SCH4U
Grade 12
University Chemistry
Lesson 17 – Introducing Equilibrium
SCH4U – Chemistry
Lesson 17
Unit 5: Chemical Systems and Equilibrium
Most of the reactions that you have studied in this and other chemistry courses have
gone to completion from reactants to products. In many chemical reactions the reaction
can proceed in both directions. The steady state of reversible reactions is called
equilibrium and it is the focus of this unit.
Overall Expectations:
•
•
•
demonstrate an understanding of the concept of chemical equilibrium, Le Chatelier’s
principle, and solution equilibria;
investigate the behaviour of different equilibrium systems, and solve problems
involving the law of chemical equilibrium;
explain the importance of chemical equilibrium in various systems, including
ecological, biological, and technological systems.
This unit consists of four lessons:
Lesson 17
Lesson 18
Lesson 19
Lesson 20
Introducing Equilibrium
The Equilibrium Constant
Acid and Bases Equilibrium
Solubility Equilibriums
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SCH4U – Chemistry
Lesson 17
Lesson 17: Introducing Equilibrium
A very important question concerning any reaction is the EXTENT to which it proceeds
under some given conditions. Does it proceed to only a limited extent? Or, does it go
to completion? Can the reaction be reversed?
We can consider the reaction between gaseous hydrogen and chlorine. If hydrogen gas
and chlorine gas are heated together, an explosive reaction occurs in which hydrogen
combine completely with chlorine to give a quantitative yield (98%) of hydrogen
chloride. Thus, this reaction is said to go to completion.
H2(g) + Cl2 (g) Æ2HCI(g).
On the other hand, if gaseous hydrogen and iodine are heated together, the reaction is
much less vigorous and does not go to completion. The yield of hydrogen iodide (HI) is
not as quantitative as some H2 and I2 remain unreacted.
H2(g) + I2 (g) ↔ 2HI(g).
We say that the system has attained a state of Equilibrium. Equilibrium will be the
focus of this unit.
What You Will Learn
After completing this lesson, you will be able to
•
•
•
illustrate the concept of dynamic equilibrium with reference to systems such as
liquid-vapour equilibrium, weak electrolytes in solution, and chemical reactions;
identify, in qualitative terms, entropy changes associated with chemical and physical
processes;
describe the tendency of reactions to achieve minimum energy and maximum
entropy;
Recognizing Dynamic Equilibrium
Many chemical reactions can easily run in both forward (towards products) and
reverse (towards reactant) directions and are called reversible reactions. This
reversible behaviour of the chemical reactions has become known as a reaction
"coming to equilibrium."
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SCH4U – Chemistry
Lesson 17
In a chemical system that can come to equilibrium, both the forward reaction direction
and the reverse reaction direction will run all the time. This is the meaning of the word
"dynamic" in the title.
The exact moment of equilibrium happens when the rate of the forward reaction
equals the rate of the reverse reaction.
When a chemical system is at equilibrium, there are no visible changes in the system.
The concentrations of every substance in the reaction will remain constant at
equilibrium.
Equilibrium occurs when opposing changes to a closed chemical system occur
simultaneously at the same rate. A closed system is a system that may exchange
energy but not matter with its surroundings.
Conditions that Apply to All Equilibrium Systems
There are four conditions that all equilibrium systems:
1. Equilibrium is achieved in reversible process when the rates of opposing changes
are equal. A double ended arrow (↔) indicates reversible changes. For example:
H2O (l) ↔H2O (g)
2. The observable (macroscopic) properties of a system at equilibrium are constant (for
example, colour, pressure, concentration, pH)
3. Equilibrium can be approached from either direction
4. Equilibrium can only be reached in a closed system. For example, carbon dioxide in
a soda drink is in equilibrium when the bottle is closed.
H2O(l)
'
H2(g) + Cl2(g)
H2O(g)
'
2HCl(g)
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SCH4U – Chemistry
Lesson 17
The effect of temperature on the
endothermic, aqueous equilibrium:
Co(H2O)62+ + 4ClCoCl42- + 6 H2O
Pink
Blue anhydrous
cobalt(II) chloride
and pink
hydrated
Blue
The violet solution in the center is at 25°C
and contains significant quantities of both
pink Co(H2O)62+ and blue CoCl42-. When
the solution is cooled, it turns pink because
the equilibrium is shifted to the left. Heating
the solution favors the blue CoCl42- ions.
Solubility Equilibrium is a dynamic equilibrium between a solute and a solvent in a
saturated solution in a closed system.
Dissolution is the process of dissolving. Consider orange drink crystals dissolving in
water. The orange crystals will dissolve into solution as the solute molecules collide
with the water molecules. As the orange crystals dissolve, the solution becomes more
and more orange. Even when the solution is saturated, the crystals and dissolved ions
are still moving. As the dissolved ions hit the solid drink crystals, they may form ionic
bonds and crystallize out of solution. Nearing equilibrium, the rate of dissolution and
crystallization approach one another and no visible changes are present.
Phase Equilibrium - A dynamic equilibrium called phase equilibrium can exist
between different states of a substance in a closed system. In a closed system,
molecules in the liquid phase can gain enough energy from collisions to move in the
gaseous phase. Similarly as the number of gaseous molecules increases, these
molecules can move back into the liquid phase.
Chemical Reaction Equilibrium - Some chemical reactions called quantitative
reactions proceed completely, meaning all of the reactants are converted into products
based on the balanced chemical equation.
Consider the decomposition of calcium carbonate in an open system:
CaCO3(s) Æ CaO(s) + CO2(g)
In this open system, equilibrium cannot be established since the carbon dioxide gas
escapes into the external environment as the calcium carbonate decomposes.
If the same reaction was placed in a closed container and allowed to proceed, both
reactants and products are present after a given time frame:
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SCH4U – Chemistry
Lesson 17
CaCO3(s) ↔ CaO(s) + CO2(g)
(Notice the arrow indicates both the forward and reverse direction)
Allowing the reaction to reach equilibrium limits the amount of product produced.
Let’s consider another reaction in a closed system that can also reach equilibrium
N2O4(g) ↔2NO2(g)
Suppose you conducted two experiments, one which began with N2O4(g) molecules in a
closed reaction vessel, and another that began with NO2(g), in a reaction vessel. Would
the system still reach equilibrium in both cases?
Table 17.1: N2O4/NO2 equilibrium concentrations
Initial Concentrations (mol/L)
N2O4(g
NO2
Experiment 1
0.75
0
Experiment 2
0
1.50
Final Concentrations (mol/L)
N2O4(g
NO2
.721
0.058
.721
0.058
According to the data above, equal ratios of N2O4(g) and NO2(g) were present regardless
of whether the reaction started in the forward or the reverse direction. Also note that
equilibrium does not mean that the concentrations are equal, rather that the rates of the
forward and reverse reactions are equal.
