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Resources
Chapter Presentation
Bellringer
Transparencies
Sample Problems
Visual Concepts
Standardized Test Prep
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Chapter 14
Chemical Equilibrium
Table of Contents
Section 1 Reversible Reactions and Equilibrium
Section 2 Systems at Equilibrium
Section 3 Equilibrium Systems and Stress
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Bellringer
• Describe what reversible means.
• Find a synonym for reversible.
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Objectives
• Contrast reactions that go to completion with
reversible ones.
• Describe chemical equilibrium.
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Completion Reactions and Reversible
Reactions
• If enough oxygen gas is provided for the following
reaction, almost all of the sulfur will react:
S8(s) + 8O2(g) → 8SO2(g)
• Reactions such as this one, in which almost all of the
reactants react, are called completion reactions.
• In other reactions, called reversible reactions, the
products can re-form reactants.
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Completion Reactions and Reversible
Reactions, continued
Reversible Reactions Reach Equilibrium
• One reversible reaction occurs when you mix
solutions of calcium chloride and sodium sulfate.
CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + 2NaCl(aq)
• The net ionic equation best describes what happens.
Ca2 (aq)  SO24 (aq)  CaSO4 (s)
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Completion Reactions and Reversible
Reactions, continued
Reversible Reactions Reach Equilibrium, continued
• Solid calcium sulfate, the product, can break down to
make calcium ions and sulfate ions in a reaction that
is the reverse of the previous one.
CaSO4 (s)  Ca2 (aq)  SO24 (aq)
• Use arrows that point in opposite directions when
writing a chemical equation for a reversible reaction.


Ca2 (aq )  SO2(
aq
)
CaSO 4 (s )


4
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Completion Reactions and Reversible
Reactions, continued
Reversible Reactions Reach Equilibrium, continued
• The reactions occur at the same rate after the initial
mixing of CaCl2 and Na2SO4.
• The amounts of the products and reactants do not
change.
• Chemical equilibrium is a state of balance in which
the rate of a forward reaction equals the rate of the
reverse reaction and the concentrations of products
and reactants remain unchanged.
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Completion Reactions and Reversible
Reactions, continued
Opposing Reaction Rates Are Equal at Equilibrium
• The reaction of hydrogen, H2, and iodine, I2, to form
hydrogen iodide, HI, reaches chemical equilibrium.

 2HI(g )
H2 (g )  I2 (g ) 

• Only a very small fraction of the collisions between H2
and I2 result in the formation of HI.
H2(g) + I2(g) → 2HI(g)
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Completion Reactions and Reversible
Reactions, continued
Opposing Reaction Rates Are Equal at Equilibrium,
continued
• After some time, the concentration of HI goes up.
• As a result, fewer collisions occur between H2 and I2
molecules, and the rate of the forward reaction drops.
• Similarly, in the beginning, few HI molecules exist in
the system, so they rarely collide with each other.
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Completion Reactions and Reversible
Reactions, continued
Opposing Reaction Rates Are Equal at Equilibrium,
continued
• As more HI molecules are made, they collide more
often and form H2 and I2 by the reverse reaction.
2HI(g) → H2(g) + I2(g)
• The greater the number of HI molecules that form, the
more often the reverse reaction occurs.
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Rate Comparison for H2(g) + I2(g)  2HI(g)
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Completion Reactions and Reversible
Reactions, continued
Opposing Reaction Rates Are Equal at Equilibrium,
continued
• When the forward rate and the reverse rate are equal,
the system is at chemical equilibrium.
• If you repeated this experiment at the same
temperature, starting with a similar amount of pure HI
instead of the H2 and I2, the reaction would reach
chemical equilibrium again and produce the same
concentrations of each substance.
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Chapter 14
Visual Concepts
Chemical Equilibrium
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Reaching an equilibrium state.
In most reversible reactions balance points exist
between the forward and backward reactions.
•Reactants and products appear together
•The reaction appears to have stopped
•Neither forward or backward reaction is complete
•This is chemical equilibrium
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Chemical Equilibria Are Dynamic
• If you drop a ball into a bowl, it will bounce.
• When the ball comes to a stop it has reached static
equilibrium, a state in which nothing changes.
• Chemical equilibrium is different from static
equilibrium because it is dynamic.
• In a dynamic equilibrium, there is no net change in
the system.
• Two opposite changes occur at the same time.
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Chemical Equilibria Are Dynamic, continued
• In equilibrium, an atom may change from being part
of the products to part of the reactants many times.
• But the overall concentrations of products and
reactants stay the same.
• For chemical equilibrium to be maintained, the rates
of the forward and reverse reactions must be equal.
• Arrows of equal length also show equilibrium.

