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Topic
22
Table of Contents
Topic
22
Topic 22: Chemical Equilibrium
Basic Concepts
Additional Concepts
Chemical Equilibrium: Basic Concepts
Topic
22
What is equilibrium?
• Consider the reaction for the formation of
ammonia from nitrogen and hydrogen.
Chemical Equilibrium: Basic Concepts
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22
What is equilibrium?
• Note that the equation for the production of
ammonia has a negative standard free
energy, ∆G°.
• Recall that a negative sign for ∆G° indicates
that the reaction is spontaneous under
standard conditions.
• Standard conditions are defined as 298 K and
one atmosphere pressure.
Chemical Equilibrium: Basic Concepts
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What is equilibrium?
• But spontaneous reactions are not always fast.
• When carried out under standard
conditions, this ammonia-forming reaction
is much too slow.
• To produce ammonia at a rate that is
practical, the reaction must be carried out at
a much higher temperature than 298 K and a
higher pressure than one atmosphere.
Chemical Equilibrium: Basic Concepts
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What is equilibrium?
• What happens when one mole of nitrogen
and three moles of hydrogen, the amounts
shown in the equation, are placed in a closed
reaction vessel at 723 K?
• Because the reaction is spontaneous, nitrogen
and hydrogen begin to react.
Chemical Equilibrium: Basic Concepts
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What is equilibrium?
• The concentrations of the reactants (H2 and
N2) decrease at first while the concentration
of the product (NH3) increases.
• Then, before
the reactants
are used up, all
concentrations
become
constant.
Chemical Equilibrium: Basic Concepts
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What is equilibrium?
• The reactants, H2 and N2, are consumed in
the reaction, so their concentrations gradually
decrease.
• After a period of time, however, the
concentrations of H2, N2, and NH3 no
longer change.
Chemical Equilibrium: Basic Concepts
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What is equilibrium?
• All concentrations become constant, as
shown by the horizontal lines on the right
side of the diagram.
• The concentrations of H2 and N2 are not
zero, so not all of the reactants were
converted to product even though ∆G for
this reaction is negative.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• When a reaction results in almost complete
conversion of reactants to products, chemists
say that the reaction goes to completion.
• But most reactions, including the ammoniaforming reaction, do not go to completion.
They appear to stop.
• The reason is that these reactions are
reversible.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• A reversible reaction is one that can occur in
both the forward and the reverse directions.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• Chemists combine these
two equations into a
single equation that uses
a double arrow to show
that both reactions occur.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• When you read the
equation, the reactants
in the forward reaction
are on the left.
• In the reverse reaction,
the reactants are on the
right.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• In the forward reaction,
hydrogen and nitrogen
combine to form the
product ammonia.
• In the reverse reaction,
ammonia decomposes
into the products
hydrogen and nitrogen.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• Just as the reaction
begins at a definite,
initial rate; no ammonia
is present so only the
forward reaction can
occur.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• As hydrogen and nitrogen combine to form
ammonia, their concentrations decrease.
• The rate of a reaction depends upon the
concentration of the reactants.
• The decrease in the
concentration of the
reactants causes the rate
of the forward reaction
to decrease.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• As soon as ammonia is present, the reverse
reaction can occur, slowly at first, but at an
increasing rate as the concentration of
ammonia increases.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• As the reaction proceeds, the rate of the
forward reaction continues to decrease and
the rate of the reverse reaction continues to
increase until the two rates are equal.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• At that point, ammonia is being produced as
fast as it is being decomposed, so the
concentrations of nitrogen, hydrogen, and
ammonia remain constant.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• The system has reached a state of balance or
equilibrium.
• The word equilibrium means that opposing
processes are in balance.
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• Chemical equilibrium is a state in which
the forward and reverse reactions balance
each other because they take place at equal
rates.
Rateforward reaction = Ratereverse reaction
Chemical Equilibrium: Basic Concepts
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Reversible reactions
• You can recognize that the ammonia-forming
reaction reaches a state of chemical
equilibrium because its chemical equation is
written with a double arrow like this.