Percent Reaction at Chemical Equilibrium
Example 1:
Consider the following equation for the formation of hydrogen fluoride from its elements
at SATP
H2(g) + F2(g) ↔2HF(g)
If the reaction begins with 1.00 mol/L concentrations of H2(g) and F2(g) and no HF(g),
calculate the concentrations of H2(g) and HF(g) at equilibrium if the equilibrium
concentration of F2(g) is measured to be 0.24mol/L.
Solution 1:
Let’s start off by stating our givens:
[H2(g)]initial
= 1.00 mol/L
[F2(g)]initial
= 1.00 mol/L
= 0.00 mol/L
[HF(g)]initial
[F2(g)]equilibrium = 0.24 mol/L
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SCH4U – Chemistry
Lesson 17
One convenient method for solving equilibrium problems is by using an I.C.E table.
I - Initial concentration
C - Change in concentration
E - Equilibrium concentration
So we will begin by setting up an ICE table for this reaction.
H2(g)
1.00
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
+
F2(g)
1.00
↔
2HF(g)
0.00
By the time equilibrium has been reached, a certain amount of H2(g)and F2(g) will have
formed product (HF). However since an equilibrium is indicated by the arrow (↔), the
reaction will not go to completion (completely to the right). Rather at equilibrium, there
will be amounts of each reactant and product available.
We will assign the variable “x” to represent the change in concentrations of reactants
and products. Assuming that some of the H2(g) and F2(g) we start with will be converted
(or lost in the formation of HF), we will assign them the value of –x, and the amount of
HF formed will be +2x. Notice how the co-efficients for the balanced equation are
included. Let’s fill in our ICE table:
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
H2(g)
1.00
-x
1.00-x
+
F2(g)
1.00
-x
1.00-x
↔
2HF(g)
0.00
+2x
+2x
Knowing that the equilibrium concentration of F2(g) is 0.24mol/L, you can determine the
value of x.
1.0mol / L − x = 0.24mol / L
− x = −0.76mol / L
x = 0.76mol / L
Now substitute the value of x to calculate the equilibrium concentrations of H2(g) and HF.
[H2(g)]equilibrium = 1.00 mol/L –x
= 1.00 mol/L – 0.76 mol/L
= 0.24 mol/L
[HF(g)]equilibrium =2x
= 2(0.76mol/L)
= 1.52 mol/L
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SCH4U – Chemistry
Lesson 17
Example 2:
In a gaseous reaction system, 0.200 mol of hydrogen gas, H2(g) is added to 2.00 mol of
iodine vapour, I2(g) in a 2.0L container at 448oC. At equilibrium the system contains
0.040 mol of hydrogen gas, H2(g). Determine the equilibrium concentrations of H2(g) and
HI(g).
Solution 2:
First write out the balanced equation for this equilibrium
H2(g) + I2(g) ↔ 2HI(g)
Then we can calculate the concentrations of the given values:
ηH2initial
ηI2initial
ηH2equilibrium
volume
= 0.200 mol
= 2.00 mol
= 0.040 mol
= 2L
[H2]initial
=
0.200mol
2.00L
= 0.100 mol/L
[I2]initial
=
2.00mol
2.00L
= 1.00 mol/L
[H2]equilibrium
=
0.040mol
2.00L
= 0.020 mol/L
Next, set up your ICE table
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
H2(g)
0.10
-x
0.020
+
I2(g) ↔
1.00
-x
1.00-x
2HI(g)
0.00
+2x
+2x
At equilibrium:
[H2 ] = 0.100 − x
0.100mol / L − x = 0.020
x = 0.080mol / L
= 0.020
[I2 ] = 0.100mol / L − x
= 0.100mol / L − 0.080mol / L
= 0.020mol / L
[HI] = 2x
= 2 ( 0.080mol / L )
= 0.160mol / L
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SCH4U – Chemistry
Lesson 17
Support Questions
1. What are the conditions necessary for equilibrium?
2. What is a forward reaction versus a reverse reaction?
3. What are the characteristics of equilibrium?
4. When ammonia is heated, it decomposes into nitrogen gas and hydrogen gas
according to the following equation;
2NH3(g) ↔ N2(g) + 3H2(g)
When 4.0 mol of NH3(g) is introduced into a 2L container and heated to a particular
temperature, the amount of ammonia changes to 2.0 mol. Determine the
equilibrium concentrations of the other two entities.
5. When carbon dioxide is heated in a closed container, it decomposes into carbon
monoxide and oxygen according to the following equilibrium equation:
2 CO2(g) ↔2CO(g) + O2(g)
When 2 mol of CO2(g) is placed in a 5.0L container and heated to a particular
temperature, the equilibrium concentration of CO2(g) is measured to be 0.39mol/L.
Determine the equilibrium concentrations of CO(g) and O2(g).
6. At 400C, 2.0 mol of pure NOCl(g) is introduced into a 2.0L flask. The NOCl(g) partially
decomposes according to the following equilibrium equation:
2NOCl(g) ↔ 2NO(g) + Cl2(g)
At equilibrium, the concentration of NO(g) is 0.032 mol/L. Determine the equilibrium
concentrations of NOCl(g) and Cl2(g) at this temperature.
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SCH4U – Chemistry
Lesson 17
Thermodynamics and Equilibrium
A favourable (spontaneous) change is a change that has a natural tendency to
happen under certain conditions.
Favourable Changes
Enthalpy
• many favourable physical and chemical changes are exothermic since the
products will have less energy than the reactants
• some favourable changes release no energy, and others are endothermic
Temperature
Hg(l) + ½ O2(g)
•
•
•
•
•
'
HgO(s)
∆H = ±90.8 kJ
the synthesis reaction is exothermic (enthalpy change, ∆H, is negative)
the synthesis reaction only takes place at relatively moderate temperatures
above 400oC, the reverse reaction (decomposition rxn) is favourable
the direction in which this reaction proceeds depends on temperature
both the rate of reaction and direction of reaction change with temperature
Entropy
•
•
•
•
Entropy, S, is the tendency towards randomness or disorder in a system
entropy of a system increases (more disordered) with temperature because the
motion of the particles becomes more chaotic at higher temperatures
2nd Law of Thermodynamics: the total entropy of a system is increasing
all favourable changes involve an increase in the total amount of entropy
Consider:
Ba(OH)2y8H2O(s) + 2NH4SCN(s) Æ Ba(SCN)2(aq) + 10 H2O(l) + 2NH3(g)
kJ
9
9
∆H = +170
two solids become a solution, water and a gas
3 molecules of reactants produce 13 molecules of product
Gibbs Free Energy & Equilibrium
•
Free energy, G, is available energy, a measure of useful work that can be
obtained from a reaction
ΔG = ΔH − T ΔS
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∆G= -ve :fwd rxn is favorable
∆G= +ve :rev rxn is favorable
∆G= 0 :rxn is at equilibrium
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SCH4U – Chemistry
Lesson 17
Support Questions
7. In each process, how does the entropy of the system change?
a)
b)
c)
d)
e)
Ice melting
water vapour condensing
sugar dissolving in water
HCl(g) + NH3(g) Æ 2NH4Cl(s)
CaCO3(s) Æ CaO(s) + CO2(g)
Key Question #17
1. After 4.00 mol of C2H4(g) and 2.50 mol of Br2(g) are placed in a sealed 1.0L container,
the reaction reaches equilibrium and is written following.
