Hydrolysis
A reaction between a salt and water to produce an
acidic or basic solution.
Chemistry 12
In hydrolysis we only care about the ions making up
the salt, not the salt itself.
Spectator ions will not react with water to produce an
acidic or basic solution.
SPECTATOR CATIONS will be:
Ions of the alkali metals (column I of the periodic table).
Ions of the alkaline earth metals (column II of the
periodic table).
How do we determine the
behavior of a salt in water?
Determine the ions produced when the salt
dissociates.
Disregard any spectator ions.
SPECTATOR ANIONS will be:
The first five anions found at the top right of the table of
Relative Acid Strengths:
If any ions remain, they will act as an ACID if they are
found on the acid (left) side of the table, or as a BASE
if they are on the base (right) side of the table.
ClO4-, I-, Br-, Cl- and NO3-
HSO4- is not a spectator ion since it is a weak acid
1
For Example:
Homework:
NH4NO3(s)
Read:
Pages 148 - 152
Do:
#69 - 73
LiCN(s)
KHCO3(s)
NH4F(s)
Fe(NO3)3(s)
Calculations Involving Ka
When a weak acid, HA, is put into water, some of the
acid ionizes and a certain amount of H3O+ will be
produced.
For the purpose of all our calculations, consider that
only the first proton of a polyprotic acid will be given
off during the ionization process.
The smaller the Ka value, the less H3O+ formed.
For Example:
H2CO3(aq)
H2CO3(aq) + H2O(l) ↔ HCO3-(aq) + H3O+(aq)
None of the acid ionizes until it reacts with the water,
therefore, we can use an “ICE” diagram to incorporate
the idea of time passing as the acid ionizes and reaches
equilibrium.
2
Homework:
Read:
Pages 152 - 153
Do:
#74 - 83
Calculations Involving Kb
Calculations involving weak bases are very similar to
calculations we have previously done with weak acids,
with two important changes:
We have to calculate the Kb value, not just take it directly
from the table.
The resulting solution will be basic, not acidic, which
means we will be using [OH-] rather than [H3O+].
3
Homework:
Read:
Pages 154 - 159
Do:
#84 – 93
Study for your quiz!!!
The equivalence point occurs when the mole ratios
involved in the reaction are exactly equal to the mole
ratios required by the stoichiometry of the reaction
equation.
Acid/Base Titrations (REVIEW)
A process in which a measured
amount of a solution is reacted with
a known volume of another solution
of an unknown concentration, until
a desired EQUIVALENCE POINT or
STOICHIOMETRIC POINT is
reached.
You must always repeat the
titration as a check on your
accuracy.
For Example:
H2SO4(aq) + 2NaOH(aq) 2H2O(l) + Na2SO4(aq)
One mole of H2SO4(aq) reacts completely
with two moles of NaOH(aq).
If the volumes of the solution added from the burette
are in agreement with each other to within ±0.02 mL
(or ±0.1mL for beginning students), then we have a
satisfactory value for the equivalence point.
4
Indicators
Homework:
A weak organic acid or base having different colours
for its conjugate acid and conjugate base forms (see
page 7 in the data booklet).
Read:
Pages 159 – 165
Read the Lab
Do:
#94 – 107
Typically very complex molecules.
∴Formula represented by an abbreviation.
For Example:
Consider Phenolphthalein:
HIn(aq) + H2O(l) ↔In-(aq) + H3O+(aq)
Conjugate Acid
Form
Conjugate Base
Form
When we place Phenolphthalein in an acidic solution,
there is an excess of H3O+(aq) ions:
HIn(aq) + H2O(l) ↔In-(aq) + H3O+(aq)
…and the solution
turns CLEAR
Causes equilibrium
to shift left…
When we place Phenolphthalein in a basic solution,
there is an excess of OH-(aq) ions:
An indicator will always be in its
CONJUGATE ACID form when in ACIDIC
solutions, and it’s CONJUGATE BASE form
in BASIC solutions.
HIn(aq) + H2O(l) ↔In-(aq) + H3O+(aq)
Causes equilibrium
to shift right…
…and the solution
turns PINK
5
Why does the colour change?