• At equilibrium, the concentrations of reactants
and products are constant.
• However, that does not mean that the
amounts or concentrations of reactants and
products are equal.
Chemical Equilibrium: Basic Concepts
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Equilibrium Expressions and Constants
• You have learned that some chemical
systems have little tendency to react and
others go readily to completion.
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Chemical Equilibrium: Basic Concepts
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Equilibrium Expressions and Constants
• In between these two extremes are the
majority of reactions that reach a state of
equilibrium with varying amounts of
reactants unconsumed.
• If the reactants are not consumed, then not
all the product predicted by the balanced
chemical equation will be produced.
Chemical Equilibrium: Basic Concepts
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Equilibrium Expressions and Constants
• According to the equation for the ammoniaproducing reaction, two moles of ammonia
should be produced when one mole of
nitrogen and three moles of hydrogen react.
• Because the reaction reaches a state of
equilibrium, however, fewer than two moles
of ammonia will actually be obtained.
• Chemists need to be able to predict the yield
of a reaction.
Chemical Equilibrium: Basic Concepts
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Equilibrium Expressions and Constants
• In 1864, the Norwegian chemists Cato
Maximilian Guldberg and Peter Waage
proposed the law of chemical equilibrium,
which states that at a given temperature, a
chemical system may reach a state in which
a particular ratio of reactant and product
concentrations has a constant value.
Chemical Equilibrium: Basic Concepts
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Equilibrium Expressions and Constants
• For example, the general equation for a
reaction at equilibrium can be written as
follows.
• A and B are the reactants; C and D the
products.
Chemical Equilibrium: Basic Concepts
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Equilibrium Expressions and Constants
• The coefficients in the balanced equation
are a, b, c, and d.
• If the law of chemical equilibrium is applied
to this reaction, the following ratio is
obtained.
Chemical Equilibrium: Basic Concepts
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Equilibrium Expressions and Constants
• This ratio is called the equilibrium constant
expression.
• The square brackets indicate the molar
concentrations of the reactants and products
at equilibrium in mol/L.
Chemical Equilibrium: Basic Concepts
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Equilibrium Expressions and Constants
• The equilibrium constant, Keq, is the
numerical value of the ratio of product
concentrations to reactant concentrations,
with each concentration raised to the power
corresponding to its coefficient in the
balanced equation.
• The value of Keq is constant only at a
specified temperature.
Chemical Equilibrium: Basic Concepts
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Equilibrium Expressions and Constants
• Keq > 1: More products than reactants at
equilibrium.
• Keq < 1: More reactants than products at
equilibrium.
Chemical Equilibrium: Basic Concepts
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Constants for homogeneous equilibria
• How would you write the equilibrium
constant expression for this reaction in
which hydrogen and iodine react to form
hydrogen iodide?
• This reaction is a homogeneous
equilibrium, which means that all the
reactants and products are in the same
physical state.
Chemical Equilibrium: Basic Concepts
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Constants for homogeneous equilibria
• All participants are gases.
• To begin writing the equilibrium constant
expression, place the product concentration
in the numerator and the reactant
concentrations in the denominator.
• The expression becomes
equal to Keq when you add
the coefficients from the
balanced chemical equation
as exponents.
Chemical Equilibrium: Basic Concepts
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Constants for homogeneous equilibria
• Keq for this homogeneous equilibrium at
731 K is 49.7.
• Note that 49.7 has no units. In writing
equilibrium constant expressions, it’s
customary to omit units.
Chemical Equilibrium: Basic Concepts
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Equilibrium Constant Expressions
for Homogeneous Equilibria
• Write the equilibrium constant expression
for the reaction in which ammonia gas is
produced from hydrogen and nitrogen.