C2H4(g) + Br2(g) ↔ C2H4Br2(g)
The graph on the right shows the
concentration of C2H4(g) as it changes
over time at a fixed temperature until
equilibrium is reached.
Calculate the equilibrium concentrations
of all three substances. (6 marks)
2. In a gaseous 2L reaction system, 2.00
mol of methane, CH4(g) is added to 10.00
mol of chlorine, Cl2(g). At equilibrium, the system contains 1.40 mol of
chloromethane, CH3Cl(g), and some hydrogen chloride, HCl(g).
a) Write a balanced chemical equation for this equilibrium (1 mark)
b) Calculate the amount of each substance at equilibrium. (5 marks).
3. Write a short paragraph, or use a graphic organizer, to show the relationship among
the following concepts: favourable chemical change, temperature, enthalpy, entropy
and free energy. (5 marks)
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SCH4U
Grade 12
University Chemistry
Lesson 18 – The Equilibrium Constant
SCH4U – Chemistry
Lesson 18
Lesson 18: The Equilibrium Constant
In 1888, Le Chatelier gave a succinct statement of the principle he had announced 4
years prior. It is:
Every change of one of the factors of equilibrium occasions a rearrangement of the
system in such a direction that the factor in question experiences a change in a sense
opposite to the original change.
Le Chateliers Principle was applied by Fritz Haber to enable the production of synthetic
ammonia during World War I. Le Chateliers principle, and further knowledge of
equilibrium concepts will be taught in this lesson.
What You Will Learn
After completing this lesson, you will be able to
•
•
•
•
demonstrate an understanding of the law of chemical equilibrium as it applies to the
concentrations of the reactants and products at equilibrium;
demonstrate an understanding of how Le Chatelier’s principle can predict the
direction in which a system at equilibrium will shift when volume, pressure,
concentration, or temperature is changed;
apply Le Chatelier’s principle to predict how various factors affect a chemical system
at equilibrium, and confirm their predictions through experimentation;
explain how equilibrium principles may be applied to optimize the production of
industrial chemicals (e.g., production of sulphuric acid, ammonia);
The law of chemical equilibrium states that, at equilibrium, there is a constant ratio
between the concentrations of the products and reactants in any change.
Consider:
N2O4(g)
'
colourless
2NO2(g)
brown
Forward reaction: N2O4(g) Æ 2NO2(g)
Forward rate: kf[N2O4]
At equilibrium,
Reverse reaction: N2O4(g) Å 2NO2(g)
Reverse rate:
kr[NO2]2
Forward rate = Reverse rate
kf[N2O4] = kr[NO2]2
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SCH4U – Chemistry
Lesson 18
The equilibrium constant, Keq, is the ratio of the forward rate constant, (kf), divided by
the reverse rate constant, (kr). Rearrange the above relationship to obtain:
Keq
kf [NO2 ]2
=
=
k r [N2O4 ]
A more general equation can be applied using the following reaction:
aP + bQ ⇔ cR + dS
Kc =
[R ]c [S ]d
[P ]a [Q ]b
Helpful Tips
9 Kc is used instead of Keq to express molar
concentration (mol/L)
9 Products in numerator (top)
9 Reactants in denominator (bottom)
9 [concentration]coefficient
9 multiply terms, never add!
For a given system at equilibrium, the value of the equilibrium constant depends only on
temperature. Changing the temperature of a reacting mixture changes the rate of the
forward and reverse reactions by different amounts, because the forward and reverse
reactions have different activation energies, Ea.
Example 1:
One of the steps in the production of sulphuric acid involves the catalytic oxidation of
sulphur dioxide.
2SO2(g) + O2(g) ↔ 2SO3
Solution 1:
Write the equilibrium expression
[SO3 ]2
Kc =
[SO2 ]2 [O2 ]
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SCH4U – Chemistry
Lesson 18
Example 2:
The reaction between ethanol and ethanoic acid to form ethyl ethanoate and water
CH3CH2OH(l) + CH3COOH(l) ↔ CH3CHOOCH2CH3(l) + H2O(l)
Write the equilibrium expression.
Solution 2:
Kc =
[CH3CHOOCH2CH3 ][H2O ]
[CH3CH2OH ][CH3COOH ]
Support Questions
8. Write the equilibrium expression for each reaction;
a) The reaction between nitrogen gas and oxygen gas at high temperatures:
N2(g) + O2(g) ↔ 2NO(g)
b) The reaction between hydrogen gas and oxygen gas to form water vapour:
H2(g) + O2(g) ↔ H2O(g)
c) The REDOX equilibrium of iron and iodine ions in aqueous solution:
2Fe3+(aq) + 2I-(aq) ↔ 2Fe2+(aq) + I2(aq)
d) The oxidation of ammonia.
4NH3(g) + 5O2(g) ↔4NO(g) + 6 H2O(g)
Measuring Equilibrium Concentrations
The equilibrium constant, Kc is calculated by substituting equilibrium concentrations into
the equilibrium expression. There are different strategies to solving these types of
problems, so we use a couple of examples to illustrate these strategies.
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SCH4U – Chemistry
Lesson 18
Example 3: (Perfect Square Method)
The following reaction increases the proportion of hydrogen gas for use as fuel.
CO(g) + H2O(g) ↔H2(g) + CO2(g)
At 700K the equilibrium constant is 0.83. Suppose that you start with 1.0 mol of CO(g)
and 1.0 mol of H2O(g) in a 5.0L container. What amount of each substance will be
present in the container when the gases are at equilibrium?
Solution 3:
First, calculate your given concentrations
1.0mol
=
[CO]initial
= 0.20 mol/L
5.0L
1.0mol
=
= 0.20 mol/L
[H2O]initial
5.0L
Next, set up an ICE table.
