Unit 2: Equilibrium
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R
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1
Unit 4
(formerly Module 5)
Equilibrium
2
11.0
Equilibrium
• Overview:
Equilibrium is the concept of a system remaining “in
balance”. A system in equilibrium does not change
at the macroscopic level (the level that we can
detect with our senses).
True equilibrium should not be confused with
homeostasis or “steady-state”, a process by which
living organisms attempt to maintain a consistent
internal conditions by absorbing or excreting
materials from and to their environment.
3
Equilibrium vs. Homeostasis
• Equilibrium usually exists
in a closed system, where
materials cannot easily
enter or leave.
• Examples:
• A reversible chemical
reaction occurs inside a
closed container. The
reaction appears to have
stopped, but at a
molecular level changes
are still going on.
• A liquid in a sealed bottle
does not appear to
evaporate.
• Homeostasis usually
exists in an open system,
where materials can
enter and leave.
• Examples:
• A dog lives in a kennel. It
eats and excretes roughly
equal amounts, and
therefore maintains a
steady weight and internal
conditions.
• A cell in a human body
maintains a balance of
nutrients.
4
The Temporary Nature of Equilibrium
• Although we study equilibrium as if it is an
unchanging state, in reality dynamic processes
as well as static forces may act upon the
equilibrium.
• Eventually something will upset the
equilibrium, temporarily or permanently
throwing it “out of balance”
• Often a new equilibrium will be re-established
after the original equilibrium is upset.
5
Examples
• A precariously balanced rock formation can
endure in “equilibrium” for centuries. Suddenly
the balanced rock falls, temporarily disturbing the
equilibrium.
• A reversible chemical reaction inside a beaker has
reached a state of equilibrium and appears to
have stopped. A researcher adds more of one of
the reactants, upsetting the equilibrium. The
reaction temporarily resumes until a new
equilibrium is established.
6
The Old Man of the Mountain
For two hundred years a
precarious rock formation
in New Hampshire was
said to resemble the face
of an old man. It had
become a symbol of New
Hampshire, appearing on
postcards, road signs and
coins
On May 3, 2003 The rock face
collapsed. In terms of equilibrium, we
could say that this was a static
equilibrium that endured for centuries,
until it was disturbed by a spring storm.
Afterwards a new equilibrium was
established, that unfortunately no
longer resembled a face.
7
8
Chapter 11
Qualitative Aspects of Chemical
Equilibrium
9
11.1
Qualitative Aspects of Equilibrium
• What is an equilibrium? What properties
does it have? How can we distinguish
dynamic, and static equilibria and tell them
apart from simple steady states?
10
Static Equilibrium
• One rock sitting on another is at
static equilibrium, even if it is
balanced precariously. No dynamic
processes are acting on it, so it
remains unchanged.
• Static equilibrium is a bit boring. We
seldom deal with it in chemistry.
Note: Gravity is not
considered to be a
dynamic force.
It is a static force.
11
Dynamic Equilibrium
• There are several types of dynamic equilibrium of
interest to chemists, which are described on the
following slides including:
• Phase Equilibrium
• Solubility Equilibrium
• Chemical Equilibrium
12
19
11
• A dynamic
equilibrium is a bit
like a hockey game.
• Barring penalties,
there is always the
same number of
players on the ice,
but some players
are constantly
leaving the bench
as others return to
it.
13
Phase Equilibrium
(1st type of dynamic equilibrium)
• Example:
• In a closed bottle a water may exist as both a
liquid and a gas at the same time (eg. Water
vapour above liquid water). As water molecules
evaporate from the liquid phase, other water
molecules condense from the gaseous phase.
14
Solubility Equilibrium
(2nd type of dynamic equilibrium)
• Example: If you add too much sugar to a cup
of tea, the tea becomes saturated with sugar
and no more appears to dissolve. In fact,
some molecules of sugar are dissolving as
other molecules recrystallize back into solid
sugar.
15
Chemical Equilibrium
(3rd type of dynamic equilibrium)
• Chemical equilibrium occurs when two
opposing chemical reactions occur at the same
rate, leaving the composition of the system
unchanged.
• Example: Dinitrogen tetroxide (N2O4) and
nitrogen dioxide (NO2) can exist in the same
container. Each can change into the other, and
at equilibrium they do so at the same rate.
N2O4  2 NO2
This is the most important type of equilibrium in chemistry!!
16
11.2
Irreversible and Reversible
Reactions
• Some chemical reactions are easily reversed,
like the electrolysis of water.
• Others, such as the burning of wood are
impossible to reverse under laboratory
conditions.
17
Irreversible Chemical Reactions
• This definition is assumed to refer to reactions
occurring under normal laboratory conditions.
• In a lab, it is easy to burn a piece of wood. It is
impossible, under laboratory conditions, to turn ash,
smoke, carbon dioxide and water back into wood.
The growth of a tree does allow wood to be produced from
materials that might include wood ashes, but growing a tree takes
decades, and requires countless changes involving many complex
chemical mechanisms and multiple organic catalysts (enzyme systems).
Burning is therefore NOT considered reversible!
18
Reactants  Products
Reactants can become products, but products cannot
turn back into reactants.
19
Reversible Chemical Reactions
• Some reactions are easily reversed
using common laboratory procedures.
• For example, it is possible to
decompose water into hydrogen and
oxygen in a electrolytic cell, and
equally possible to synthesize water
from hydrogen and oxygen in a fuel
cell:
2H2O + electrical energy  2 H2 + O2
2H2 +O2  2H2O + electrical energy
20
Reactants
Products
Reactants can become products, and products can also turn
back into reactants. Also show as: reactants  products
21
Reversibility and Equilibrium
Only reversible
reactions can
produce a true
dynamic chemical
EQUILIBRIUM
22
A Chemical System is at Equilibrium if it
meets these Criteria:
1. The System is Closed, for example by being sealed
inside a container so material cannot enter or leave.
2. The Change is reversible. The reaction or change can
proceed in both direct and reverse directions.
3. There is no Macroscopic Activity. Nothing seems
to be happening – the properties of the system are constant.
These unchanging properties can include: colour, amount of undissolved
solute, concentration, pressure etc.
4. There is Molecular Activity.
Reactions continue at
the microscopic or molecular level
23
Definitions
of some easily confused terms
• Macroscopic: Occurring at the level we can detect with
our senses, as opposed to microscopic. Observable
changes.
• Microscopic: Occurring at a level below what we can
see. Too small to observe without instruments.
• Dynamic Equilibrium: A balance that involves two
opposing active processes that are occurring at the
same rate. This contrasts with Static Equilibrium and
Steady State (Homeostasis).
• Static Equilibrium: A balance that does not involve
active processes.
• Steady State (including Homeostasis): An apparent
balance that occurs in an open system. An unchanging
set of properties is maintained, but materials enter and
leave the system.
24
• Page 287
• Read all the questions, make sure you
understand them, be prepared to answer
them verbally next class.