Consider Phenol Red (pH change 6.6-8.0):
HIn(aq) + H2O(l) ↔In-(aq) + H3O+(aq)
Conjugate Acid
Form
Conjugate Base
Form
In an pH < 6.6 solution:
In an pH > 8.0 solution:
Hin > InSolution turns YELLOW
Hin < InSolution turns RED
At some point it is possible to have Hin = In-.
An equal number of YELLOW and RED molecules
give an ORANGE solution.
This point at which an indicator is half way through its
colour change is called the END POINT or
TRANSITION POINT.
At the end point we have:
[In ][H O ]= [H O ]and… -log ka = -log [H3O+]
=
−
Don’t get the terms endpoint and equivalence point
confused!!
ENDPOINT:
The point in the titration where the colour of the indicator
changes.
EQUIVALENCE POINT:
The point of the titration where the stoichiometry of the
reaction is exactly satisfied.
If we choose our indicator correctly, it
should change colour at, or very close
to the equivalence point.
ka
+
3
[HIn]
+
3
pka = pH
Indicators do not change colour at exactly some pH
value.
The colour change takes place over a range of about 2
pH units.
For Example:
Consider Thymol blue:
HIn(aq) + H2O(l) ↔ In-(aq) + H3O(aq)
ph < 8.0
ph > 9.6
Solution turns
(8.0 + 9.6) ÷ 2 = 8.8
GREEN
6
Examples:
Universal Indicator:
An indicator solution which changes colour several
times over a range of pH values.
Homework:
Practical Aspects of Titrations
Read:
Pages 166 – 176
Do:
#108 - 123
Standard Solutions:
A solution of accurately known concentration.
Once we have a standard solution, we can use it to
titrate other solutions and determine their
concentrations.
There are two main ways in which we can prepare a
standard solution.
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Use a substance which can be obtained in a pure and stable form
that does not absorb water or Carbon dioxide from the air.
It must have a known molar mass so it can be used to
prepare a solution of exactly known concentration.
Such a substance is called primary standard.
For Example:
Potassium hydrogen phthalate
(KHC8H4O4; molar mass = 204.23)
Titrate a base with an acidic primary standard.
Once the concentration of the base is known, it can
then be used as a “standard” for titrating other
unknown acid solutions.
For Example:
NaOH cannot easily be used as a primary standard
because solid NaOH is about 95-98% pure and rapidly
absorbs water from the atmosphere.
We make up solutions of NaOH which are known to be
close to 0.1 M.
Types of Titration Curves
A primary standard of 0.1000 M potassium hydrogen
phthalate can be titrated against the NaOH to find the
exact [NaOH].
The [NaOH] can then be calculated and we say that we
have a “standardized solution of NaOH”.
The NaOH solution can now be used to titrate several
different HCl solutions and find their concentrations.
In addition to looking at the details of the shapes of
titration curves, we will also come up with some “rules
of thumb” which will allow us to predict the
approximate pKa values of the indicators needed to
help us find the equivalence point of titrations.
8
The titration of a STRONG ACID with a
STRONG BASE
Special features:
VB is the volume of base
required to get to the
stoichiometric point.
pH rises almost vertically
around the value of VB
The corners before and
after the stoichiometric
point are fairly sharp.
The titration of a WEAK ACID with a STRONG
BASE
Special features:
There is an initial upswing in
the pH at the start of the
titration.
The curve has a fairly sharp
corner after the
stoichiometric point and a
somewhat rounded corner
just before the stoichiometric
point.
VB is the volume of base
required to get to the
stoichiometric point and
V
V1 2 = B .
2
Choosing an Indicator?
The salt of a strong acid and a strong base is neutral in
solution.
For example:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
We require an indicator which changes colour around
pH = 7; that is, pKa = 7.
As long as the indicator changes at a pH > 7, and in the
vertical region of the curve, it is acceptable.
