Acid-Base Equilibria

1
Acid-Base Equilibria
Aqueous solutions can be classified as acidic,
basic, or neutral. This classification scheme is
based on the quantities of 2 ions, hydronium ion,
H301 and hydroxide ion, OH1-. Where do these
ions come from in solutions of pure water?
Water molecules in motion will randomly collide
with one another. When this happens occasionally
a hydrogen nucleus from one molecule will be
transferred from one molecule to the other.
This can be illustrated.
2
Notice the nucleus of one hydrogen atom, a
proton, was transferred, but the electron pair
was left behind. This produces the H301 ion
(hydronium) and the OH1- ion (hydroxide)
3
water ionizing
the equilibrium expression is
Which is usually shortened to
4
The equilibrium constant for
is
H1OH1-
Ke
H2O
H2O in mol/L is mass of 1 L of water is 1000
g n (1000g)/(18g/mol) 55.6 molL-1
5
To simplify this equation
Here is a new constant H2OKe Kw the ion
product constant of water Kw H1OH1- 1.0
x 10-14 _at_ 25oC at higher temperatures Kw
increases
6
Kw H1OH1- 1.0 x 10-14 _at_ 25oC if
H1OH1- then H1OH1- 1.0 x 10-7 which
is the situation for neutral water. A scale
which simplifies this very small number is the pH
(potency of H1) scale. It is based on powers of
ten. Simply take the exponent from 10-7 and then
multiply it by -1. This produces the number 7.
7
In mathematical terms pOH -logOH1- so if in
an aqueous solution the OH1- 2.4 x 10-4, the
pOH is 3.62 If the pOH 4.56 then the OH1-
2.8 x 10-5 This is calculated using the
equation OH1- 10-pOH , similarly
H1 10-pH.
8
In mathematical terms pH -logH1 so if in an
aqueous solution the H1 2.4 x 10-8, the pH
is 7.62 Remember the whole number portion of a pH
doesnt count as a significant digit (SD), just
like in the number 2.4 x 10-8 the exponent -8
doesnt count as a SD.
9
If the H1OH1- 1.0 x 10-14 then taking the
log of each side pH pOH 14 so if the pH of a
solution is 2.3 the pOH is 11.7 This means for
these 4 values H1, OH1-, pH, pOH if one of
them is given the other 3 can be calculated.
Here is an example
10
If the pH of a solution is 1.45 find the H1,
OH1-, and the pOH. Remember here are the
equations needed.
If pH 1.45 6. pOH 14 - pH pOH 12.55 3.
OH1- 10-12.55 OH1- 2.8 x 10-13M 4. H1
10-1.45 H1 0.035 M
1. pH -logH1 2. pOH -logOH1- 3. OH1-
10-pOH 4. H1 10-pH. 5. H1OH1- 1.0 x
10-14 6. pH pOH 14
11
Take note the equations can be applied in varying
orders and there are choices as to exactly which
equations are used. These decisions are yours to
make.
12
If the OH1- of a solution is 9.4 x 10-9 M find
the H1, pH, and the pOH.
If OH1- 9.4 x 10-9 2. pOH -log 9.4 x
10-9 pOH 8.03 6. pH 14 - 8.03 5.97 4. H1
10-5.97 1.1 x 10-6 M
1. pH -logH1 2. pOH -logOH1- 3. OH1-
10-pOH 4. H1 10-pH. 5. H1OH1- 1.0 x
10-14 6. pH pOH 14
13
If the H1 of a solution is 6.2 x 10-2 M find
the OH1-, pH, and the pOH.
If H1 6.2 x 10-2M 1. pH -log 6.2 x 10-2
pH 1.21 6. pOH 14 - 1.21 pOH
12.79 3.OH1- 10-12.79 OH1- 1.6 x
10-13M
1. pH -logH1 2. pOH -logOH1- 3. OH1-
10-pOH 4. H1 10-pH. 5. H1OH1- 1.0 x
10-14 6. pH pOH 14
14
pH Scale
-1 0 1 2 3 4 5 6 7 8 9 10 11
12 13 14 15
N e u t r a l
Since this is a logarithmic scale pH 9 is 10xs
more basic than pH 8, pH 12 is 1000 xs more
basic than pH 9
How much more acidic is pH 1 than pH 5? 10 000
xs What pH is 1000xs more acidic than pH 2?
pH of -1
15
Conceptual Definitions of Acids and
Bases Arrheniuss Concept - acids are substances
which react in water and produce hydronium
ions. HCl(g) H20 -------gt H301(aq)
Cl1-(aq) Bases are substances which react with
water and produce hydroxide ions. NH3(g) H20
------gt NH41(aq) OH1-(aq)
16
This concept has its limitations however. Cant
substances be classified as acids or bases
without the involvement of water?