25
11.4
Le Châtelier’s Principle
• Henri Louis Le Châtelier (1850-1937) was a French
chemist who is most famous for his studies of
chemical equilibrium.
• In addition he studied metal alloys and, with his
father, was involved in the development of methods
of purifying aluminum
26
Equilibrium is not eternal
• An equilibrium can exist for a long time, only
to change (be upset) when certain conditions
change.
• After it is upset, there is a period of
adjustment, then a new equilibrium is
established.
New Equilibrium
Original Equilibrium
27
Some factors that might upset an
equilibrium.
• Which of these factors do you think might
affect the amount of reactant and product
at equilibrium?
• Maybe Temperature?
• Maybe Pressure?
• Maybe Concentration of reactants and
products?
• Maybe Catalyst?
• Most of them do, but one does not.
• We’ll see which one doesn’t a bit later!
28
• Henri LeChâtelier studied many of these
factors to see how they could effect a system
at equilibrium.
• He found some factors could favour the direct
reaction, increasing the amount of product.
• Others could favour the reverse reaction, increasing
the amount of reactant.
• Regardless, eventually an equilibrium was reestablished, but with new amounts of product and
reactant.
29
When an Equilibrium is Upset...
• Henri LeChatelier stated the following
generalization:
• “If the conditions of a system in
equilibrium change, the system will react
to partially oppose this change until it
attains a new state of equilibrium.”
• In other words: The system will establish a new
equilibrium.
.
30
• Will any reversible reaction will eventually
reach equilibrium?
• Answer: yes, as long as it in a closed system.
• Will the amount of product always equal the
amount of reactant at equilibrium?
• Short answer: NO! they are not always equal.
• At equilibrium the amount of “reactant” and
“product” may vary depending on several
factors.
31
The Effect of a Catalyst
• Adding a catalyst to a system already at
equilibrium will have NO EFFECT.
• Adding a catalyst to a system that has not yet
reached equilibrium will cause it to reach
equilibrium faster.
• Why? A catalyst increases both forward and
backward rates equally, so the final result will be
the same, but the process of reaching equilibrium
will be faster.
32
Effect of Temperature
• Increasing the temperature will favour the
endothermic reaction.
• Decreasing the temperature will favour the
exothermic reaction.
33
Effect of Pressure
• Increasing the pressure may favour the
reaction that produces fewer gas particles
• Decreasing the pressure may favour the
reaction that produces more gas particles.
• Note: only reactants or products that are in
the gaseous state are counted towards the
effects of pressure.
34
Effects of Concentration
• The effects of concentration of the reactants
and products are most important of all.
Increasing or decreasing the concentration of
Reactants WILL have an effect.
• Remember: Pure solids(s) and liquids(l) do NOT
have a variable concentration.
• Before we can see the effects of changing a
concentration we should remember what
LeChatelier said...
35
Remember LeChatelier’s Principle:
• “If the conditions of a system in equilibrium
change, the system will react to partially
oppose this change until it attains a new state
of equilibrium.”
• In other words, any “stress” or change that
you make to the system will cause it to react in
a way that tries to (partially) undo the change
that you made.
36
Changes in Concentration
• If you increase the concentration of a reactant
(gas or aqueous), the equilibrium will shift to use up
some of the reactant you added.
• If you increase a product (gas or aqueous), the
equilibrium will shift to reduce the product and
make more reactant
H2(g) + I2(g)
2HI(g)
If you increase the
concentration of
Hydrogen…
…The system will react to reduce
the amount of Hydrogen…
…by shifting the
reaction towards the
product
37
H2 + I2  2HI
 Sudden increase
in[H2].
 Theequilibrium
The
equilibrium
changes:
[HI] goes way up,
[I2] goes down.
This causes [H2] to
adjust towards its
original level
38
Note:
• Adding more of a undissolvable solid or pure
liquid normally has NO EFFECT on a system at
equilibrium.
• It can only affect the equilibrium if the
concentration changes, and usually only gases
and aqueous solutions have variable
concentration.
• However:
• adding water to an aqueous solution can change its
concentration by dilution.
• Adding solid to a solution that is not saturated MIGHT
increase its concentration (if it dissolves!).
39
Changes in Temperature
• Increasing the temperature causes the
equilibrium to shift in the direction that
absorbs heat (the endothermic direction).
• Decreasing the temperature shifts the
equilibrium in the exothermic direction.
2SO2 +
O2
2SO3 +
If you increase the
temperature…
…The system will react to reduce
the temperature…
Heat
…by shifting the
reaction towards
endothermic side
40
2SO2 + O2  2SO3 + heat
 Sudden increase in
temperature.
 The endothermic
reaction kicks in, getting
rid of some SO3, and
creating more SO2 and
more O2 and cooling
things off
 This causes a lowering
of the temperature,
moving it towards its
original level
41
Changes in Pressure
(only affects systems where one or more materials are gases)
• Increasing the pressure causes the equilibrium
to shift in the direction that has the fewest gas
molecules.
• Decreasing the pressure shifts the equilibrium
in the direction that produces more gas
molecules.
8 molecules
If you increase the pressure…
4 molecules
…the system will create fewer molecules…
…to reduce the pressure.
42
N2 + 3 H2  2 NH3
Sudden increase in
pressure.
Pressure partially
adjusts towards the
original level..
When the pressure
increases, the
equilibrium adjusts,
making more NH3
(since 2 molecules NH3 are
fewer particles than 3
molecules of H2 plus 1
molecule of N2)
43
Special Note Re. Gases
• Only gases can be affected by pressure.
• If only one side of an equation has gases,
then...
• Increasing pressure will favour the side with no
gases.
• Decreasing the pressure will favour the side with
gases.
I2(s)
Decreased pressure
Increased pressure
I2(g)
44
• Page 304, Questions 1 to 4
45
Chapter 12
The Quantitative Aspects of
Equilibrium
46
12.1
Chapter 12
In this section we will explore the
mathematical aspects of equilibrium,
including:
• The Equilibrium Constant (Kc)
• The Equilibrium Law
47
The Equilibrium Constant
and Equilibrium Law Expressions
• The equilibrium constant (Kc) is:
• A number
• derived from an equilibrium law expression.
• a ratio between the concentration of products and
the concentration of reactants of a reversible
reaction at equilibrium, but…
• With each aqueous or gaseous product and reactant
raised to the power of its corresponding coefficient
48
Deriving the Equilibrium Law
• The next two slides show how the equilibrium law
was derived from the rate law. If you want to
understand the relationship between rates and
equilibrium you should follow this.
• If you just want to use the equilibrium law to find Kc,
you can skip forward three slides.