 products
reac tants 

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Dynamic Equilibrium
In dynamic equilibrium the forward and
backwards reactions continue at equal rates so
the overall effect does not change. On a molecular
scale there is continuous change. On the
macroscopic scale nothing appears to be
happening. The system needs to be closed –
isolated from the outside world.
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Chemical Equilibria Are Dynamic, continued
• In some cases, the equilibrium has a higher
concentration of products than reactants.
• This type of equilibrium is also shown by using two
arrows.
reac tants 

 products
• The forward reaction has a longer arrow to show
that the products are favored.
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
More Examples of Equilibria
• Even when systems are not in equilibrium, they are
continuously changing in order to reach equilibrium.
• For example, combustion produces carbon dioxide,
CO2, and poisonous carbon monoxide, CO. After
combustion, a reversible reaction produces soot.

 C(s )  CO2 (g )
2CO(g ) 

• This reaction of gases and a solid will reach chemical
equilibrium.
• Equilibria can involve any state of matter, including
aqueous solutions.
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Chapter 14
Section 1 Reversible Reactions and
Equilibrium
Homework
1. Read pgs.496-499
2. Solve Section Review
pg.501, Qs. 1-4
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Section Review, pg.501
1. Which of the following equations best represents
a reaction that goes to completion?
a. Reactants
products
b. reactants
products
c. Reactants
products
2. At equilibrium, what is the relationship between the
rates of the forward and reverse reactions?
The rate of the reactions are equal.
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3. Explain what each reaction below shows. Describe
the relationship between the amounts of products and
reactants for each case.
a. reactants
products
b. Reactants
products
a.The reaction is reversible and at equilibrium. The relative
amounts of products and reactants cannot be determined.
b. This reversible reaction is also at equilibrium; however,
there are more products than reactants.
4. What is the difference between dynamic and static
equilibria?
During static equilibrium, nothing changes. During dynamic
equilibrium, there are no net changes.
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Chapter 14
Section 2 Systems at Equilibrium
Bellringer
• Make a list of numbers that are “constants” under
constant conditions.
• An example is the speed of light.
• What do these constants have in common?
• Answer: Each constant is always the same number
for a certain and constant set of conditions.
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Chapter 14
Section 2 Systems at Equilibrium
Objectives
• Write Keq expressions for reactions in equilibrium,
and perform calculations with them.
• Write Ksp expressions for the solubility of slightly
soluble salts, and perform calculations with them.
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Chapter 14
Section 2 Systems at Equilibrium
The Equilibrium Constant, Keq
• Limestone caverns form as rainwater, slightly acidified
by H3O+, dissolves calcium carbonate.
• The reverse reaction also takes place, depositing
calcium carbonate and forming stalactites and
stalagmites.

 Ca2 (aq )  CO2 (g )  3H2O(l )
CaCO3 (s )  2H3O (aq ) 

• When the rates of the forward and reverse reactions
become equal, the reaction reaches chemical
equilibrium.
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Chapter 14
Section 2 Systems at Equilibrium
The Equilibrium Constant, Keq, continued
• There is a mathematical relationship between product
and reactant concentrations at equilibrium.
• For limestone reacting with acidified water at 25°C:
[Ca2 ][CO2 ]
-9
Keq 