17
Bronsteads Definition of Acids and Bases Acids
are substances which donate protons and bases are
substances which accept protons. In the
examples above HCl(g) is an acid because it
donates protons to H2O molecules and NH3 is a
base because it accepts protons from H2O
molecules.
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Strength of Acids and Bases is determined by the
degree to which a substance produces ions in
solution. A strong acid or base is a substance
which completely ionizes. In other words if 100
molecules of a strong acid like HCl are placed in
water all 100 of them will react with H2O
producing 100 H3O1 ions and 100 Cl1- ions. Weak
acids and bases only partially ionize. Strong
Acid - the reaction below goes to
completion. HCl(g) H20 --------gt H301(aq)
Cl1-(aq)
19
Weak Acid - the reaction occurs to a limited
extent. In the example below if 100 acetic acid
molecules are placed in water only a few of them
will successfully react with water molecules
producing hydronium ions. Most CH3COOH
molecules remain intact. CH3COOH H20
H301(aq) CH3COO1-(aq)
20
Strong Acids in order of decreasing strength are
HClO4, HI, HBr, H2SO4, HCl, HNO3 A table with
the remaining moderate and weak acids can be
found on page 803. Acid strength has to do with
the ease with which an acid can lose a proton.
If the binary acid strengths (HI, HBr, HCl) are
compared it can be seen that HI is the strongest
acid of this group because its iodide ion is the
largest of the group so the force between the
hydrogen ion and the iodide ion is the weakest so
it loses its proton most easily.
21
Remember the weaker the force the stronger the
acid
I1-
22
Strong Bases include hydroxides of group 1A and
Ca2, Ba2, and Sr2. A table with the
remaining moderate and weak bases can be found on
page 803. As with acids the weaker the bonds,
the stronger the base since liberation of OH1-
ions is easiest when the bonds are weakest.
23
Polyprotic Acids donate protons in steps. For
instance carbonic acid, H2CO3 has two protons to
donate and it does this in two steps step
1 H2CO3 H20 HCO31- H301 step
2 HCO31- H20 CO32- H301 note The
arrows are constructed in this manner to show the
reverse reaction has a greater tendency than the
forward reaction.
24
Conjugate Acid - Base Pairs - When using the
Bronsted concept for acids and bases it is
convenient to consider all acid - base reactions
as reversible equilibria. For instance when
sulfurous acid, H2SO3 reacts with water the
following equilibrium is established H2SO3
H2O H301 HSO31-
acid base acid
base
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acid base acid
base
H2SO3 H2O H301 HSO31-
In the forward direction the H2SO3 is the proton
donor so its the acid and the H2O is the proton
acceptor so its the base. In the reverse
direction the H301 is the proton donor so its
the acid and the HSO31- is the proton acceptor so
its a base.
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acid base acid
base H2SO3 H2O H301 HSO31-
When looking at both forward and reverse
reactions it is easy to pick out a pair of
molecules which differ by a single proton (H atom
without its electron). These pairs are called
conjugate acid-base pairs.
27
Identify the acid base conjugate pairs in the
equilibrium below
PO43-(aq) H2O(aq) HPO42-(aq)
OH1-(aq)
acid
base
acid
base
Show how HPO4-2 can act as both an acid and a
base.
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Amphoteric (Amphiprotic) Substances can behave as
both acids or bases dependent on the
circumstances. Water molecules, for instance,
can sometimes accept protons and behave as bases
or donate protons and behave as acids.
HBr(g) H2O H301(aq)
Br1-(aq) base NH3(g) H2O 0H1-(aq)
NH41(aq) acid
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Any substance which alters the pH of an aqueous
solution is regarded as an acid or a base.
Acids and Bases are classified into 2
categories, strong and weak. Strong Acids and
Bases are completely ionized. This means in
solution no molecules are present. All the
molecules break apart (dissociate), to form ions.
Weak acids and bases only partially dissociate
and ionize and so are involved in equilibria.
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Strong Acid
Weak Acid
31
Is this a Strong or Weak Acid?
32
Is this a Strong or Weak Acid?