49
A Generalized Reversible Reaction with
reactants and products
Forward reaction = dir
Reverse reaction = rev
aA + bB ↔ cC + dD
reactants
products
Forward rate: rdir = kdir [A]a [B]b
Reverse rate: rrev = krev [C]c [D]d
At equilibrium, rdir = rrev
so, through the magic
k dir
of algebra…
k rev
Reactants on bottom
Products on top
c
d
[C ] [ D]

a
b
[ A] [ B]
50
Simplifying the Rate Constant
• The two separate rate constants (kdir and krev) are
often replaced by a single equilibrium constant,
Kc:
c
c
d
k dir [C ] [ D]

a
b
k rev [ A] [ B]
Becomes:
d
[C ] [ D]
Kc 
a
b
[ A] [ B]
51
The Equilibrium Law
For a chemical equation of the type:
aA + bB ↔ cC + dD
The equilibrium law expression is:
c
d
[C ] [ D]
Kc  a b
[ A] [ B]
• The lowercase letters represent
coefficients,
• the uppercase letters are
chemical formulas,
• the square brackets mean
concentration.
• Kc is the equilibrium constant
“Products” always go on the top!
“Reactants” always go on the bottom!
52
c
d
[C ] [ D]
Kc 
a
b
[ A] [ B]
keq
Kc
Ka
Kb
Ksp
Variants of the
Equilibrium Constant
These symbols are all used for
Equilibrium Constants in different text books
or for different types of reaction.
I usually use Kc, since your text book and the
study guide both use it. The old textbook
used keq.
In problems involving acids and bases, Ka and
Kb are often used.
Ksp is used for problems involving the
solubility product.
53
What does the Equilibrium Constant
mean?
• It is a ratio between the amount of reactant and product
that exist at equilibrium.
• If Kc is greater than 1, then the direct reaction is
favoured, and there is more product than reactant at
equilibrium*
• If Kc is less than 1, then the reverse reaction is favoured,
and there is more reactant than product at equilibrium*
• If Kc is 0, the reaction is impossible.
• if Kc is infinite (∞) the reaction is spontaneous and
irreversible so as much reactant as possible will change
to product
• Reactions we call irreversible have high Kc value (>1010)
• Reactions that don’t normally occur have Kc values near 0 (<10-10)
*this is a slight over-simplification, since the formula can be complicated, but it
is generally true.
54
Effect of Temperature on an
Equilibrium Constant
• An equilibrium constant relates to
concentrations, and only remains constant if the
other conditions, such as temperature, remain
fixed.
• If the temperature were to change, so would the
value of Kc
• How the value of Kc might change depends on
the type of reaction (exothermic or endothermic)
and how the temperature changed (increased or
decreased)
55
Table Showing Effects of Temperature
on Equilibrium Constants
Type of reaction
Temperature
change
Favoured Reaction
Change in Kc
Exothermic*
Exothermic*
Endothermic
Endothermic
Increase
Decrease
Increase
Decrease
Reverse ()
Direct ()
Direct ()
Reverse ()
Decrease
Increase
Increase
Decrease
*Exothermic means either ΔH < 0 or that Reactants  Products + Energy
Notice that any change in temperature that favours the
direct reaction will cause the value of Kc to increase.
56
Fixed Concentrations
• Some materials have “fixed” concentrations,
ie. Their concentrations cannot change in an
equilibrium.
• Examples:
• An undissolved solid, (it can’t have a concentration
unless it dissolves.)
• A pure liquid. (a pure substance always has its
maximum concentration)
• These cases, which include all substances with
the (s) and (l) phase markers, are not included
in equilibrium calculations.
57
Example
• What is the equilibrium expression for the following
equation:
CN1-(aq) + H2O(l)  HCN(aq) + OH1- (aq)
Answer:
1
[ HCN ][OH ]
Kc 
1
[CN ]
1
Not this:
[ HCN ][OH ]
Kc 
1
[CN ][ H 2O]
Why? Because H2O (liquid water) is a pure substance and therefore has a fixed
concentration. Substances with fixed concentration are not included in an
equilibrium expression.
58
• Copy the following equations, and write the
equilibrium law expression for each
1) H2(g) + I2(g)  2 HI(g)
2) 2 BrCl(g) Cl2(g) + Br2(g)
3) CO2(g) + H2O(l) H2CO3(aq)
59
12.1.4
Calculating Equilibrium
Concentrations
• To calculate the concentrations of reactants
and products at equilibrium we sometimes
use a table that records the initial
concentration, the change in concentration
and the final equilibrium concentration of
each reactant or product.
• We call such a table an I.C.E. table
60
• The I.C.E. method is a technique that can help
solve some equilibrium problems
• I.C.E. stands for:
• Initial concentration
• Change in concentration
• Equilibrium concentration
[A]I
Δ[A]
[A]E
61
Write thethe
Identify
chemical
molar ratios
equation
(based
of the
on the
reaction.
coefficients
Show
of all
thereactants
equation) and Products
Eliminate any unused ratios, such as those based on liquids and undissolved solids.
Reactant + Reactant  Product + Product
Molar
ratio
Molar
ratio
Molar
ratio
Reactant 1 Reactant 2 Product 1
(if needed)
I
(Initial)
C
(Change)
E
(Equilibrium)
Molar
ratio
Product 2
(if needed)
Concentration Concentration Concentration Concentration
Before
Before
Before
Before
Negative
Ratio
(reactant)
Negative
Ratio
(reactant)
Sum
Sum
Positive
Ratio
(product)
Sum
Concentration
Difference
(Δ[C])
Concentration
After
Fill
inthe
the
missing
squares
in “C”
row. know
They
benegative
ratios
toenough
the
one
you
know
Use
Equilibrium
to
any
further
Make
afind
table
3Concentrations
rows
labelled
I, answer
C,will
and
E,
and
columns
for
all the
Enter
the
information
you
already
into
the
table.
If questions:
no values
are
given
for
Add
to
thewith
equilibrium
values
(adding
a
number
is like
subtracting)
such
as
Kbe
Kthat
or
Kcan
) complete!
the
initial
concentrations,
it is positive
usually
safe
to assume they are zero.
Look
forfinding
aproduct
column
The
reactants
numbers
and
willproducts
you
for
reactants,
interested
in for
products.
c , negative
a Kbyou
spare
62
ICE method: Step by Step
Write the chemical equation of the reaction. Show all reactants and Products
Identify the molar ratios (based on the coefficients of the equation)
Eliminate any unused ratios, such as those based on liquids and undissolved solids
.
Make a table with 3 rows labelled I, C, and E, and enough columns for all the
reactants and products you are interested in
Enter the information you already know into the table. If concentrations are not
given as mol/L you may have to convert them. If not told otherwise, you may
assume that the initial product concentrations are zero.
Look for a column that you can complete!
Fill in the missing squares in “C” row. They will be ratios to the one you know
The numbers will be negative for reactants, positive for products.
Add to find the equilibrium values
(adding a negative number is like subtracting)
Use the Equilibrium Concentrations to answer any further questions (such as finding
Kc , Ka or Kb)
63
Copy this work!