1.4

10
[H3O ]2
• Keq is the equilibrium constant of the reaction.
• Keq for a reaction is unitless, applies only to systems
in equilibrium, and depends on temperature and must
be found experimentally or from tables.
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Chapter 14
Visual Concepts
Equilibrium Constant
The equilibrium constant always
has the same value (provided
you don't change the
temperature), irrespective of the
amounts of A, B, C and D you
started with.
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Chapter 14
Section 2 Systems at Equilibrium
The Equilibrium Constant, Keq, continued
Determining Keq for Reactions at Chemical
Equilibrium
1. Write a balanced chemical equation.
• Make sure that the reaction is at equilibrium
before you write a chemical equation.
2. Write an equilibrium expression.
• To write the expression, place the product
concentrations in the numerator and the reactant
concentrations in the denominator.
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Chapter 14
Section 2 Systems at Equilibrium
The Equilibrium Constant, Keq, continued
Determining Keq for Reactions at Chemical
Equilibrium, continued
• The concentration of any solid or a pure liquid that
takes part in the reaction is left out.
• For a reaction occurring in aqueous solution,
water is omitted.
3. Complete the equilibrium expression.
• Finally, raise each substance’s concentration to
the power equal to the substance’s coefficient
in the balanced chemical equation.
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Chapter 14
Visual Concepts
Equilibrium Constant
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Chapter 14
Section 2 Systems at Equilibrium
Calculating Keq from Concentrations of
Reactants and Products
Sample Problem A
An aqueous solution of carbonic acid reacts to reach
equilibrium as described below.
–



H2CO3 (aq )  H2O(l ) 
HCO
(
aq
)

H
O
(aq )

3
3
The solution contains the following solution
concentrations: carbonic acid, 3.3 × 10−2 mol/L;
bicarbonate ion, 1.19 × 10−4 mol/L; and hydronium ion,
1.19 × 10−4 mol/L. Determine the Keq.
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Section 2 Systems at Equilibrium
Chapter 14
Calculating Keq from Concentrations of
Reactants and Products
Sample Problem A Solution
[H2CO3 ]  3.3 10-2
[HCO3– ]  [H3O ]  1.19 10-4
For this reaction, the equilibrium constant expression is
Keq
[HCO3– ][H3O ]

[H2CO3 ]
Substitute the concentrations into the expression.
Keq
[HCO3– ][H3O ] (1.19  10-4 )  (1.19  10-4 )
-7



4.3

10
[H2CO3 ]
(3.3  10-2 )
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H.W. #2
Sample Problem A, practice pg. 506
1. For the system involving N O and NO at
equilibrium at a temperature of 100C, the product
concentration of N O is 4.0  10-2 mol/L and the
reactant concentration of NO is 1.4  10-1 mol/L. What
is the K value for this reaction?
2
2
4
2
4
2
eq
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Chapter 14
Section 2 Systems at Equilibrium
The Equilibrium Constant, Keq, continued
Keq Shows If the Reaction Is Favorable
• When Keq is large, the numerator of the equilibrium
constant expression is larger than the denominator.
• Thus, the concentrations of the products will
usually be greater than those of the reactants.
• In other words, when a reaction that has a large
Keq reaches equilibrium, there will be mostly
products.
 Reactions in which more products form than
reactants form are said to be “favorable.”
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Chapter 14
Section 2 Systems at Equilibrium
The Equilibrium Constant, Keq, continued
Keq Shows If the Reaction Is Favorable, continued
• The synthesis of ammonia is very favorable at 25°C
and has a large Keq value.

 2NH3 ( g )
N2 (g )  3H2 (g ) 

K eq
[NH3 ]2
8


3.3

10
[N2 ][H2 ]3
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at 25 C
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Chapter 14
Section 2 Systems at Equilibrium
The Equilibrium Constant, Keq, continued
Keq Shows If the Reaction Is Favorable, continued
• However, the reaction of oxygen and nitrogen to give
nitrogen monoxide is not favorable at 25°C.