• The form of the equilibrium constant
expression is
Chemical Equilibrium: Basic Concepts
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Equilibrium Constant Expressions
for Homogeneous Equilibria
Chemical Equilibrium: Basic Concepts
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Equilibrium Constant Expressions
for Homogeneous Equilibria
• Place the product concentration in
the numerator and the reactant
concentrations in the denominator.
• Raise the concentration of
each reactant and product
to a power equal to its
coefficient in the balanced
chemical equation and set
the ratio equal to Keq.
Chemical Equilibrium: Basic Concepts
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Constants for heterogeneous equilibria
• When the reactants and products of a reaction
are present in more than one physical state,
the equilibrium is called a heterogeneous
equilibrium.
• When ethanol is placed in a
closed flask, a liquid-vapor
equilibrium is established.
Chemical Equilibrium: Basic Concepts
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Constants for heterogeneous equilibria
• To write the equilibrium constant expression
for this process, you would form a ratio of
the product to the reactant.
• At a given temperature, the ratio would have
a constant value K.
Chemical Equilibrium: Basic Concepts
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Constants for heterogeneous equilibria
• Note that the term in the denominator is the
concentration of liquid ethanol.
• Because liquid ethanol is a pure substance,
its concentration is constant at a given
temperature.
• That’s because the concentration of a pure
substance is its density in moles per liter.
Chemical Equilibrium: Basic Concepts
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Constants for heterogeneous equilibria
• At any given temperature, density does
not change.
• No matter how much or how little C2H5OH
is present, its concentration remains constant.
• Therefore, the term in the denominator is a
constant and can be combined with K.
Chemical Equilibrium: Basic Concepts
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Constants for heterogeneous equilibria
• The equilibrium constant expression for this
phase change is
Chemical Equilibrium: Basic Concepts
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Constants for heterogeneous equilibria
• Solids also are pure substances with
unchanging concentrations, so equilibria
involving solids can be simplified in the
same way.
• For example, notice the
experiment involving the
sublimation of iodine
crystals.
• The equilibrium depends only on the
concentration of gaseous iodine in the system.
Chemical Equilibrium: Basic Concepts
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Equilibrium Constant Expressions
for Heterogeneous Equilibria
• Write the equilibrium constant expression
for the decomposition of baking soda
(sodium hydrogen carbonate).
Chemical Equilibrium: Basic Concepts
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Equilibrium Constant Expressions
for Heterogeneous Equilibria
• You are given a heterogeneous equilibrium
involving gases and solids.
• The general form of the
equilibrium constant
expression for this reaction is
• Because the reactant and one of the products
are solids with constant concentrations, they
can be omitted from the equilibrium
constant expression.
Chemical Equilibrium: Basic Concepts
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Equilibrium Constant Expressions
for Heterogeneous Equilibria
• Known
• Unknown
• equilibrium constant expression = ?
Chemical Equilibrium: Basic Concepts
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Equilibrium Constant Expressions
for Heterogeneous Equilibria
• Write a ratio with the concentrations of
the products in the numerator and the
concentration of the reactant in the
denominator.
• Leave out [NaHCO3] and
[Na2CO3] because they are
solids.
Chemical Equilibrium: Basic Concepts
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Equilibrium Constant Expressions
for Heterogeneous Equilibria
• Because the coefficients of [CO2] and [H2O]
are 1, the expression is complete.
Chemical Equilibrium: Basic Concepts
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Determining the Value of
Equilibrium Constants
• When equilibrium is established, the
concentration of each substance is
determined experimentally.
• Although an equilibrium system has only
one value for Keq at a particular temperature,
it has an unlimited number of equilibrium
positions.
• Equilibrium positions depend upon the initial
concentrations of the reactants and products.
Chemical Equilibrium: Basic Concepts
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Calculating the Value
of Equilibrium Constants
• Calculate the value of Keq for the equilibrium
constant expression
given concentration data at one equilibrium
position: [NH3] = 0.933 mol/L, [N2] = 0.533
mol/L, [H2] = 1.600 mol/L.
Chemical Equilibrium: Basic Concepts
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Calculating the Value
of Equilibrium Constants
• You have been given the equilibrium
constant expression and the concentration
of each reactant and product.