Sample Problem
(see p. 316 and 317 in Text Book)
At a given temperature, 10 moles of nitrogen oxide (NO)
and 8 moles of Oxygen (O2) are placed in a 2 litre
container. After a given period of time, the following
equilibrium is obtained:
2 NO(g) + O2(g)  2NO2(g)
Once equilibrium is attained, 8 moles of NO2 have been
produced. Calculate the equilibrium constant.
Preliminary work: It is necessary to convert to moles per litre!
assume:
[NO]I = 10 mol in 2 L = 5 mol/L
[O2]I = 8 mol in 2L = 4 mol/L
[NO2] I = 0 mol in 2L = 0 mol/L
Final Concentration:
[NO2]E = 8 mol in 2L = 4 mol/L
Initial concentrations
64
Fill in the missing squares in “C” row. They will be ratios to the one you know
Look
for
athe
column
that
you
can
complete!
Make
athe
table
with
3you
rows
labelled
I,
and
E,table
and
enough
columns
information
already
know
the
AddEnter
to
find
equilibrium
values
(adding
ainto
negative
number
is equation)
like
subtracting)
the
molar
ratios
onC,
the
coefficients
of the
Write Identify
the chemical
equation
of(based
the reaction.
Show all reactants
and Products
The numbers
willreactants
be negative
reactants,
for products.
for all the
andfor
products
youpositive
are interested
in
2 NO(g) +
I
(Initial C)
C
(Change)
E
(Equilibrium)
O2(g) 
2NO2(g)
2
NO
1
O2(g)
2
NO2
5 mol/L
4 mol/L
0 mol/L
[NO]i
i[O2]
[NO2]i
-4 mol/L
-2 mol/L
(-ratio)
(-ratiot)
1 mol/L
2 mol/L
+4 mol/L
Δ [NO2]
4 mol/L
[NO2]e
65
Finishing the Problem
Now you need to find the Kc value. To do this, use
the values from the “E” line of your table (the
equilibrium concentrations) and substitute them
into the Kc formula:
2
2
[ NO2 ]
4
16
Kc 



8
2
2
[ NO]  [O2 ] 1  2 2
The value of the equilibrium constant is 8
(although you do not need to give the units of an equilibrium constant,
since our concentrations were in moles/Litre, and our time is assumed
to be in seconds it would be safe to call it mol/(L∙s))
66
A Tougher Sample Problem
(see p. 316 and 317 in Text Book)
At 1100K the equilibrium constant of the following reaction is 25
H2(g) + I2(g)  2HI(g)
2 moles of hydrogen and 3 moles of iodine are placed in a
1 L container. What is the concentration of each
substance when the reaction attains equilibrium..
Initial concentrations: [H2] = 2 mol/L, [I2] = 3 mol/L
This time we have no numbers for equilibrium concentrations, we will
have to use algebraic variable for some of the values.
We can use
x to represent the change in Hydrogen Concentration
Let: Δ[H2] = -x
(why negative? Because hydrogen is a reactant!)
67
Add
to
find
the
equilibrium
values
(adding
awill
negative
number
isone
like
subtracting)
Fill
inMake
the
squares
in
“C”
row.
They
bethe
to
you
know
athe
table
withequation
3ratios
rows
labelled
I,theC,
and
E,ratios
and
enough
columns
Look
Write
Enter
for
Identify
the
amissing
column
chemical
the
information
molar
that
you
you
can
(based
already
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complete!
the
on
reaction.
know
coefficients
into
Show
table
all
ofreactants
thethe
equation)
and
Products
for all the
reactants
andfor
products
youpositive
are interested
in
The numbers
will
be negative
reactants,
for products.
H2(g) +
I
(Initial C)
C
(Change)
E
(Equilibrium)
I2(g) 
2HI(g)
1
H2
1
I2(g)
2
HI
2 mol/L
3 mol/L
0 mol/L
-x mol/L
-x mol/L
+2x mol/L
2-x mol/L
3-x mol/L
2x mol/L
68
Continuing the Problem
Set up the Kc formula, substituting in the expression from line “E”
[ prod .]
Kc 
[react .]
2
So..
( 2 x)
25 
(2  x)(3  x)
This equation can be rearranged in the following steps:
25(2  x)(3  x)  (2 x)
25(6  5x  x )  4 x
2
2
2
150  125 x  25 x  4 x
2
21x  125 x  150  0
2
2
 Carried over to next slide
69
Using the Quadratic formula
21x 2  125 x  150  0
 b  b  4ac
x
2a
2
 Equation to be solved
Quadratic formula, where:
a=21, b=125, c=150
 (125)  (125)  4(21)(150)
x
2(21)
2
x  4.29
Two solutions to formula
substitute
x  1.67
But only one of the solutions will give a real answer!
70
Choosing the Best Answers
x  4.29
Two solutions to formula
x  1.67
Try finding the correct concentrations using the solutions
[ H 2 ]E  2  x [H2 ]  2  4.29  2.29
[ I 2 ]E  3  x
[ HI ]E  2 x
Solve using 4.29
Concentration cannot
be negative!
[ H 2 ]  2 1.67  0.33
1.67
[Solve
I 2 ]  3using
1.67
 1.33
[ HI ]  2(1.67)  3.34
At equilibrium:
The concentration of hydrogen will be 0.33 mol/L,
The concentration of iodine will be 1.33 mol/L and
The concentration of hydrogen iodide will be 3.34 mol/L
71
Simplifying I.C.E. tables
(the 5% rule)
Sometimes, when doing an ICE table you may have
to subtract a very small value from a relatively large
value, for example 2.0 mol/L – 1.0x10-4 mol/L. In
this case, don’t bother doing the subtraction, since
by the time you change it to show significant digits,
the result will be the same:
2.0 mol/L – 0.0001 mol/L = 1.9999 mol/L ≈ 2.0 mol/L.
In fact, as a rule of thumb, if the number you
subtract is less than 5% of the original number, you
can skip the subtraction.
72
• Page 318
• Questions #1 to 8, (Equilibrium Law and Kc)
• Questions #19 to 21, (I.C.E. method)
• Optional Assignments:
• Link to another worksheet
73
12.2
Acids and Bases
• The concept of acids and bases has been with us for a
long time, but there has always been difficulty with the
exact definitions.
• In this section we will look at four theories about acids
and bases (Don’t worry, you only need to remember 2
of them)
• We will find out the difference between “strength”,
“concentration” and “acidity” of an acid.
• We will also explore the concepts of pH and pOH a bit
deeper than you did in grade 9 or 10.
• We will explore the mathematical relationships
between concentration and pH
74
Properties of Acids & Bases
• Acids are solutions that:
• Turn litmus red, but leave phenolphthalein clear.
•
•
•
•
React with active metals to give off H2 gas
React with carbonate salts to give off CO2 gas
Taste sour (if safe to taste)
Have low pH numbers (below 7)
• Bases are solutions that:
• Turn litmus blue, and turn phenolphthalein purple.