Bromothymol blue, Phenol red, Neutral red,
Thymol blue, or Phenolphthalein
Calculating Ka
Assume we are adding NaOH to CH3COOH. The value
of Ka can be found in two different ways:
From point “pHinit”
(the pH of the acid before the titration starts):
The acid ionizes:
CH3COOH(aq) + H2O(l) ↔ CH3COO-(aq) + H3O+(aq)
If we measure the pHinit, we can calculate the equilibrium
[H3O+] since the ionization of CH3COOH produces a
solution having [H3O+] = [CH3COO-].
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From point “pH1/2”
We can write:
−
ka
+ 2
[CH COO ][H O ]= [H O ]
=
3
+
3
3
[CH 3COOH]
[CH 3COOH]
−pH 2
ka =
[10 ]
[CH COOH]
(the pH corresponding to V1/2):
We know that [H3O+], [CH3COO-], and [CH3COOH] must be
related by means of the Ka expression.
When we are exactly half-way to the stoichiometric point, half
of the initial CH3COOH will have been neutralized and
converted to CH3COO-:
3
If we know the starting concentration of the acid and
measure the pHinit of the acidic solution, we can solve
to find the value of ka for the acid.
CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(aq)
Therefore:
[CH3COOH] = [CH3COO-]
[CH COO ][H O ]= [H O ]
=
−
ka
3
+
3
[CH COOH]
+
k a = 10 −pH
3
3
Choosing an Indicator:
The salt of a weak acid and a strong base is basic in
solution:
CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l)
Since the Na+ is a spectator and the CH3COO- is a weak
base, the resulting solution will be BASIC.
We require an indicator which changes colour in the basic
range.
Homework:
Read:
Pages 177 – 184
Do:
#124, 126, 129
If the acid being titrated is very weak, the pKa of the
indicator should be several pH units above 7 (around 9).
The stronger the acid, the closer to 7 the pKa value of the
indicator should be.
10
The titration of a WEAK BASE with a STRONG
ACID
Special features:
This case is very similar
to the case of a weak acid
and a strong base, except
that the curve looks like
it has been flipped
upside down.
The ionization of NH3 produces a solution with:
[NH4+] = [OH-]
kb
− 2
[NH ][OH ]= [OH ]
=
−
Assume we are adding HCl to NH3. The value of Kb can
be found in two different ways:
From point “pHinit”
(the pH of the base before the titration starts):
From point “pH1/2”
4
[NH ]
[NH ]
3
kb
[10
=
The base ionizes:
NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
If we measure the pH (and convert it to pOH) we can
calculate the equilibrium [OH-].
(the pH corresponding to V1/2):
We can write:
+
Calculating Kb
3
When we are exactly half-way to the stoichiometric point, half
of the initial NH3 will have been neutralized and converted to
NH4+:
NH3(aq) + HCl(aq) → NH4Cl(aq)
−pOH 2
]
[NH ]
3
Therefore:
[NH3] = [NH4+]
If we know the starting concentration of the base and
measure the pH (to calculate pOH) of the solution, we
can solve to find the value of kb for the base.
[NH ][OH ]= [OH ]
=
+
kb
−
4
−
[NH ]
k b = 10 −pOH
3
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Choosing an Indicator:
The salt of a strong acid and a weak base is acidic in
solution:
NH3(aq) + HCl(aq) → NH4Cl(aq)
Since the Cl- is a spectator and the NH4+ is a weak acid, the
resulting solution will be acidic.
We require an indicator which changes colour in the acidic
range.
If the base being titrated is very weak, the pKa of the
indicator should be several pH units below 7 (around 5).
The stronger the base, the closer to 7 the pKa value should
be.
Buffers
A solution containing equal amounts of a weak acid
and its conjugate base, or, weak base and its conjugate
acid.
For Example:
What happens when we make
up a buffer?
If we make a solution of CH3COOH, there will be a
small amount of both CH3COO- and H3O+ present.
CH3COO-
Adding extra
in the form of NaCH3COO
causes the equilibrium to shift to the left and there
will be a large decrease in the [H3O+].
The buffer will then have a relatively large
concentration of both the weak acid and its conjugate
base, and a small concentration of H3O+.
We can make an Acetic acid buffer by mixing 1.0 L of 1.0 M
CH3COOH(aq) with 1.0 mol of NaCH3COO(s).