• You must calculate the equilibrium constant.
• Because the reactant, H2, has the largest
concentration and is raised to the third power
in the denominator, Keq is likely to be less
than 1.
Chemical Equilibrium: Basic Concepts
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Calculating the Value
of Equilibrium Constants
• Known
• [NH3] = 0.933 mol/L
• [N2] = 0.533 mol/L
• [H2] = 1.600 mol/L
• Unknown
• Keq = ?
Chemical Equilibrium: Basic Concepts
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Calculating the Value
of Equilibrium Constants
• Substitute the known values into the
equilibrium constant expression and
calculate its value.
Chemical Equilibrium: Basic Concepts
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Factors Affecting Chemical Equilibrium
• Le Châtelier’s principle states that if a
stress is applied to a system at equilibrium,
the system shifts in the direction that relieves
the stress.
• For example, consider the equilibrium system
• If an additional amount of reactant (NO or
Br2) is added to the system, the equilibrium
will shift to the right, that is, more product
(NOBr) will be formed.
Chemical Equilibrium: Basic Concepts
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Factors Affecting Chemical Equilibrium
• Conversely, adding more NOBr to the system
will result in a shift to the left, forming more
NO and Br2.
• The removal of a reactant or product also
results in a shift in the equilibrium.
• Removing a reactant causes the equilibrium
to shift to the left, forming more reactants.
• Removing the product causes a shift to the
right, forming more product.
Chemical Equilibrium: Basic Concepts
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Changes in volume
• Le Châtelier’s principle also applies to
changes in the volume of the reaction vessel
containing the equilibrium system.
• Suppose the volume of the reaction vessel for
the
system is decreased, resulting in an increase
in pressure.
• The equilibrium will shift to relieve the stress
of increased pressure.
Chemical Equilibrium: Basic Concepts
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Changes in volume
• In this case, the shift will be to the right because
three moles of reactant gas combine to form only
two moles of product gas.
• Thus, a shift toward the product will reduce the
pressure of the system.
• If the volume of the reaction vessel was
increased, the equilibrium would shift to the left,
and more of the reactants would be formed.
Chemical Equilibrium: Basic Concepts
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Changes in volume
• Note that changing the volume of the
reaction vessel causes no shift in the
equilibrium when the number of moles of
product gas equals the number of moles of
reactant gas; an example is the equilibrium
Chemical Equilibrium: Basic Concepts
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Changes in temperature
• Even though equilibrium may shift to the
right or left in response to a change in
concentration or volume, the value of the
equilibrium constant remains the same.
• A change in temperature, however, alters
both the equilibrium position and the
value of Keq.
Chemical Equilibrium: Basic Concepts
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Changes in temperature
• For example, consider the thermochemical
equation for the reversible formation of
hydrogen chloride gas from its elements.
• The forward reaction releases heat, so
you can consider heat as a product in the
forward reaction and a reactant in the
reverse reaction.
Chemical Equilibrium: Basic Concepts
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Changes in temperature
• Raising the temperature of this system
requires the addition of heat, which shifts
the equilibrium to the left and reduces the
concentration of hydrogen chloride.
• Thus, the value of Keq decreases.
• Lowering the temperature of the system
means that heat is removed, so the
equilibrium relieves the stress by shifting to
the right, increasing both the concentration
of hydrogen chloride and Keq.
Basic Assessment Questions
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Question 1
Write equilibrium constant expressions for
the following homogeneous equilibria.
Basic Assessment Questions
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Question 1a
Answer 1a
Basic Assessment Questions
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Question 1b
Answer 1b
Basic Assessment Questions
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Question 2
Write equilibrium constant expressions for
the following heterogeneous equilibria.
Basic Assessment Questions
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Question 2a
Answer 2a
Basic Assessment Questions
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Question 2b
Answer 2b
Basic Assessment Questions
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Question 3
Hydrogen and carbon disulfide react to form
methane and hydrogen sulfide according to
this equation.