 2NO(g )
N2 (g )  O2 (g ) 

K eq
[NO2 ]2

 4.5  10–31
[N2 ][O2 ]
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at 25 C
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Chapter 14
Section 2 Systems at Equilibrium
The Equilibrium Constant, Keq, continued
Keq Shows If the Reaction Is Favorable, continued
• When Keq is small, the denominator of the equilibrium
constant expression is larger than the numerator.
• The larger denominator shows that the concentrations
of reactants at chemical equilibrium may be greater
than those of products.
 A reaction that has larger concentrations of
reactants than concentrations of products is an
“unfavorable” reaction.
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Chapter 14
Section 2 Systems at Equilibrium
The Equilibrium Constant, Keq, continued
Keq Shows If the Reaction Is Favorable, continued
These pie charts show the relative
amounts of reactants and products
for three Keq values of a reaction.
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Chapter 14
Section 2 Systems at Equilibrium
Calculating Concentrations of Products
from Keq and Concentrations of Reactants
Sample Problem B
Keq for the equilibrium below is 1.8 × 10−5 at a
temperature of 25°C. Calculate [NH4 ] when
[NH3] = 6.82 × 10−3.

–


NH3 (aq )  H2O(l ) 
NH
(
aq
)

OH
(aq )

4
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Chapter 14
Section 2 Systems at Equilibrium
Calculating Concentrations of Products
from Keq and Concentrations of Reactants,
continued
Sample Problem B Solution
The equilibrium expression is
NH4and OH− ions are produced in equal numbers, so
[OH– ]  [NH4 ].
So, the numerator can be written as x2.
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Chapter 14
Section 2 Systems at Equilibrium
Calculating Concentrations of Products
from Keq and Concentrations of Reactants,
continued
Sample Problem B Solution, continued
Keq and [NH3] are known and can be put into the
expression.
1.8  10–5  Keq
[NH4 ][OH– ]
x2


[NH3 ]
6.82  10-3
x2 = (1.8  10−5)  (6.82  10−3) = 1.2 × 10−7
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Chapter 14
Section 2 Systems at Equilibrium
Calculating Concentrations of Products
from Keq and Concentrations of Reactants,
continued
Sample Problem B Solution, continued
Take the square root of x2.
[NH4 ]  1.2  10-7  3.5  10-4
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Chapter 14
HW. #2
Sample Problem B, practice pg. 506
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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp
• The maximum concentration of a salt in an aqueous
solution is called the solubility of the salt in water.
• Solubilities can be expressed in moles of solute per
liter of solution (mol/L or M).
• For example, the solubility of calcium fluoride in
water is 3.4 × 10−4 mol/L.
• So, 0.00034 mol of CaF2 will dissolve in 1 L of
water to give a saturated solution.
• If you try to dissolve 0.00100 mol of CaF2 in 1 L
of water, 0.00066 mol of CaF2 will remain
undissolved.
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Chapter 14
Visual Concepts
Solution Equilibrium
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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp, continued
• Calcium fluoride is one of a large class of salts that are
said to be slightly soluble in water.
• The ions in solution and any solid salt are at
equilibrium.
2
–


CaF2 (s ) 
Ca
(
aq
)

2F
(aq )

• Solids are not a part of equilibrium constant
expressions, so Keq for this reaction is the product
of [Ca2+] and [F−]2, which is equal to a constant.
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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp,
continued
• Equilibrium constants for the dissolution of slightly
soluble salts are called solubility product
constants, Ksp, and have no units.
• The Ksp for calcium fluoride at 25°C is 1.6  10−10.
Ksp = [Ca2+][F−]2 = 1.6  10−10
• This relationship is true whenever calcium ions and
fluoride ions are in equilibrium with calcium fluoride,
not just when the salt dissolves.
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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp,
continued
• For example, if you mix solutions of calcium nitrate
and sodium fluoride, calcium fluoride precipitates.
• The net ionic equation for this precipitation is the
reverse of the dissolution.