33
Problems Involving Strong Acids and Bases
Find the pH and pOH of a 0.28 molL-1 solution of
HClO4. Since this is a strong acid the ionization
is complete. This can be represented by
HClO4
H1(aq) ClO41-(aq)
(molL-1) initial
0.28
0.0
0.0
-0.28
0.28
0.28
shift
0.00
0.28
0.28
final
34
Remember pH -logH1 so pH - log 0.28 pH
0.55 pH pOH 14 so pOH 14 - 0.55 pOH 13.45
35
If the pOH of a solution of HCl is found to be
12.9 what is the concentration of this
solution. Since this is a strong acid the
ionization is complete. This can be represented
by
H1(aq) Cl1-(aq)
HCl
(molL-1) initial
0.08
0.0
0.0
-0.079
0.079
0.079
shift
0
0.079
0.079
final
In this case the pOH can be used to determine the
final H1, pH 14 - 12.9, H1 10-1.1
36
Find the pH and pOH of a 0.12 molL-1 solution of
Sr(OH)2. Since this is a strong base the
ionization is complete. This can be represented
by
Sr2(aq) 2 OH1-(aq)
Sr(OH)2
(molL-1) initial
0.12
0.0
0.0
-0.12
0.12
0.24
shift
0.00
0.12
0.24
final
Remember pOH -logOH1- so pOH - log 0.24
0.62 pH 14 - 0.62 13.38
37
Problems Involving Weak Acids and Bases
Find the pH and pOH of a 1.68 molL-1 solution of
HCN (Ka 6.2 x 10-10. Since this is a weak acid
the ionization is incomplete. This can be
represented by
HCN
H1(aq) CN1-(aq)
(molL-1) initial
1.68
0.0
0.0
- x
x
x
shift
1.68 - x
x
x
final
38
To determine the value of x the Ke for this
reaction must be known. It is found in tables
called Ka (equilibrium constants for acids).
For HCN Ka 6.2 x 10-10.
H1CN1-
Ka
HCN
6.2 x 10-10 x2
1.68
x 3.2 x 10-5, pH 4.49, pOH 9.51
39
What is the percentage ionization of the 1.68
mol/L HCN solution?
40
If an unknown 0.34 mol/L monoprotic acid is 1.3 x
10-5 ionized what is its Ka?
41
If the pH of a solution of HOCl is found to be
4.9 what is the concentration of this solution.
Ka 3.0 x 10-8 Since this is a weak acid the
ionization is incomplete. This can be
represented by
H1(aq) OCl1-(aq)
HOCl
(molL-1) initial
x
0.0
0.0
1.26 x 10-5
1.26 x 10-5
-1.26 x 10-5
shift
1.26 x 10-5
1.26 x 10-5
x - 1.26 x 10-5
final
In this case the pH can be used to determine the
final H1, H1 10-4.9 1.26 x 10-5.
42
H1OCl1-
Ka

HOCl
3.0 x 10-8 1.58 x 10-10
x
x 5 x 10-3 molL-1
43
Negative ions Which Act Like Weak Bases
Strong acids totally ionize because the force of
attraction between the two ions is relatively
weak and water molecules pulling on them can
separate them.
NO31
This means NO31- has a weak pull on H1 ions
44
Conversely weak acids have negative ions with
relatively strong attractions for H1 ions
If a salt like NaNO2 is added to water it is
easily pulled apart since sodium ions are quite
large compared to H1 ions making the attraction
between Na1 and NO21- fairly weak.
45
NO21-
All negative ions except those found in strong
acids, Cl1-, Br1-, I1-, NO31-, ClO41-, HSO41-,
are capable of creating OH1- ions, this means
they act like bases because they raise pH
46
The equilibrium for NO21- where it alters pH is
NO21-(aq)
H20
HNO2 (aq) 0H1-(aq)
Show how K2CO3 alters pH
CO32-(aq)
H20
HCO31-(aq) 0H1-(aq)
Show how NaF alters pH
F1-(aq)
H20
HF(aq) 0H1-(aq)
Show how NaCl alters pH it doesnt, Cl1- has too
weak a pull on H1 remember, Cl1- is part of the
strong acid HCl
47
What is the pH of a 0.24 molL-1 solution of
potassium sulfite Kb of sulfite ion is 1.5 x
10-7? When potassium sulfite is placed in water
it dissociates into ions. This can be shown by
K2SO3(s)
2K1(aq) SO32-(aq)
Since SO32- is not part of a strong acid it is
capable of pulling apart (hydrolyzing) water
molecules. This can be shown by
SO32-(aq) H20
HSO31-(aq) 0H1-
molL-1 initial
0.24
0.00
0.00
-x
x
x
shift
0.24-x
x
x
_at_E
48
The Kb of SO32- is 1.5 x 10-7. So to find x
x2
1.5 x 10-7
0.24 -x
X is so small compared to 0.24 it can be regarded
as negligible
x
1.9 x 10-4 OH1-
pOH -log (1.9 x 10-4) pOH 3.72 pH 14 -
3.72 10.28
49
Positive Ions Which Act Like Acids
Positive ions like NH41, can donate H1 ions to
water molecules. This increases the amount of
hydronium (H301) creating an acidic affect.