CO(g)
Initial concentration
(mol/L)
Change in
concentration (mol/L)
Equilibrium
concentration (mol/L)
+
H2O(g)
↔
H2(g)
+
CO2(g)
.20
.20
0
0
-x
-x
+x
+x
.20-x
.20-x
x
x
Write the equilibrium expression and then substitute the equilibrium concentrations into
the expression.
Kc =
[H2 ][CO2 ]
[CO ][H2O ]
0.83 =
[ x ][ x ]
[0.20 − x ][0.20 − x ]
This type of problem is called a perfect square, since taking the square root makes the
problem easier to solve. Recall that the square root is plus or minus (±).
0.83 =
±0.911 =
[x2 ]
[0.20 − x ]2
x
0.20 − x
x = 0.148 or x = 0.306
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SCH4U – Chemistry
Lesson 18
The value x = 0.306 is impossible because it would result in a negative concentration of
both CO and H2O at equilibrium
Using x = 0.148 mol/L, solve for the equilibrium concentrations
[H2]equilibrium = [CO2]equilibrium= 0.15 mol/L
[CO]equilibrium = [H2O]equilibrium = 0.25 mol/L
To find the amount of each gas, multiply the concentration of each gas by the volume of
the container (5L)
Amount of H2(g) = CO2(g) = 0.75mol
Amount of CO(g) = H2O(g) = 0.25 mol
Example 4: (Quadratic Equation Method)
The following reaction has an equilibrium constant of 25.0 at 1100K.
H2(g) + I2(g) ↔ 2HI(g)
2.00 mol of H2(g) and 3.00 mol of I2(g) are placed in a 1.00L reaction vessel at 1100K.
What is the equilibrium concentration of each gas?
Solution 4:
Since the container has a volume of 1.0 L, the concentrations are:
[H2]initial = 2.0mol/L
[I2]initial = 3.00mol/L
Next, set up an ICE table
H2(g)
2.00
-x
2.00-x
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
+
I2(g)
3.00
-x
3.00-x
↔
2HI(g)
0.00
+2x
+2x
Then write and substitute into the equilibrium expression.
[HI ]2
[2 x ]2
Kc =
25.0 =
[H2 ][I2 ]
[2.00 − x ][3.00 − x ]
This equation cannot be solved using the perfect square method. The quadratic
equation must be used:
First rearrange the equation into quadratic form:
ax 2 + bx + c = 0
0.840x2 - 5.00x + 6 = 0
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SCH4U – Chemistry
Lesson 18
Then sub into the quadratic formula:
x=
=
=
−b ±
2
b − 4ac
2a
−( −5) ±
25 − 20.16
1.68
5.0 ± 2.2
1.68
x = 4.3 and x =1.7
The value of x = 4.3 is not possible.
Substituting x =1.7 into the last line of the ICE table
Equilibrium concentration (mol/L)
[H2] =.30mol/L
2.00-1.7
[I2] =1.3mol/L
3.00-1.7
+2(1.7)
[HI] =3.4 mol/L
Note, you can check your solution by substituting the equilibrium concentrations into the
equilibrium expression. If you have calculated everything correctly, you should get a
value of 25 for the equilibrium constant.
Summary: Equilibrium Calculations
•
•
•
•
organize data with an ICE table (initial, change, equilibrium)
letting x represent the substance with the smallest coefficient in the chemical
equation avoids fractional values of x
look for perfect squares when solving an equilibrium expression
9 take the square root of each side to solve the equation
9 remember that the square root is ± x
many problems do not involve perfect squares
9
you may need to use the quadratic equation to solve the expression
Recall that a quadratic equation of the form:
ax 2 + bx + c = 0
Has the solution:
−b ± b 2 − 4ac
x=
2a
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SCH4U – Chemistry
Lesson 18
Support Questions
9. At 25oC, the value of Kc for the following reaction is 82.
I2(g) + Cl2 ↔ 2ICl(g)
0.83 mol of I2(g) and 0.83 mol of Cl2(g) are placed in a 1.0L container. What are the
concentrations of the gases at equilibrium?
10. At a certain temperature, Kc =4.0 for the following reaction:
2HF(g)↔H2(g) + F2(g)
A 1.0L reaction vessel contained 0.045 mol of F2(g) at equilibrium. What was the
initial amount of HF in the reaction vessel?
11. Consider the following reaction:
SO2(g) + NO2(g) ↔NO(g) + SO3(g)
In a 1.0L container, 0.017 mol of SO2(g) and 0.011 mol of NO2(g) were added. The
value of Kc for the reaction at 200K is 4.8. What is the equilibrium concentration of
SO3(g) at this temperature?
12. Consider the following reaction
CO(g) + Cl2(g) ↔COCl2(g)
0.055 mol of CO(g) and ) and .072 mol of Cl2(g) are placed in a 5.0L container. At
870K, the equilibrium constant is 0.20. What are the equilibrium concentrations of
the mixture at 870K?
Qualitative Interpretation
The following assumptions can be made when interpreting the equilibrium constant:
•
when K›1, products are favoured. The equilibrium lies far to the right
•
when K≈1, there are approximately equal concentrations of reactants and
products at equilibrium
when K‹1, reactants are favoured. The equilibrium lies far to the left
•
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SCH4U – Chemistry
Lesson 18
Example 4:
Consider the reaction of carbon monoxide and chlorine to form phosgene, COCl2(g).
CO(g) + Cl2(g) ↔ COCl2(g)
At 870K, the value of Kc is 0.20. At 370K, the value of Kc is 4.5 x 107. Based on only the
values of Kc, is the production of COCl2(g) more favourable at higher or lower
temperature?
Solution 4:
At 870K Kc =0.020
At 370K Kc =4.5 x 107
The value of Kc is large at the lower temperature, 370K. Therefore a lower temperature
is more favourable.
The Meaning of a Small Equilibrium Constant
When Kc is small compared with the initial concentration, the value of the initial
concentration minus x is approximately equal to the initial concentration. Thus, you can
ignore x. If the initial concentration of a substance is zero, any equilibrium
concentration of the substance, no matter how small, is significant.
In general, values of Kc are not measured with accuracy better than 5%. Therefore,
making approximations is justified if the calculation error you introduce in less than 5%
Example 5: (Using Approximation for Small Equilibrium Constant)
A chemist is studying the following equilibrium reaction.
N2(g) + O2(g) ↔ 2NO(g)
The chemist puts 0.085 mol of N2(g) and 0.038 mol of O2(g) in a 1.0L container. At a
given temperature the value of Kc is 4.2 x 10-8. What is the concentration of NO(g) in the
mixture at equilibrium?