 CaF2 (s )
Ca2 (aq )  2F– (aq ) 

• This equation is the same equilibrium. So, the Ksp for
the dissolution of CaF2 in this system is the same and
is 1.6 × 10−10.
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Chapter 14
Visual Concepts
Solubility Product Constant
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Chapter 14
Visual Concepts
Solubility Product Constant
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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp, continued
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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp, continued
Determining Ksp for Reactions at Chemical Equilibrium
1. Write a balanced chemical equation.
• Solubility product is only for salts that have low
solubility. Soluble salts do not have Ksp values.
• Make sure that the reaction is at equilibrium.
• Equations are always written so that the solid salt
is the reactant and the ions are products.
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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp, continued
Determining Ksp for Reactions at Chemical Equilibrium
2. Write a solubility product expression.
•
Write the product of the ion concentrations.
•
Concentrations of solids or liquids are omitted.
3. Complete the solubility product expression.
•
Raise each concentration to a power equal to
the substance’s coefficient in the balanced
chemical equation.
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Chapter 14
Section 2 Systems at Equilibrium
Calculating Ksp from Solubility
Sample Problem C
Most parts of the oceans are nearly saturated with
CaF2.The mineral fluorite, CaF2, may precipitate when
ocean water evaporates. A saturated solution of CaF2
at 25°C has a solubility of 3.4 × 10−4 M. Calculate the
solubility product constant for CaF2.
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Chapter 14
Section 2 Systems at Equilibrium
Calculating Ksp from Solubility, continued
Sample Problem C Solution
2
–


CaF2 (s ) 
Ca
(
aq
)

2F
(aq )

[CaF2] = 3.4  10–4, [F−] = 2[Ca2+]
Ksp = [Ca2+][F−]2
Because 3.4 × 10−4 mol CaF2 dissolves in each liter of
solution, you know from the balanced equation that
every liter of solution will contain 3.4 × 10−4 mol Ca2+
and 6.8 × 10−4 mol F−. Thus, the Ksp is:
[Ca2+][F−]2 = (3.4 10−4)(6.8  10−4)2 =
1.6 × 10−10
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HW. # 3
Sample Problem C, practice pg. 509
1) Copper(I) bromide is dissolved in water to saturation
at 25C. The concentration of Cu ions in solution is
7.9  10-5 mol/L. Calculate the Ksp for copper(I)
bromide at this temperature. (ans. 6.2E-9)

2) What is the Ksp value for Ca (PO ) at 298 K if the
concentrations in a solution at equilibrium with excess
solid are 3.42  10-7 M for Ca ions and 2.28  10 M for
PO -3 ions? (ans.2.08E-33)
3
4 2
-7
2
4
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Chapter 14
Section 2 Systems at Equilibrium
Calculating Ionic Concentrations Using Ksp,
Sample Problem D
Copper(I) chloride has a solubility product constant of
1.2 × 10−6 and dissolves according to the equation
below. Calculate the solubility of this salt in ocean
water in which the [Cl−] = 0.55.

–


CuCl(s ) 
Cu
(
aq
)

Cl
(aq )

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Chapter 14
Section 2 Systems at Equilibrium
Calculating Ionic Concentrations Using Ksp,
continued
Sample Problem D Solution
The product of [Cu+][Cl−] must equal Ksp = 1.2 × 10−6.
[Cl−] = 0.55
Ksp = [Cu+][Cl−] = 1.2 × 10−6
–6
K
1.2