•
•
•
•
Taste bitter (if safe to taste)
Seldom react with metals or carbonates
Have High pH numbers (above 7)
Emulsify fats and oils
75
Theories of Acids and Bases
Hypothesis#1: Lavoisier’s Mistake
(optional item)
Lavoisier (1776) dealt mostly with strong
oxyacids, like HNO3 and H2SO4. He claimed
that:
• An acid is a substance that contains oxygen.
• This hypothesis is now known to be completely wrong,
but it did give oxygen its name: “oxy” (meaning acid)+
“gen” (former or creator)
76
Theories of Acids and Bases
(Hypothesis#2: Arrhenius’ Theory)
Arrhenius (c. 1884) claimed that:
• An acid is a substance that dissociates in water
to produce H+ ions
• A base is a substance that dissociates in water
to produce OH- ions
Arrhenius’ theory is still used to this day as a “simplified” way of
explaining acids and bases. It explains most of the properties of
acids... Why their formulas usually begin with H, why they give off
hydrogen when reacting with metals, etc. But this theory does not
account for acidic and basic salts– substances that act like acids and
bases, but don’t have an H or OH in their formula, or for anhydrous
acids, or for reactions that occur outside of water.
77
Theories of Acids and Bases
(Hypothesis#3: Brønsted-Lowry)
Brønsted and Lowry (1923) proposed:
• An acid is a substance from which a proton
(H+) can be removed. An acid is a proton
donor.
• A base is a substance that can cause a proton
to be removed from an acid. A base is a
proton acceptor.
Although a bit complicated for explaining simple acid/base reactions,
this is the main theory in use today.
78
Brønsted-Lowry and Conjugates
• One of the results of the Brønsted-Lowry theory
is that in a reaction, each acid has a
corresponding base and each base has a
corresponding acid (called their conjugates)
H+
H+
HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
Becomes
Becomes
79
Simple way of finding conjugates
• The formula of a conjugate base is the formula of
the acid with one H removed, and more negative
charge.
•
•
•
•
The conjugate base of HBr is BrThe conjugate base of HCN is CNThe conjugate base of water (H2O) is OHThe first conjugate base of H3PO4 is H2PO4-
• H3PO4 can have other conjugates, such as HPO42-, PO43-
• The formula of a conjugate acid is the usually the
negative ion, with an H added in front of it and
one step more positive:
• The conjugate acid of F- is HF
• The conjugate acid of water (H2O) is H3O+
80
Theories of Acids and Bases
(Hypothesis#4: Lewis Acids)
Gilbert Lewis (c.1940) proposed:
• An acid is a substance that can accept an electron
pair to form a covalent bond.
• A base is a substance that can donate a pair of
electrons to form a covalent bond.
This theory is not explained in the new textbook, and will probably not be
required for examinations. It is given here as optional enrichment material.
The Lewis theory is the only one that can explain the properties of acidic &
basic salts.
81
Comparing the Hypotheses
Theory
An Acid is...
A Base is...
Details
Lavoisier
A substance with
oxygen
Arrhenius
A substance that
dissociates in water to
produce H+ ions
A substance that
dissociates in water to
produce OH- ions.
All acids have H
All bases have OH
BronstedLowry
A substance from
which an H+ ion can be
removed (a proton
donor)
A substance which can
remove the H+ from an
acid (a proton acceptor)
All acids have H
Bases can have any
negative ion.
Lewis
A substance that can
accept a pair of
electrons to form a
covalent bond
A substance that can
donate a pair of electrons
to form a covalent bond
Acids may have H+ or
some other + ions, a
base may have OH- or
some other – ions
Incorrect Hypothesis
Incorrect hypothesis
Optional Hypothesis
82
Assignment
• Find the Conjugate base of each of these
acids:
1) HCl
2) HNO3
3) HBr
Find the conjugate acid of each of these bases:
1) CN2) HCO33) H2O
83
Dissociation
• Both theories of acids tell us that acids are compounds
that “dissociate” in water to give off H+ ions (protons),
which immediately attach to water molecules to
become H3O+ ions
• Eg. HCl added to water breaks up into:
• H+ and Cl- ions, the H+ ions join H2O as shown by:
(monoproteic acid)
• HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
• Different acids can dissociate differently:
• H2SO4(aq) + 2H2O(l)  2H3O+(aq) + SO4-(aq)
• H3PO4 (aq) + 3H2O  3H3O+(aq) + PO4-(aq)
(diproteic acid)
(triproteic acid)
84
Dissociation of Water
• In distilled water, at any given moment about
2 molecules out of every billion
(0.00000002%) exist as ions: H+ and OH• This means there is an equilibrium reaction
occurring.
H2O  H+ + OH•
Since so few of the molecules are ionized, the K value of this
equilibrium must be tiny! (see calculation 3 slides from now)
• This also means that water can act as either a very weak acid
or a very weak base! (it can give off H+ or OH-)
85
Equivalence of H+ and H3O+
• Some textbooks use H+ and other books use
H3O+ to represent the hydrogen ion content of
acids.
• Although H3O+ better represents the actual state of
the ions in a water solution, H+ is simpler to write.
• As long as the solvent is water, the two are the same
In most
cases,
when
water is
present
+
[H ]=[H
+
O
]
3
86
Similarity of Ka Kb Kw and Kc
• Ka is called the acidity constant
• Kb is called the basicity constant
• Kw is the ionization constant of water
• but all three are calculated the same way as Kc and are
used for similar purposes.
We usually use Ka or Kb in acid-base reactions
Tables of Ka values help us compare the “natural
strength” of various acids.
87
The Ionization Constant of Water (Kw)
• Water is H2O, but at any given temperature a
tiny fraction (≈ 2/ 109) of water molecules
dissociate into ions:
• H2O(l)  H+(aq) + OH-(aq)
• Kw represents a ratio between the ionized and
unionized molecules:


[ H ][OH ]
Kw 
[ H 2O ]


K w  [ H ][OH ]
88
• Because of the equivalence of H+ and H3O+ in
water solutions, you could also write:


K w  [ H 3O ][OH ]
• At room temperature [H3O+] and [OH-] in pure
water are both about 1.0×10-7 mol/L
So...
7
7
K w  [10 ][10 ]
• The value of Kw varies a bit with temperature
(see p 328), but has an average value of 1.0 ×
10-14 at 25°C.
14
K w 1.0 10
at 25C
89
Effect of Temperature on Kw
• At lower temperatures, Kw becomes smaller.
Less water molecules dissociate when it is cold
• At higher temperatures, Kw becomes larger.
More water molecules dissociate when it is
hot.
• Kw affects the pH of water. We say that the pH of
pure water is 7, but in reality it varies by
temperature:
• At the boiling point (100°C) , the pH of pure water
will be close to 6!
• At the freezing point (0°C) it will be about 7.5!