Calculate Keq if the equilibrium concentrations
are [H2] = 0.205 mol/L, [CS2] = 0.0664 mol/L,
[CH4] = 0.0196 mol/L, and [H2S] = 0.0392
mol/L.
Basic Assessment Questions
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0.257
Answer
Basic Assessment Questions
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Question 4
How would decreasing the volume of the
reaction vessel affect these equilibria?
Basic Assessment Questions
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Question 4a
Answer 4a
shift to the right
Basic Assessment Questions
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Question 4b
Answer 4b
shift to the left
Chemical Equilibrium: Additional Concepts
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Additional Concepts
Chemical Equilibrium: Additional Concepts
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Using Equilibrium Constants
• When Keq is known, the equilibrium
concentration of a substance can be
calculated if you know the concentrations
of all other reactants and products.
• The following example problem shows
you how to determine an equilibrium
concentration.
Chemical Equilibrium: Additional Concepts
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Calculating Equilibrium Concentrations
• At 350C, Keq = 66.9 for the formation of
hydrogen iodide from its elements.
• What is the concentration of HI if [H2] =
0.0295 mol/L and [I2] = 0.0174 mol/L?
Chemical Equilibrium: Additional Concepts
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Calculating Equilibrium Concentrations
• Write the equilibrium constant expression.
• Multiply both sides of the equation by
[H2] [I2].
Chemical Equilibrium: Additional Concepts
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Calculating Equilibrium Concentrations
• Substitute the known quantities into the
equation and solve for [HI].
• An extra digit is retained here for accuracy,
but the final answer will be rounded to three
digits.
• The equilibrium concentration of HI is 0.185
mol/L.
Chemical Equilibrium: Additional Concepts
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Solubility equilibria
• The solubility product constant (Ksp) is
an equilibrium constant for the dissolving
of a sparingly soluble ionic compound in
water.
Chemical Equilibrium: Additional Concepts
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Solubility equilibria
• The solubility product constant expression
is the product of the concentrations of the
ions with each concentration raised to a
power equal to
the coefficient
of the ion in
the chemical
equation.
Chemical Equilibrium: Additional Concepts
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Solubility equilibria
• For example, copper (II) hydroxide dissolves
in water according to this equation.
• The coefficient of Cu2+ is 1, and the
coefficient of OH– is 2, so the following is
the solubility product constant expression.
Chemical Equilibrium: Additional Concepts
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Solubility equilibria
• Tabulated Ksp values may be used to
calculate the molar solubility of a sparingly
soluble ionic compound and also to calculate
ion concentrations in a saturated solution.
Chemical Equilibrium: Additional Concepts
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Calculating Molar Solubility
and Ion Concentration from Ksp
• The Ksp for lead(II) fluoride (PbF2) is
–8
3.3 x 10 at 25C. Use this Ks value to
calculate the following.
a. The solubility in mol/L of PbF2
b.The fluoride ion concentration in a saturated
solution of PbF2.
Chemical Equilibrium: Additional Concepts
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Calculating Molar Solubility
and Ion Concentration from Ksp
a. Write the balanced equation for the solubility
equilibrium, and write the Ksp expression.
• The moles of Pb2+ ions in solution equal the
moles of PbF2 that dissolved.
Chemical Equilibrium: Additional Concepts
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Calculating Molar Solubility
and Ion Concentration from Ksp
• Therefore, let [Pb2+] equal s, where s
represents the solubility of PbF2.
• Because there are two F– ions for every Pb2+
ion, [F–] = 2s.
Chemical Equilibrium: Additional Concepts
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Calculating Molar Solubility
and Ion Concentration from Ksp
• Substitute these terms into the Ksp expression
and solve for s.
• Here three digits are retained for accuracy,
but the final answer will be rounded to two
digits.
Chemical Equilibrium: Additional Concepts
Topic
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Calculating Molar Solubility
and Ion Concentration from Ksp
• The molar solubility of PbF2 in water at
25°C is 2.0 x 10–3 mol/L.