Solution 5:
Start by dividing the smallest initial concentration by Kc to determine whether you can
ignore the changes in concentration.
Smallest initial concentration
0.038
=
= 9.0 × 105
Kc
4.2 × 10 −8
Because this value is well above 500, you can ignore the changes in [N2] and [O2]
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SCH4U – Chemistry
Lesson 18
Next set up your ICE table
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
N2(g)
+
0.085
-x
0.085-x
~0.085
O2(g)
↔
0.038
-x
0.038- x
~0.038
2NO(g)
0.00
+2x
+2x
Now set up your equilibrium expression, and substitute in the equilibrium values from
your ICE table.
Kc =
[NO ]2
[N2 ][O2 ]
4.2 × 10−8 =
[2 x ]2
[0.085][0.038]
x = 5.82 × 10 −6
[NO] = 2x
[NO] = 1.2 x 10-5 mol/L
Support Questions
13. The following equation represents the equilibrium reaction for the dissociation of
phosphene gas.
COCl2(g) ↔CO(g) + Cl2(g)
At 100oC, the value of Kc for this reaction is 2.2 x 10-8. The initial concentration of
COCl2 in a closed container at 100oC is 1.5 mol/L. What are the equilibrium
concentrations of CO(g) and Cl2(g)?
Predicting the Direction of a Reaction
The reaction quotient, Qc, is an expression that is identical to the equilibrium constant
expression, but its value is calculated using concentrations that are not exactly at
equilibrium. It is used to predict the direction in which a reaction proceeds to reach
equilibrium.
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SCH4U – Chemistry
Lesson 18
aP + bQ ⇔ cR + dS
Qc =
[R ]c [S ]d
[P ]a [Q ]b
Consider:
If Qc = Kc the system must be at equilibrium
If Qc › Kc the numerator must be very large ([right side] greater than at equilibrium) and
the system must attain equilibrium by moving to the left
If Qc ‹ Kc the system must attain equilibrium by moving to the right
Example: 6
Consider the reaction for the production of ammonia (NH3), also known as the Haber
Process.
N2(g) + 3H2(g) ↔2NH3(g)
At 500oC, the value for Kc for this reaction is 0.40. The following concentrations of
gases are present in a container at 500EC. [N2(g)] = 0.10mol/L, [H2(g)] = 0.30 mol/L, and
[NH3] = 0.20 mol/L. Is this mixture of gases at equilibrium? If not, which direction will
the reaction go to reach equilibrium?
Solution 6:
Calculate the reaction Qc
[NH3 ]2
Qc =
[N2 ][H2 ]3
[0.20]2
[0.10][0.30]3
= 14.8
=
Now compare the Qc to the given value for Kc, 0.40
Qc > Kc
Therefore the system is not at equilibrium. The reaction will proceed by moving to the
left.
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SCH4U – Chemistry
Lesson 18
Support Questions
14. At 448°C the equilibrium constant, Kc, for the reaction, H2(g) + I2(g) ↔ 2HI(g) is 51.
Predict how the reaction will proceed to reach equilibrium at 448°C if we start with
2.0 × 10-2 mol of HI, 1.0 × 10-2 mol of H2, and 3.0 × 10-2 mol of I2 in a 2.0L container.
Le Châtelier’s Principle
This principle states:
A dynamic equilibrium tends to respond so as to relieve the effect of any change in the
conditions that affect the equilibrium
•
•
•
qualitatively predicts what can be shown quantitatively by evaluating the Qc
predicts the way that an equilibrium system responds to change in concentration,
volume, or temperature
can be applied to many industrial processes, such as the Haber-Bosch synthesis of
ammonia
Figure 18.1: The Synthesis of Ammonia (Haber Process)
The variables affecting equilibrium are summarized in the table 18.1 on the following
page.
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SCH4U – Chemistry
Lesson 18
Table 18.1: Variables that affect equilibrium
Variable
Type of Change
Concentration
Increase
Decrease
Temperature
Increase
Decrease
Volume
Increase
(decrease in pressure)
Decrease
(increase in pressure)
Variables that do not effect equilibrium
Catalysts
Inert Gases
-
Response of System
Shifts to consume some of
the added reactant or
product
Shifts to replace some of
the decreased reactant or
product
Shifts to consume some of
the added thermal energy
Shifts to compensate some
of the decreased thermal
energy
Shifts towards the side with
the larger total amount of
gaseous entities
Shifts towards the side with
the smaller total amount of
gaseous entities
No effect
No effect
Support Question
15. Consider the following equilibrium:
N2O4(g)
2NO2(g)
H° = 58.0 kJ
In what direction will the equilibrium shift when each of the following changes is
made to a system at equilibrium:
a)
b)
c)
d)
e)
add N2O4
remove NO2
increase the total pressure by adding N2(g)
increase the volume
decrease the temperature?
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SCH4U – Chemistry
Lesson 18
Key Question #18
1. Write equilibrium expressions for each homogeneous reaction. (4 marks)
a) SbCl5(g) ↔ SbCl3(g) + Cl2(g)
b) 2H2(g) + 2NO(g) ↔N2(g) + 2H2O(g)
2. When 1.0 mol of ammonia gas is injected into a 0.50L flask, the following reaction
proceeds to equilibrium.
2NH3(g) ↔ N2(g) + 3H2(g)
At equilibrium, 0.30mol of hydrogen gas is present.
a) Calculate the equilibrium concentrations of N2(g) and NH3(g). (3 marks)
b) What is the value of Kc? (2 marks)
3. At a certain temperature, Kc for the following reaction between sulphur dioxide and
nitrogen dioxide is 4.8. (6 marks)
SO2(g) + NO2(g) ↔ NO(g) + SO3(g)
SO2(g) and NO2(g) have the same initial concentration: 0.36 mol/L. What amount of
SO3(g) is present in a 5.0L container at equilibrium?
4. Phosphorus trichloride reacts with chlorine to form phosphorus pentachloride.
PCl3(g) + Cl2(g) ↔ PCl5(g)
0.75 mol of PCl3 and 0.75 mol of Cl2 are placed in a 8.0L reaction vessel at 500K.
What is the equilibrium concentration of the mixture? The value of Kc at 500K is 49.