This can be illustrated by the following
50
(No Transcript)
51
Multivalent metal ions with a large charge
density can also donate H1 . They do this by
attracting water molecules. Once the water
molecules are attached to these highly charged
ions the electron pairs shared by the hydrogen
and oxygen atoms are pulled toward the oxygen end
of the water molecule making the H1 ions
vulnerable to removal by other colliding water
molecules. This can be illustrated by the
following
52
Notice the electron pairs moving closer to the
Al3
Only small trivalent ions and Be2 act in this way
53
If the pH of a solution of NH4Cl is found to be
5.23 what is the concentration of this solution
Ka NH41 5.7 x 10-10 ? Since the pH is below
7 this substance is acting like an acid. It is
not among the strong acids so it must be a weak
acid. Since it has no H1 to donate it must be
acting like an acid by hydrolyzing water.
Positive ions can behave this way. When NH4Cl is
dissolved it dissociates into ions. This is
shown by
NH4Cl(s)
NH41(aq) Cl1-(aq)
The NH41 acts like a weak acid so it must .
54
NH41 H2O
NH3 H3O1
(molL-1) initial
x
0.0
0.0
- 5.9 x 10-6
5.9 x 10-6
5.9 x 10-6
shift
x - 5.9 x 10-6
5.9 x 10-6
5.9 x 10-6
final
In this case the pH can be used to determine the
final H1, pH 5.23, H1 10-5.23 5.9 x
10-6
H1NH3
Ka

NH41
5.7 x 10-10 3.47 x 10-11
x 0.061 molL-1
x
55
Buffered Solutions
There are experimental situations where real life
situations are simulated. Living organisms have
fluids which are maintained within very narrow pH
ranges so these simulated experiments require
solutions in which pH changes are minimal despite
the introduction of acids and bases. Mixtures
which resist changes in pH when acids or bases
are added are called buffers. Buffers are
equilibrium mixtures which shift when acids or
bases are added.
56
Buffers contain 1. a weak acid and a soluble
salt containing the conjugate base of the weak
acid OR 2. A
weak base and a soluble salt containing the
conjugate acid of the weak base
57
If the weak acid is HF then a soluble salt
containing its conjugate base partner
is NaF. Remember that conjugate acid - base
partnerships are different by a single H, bases
gain Hs, acids lose Hs. If the weak acid is
HC2H3O2 then a soluble salt containing its
conjugate base partner is KC2H3O2 . If the weak
base is NH3, then a soluble salt containing its
conjugate acid partner is NH4Cl.
58
Lets now look at how buffers work. If the
buffer is HF and NaF the equilibrium mixture is
If the buffer is HC2H3O2 and NaC2H3O2 the
equilibrium mixture is
If the buffer is H2CO3 and NaHCO3 the equilibrium
mixture is
59
Why do buffers require a salt? If you can recall
the definition of a weak acid they only ionize a
small amount, so there is only a small amount of
HCO31- present at equilibrium. If the
equilibrium is going to resist a change in pH
when acids or bases are added there must be
sufficient quantities of substances on both sides
of this equilibrium. When acids are added (a
source of H1, the equilibrium shifts
60
to get rid of it. If a base is added, the OH1-
produced reacts with the H1 forming water. This
means the equilibrium shifts
to replace the H1 removed by the addition of the
OH1-. A buffers capacity to resist large changes
in pH is determined by the quantity of weak acid
and conjugate base present. When either of these
quantities is used up the solution can no longer
act as a buffer.
61
Buffer Problems 1. 1.0 L of a buffer is prepared
by mixing acetic acid and sodium acetate. If
both of these substances are 1.0 molL-1 at
equilibrium find the pH if the Ka for acetic acid
is 1.8 x 10-5? Equilibrium is
pH -log 1.8 x 10-5 pH 4.74
62
1. If 1.0 mL of 2.0 M HCl is now added to this
buffer what is the new pH?(assume no volume
change of 1.0L.)
_at_E
1.0
1.0
1.8 x 10-5
upset
2.0 M x 0.0010L0.0020
shift
0.0020
-0.0020
-0.0020
_at_E
1.002
0.998
x
Assume all the HCl is consumed by the equilibrium
shifting right to left
63
1. If 5.0 mL of 2.0 M NaOH is now added to this
buffer what is the new pH?(assume no volume
change.)
_at_E
1.0
1.0
1.8 x 10-5
upset
2.0 M x 0.0050 L - 0.010
shift
-0.010
0.0100
0.010
_at_E
0.99
1.010
x
Assume all the OH1- is consumed by the H1
equilibrium shifts left to right to replace the
H1
64
pH -log x 4.75