Chemical Equilibrium: Additional Concepts
Topic
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Calculating Molar Solubility
and Ion Concentration from Ksp
b. As shown above, the fluoride ion
concentration in a saturated solution of
PbF2 at 25C is as follows.
Chemical Equilibrium: Additional Concepts
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Predicting precipitates
• The solubility product constant expression can
also be used to predict whether a precipitate
will form when two solutions of ionic
compounds are mixed.
• The molar concentrations of the ions in a
solution are used to calculate the ion product,
Qsp.
Chemical Equilibrium: Additional Concepts
Topic
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Predicting precipitates
• If Qsp < Ksp, a precipitate will form, reducing
the ion concentrations until the system reaches
equilibrium and the solution is saturated.
• If Qsp < Ksp, no precipitate forms.
Chemical Equilibrium: Additional Concepts
Topic
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Predicting precipitates
• The following example problem
demonstrates how to use Qsp and Ksp to
determine whether a precipitate will form.
Predicting a Precipitate
• Predict whether a precipitate will form if 200
mL of 0.030M CaCl2 is added to 200 mL of
0.080M NaOH.
Chemical Equilibrium: Additional Concepts
Topic
22
Predicting a Precipitate
• A double-replacement reaction might
occur according to this equation.
• You know that NaCl is a soluble compound
and will not form a precipitate.
• However, Ca(OH)2 is sparingly soluble with
Ksp = 5.0 x 10–6, so it might precipitate if the
concentrations of its ions are high enough.
Chemical Equilibrium: Additional Concepts
Topic
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Predicting a Precipitate
• Write the equation for the dissolving of
Ca(OH)2.
• The ion product expression is as follows.
• Qsp is a trial value that will be compared
to Ksp.
Chemical Equilibrium: Additional Concepts
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Predicting a Precipitate
• Next, find the concentrations of the Ca2+
and OH– ions.
• Divide the initial concentrations in half
because the volume doubles on mixing.
Chemical Equilibrium: Additional Concepts
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Predicting a Precipitate
• Calculate Qsp.
• Calculate Qsp with Ksp.
–
• The concentrations of
and OH are
high enough to cause a precipitate of
Ca(OH)2 to form.
Ca2+
Chemical Equilibrium: Additional Concepts
Topic
22
Common ion effect
• The solubility of a substance is reduced when
the substance is dissolved in a solution
containing a common ion. This is called the
common ion effect.
• For example, PbI2 is less soluble in an
aqueous solution of NaI than in pure water
because the common ion I–, already present
in the NaI solution, reduces the maximum
possible concentration of Pb2+ and thus
reduces the solubility of PbI2.
Additional Assessment Questions
Topic
22
Question 1
At a certain temperature, Keq = 0.118 for the
following reaction.
Calculate the concentration of [H2] in an
equilibrium mixture with [CH4] = 0.0492
mol/L and [C2H2] = 0.0755 mol/L.
Additional Assessment Questions
Topic
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Answer
0.156 mol/L
Additional Assessment Questions
Topic
22
Question 2
Use the data
in this table to
calculate the
solubility in
mol/L of
these ionic
compounds at
298 K.
Additional Assessment Questions
Topic
22
Question 2a
MgCO3
Answer 2a
2.6 x 10–3
mol/L
Additional Assessment Questions
Topic
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Question 2b
AlPO4
Answer 2b
9.9 x 10–11
mol/L
Additional Assessment Questions
Topic
22
Question 3
Will a precipitate form when 125 mL of
0.010M K2SO4 is mixed with 250 mol of
0.015 M CaBr2? (Hint: Note that the volumes
of the two solutions are not equal, and the
volume after mixing is 375 mL.)
Additional Assessment Questions
Topic
22
Answer
For CaSO4 : Qsp (3.3 x 10–5) < Ksp (4.9 x 10–5);
no precipitate forms
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