(6 marks)
5. Consider the following reaction:
2H2O(l) ↔2H2(g) + O2(g) Kc = 7.3 x10-18 at 1000oC
The initial concentration of water in a reaction vessel is 0.055 mol/L. What is the
equilibrium concentration of H2(g) at 1000oC? (6 marks)
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SCH4U – Chemistry
Lesson 18
Key Question #18 (continued)
6. A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value
of the equilibrium constant, Kc, for the reaction,
H2(g) + I2(g) ↔ 2HI(g)
at 448°C is 50.5. What are the concentrations of H2, I2, and HI in the flask at
equilibrium? (6 marks)
7. Consider the reaction for the synthesis of the ester, ethyl acetate
CH3COOH(l) + CH3CH2OH(l) ↔ CH3COOCH2CH3(l) + H2O(l).
Various samples were analyzed. The concentrations are given in the table below.
Describe whether each sample is at equilibrium. If it is not at equilibrium, predict the
direction in which the reaction will proceed to establish equilibrium. (8 marks)
Sample
a)
b)
c)
d)
CH3COOH(l)
0.10
0.084
0.14
0.063
CH3CH2OH(l)
0.10
0.13
0.21
0.11
CH3COOCH2CH3(l) H2O(l)
0.10
0.10
0.16
0.28
0.33
0.20
0.15
0.17
8. For the reaction.
PCl5(g) ↔PCl3(g) + Cl2(g)
H° = 87.9 kJ
in what direction will the equilibrium shift when; (3 marks)
a) Cl2(g) is added
b) the temperature is increased
c) the volume of the reaction system is decreased
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SCH4U
Grade 12
University Chemistry
Lesson 19 – Acid and Bases Equilibrium
SCH4U – Chemistry
Lesson 19
Lesson 19: Acid and Bases Equilibrium
In a previous chemistry course you likely learned about the properties of acids and
bases. For example acids are sour tasting, turn litmus paper red, and have a pH lower
than 7. Bases, on the other hand, are slippery, bitter tasting, turn litmus paper blue, and
have a pH above 7. In this lesson you will learn more about acid and bases, including
the equilibriums that can be established in weak acid and base solutions.
What You Will Learn
After completing this lesson, you will be able to
•
•
•
•
•
define constant expressions, such as Ksp, Kw, Ka, and Kb;
compare strong and weak acids and bases using the concept of equilibrium;
describe the characteristics and components of a buffer solution.
solve equilibrium problems involving concentrations of reactants and products and
the following quantities: Keq, Ka, Kb, pH, pOH;
explain how buffering action affects our daily lives, using examples (e.g., the
components in blood that help it to maintain a constant pH level; buffered
medications).
Explaining the Properties of Acids & Bases
You have learned about acids and bases in previous chemistry courses. Some of the
general properties of acids and bases are summarized in Table 19.1 below
Table 19.1: Properties of Acids and Bases
Property
Acids
pH
0 - 6.9
taste
sour
Texture of solution
No characteristic texture
Reaction with litmus paper
Turn blue litmus red
Red litmus remains red
Reaction with phenothalein colourless
Bases
7.1 - 14
Bitter
Slippery
Turn red litmus blue
Blue litmus remains blue
Pink
The Arrhenius Theory
•
•
•
an acid dissociates in water to form H+(aq)
(ex: HCl, H2SO4)
(ex: NaOH, KOH)
a base dissociates in water to form OH (aq)
different combinations of strong acids and strong bases react with the same
exothermic result
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SCH4U – Chemistry
Consider:
Lesson 19
HCl(aq) + NaOH(aq) Æ NaCl(aq) + H2O(l)
Total Ionic Equation:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Æ Na+(aq) + Cl-(aq) + H2O(l)
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Æ Na+(aq) + Cl-(aq) + H2O(l)
H+(aq) + OH-(aq) Æ H2O(l)
Net Ionic Equation:
∆H = -56 kJ
The Brønsted-Lowry Theory
•
•
(H+(aq) does not exist in water; rather H3O+(aq))
an acid is an “proton-donor”
a base is a “proton-acceptor”
CH3COOH(aq)
Acid
+
H2O(l)
'
Base
H3O+(aq)
Conj. Acid
+
CH3COO-(aq)
Conj. Base
Conjugate pair
Conjugate pair
Strength of Acids & Bases
• a strong acid or base completely dissociates in water, therefore [H3O+(aq)] or
[OH-(aq)] is equal to the [strong acid] or [strong base], respectively
9 Strong Acids: HCl, HBr, HI, H2SO4, HNO3, HClO3, HClO4
9 Strong Bases: NaOH, KOH, Ca(OH)2, Ba(OH)2
• a weak acid or base only slightly dissociates in water
Calculating Ion Concentrations in Acidic and Basic Solutions
For strong acids and bases, the calculations are simple. Let’s try an example
Example 1:
During an experiment, a student pours 25.0mL of 1.40 mol/L nitric acid into a beaker
that contains 15.0mL of 2.00mol/L sodium hydroxide. What is the concentration of the
ion that causes the solution to acidic or basic?
Solution 1:
Given:
Volume of HNO3
Volume of NaOH = 15.0 mL
[HNO3] = 1.40mol/L
[NaOH] = 2.0 mol/L
Begin by writing the balanced chemical equation for the reaction;
HNO3(aq) + NaOH(aq) Æ NaNO3(aq) + H2O (l)
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SCH4U – Chemistry
Solution:
Lesson 19
ηHNO3 = 1.40 mol/L x 0.0250L
η = 0.0350 mol
Amount of NaOH = 2.00 mol/L x 0.0150L
= 0.0300 mol
The reactants combine in a 1:1 ratio. There is less of the sodium hydroxide, therefore it
is the limiting reactant.
Amount of excess HNO3 = 0.0350 mol – 0.0300 mol = 0.005 mol
Therefore the amount of excess acid (H3O+) is 0.005 mol
Total volume of solution = 25mL + 15mL = 40mL
[H3O+] = 0.005 mol/0.0400L
= 0.12 mol/L
Support Questions
16. Calculate the concentration of hydronium ions in 4.5 mol/L of HCl
17. Calculate the concentration of hydroxide ions in 3.1 mol/L KOH.
The Equilibrium of Weak Acids & Bases
Water dissociates according to the following equation:
2H2O(l) ' H3O+(aq) + OH-(aq)
Pure water is a poor conductor of electricity because very few ions dissociate. At 25oC,
[H3O+] = [OH-] = 1.0 x 10-7 mol/L
The equilibrium constant, Kc, for the dissociation of water is given by the following
expression:
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SCH4U – Chemistry
Kc =
[H3O + ][OH − ]
[H2O ]2
K c [H2O ]2 = [H3O + ][OH − ]
= 1.0 x10 −7 mol / L × 1.0 x10 −7 mol / L
= 1.0 x10 −14
= Kw
Lesson 19
9
9
[H3O+] of a strong acid is equal to the
[dissolved acid].