10
[Cu ]  sp– 
 2.2  10–6
[Cl ]
0.55
This is the solubility of copper(I) chloride because the
dissolution of 1 mol of CuCl produces 1 mol of Cu+.
Therefore, the solubility of CuCl is 2.2 × 10−6 mol/L.
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HW. # 1,2
Sample Problem D , practice pg. 510
3) Calculate the concentration of Pb ions in solution
when PbCl is dissolved in water.The concentration of Clions in this solution is found to be 2.86  10 mol/L. At
25C, the K of PbCl is 1.17  10 . ans. 0.0143
2
2
-2
sp
2
-5
4) What is the concentration of Cu ions in a saturated
solution of copper(I) chloride given that the K of CuCl is
1.72  10 at 25C? ans. 4.15E-4
+
sp
-7
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Sample Problem D , practice pg. 510
1)The K for silver carbonate is 8.4  10-12 at 298 K.
The concentration of carbonate ions in a saturated
solution is 1.28  10-4 M. What is the concentration of
silver ions? (2.6E-4)
sp
2) Lead-acid batteries employ lead(II) sulfate plates in
a solution of sulfuric acid. Use data from Table 3 to
calculate the solubility of PbSO4 in a battery acid that
has an SO4 -2concentration of 1.0 M. (1.8E-8)
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HW.
Section Review,
pg. 511, # 1-7
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Section 14.2 Review, pg. 511
1. Giving an example, explain how to write an expression
for Keq from a chemical equation.
2. Which
species are left out from the Keq expression, and
why?
3. To
which chemical systems can a Ksp be assigned?
4. When
does Ksp not apply?
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Section 14.2 Review, pg. 511
5) For the reaction in which hydrogen iodide is made at
425C in the gas phase from its elements, calculate [HI],
given that [H ]  [I ] 4.79  10-4 and Keq  54.3.(ans. 3.53E-3)
2
2
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Section 14.2 Review, pg. 511
6) Given that the Ksp value of CuS is 1.3  10-36,
what is [Cu 2] in a saturated solution?(ans. 1.14E-18)
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Section 14.2 Review, pg. 511
7) Write
the equation for the reaction in which solid carbon
reacts with gaseous carbon dioxide to form gaseous
carbon monoxide. At equilibrium, a 2.0 L reaction vessel is
found to contain 0.40 mol of C, 0.20 mol of CO , and 0.10
mol of CO. Find the Keq. (ans. 2.5E-2)
2
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Chapter 14
Standardized Test Preparation
Understanding Concepts
1. In which of these reactions is the formation of the
products favored by an increase in pressure?
A. 2O3(g)  3O2(g)
B. C(s) + O2(g)  CO2(g)
C. 2NO(g) + O2(g)  2NO2(g)
D. CO2 (aq) + 2H2O(l )  H3O (aq) + HCO3– (aq)
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Understanding Concepts
2. What is the effect of an increase in temperature on an
exothermic reaction at equilibrium?
F. It has no effect on the equilibrium.
G. It shifts the equilibrium in favor of the forward
reaction.
H. It shifts the equilibrium in favor of the reverse
reaction.
I. It shifts the equilibrium in favor of both the forward
and reverse reactions.
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Chapter 14
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Understanding Concepts
2. What is the effect of an increase in temperature on an
exothermic reaction at equilibrium?
F. It has no effect on the equilibrium.
G. It shifts the equilibrium in favor of the forward
reaction.
H. It shifts the equilibrium in favor of the reverse
reaction.
I. It shifts the equilibrium in favor of both the forward
and reverse reactions.
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Understanding Concepts
3. Which of the following properties of a reactant is
included in a Keq equation?
A. charge
B. concentration
C. mass
D. volume
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Understanding Concepts
3. Which of the following properties of a reactant is
included in a Keq equation?
A. charge
B. concentration
C. mass
D. volume
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Chapter 14
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Understanding Concepts
4. Explain how the common ion effect is involved in the
addition of NaIO3 to a solution of Cd(IO3)2, which is
much less soluble than NaIO3.
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Understanding Concepts
4. Explain how the common ion effect is involved in the
addition of NaIO3 to a solution of Cd(IO3)2, which is
much less soluble than NaIO3.
Answer: Some of the Cd(IO3)2 would precipitate out of
the solution, reducing the cadmium ion concentration.
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Chapter 14
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Understanding Concepts
5. Explain how pressure can be used to maximize the
production of carbon dioxide in the reaction

 2CO2 (g )?
2CO(g )  O2 (g ) 

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Chapter 14
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Understanding Concepts
5. Explain how pressure can be used to maximize the
production of carbon dioxide in the reaction

 2CO2 (g )?
2CO(g )  O2 (g ) 

