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
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
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
K eq
[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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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
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
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
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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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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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp, continued
Using Ksp to Make Magnesium
• Though slightly soluble hydroxides are not salts, they
have solubility product constants.
• Magnesium hydroxide is an example.
2
–


Mg(OH)2 (s ) 
Mg
(
aq
)

2OH
(aq )

[Mg2+][OH−]2 = Ksp = 1.8 × 10−11
• This equilibrium is the basis for obtaining
magnesium.
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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp, continued
Using Ksp to Make Magnesium, continued
• The table at right lists
the most abundant ions
in ocean water and
their concentrations.
• Mg2+ is the third most
abundant ion in the
ocean.
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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp, continued
Using Ksp to Make Magnesium, continued
• To get magnesium, calcium hydroxide is added to sea
water, which raises the hydroxide ion concentration to a
large value so that [Mg2+][OH−]2 would be greater than
1.8 × 10−11.
• Magnesium hydroxide precipitates.
• Magnesium hydroxide is treated with hydrochloric acid
to make magnesium chloride, MgCl2.
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Chapter 14
Section 2 Systems at Equilibrium
The Solubility Product Constant, Ksp, continued
Using Ksp to Make Magnesium, continued
• Finally, magnesium is obtained by the electrolysis of
MgCl2 in the molten state.
• One cubic meter of sea water yields 1 kg of
magnesium metal.
• Because of magnesium’s low density and rigidity,
alloys of magnesium are used when light weight and
strength are needed.
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