If dealing with a very dilute acid
solution (about 1x10-7mol/L), the
dissociation of water molecules can be
ignored when determining [H3O+] of a
strong acid
So few ions form that the concentration of water is essentially constant. The ion
product constant, Kw, is the product of [H3O+] and [OH-] ions.
[H3O+] and [OH-] in Aqueous Solutions at 25oC
Acidic Solutions: [H3O+] is greater than 1.0 x 10-7 mol/L and [OH-] is less than 1.0 x 10-7
Basic Solutions: [H3O+] is less than 1.0 x 10-7 mol/L and [OH-] is greater than 1.0 x 10-7
Example 2:
Find the [H3O+] and [OH-] in each solution
a) 2.5 mol/L nitric acid
b) 0.16 mol/L barium hydroxide
Solution 2:
The Kw expression will be used to solve both of these problems.
Kw = 1.0 x 10-14
= [H3O+][OH-]
a) Begin by writing the dissociation equation for nitric acid.
HNO3 Æ H+(aq) + NO3-(aq)
[HNO3] = 2.5 mol/L and since nitric acid is a strong acid [H3O+] = 2.5 mol/L
Rearrange the Kw expression to solve for the [OH-]
1.0 × 10−14
2.5
[OH ] = 4.0 x 10-15 mol/L
[OH-] =
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SCH4U – Chemistry
Lesson 19
b) Begin by writing the dissociation equation for barium hydroxide.
Ba(OH)2 Æ Ba2+(aq) + 2OH-(aq)
Each mole of Ba(OH)2 produces two moles of OH- ions.
Therefore, [OH-] = 2 x 0.16 mol/L = 0.32 mol/L
Rearrange the Kw expression to solve for the [OH-]
[H3O+] =
1.0 × 10−14
0.32
[H3O+] = 3.1 x 10-14 mol/L
pH and pOH
The pH of a solution is the exponential power of hydrogen. It can be expressed as
follows
pH = -log[H3O+]
The range of the pH scale is from 1 - 14.
9 Calculate [H3O+] or [OH-] using
antilog (inverse log, 10x)
[H3O+] = 10-pH
[OH-] = 10-pOH
You can also calculate the pOH (the power of hydroxide ions) of a solution from the
[OH-]
pOH = -log[OH-]
pH + pOH = 14
Example 3:
A liquid shampoo has a hydroxide concentration of 6.8 x 10-5 mol/L
a) Is the shampoo acidic, basic, or neutral?
b) Calculate the hydronium ion concentration.
c) What is the pH and the pOH of the shampoo?
Solution 3:
a) [OH-] = 6.8 x 10-5 mol/L > 1.0 x 10-7. Therefore, the shampoo is basic/
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SCH4U – Chemistry
Lesson 19
b) Use the Kw expression to find the hydronium ion concentration.
1.0 × 10−14
[H3O ] =
6.8 × 10−5
+
[H3O+] = 1.5 x 10-10 mol/L
c) Substitute hydronium and hydroxide concentration into the pH and pOH equations,
respectively.
pH = -log[H3O+]
pH = -log[1.0 x 10-10]
pH = 9.83
pOH = -log[OH-]
pOH = -log[6.8 x10-5]
pOH = 4.17
You can check your answer using pH + pOH =14, 9.83 + 4.17 = 14 (check!)
Example 4:
Suppose that the pH of a urine sample was measured at 5.53 at 250EC. Calculate the
pOH, [H3O+] and [OH-].
Solution 4:
pOH = 14-pH
pOH = 8.47
[H3O+] = 10-5.53
[H3O+] = 3.0 x 10-6 mol/L
[OH-] = 10-8.47
[OH-] = 3.4 x 10-9 mol/L
You can check your answer using [H3O+][OH-] =1.0 x 10-14
Support Questions
18. Phenol. C6H5OH is used as a disinfectant. An aqueous solution of phenol was found
to have a pH of 4.72. Is phenol acidic, neutral or basic? Calculate [H3O+] [OH-] and
pOH of the solution.
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SCH4U – Chemistry
Lesson 19
Acid Dissociation Constant, Ka
Weak acids (example include citric acid, vitamins such as niacin and vitamin B3) do not
completely dissociate in water
For monoprotic acids, the [H3O+] depends on the [HA]initial and the amount that
dissociates:
HA(aq) + H2O(l) ' H3O+(aq) + A-(aq)
Kc =
K c [H2O ] =
[H3O + ][ A− ]
[HA][H2O ]
[H3O + ][ A− ]
= Ka
[HA]
Ka is the acid dissociation constant, and it will be used extensively in this section.
Table 19.1 below lists some common Ka values at 25oC (these values are temperature
dependent). However table 1 in Appendix A will have an extensive list. The Ka of weak
acids have values that are between 1 and 1 x 10-16.
Table 19.1: Some Acid Dissociation Constants for Weak Acids at 25oC
Acid
Formula
Acid Dissociation
constant, Ka
Acetic acid
CH3COOH
1.8 x 10-5
Chlorous acid
HClO2
1.1 x 10-2
Formic acid
HCOOH
1.8 x 10-4
Hydrocyanic acid
HCN
6.2 x 10-10
Hydrofluoric acid
HF
6.6 x 10-4
Hydrogen oxide (water)
H 2O
1.0 x 10-14
Lactic acid
CH3CHOHCOOH
1.4 x 10-4
Nitrous acid
HNO2
7.2 x 10-4
phenol
C6H5OH
1.3 x 10-10
The Equilibrium of Weak Acids & Bases
The percent dissociation of a weak acid is the fraction of acid molecules that
dissociate compared with the initial concentration of the acid. It depends on:
9 Ka for the weak acid
9 initial concentration of the weak acid
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SCH4U – Chemistry
Lesson 19
Polyprotic Acids have more than one hydrogen atom that dissociates. Each
dissociation has its own Ka.
9 all polyprotic acids, except sulfuric acid, are weak acids. Ex H3PO4 is a weak acid
that is polyprotic
9 their 2nd dissociation is much weaker than their 1st
9 consider only their 1st dissociation when calculating [H3O+] and pH
NOTE: Refer to Table 1 (Appendix A) for Ionization Constants for Polyprotic Acids
If
[HA]
〉500
Ka
The change in the initial concentration, x, is negligible and can be ignored.
If
[HA]
〈500
Ka
The change in the initial concentration, x, may not be negligible. The
equilibrium equation will be more complex, possibly requiring the solution
of a quadratic equation.
Example 5:
Formic Acid, HCOOH is present in the sting of certain ants. What is the pH of a 0.025
mol/L solution of formic acid.