90
The Equilibrium Constant of an Acid
• The acidity constant of an acid (Ka) represents
the fraction of the acid which will dissociate
and release H+ (ie. H3O+) ions.
• It is one of two factors that affect what the pH
of an acid solution will be.
• (the other factor is the concentration of the acid)
• Acids with low Ka values are considered
naturally “weak” acids. Acids with high Ka
values are considered naturally “strong” acids.
91
If we consider an acid to have the formula
HA then its dissociation can be represented
by HA(aq) + H2O(l)  H3O+(aq) + A-(aq)
In this case, we can calculate the Ka value
this way:


( aq )
( aq )
3
a
( aq )
[H O
][ A
K 
[ HA ]
[H+]
NOTE: Because of the equivalence of
and
[H3O+] in some texts, this could be shown as 
Ka 
]
[H


][ A ( aq ) ]
[ HA( aq ) ] 92
( aq )
The Ionization Percentage of an Acid
If you ever want to work out the
percentage of an acid that is ionized
(like if you had too much time on your
hands, or were bored or something) you
can use this formula:

[ H 3O ( aq) ]
% ionization 
 100
[ HA( aq) ]
93
• Copy the problem and solve:
Johnny dissolves 3.4 mols of dried weak acid
powder in 500 mL of water.
HA(aq) + H2O(l)  H3O+(aq) + A-(aq)
When he measures the acidity he discovers that
the acid has 4.5x10-2 mol/L of H3O+ ions.
What is the Ka of this acid?
What is the percentage ionization of this acid?
94
12.
pH, [H3O+] and Ka of Acids
There are three different factors that affect the
apparent strength of an acid. In this section we
will discover some of the ways of measuring acids
and bases, including:
• Strength vs. concentration vs. acidity
• What are pH and pOH?
• pH (by the chart)
• pH (by calculation)
• Applying the I.C.E. Method to Acids and Bases
• For finding [H+], [OH-]concentrations, Ka, Kb.
95
Measures of an Acid
• There are three different ways of
measuring acids:
1) The natural strength
2) The concentration
3) The degree of Acidity
Ka
mol/L
pH
96
Natural Strength of an Acid
• A “strong” acid is one that dissociates completely
(100%) when dissolved in water. Strong acids
have Ka values close to infinity (huge numbers >
1010). Strong acids dissociate irreversibly.
Typical Reaction: HA + H2O  H3O+ + A• A “weak” acid is one that does NOT dissociate
completely when dissolved in water. Weak acids
have small Ka values, usually less than 1. Weak
acids dissociate reversibly, creating a possible
equilibrium.
Typical reaction: HA + H2O  H3O+ + A97
“Strong” Acids and “Weak” Acids
Acids shown in red, Conjugate Base in blue
Strong Acid Name
Reaction
Hydrochloric Acid
HCl(aq) +H2O H3O+ +
Ka (>>1)
Cl-
∞
Hydrobromic Acid
HBr(aq) +H2O  H3O+ + Br-
∞
Hydroiotic Acid
HI(aq) +H2O H3O+ +
Nitric Acid
HNO3(aq) +H2O  H3O+ + NO3-
∞
Sulphuric Acid
H2SO4+2H2O  2H3O+ + SO4-
∞
I-
∞
Weak Acid Name
Reaction
Ka (<1)
Iodic
HIO3+H2O  H3O++IO3-
1.7x10-1
Oxalic Acid
H2C2O4+H2O H3O++HC2O4
5.8x10-2
Formic Acid
HCOOH +H2O H3O++HCOO-
1.8x10-4
Acetic Acid
CH3COOH +H2O H3O++CH3COO-
1.7x10-5
Citric Acid
H3C6H5O7+H2O H3O++H2C6H5O7
3.2x10-7
Hydrogen Peroxide H2O2 + H2O  H3O+ + HO2-
Water
H2O + H2O  H3O+ OH-
2.4x10-12
1.0x10-14
98
Concentration of an Acid
• The concentration of an acid is the number of
moles of an acidic substance that have been
dissolved in a volume of water. This is
determined by the concentration formula:
n
C
V
Where:
n = number of moles of acidic solute
V= volume of solution in Litres
6 mol/L is considered a very concentrated acid
0.1 mol/L is considered a fairly dilute acid
99
Strength, Concentration and Acidity
The strength and concentration of an acid
together determine its degree of acidity
Concentration
STRENGTH
The natural strength of an
acidic compound,
determined by its Ka value
The number of moles per
litre of the acidic compound
in solution.
Degree of Acidity
The effective strength of an acid,
based on its [H3O+] concentration and
Usually recorded as its pH
100
Degree of Acidity
The degree of acidity is the EFFECTIVE
strength of the acid, that is, how effective
the acid is at reacting with other materials.
Acidity can be measured using:
a) the H3O+ concentration, or more commonly
b) a measure called the pH
101
Who invented pH and pOH?
(optional background information)
The degree of acidity of a solution is directly related
to its [H3O+] concentration. Unfortunately this is
often a small number expressed in scientific
notation, such as 1.3x10-5 mol/L or 2.3x10-13 mol/L.
Not easy numbers to remember, compare or write.
In 1909 Søren Sørensen suggested an easier way to
record the degree of acidity of solutions. It was a
logarithmic scale called pH. Each level of the pH
scale represents a 10 fold difference in H3O+ (or H+)
concentration.
ie. A acid of pH 2 has 10 times more [H+] than one that’s pH3.
102
pH and pOH
pH is a measure of the degree of acidity,
given by the following formula:
pH = - log [H3O+]
Or: (since [H3O+] is equivalent to [H+] in water solutions)
pH = - log [H+]
Although less used, there is also a measure
of the degree of alkalinity, called pOH:
pOH = - log [OH-]
103
pH and Concentration of H+ ions
For solutions of an exact pH you may use the chart below:
For simplicity I used
[H+] instead of [H3O+]
Acids
pH [H+]* [OH-]* pOH
0
1
2
3
4
5
6
7
10 0
10-1
10-2
10-3
10-4
10-5
10-6
10-7
10-14
10-13
10-12
10-11
10-10
10-9
10-8
10-7
14
13
12
11
10
9
8
7
Bases
pH [H+]* [OH-]* pOH
8
9
10
11
12
13
14
10-8
10-9
10-10
10-11
10-12
10-13
10-14
10-6
10-5
10-4
10-3
10-2
10-1
10-0
6
5
4
3
2
1
0
*Concentration in mol/L
Temperature = 25°C
104
What about “in-between” pH values?
• The formula for pH is:
• pH = -log [H+]
• On your calculator you must find out how to calculate the
negative logarithm of a number!
• The formula for [H+] concentration is:
• [H+]=10 –pH
or…
[H+]=log-1(-pH)
• Sometimes log-1 is called antilog: [H+] =antilog (-pH)
• Sometimes log-1 is called inverse log:
[H+] =invlog (-pH)
• Sometimes log-1 is called 10x:
[H+] =10(-pH)
• On your calculator find out raise 10 to a negative number
or how to calculate the inverse logarithm (antilog) of a
negative number.
