(Con. Acid)
HC2O4-/C2O4-2 and CH3COOH/CH3COO- are con. acid/base pairs.
Assignment: Page 119 # 13
Page 121 # 16, 17, 18, 19
(If time permits, move onto part 2 or show the video from yesterday)
Part II: Relative Strengths of Acids and Bases
Orally: 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-
Orally: 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. Look at page 6 of your data booklet and
identify other acids that are considered to be strong.
e.g., HClO4, HBr, HNO3 or any acid that fully dissociates (i.e., equations with a single
arrow)
Weak acid – forms an equilibrium in water (i.e., does not fully dissociate).
e.g.,
HF
======
H+ + F-
Orally: 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.
Orally: 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 – A base that fully dissociates in water. Strong bases contain hydroxide and
a cation from group #1/#2 of the periodic table.
e.g.,
NaOH ------------------
e.g.,
Ca(OH)2 --------------
Na+ + OHCa+2 + 2OH-
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+ for both equations below.
HCl ----------------1.0M
HNO3 ---------------1.0M
H+ + ClH+ + NO3-
Orally: Since the molar concentrations of the acids are equal and they both fully
dissociate, both acids are equally effective at producing H+ (i.e., both acids will produce
1.0M [H+] ).
Weak Bases: Bases that form equilibriums.
Orally: Weak bases will be discussed in much greater detail later in the chapter.
Part III: Predicting reactions using the acid/base table (page 6)
Predict the products for the following reactions
1.
HCO3- + HF
=========
?
+ ?
Step #1: Determine the relative strength of each acid
Orally: 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
HCO3-
Ka = 3.5 x 10-4
Ka = 5.6 x 10-11
Since the Ka for HF is larger than HCO3-, HF is the strongest acid (i.e., HF is better at
donating protons than HCO3- is) and HCO3- is forced to accept the proton.
Step #2: Predict the products
HCO3- + HF
=========
H2CO3 + F-
Step #3: Label each substance as an acid, base, con acid or con base
HCO3- + HF
=========
(base)
(Acid)
H2CO3
+
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., H2CO3).
Ka = 3.5 x 10-4
Ka = 4.3 x 10-7
HF
H2CO3
Since the Ka for HF is greater than H2CO3, 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 + H2O
HF
H2O
=========
? + ?
Ka = 3.5 x 10-4
Ka = 1.0 x 10-14
HF +
(Acid)
H2O
(Base)
=========
F- + H3O+
(Con Base) (Con Acid)
HF
Ka = 3.5 x 10-4
+
H3O Ka = 1.0
Since the Ka for H3O+ is larger than HF, we predict that the reactants are favoured.
3.
HClO4 + H2PO4-
=========
? + ?
HClO4is a strong acid and H2PO4- is weak, so HClO4is automatically the acid and forces
H2PO4- to behave like a base.
HClO4
(Acid)
+ H2PO4(Base)
=========
ClO4- + H3PO4
(Con Base) (Con Acid)
Since HClO4 is a strong acid and H3PO4 is weak, the equilibrium will favour products.
Orally: 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?
A: Since HF is identified as the strongest acid, HCO3- must be the strongest base.
Q: In #2, which base is the weakest?
A: Since HF is the weakest acid in this system, H2O is the weakest base present.
Q: In #3, which base is the strongest?
A: Since HClO4 is the strongest acid, H2PO4- must be the strongest base.
Orally: 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
(HPO4-2 is a very weak acid, but PO4-3 is a relatively good base)
Assign: page 125 #21 – 23
page 133 #38, 39, 41, 42, 44, 45, 46
http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons2.htm