Answer: Increasing the pressure will cause the reaction
to favor the production of carbon dioxide because
there are three gas molecules in the reactants and
only two in the products.
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Chapter 14
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Reading Skills
Read the passage below. Then answer the questions.
Coral reefs are made by tiny organisms known as
polyps. They attach themselves permanently in one
place and survive by eating tiny marine animals that
swim past. The polyps secrete calcium carbonate to
make their shells or skeletons. When the polyps die,
the calcium carbonate structures remain and
accumulate over time to form a reef. This reef building
is possible because calcium carbonate is only slightly
soluble in water.
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Chapter 14
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Reading Skills
6. What is the Ksp expression for calcium carbonate?
F. Ksp  [Ca2 ][CO32– ]
G. Ksp  [Ca2 ][CO32– ]2
2 2
2–
H. Ksp  [Ca ] [CO3 ]
I.
Ksp  [Ca2 ]2 [CO32– ]2
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Chapter 14
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Reading Skills
6. What is the Ksp expression for calcium carbonate?
F. Ksp  [Ca2 ][CO32– ]
G. Ksp  [Ca2 ][CO32– ]2
2 2
2–
H. Ksp  [Ca ] [CO3 ]
I.
Ksp  [Ca2 ]2 [CO32– ]2
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Reading Skills
7. What is the most likely source of calcium used by the
polyps to build their shells?
A. calcium atoms in solution in sea water
B. calcium ions in solution in sea water
C. calcium particles that reach the water in acid rain
D. calcium containing rocks gathered from the ocean
floor
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7. What is the most likely source of calcium used by the
polyps to build their shells?
A. calcium atoms in solution in sea water
B. calcium ions in solution in sea water
C. calcium particles that reach the water in acid rain
D. calcium containing rocks gathered from the ocean
floor
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Reading Skills
8. The Ksp for calcium carbonate is 2.8  10-9 and the
Ksp value for calcium sulfate is 9.1  10-6. If coral
polyps secreted calcium sulfate rather than calcium
carbonate, how would this affect the formation of the
coral reef?
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Reading Skills
8. The Ksp for calcium carbonate is 2.8  10-9 and the
Ksp value for calcium sulfate is 9.1  10-6. If coral
polyps secreted calcium sulfate rather than calcium
carbonate, how would this affect the formation of the
coral reef?
Answer: A reef would not be built as quickly if polyps
secreted calcium sulfate, because calcium sulfate has
a higher solubility in water.
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Chapter 14
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Interpreting Graphics
Use the information from the table below to answer
questions 9 through 12.
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Chapter 14
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Interpreting Graphics
9. What is the concentration of magnesium carbonate in
a saturated aqueous solution?
F. 0.0000068 M
G. 0.0026 M
H. 0.84 M
I. 1.31 M
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Interpreting Graphics
9. What is the concentration of magnesium carbonate in
a saturated aqueous solution?
F. 0.0000068 M
G. 0.0026 M
H. 0.84 M
I. 1.31 M
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Interpreting Graphics
10. If a saturated solution of silver carbonate is mixed
with a saturated solution of zinc sulfide, a precipitate
forms. What compound precipitates?
A. Ag2CO3
B. Ag2S
C. ZnCO3
D. ZnS
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10. If a saturated solution of silver carbonate is mixed
with a saturated solution of zinc sulfide, a precipitate
forms. What compound precipitates?
A. Ag2CO3
B. Ag2S
C. ZnCO3
D. ZnS
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Interpreting Graphics
11. Calculate the concentration of S2– ions in a saturated
solution of FeS that contains 0.010 M Fe2+ ions.
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Interpreting Graphics
11. Calculate the concentration of S2– ions in a saturated
solution of FeS that contains 0.010 M Fe2+ ions.
Answer: 1.6  10–17 M
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Interpreting Graphics
12. What will happen if a solution containing 1  10–10 M
Na2CO3 is mixed with an equal volume of a solution
containing 1  10–10 M MgCl2 and 1  10–10 M ZnCl2?
F. No precipitate will form.
G. MgCO3 will precipitate out of the solution.
H. ZnCO3 will precipitate out of the solution.
I. MgCO3 and ZnCO3 will precipitate out of the
solution.
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Interpreting Graphics
12. What will happen if a solution containing 1  10–10 M
Na2CO3 is mixed with an equal volume of a solution
containing 1  10–10 M MgCl2 and 1  10–10 M ZnCl2?
F. No precipitate will form.
G. MgCO3 will precipitate out of the solution.
H. ZnCO3 will precipitate out of the solution.
I. MgCO3 and ZnCO3 will precipitate out of the
solution.
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