105
Question: Find the pH if the H3O+
concentration is 1.40x10-8 mol/L
Solution: pH = – log(1.40 x 10 – 8)
(TI 83 instructions)
log (1.40 x 10 ^^ (–)
Enter
Type: (–) log
(-) 8) Enter
The answer should be:
pH=7.85… (I’ve rounded to 3 Sig.Fig.)
Question: Find the H3O+ concentration if
the pH is 4.30
Solution: [H+] =10 – 4.30
(TI-83 instructions)
^ (-)
(–) 4.3
Enter
10 ^
The answer should be:
pH=5.01… x 10-5 mol/L(I’ve rounded to 3 Sig.Fig.)
106
Using the Windows Calculator
• Switch to scientific view
To find pH, use: 1.4 Exp 8
To find [H3O+] use: 4.30
or:
10
=
5.01…x10-5
=
4.30
7.85…
=
Finding the pH if H3O+
concentration is 1.40x10-8 mol/L
Finding [H3O+] concentration if
pH is 4.30
107
Traditional Scientific Calculators
Question: Find the pH if the H3O+
concentration is 1.40x10-8 mol/L
Solution: pH = – log(1.40 x 10 – 8)
Try:
Exp 8 +/+/- ) log
+/- =
log +/(1.4 Exp
7.85...
Or:
EE 8 +/+/- ) log
+/- =
log +/(1.4 Exp
Question: Find the H+ concentration if the
pH is 4.30
Solution: [H+] = 10 - 4.30
Try:
Or:
Or
Or:
10 yx 4.30 +/- =
4.30 +/- Inv log =
4.30 +/- 2ndF 10x =
10 ^ (-) 4.3
5.01... ×10-5
108
1. Find the [H+] and [OH-] of the following pH solutions without
using a calculator:
• A) pH=4 B) pH=12 C) pH=9 D) pH=7 E) pH=8
2. Find the pH of the following solutions without a calculator:
• A) [H+]=1.0x10-8 mol/L
• C) [H+]=1.0x10-5 mol/L
B) [OH-]=1.0x10-3 mol/L
D) [OH-]=1.0x10-9 mol/L
3. Find the [H+] of these solutions using the 10x or antilog
function of your calculator: [H+]=10-pH
• A) pH=3.7 B) pH=9.8 C) pH=6.2
D) pH=4.0 E) pH=7.1
4. Find the pH of the following solutions using the log function
of your calculator: pH= - log [H+].
• A) [H+]=4.5x10-3 mol/L
• C) [H+]=3.0x10-7mol/L
B) [H+]=3.4x10-8 mol/L
D) [H+]=2.5x10-2 mol/L
109
2
sig. digits
Problem 5.
5. Calculate the Ka value of a monoproteic, weak
acid if a 0.10 mol/L (initially) solution has a pH
of 5.5 when it reaches equilibrium.
• Assume: HA H+ + A• Hint: You must find the concentration of [H+] ions
before you do the problem.
2
sig. digits
110
Solutions to Problems 1- 4
1. Answers
• A) 1x10-4, 1x10-10 mol/L
• C) 1x10-9, 1x10-5 mol/L
• E) 1x10-8, 1x10-6 mol/L
B) 1x10-12, 1x10-2 mol/L
D) 1x10-7, 1x10-7 mol/L
2. Answers:
• A) pH=8
• C) pH=5
B) pOH-=3 pH= 11
D) pOH-=9 pH=5
3. Answers in mol/L
• A) 2.0x10-4 mol/L
• D) 1.0x10-4 mol/L
B) 1.6x10-10 mol/L C) 6.3x10-7 mol/L
E) 7.9x10-8 mol/L
4. Answers: pH= - log [H+].
• A) pH= 2.3
• C) pH=6.5
B) pH=7.5
D) pH=1.6
111
Solution to Problem 5
• Calculate the Ka value of a monoproteic weak acid if a 0.1
mol/L (initially) solution has a pH of 5.5 when it reaches
equilibrium.
• Assume: HA H+ + A• Hint: You must find the concentration of [H+] ions before
you do the problem.
• Find the [H+]:
• [H+] = log-1 (-5.5)
or
[H+] = 10 -5.5
• [H+] = 3.162x10-6 mol/L
• For an extremely accurate solution, use the I.C.E. method to
find the equilibrium concentrations… (continued on next slide)
• For a quicker solution, using the 5% rule, see later slides
112
HA  H+
1
I
(initial)
C
E
:
1
+
:
A-
Round to 2 Sig.
Digits
1
HA
H+
A-
0.10
0
0
(change)
-3.163x10-6 +3.163x10-6 +3.163x10-6
(equil.)
-6
9.9997x10-2 3.163x10-6 3.163x10
Ka =
[H+][A-]
[HA]
= (3.163x10-6 x 3.163x10-6) mol/L
9.9997x10-2 mol/L
=1.0004 x10-10
≈ 1.0x10-10
113
The Short-cut:
An alternative solution using 5% rule
Ka =
[H+][A-]
= (3.163x10-6 x 3.163x10-6) mol/L
[HA]
I
mol/L
Since 3.163 x 10-6 is less than 5% of
0.1, we can use the 5% rule and
simply write
of
HA 0.1, insteadH+
subtracting (0.1 – 0.000003163)
and using an ICE table
(initial)
C
(change)
E
(equil.)
0.10
0
assume
0.10
0.10
=1.0004 x10-10
3.163x10-6
= 1.0x10-10
A0
After rounding to the correct
number of significant digits, our
answer is the same as doing it the
-6 hard way!
3.163x10
114
A Half-Evil Example
Calculate the pH of an aqueous solution of formic
acid HCOOH at 0.20 mol/L if its acidity constant is
1.8x10-4.The equation of this reaction is as follows
HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO-(aq)
See the solution on page 333
(I told you, it’s only half-evil)
115
Assignments
Page 339, Questions 4 to 23
116
Unit 4: Chapter 12.2.5
Solubility Product Constant
Solubility
Calculating the solubility constant
Examples
117
Solubility
Dissolved ions 
Solid solute crystals
• A saturated solution contains:
• Dissolved solute (usually ions) in the solution
• Some non-dissolved solute (crystal)at the bottom
118
In a saturated solution...
• There is an equilibrium situation:
Solute(s)  ion+(aq) + ion–(aq)
Some of the solute dissociates (dissolves into
ions) and at the same time, some of the ions
crystallize back into solid.
119
Solubility
• The solubility of a substance is the maximum
amount of the substance that dissolves in a
given volume of the solvent
• Usually given in g/L or sometimes in g/100mL
• For calculation purposes you should use molar
solubility (mol/L)
• To convert g/100 mL to g/L, multiply by 10
• To convert g/L to mol/L, change grams to moles using
the mole formula:
Moles
m
n
M
mass (g)
molar mass(g/mol)
120
• The solubility of a substance partly
depends on the temperature of the water
you are dissolving it in.