We get the equilibrium:
CH3COOH(aq) + H2O(l) ↔ CH3COO-(aq) + H3O+(aq)
The job of a buffer is to prevent large
changes in pH when an acid or base is
added to the solution.
Going back to our example:
CH3COOH(aq) + H2O(l) ↔ CH3COO-(aq) + H3O+(aq)
Since the [CH3COOH] and [CH3COO-] are equal, we
get:
CH 3COO − H 3O +
ka =
= H 3O +
CH
COOH
[ 3
]
[
][
] [
]
and:
pH = −log k a = 4.74
The pH of the buffer.
12
When we add an acid or a base, the equilibrium will
shift in order to maintain the pH relatively constant at
4.74.
If we add a base:
If we add an acid:
The [H3O+]
returns to the
original amount
(almost).
CH3COOH(aq) + H2O(l)
The [H3O+]
returns to the
original amount
(almost).
↔ CH3COO-(aq)
Shifts left to
counteract the
change of H3O+(aq).
+
H3O+(aq)
The
CH3COOH(aq) + H2O(l) ↔ CH3COO-(aq) + H3O+(aq)
Shifts right to
counteract the
change of H3O+(aq).
The
[H3O+]decreases as
it is neutralized by
the addition of OH(aq).
[H3O+]increases.
RESULT: The pH stays relatively
constant.
Basic buffers also exist.
You need to make sure that the pH of the buffer will be
in the basic range.
For Example:
Consider the NH4+(aq)/NH3(aq) buffer:
NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
pH = pKa = -log ka
pH = -log (5.6 x 10-10)
pH = 9.25 (basic range)
RESULT: The pH stays relatively
constant.
Buffers are not infinite in their ability to counteract
changes in pH.
If we add too much H3O+(aq) or OH-(aq), we may
overrun the buffer.
For Example:
CH3COOH(aq) + H2O(l) ↔ CH3COO-(aq) + H3O+(aq)
buffer
will be overrun
and cannot counteract
useThe
up all
of the
we cause too
If we add too much
much
of
a
shift
to H3O+(aq).
the
addition
of
any
more
conjugate base
H3O+(aq)…
the left and…
in the buffer!
RESULT: The pH drops
significantly!
13
Whenever we titrate a weak acid or weak base we will
have a buffer solution in the middle portion of the
titration curve.
Consider the behavior of a CH3COOH/CH3COObuffer when H3O+ or OH- is added:
Applied Acid/Base Chemistry
Metal Oxides:
When a metal oxide is added to water we have a
dissociation:
Na2O(s) → 2 Na+(aq) + O2-(aq)
The metal ions are spectators and the oxide ion, O2-, is a
strong base.
The hydrolysis of the oxide ion is then given by:
O2-(aq) + H2O(l) → 2 OH-(aq)
We get a basic solution.
Homework:
Read:
Pages 184 – 188
Lab
Do:
#125, 127, 128, 130 - 143
But, the spectator ion will react with OH- to produce an
ionic compound.
Na+(aq) + OH-(aq) → NaOH(aq)
The overall reaction is:
Na2O(s) + H2O(l) → 2 NaOH(aq)
Metal oxides produce
basic solutions. (MOB)
14
Non-Metal Oxides:
When a non-metal oxide reacts with water, the water
bonds to the oxide molecule to produce an ACIDIC
solution.
For Example:
SO3(g) + H2O(l) → H2SO4(aq)
Non-metal oxides produce
acidic solutions. (NOA)
Where does acid rain come from?
Combustion of fuels that
contain Sulphur and Nitrogen:
Acid Rain
Because of dissolved CO2(g), “normal”
rainwater is naturally somewhat acidic and
has a pH of 5.6:
CO2(g) + H2O(l) ↔H3O+(aq) + HCO3-(aq)
Therefore, any precipitation with pH < 5.6
is called ACID RAIN.
Homework:
Do:
#144 - 147
SO2(g) + H2O(l) → H2SO3(aq)
Sulphurous acid
SO3(g) + H2O(l) → H2SO4(aq)
Sulphuric acid
2 NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq)
Nitric and Nitrous acid
15