• Standard tables give solubility at 25°C
• Solids generally have a higher solubility at
higher temperatures
• Gases generally have a lower solubility at
higher temperatures.
• There are a few exceptions to these
generalizations!
121
What Is Ksp
• The solubility product constant is a number
used to compare the solubilities of different
solutes.
• The higher the Ksp, the more of the solute can
be dissolved before the solution becomes
saturated. Low Ksp values mean very little of
the substance will dissolve.
• Ksp is calculated in a similar way to other
equilibrium constants, but it’s a bit easier!
122
“Soluble” and “Insoluble” Substances
• A “soluble” substance is a substance with a
high Ksp value. At room temperature, a large
amount of the substance can be dissolved in
water.
• A truly insoluble substance doesn’t exist, but
any substance with a very low Ksp is said to be
“insoluble”, since so little of it will dissolve in
water that it can barely be measured.
See table 8.11 on page 423 to find out what common
substance will be “soluble” and “insoluble”
123
General Formula for Ksp
• For a solute that dissociates like this:
XmYn(s)  mX+(aq) +nY-(aq)
• The Ksp can be calculated by this way:
Ksp = [X+]m[Y-]n
124
Calculating Ksp
• Example
1:
1
:
1
Eliminate solid.
:
1
• BaSO4(s)  Ba2+(aq) + SO42-(aq)
• At equilibrium there will be a certain concentration of
Ba2+ ions and SO42- ions
Ksp = [Ba2+][SO42-]
• Example
2: 1
1
:
:
2
• CaCl2(s) Ca2+(aq) + 2Cl-(aq)
Eliminate solid.
• At equilibrium there will be a certain concentration of
Ca2+ ions and Cl- ions
Ksp = [Ca2+][Cl-]2
125
1
:
1
Dissociates equally (1:1)
• Ksp = [Ba2+][SO32-]
• If the solubility of BaSO3 is 0.0025g/L, what is the
Ksp of this compound?
• Since it dissociates equally, the concentration of
both ions will be equal to the moles of BaSO3 that
dissolved (0.0025 g/L).
• We have to convert that to mol/L
m • MBaSO3 = 137.3 + 32.1 + 3(16.0) = 217.4 g/mol
n
-5 mol
•
n
=
0.0025
g/L
/
217.4
g/mol
=1.15x10
BaSO3
M
• So... [Ba2+] = [SO32-] = 1.15x10-5 mol/L
• Ksp
= [Ba2+][SO32-]
=(1.15x10-5)(1.15x10-5) = 1.32x10-10 mol2/L2
126
Example
• The solubility of silver carbonate (Ag2CO3) is
3.6x10-3 g/100mL at 25C. Calculate the value
of the solubility product constant of silver
carbonate.
Data:
Solubility = 3.6×10-3 g/100mL
Molar Solubility =?
Use solubility:
[+ ions] = ?
m = g (per litre)
[– ions] = ?
Ksp= ?
M = 2(107.8)+12+3(16)
Values from periodic table and
formula of Ag2CO3.
we want the solubility in g/L, so multiply by
10...
3.6×10-3 g/100mL = 3.6×10-2 g/L
Now we need it in mol/L, so...
n=
n=
3.6×10-2 g
/L
275.8 g/mol
1.3×10-4 mol/L
m
n
M
Molar solubility = 1.3x10- 4 mol/L
Carry over to the next slide.
127
Info carried over
from previous slide
The molar concentration of CO3 2-(aq)
in solution will equal the molar
solubility!
Data:
Solubility = 3.6x10-3 g/100mL
Molar Solubility= 1.3x10-4 mol/L
Equation:
1
:
2
:
1
Ag2CO3(s)  2 Ag+ + CO32[CO32-]
[Ag+]
= 1.3x10- 4 mol/L
=
2.6x10- 4
mol/L
[Ag+] is double [CO32-]. Multiply by 2 
Ksp = [Ag+]2 [CO32-]
[CO32-]=1.3×10- 4 mol/L
[Ag+] = 2.6×10- 4 mol/L
Ksp=?
Ksp=(2.6×10- 4 mol/L)2(1.3×10-4 mol/L)
Ksp =8.8×10-12 mol3/L3
128
• At the annual chemistry Christmas party, a
careless chemist spills a whole bottle of calcium
fluoride, CaF2, into a 2 litre punch bowl filled
with fruit punch. Most of the calcium fluoride
dissolves, but a little powder settles to the
bottom of the punch bowl.
•
•
•
•
The Ksp of calcium fluoride is 3.4*10-11
CaF2 dissociates: CaF2(s)  Ca2+(aq) + 2F-(aq)
The lethal dose of fluoride ions is 1g.
Will all of the eight chemists at the party die if each
drinks a cup of the punch? (1 cup ≈ 250mL or ¼ L.)
• Would one chemist die if he drank all the punch?
129
• Ksp = [Ca2+] [F-]2
• The dissociation formula is:
CaF2(s)  Ca2+(aq) + 2F-(aq)
• So there will be twice as much F- produced as
Ca2+
• Let’s choose a variable to represent the
concentration of Ca2+ ions produced at
equilibrium. Double that variable to represent
the F- ion concentration.
Let x = [Ca2+] and 2x = [F-]
Note: The units of x will be mol/L
130
• Ksp = [Ca2+] [F-]2
• Simplify x(2x)2
• Reverse the equation
and divide by 4 to
isolate the x3
• Simplify:
• Cube root of both
sides...
• ...Gives us “x”, but we
aren’t finished...
3.4 10
11
 x( 2 x)
3.4  10
11
 4x
3.4 10
x 
4
2
3
11
3
x  8.5 10
3
3
12
x  8.5 10
3
3
12
4
x  2.04 10 mol / L
131
4
x  2.04 10 mol / L
But x represents [Ca2+], and since the [F-] is twice as
high:
[ F  ]  2  2.04 104  4.08 104 mol / L
So to change this to grams per litre we must
multiply by the molar mass of F-, which is 19.0
g/mol (note: this is not F2)
[ F  ]  4.08 104 19.0  7.76 103 g / L
A cupful (250mL) is one quarter of this, so...

3
[ F ]  1.94 10 g / cup
132

3
[ F ]  1.94 10 g / cup
• Each cup contains only 0.00194 g of fluoride ions,
so nobody would die from drinking one cup.
• If somebody drank the whole punchbowl (2 litres
= 8 cups) he would still only get 0.0155 g of
fluoride ions. Still well below the deadly limit.
The chemists would survive and have sparkling
white teeth!
• Note: Although this dosage might not be lethal it could have longterm side effects. Never drink contaminated punch! The maximum
recommended concentration of fluoride in water is 1 mg/L, about one
eighth what is in the punch.
133
Exercises on Solubility
• Page 340 # 36 to 39
134