Solution 5:
Since formic acid is a weak acid, and does not ionize completely, we have to set up an
ICE Table to determine the hydronium ion concentration.
First write the equation for the dissociation of formic acid, HCOOH
HCOOH(aq) + H2O(l) ↔ H3O+(aq) + HCOO-(aq)
Notice that the equilibrium arrow is used to indicate incomplete dissociate, thus an
equilibrium calculation will be used
HCOOH(aq)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
0.025
-x
0.025-x
+ H2O(l) ↔ H3O+(aq) +
0.00
+x
+x
HCOO-(aq)
0.00
+x
+x
Note: the water is not included in the equilibrium expression since it is in liquid form.
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SCH4U – Chemistry
Ka =
1.8 × 10 −4 =
Lesson 19
[HCOO − ][H3O + ]
[HCOOH ]
( x )( x )
( 0.025 − x )
**The value for Ka for formic acid (found in table 1, appendix A) is 1.8 x 10-4**
Next check to see if the amount of acid that dissociates is negligible compared to initial
concentration of the acid.
[HCOOH ]
0.025
=
Ka
1.8 × 10−4
= 139
Since this value is < 500, the amount that dissociate is NOT negligible compared with
initial concentration. In other words we can NOT disregard “x”.
Solve the Ka expression above using the quadratic equation.
−4
x + 1.8 × 10 × −4.5 × 10
2
−6
=0
x =
−b ±
b − 4ac
2
2a
−4
x =
−( −1.8 × 10 ) ±
−4
−6
(1.8 × 10 ) − 4 × 1 × ( −4.5 × 10 )
2
2×1
x = 0.0020 and x =-0.002
The negative value is not reasonable, since a concentration term cannot be negative.
Therefore x = 0.0020 mol/L = [H3O+]
Finally, calculate the pH using pH = -log[H3O+]
pH = -log0.0020
pH = 2.70
Let’s do a sample calculation for polyprotic acids (an acid that dissociate more than one
proton)
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SCH4U – Chemistry
Lesson 19
Example 6:
Calculate the pH, [H2PO4-] and [HP042-] of a 3.5 mol/L aqueous solution of phosphoric
acid, H3PO4.
Solution 6:
Begin by writing the equation for the dissociation of phosphoric acid, H3PO4.
H3PO4(aq) + H2O(l) ↔ H2PO4-(aq) + H3O+(aq)
Given Ka1 for H3PO4 = 7.0 x 10-3, Ka2 for H2PO4 = 6.3 x 10-8
H3PO4(aq) + H2O(l) ↔ H2PO4-(aq)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Ka =
7.0 × 10−3 =
3.5
-x
3.5 -x
H3O+(aq)
+
0.00
+x
+x
0.00
+x
+x
[H2PO4 − ][H3O + ]
[H3PO4 ]
( x )( x )
3.5 − x
Next determine if the amount of dissociation of phosphoric acid is negligible or not.
[H2PO4 − ]
3.5
=
Ka
7.0 × 10−3
= 500
Therefore, x is probably negligible, so our solution becomes much simpler.
x2
3.5
x = 0.16 mol/L
7.0 × 10−3 =
Now write the expression for the dissociation of H2PO4H2PO4(aq) + H2O(l) ↔ HPO42-(aq) + H3O+(aq)
Copyright © 2008, Durham Continuing Education
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SCH4U – Chemistry
Lesson 19
H2PO4(aq) + H2O(l) ↔
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Ka =
6.3 × 10−8 =
0.16
-x
.16-x
HPO42-(aq) +
0.00
+x
+x
H3O+(aq)
0.16
+x
0.16 +x
[HPO4 2− ][H3O + ]
[H2PO4 − ]
( x )( 0.16 + x )
( 0.16 − x )
Now check to see if the value of x is negligible.
[HPO4 2− ]
0.16
=
Ka
6.3 × 10 −8
=> 500
Therefore the value of x is negligible.
x ( 0.16 )
0.16
x = 6.3 × 10 −8
6.3 × 10−8 =
pH = -log[0.16]
pH = 0.80
Support Questions
19. Calculate the pH of a sample of vinegar that contains 0.083 mol/L acetic acid. What
is the percent dissociation of the vinegar?
20. A chemist finds that 0.03% of hypochlorous acid molecules are dissociated in a 0.40
mol/L solution of the acid. What is the value of Ka for the acid?
21. Adipic acid is a diprotic acid used to manufacture nylon. Its formula can be
shortened to H2Ad. The acid dissociation constants for adipic acid are Ka1 = 3.71 x
10-5 and Ka2 = 3.87 x 10-6. What is the pH of a 0.085 mol/L solution of adipic acid?
Copyright © 2008, Durham Continuing Education
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SCH4U – Chemistry
Lesson 19
Bases and Buffers
Base Dissociation Constant, Kb
• weak bases, B, react with water to form an equilibrium solution of ions:
B(aq) + H2O(l) HB+(aq) + OH-(aq)
[HB + ][OH − ]
Kc =
[B][H2O ]
Many compounds present in plants are weak bases
9
9
[HB + ][OH − ]
= Kb
K c [H2O ] =
[B]
Acids and Their Conjugate Bases
Consider:
CH3COOH(aq) + H2O(l)
Ka =
'
Caffeine
Piperidine (black pepper)
H3O+(aq)
+
CH3COO-(aq)
[CH3COO − ][H3O + ]
[CH3COOH ]
The acetate ion is the conjugate base of acetic acid. A soluble salt of the conjugate
base, such as sodium acetate, forms acetate ions in solution and the acetate ion acts as
a base with water:
CH3COO-(aq)
Kb =
+
H2O(l)
'
CH3COOH(aq) + OH-(aq)
[CH3COOH ][OH − ]
[CH3COO − ]
The product of KaKb gives an interesting result:
K aK b =
[CH3COO − ][H3O + ] [CH3COOH ][OH − ]
×
[CH3COOH ]
[CH3COO − ]
= [H3O + ][OH − ]
The strength of an acid and its
conjugate base are inversely related.
= Kw
Table 19.2 below lists some common Kb values at 25oC (these values are temperature
dependent). However table 2 in Appendix A will have an extensive list.
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SCH4U – Chemistry
Lesson 19
Table 19.2: Some Base Dissociation Constants at 25oC
Base
Formula
Base Dissociation constant,
Kb
Ethylenediamine
NH2CH2CH2NH2
5.2 x 10-4
Dimethylamine
(CH3)2NH
5.1 x 10-4
Methylamine
CH3NH2
4.4 x 10-4
Trimethylamine
(CH3)3N
6.5 x 10-5
Ammonia
NH3
1.8 x 10-5
Hydrazine
N2H4
1.7 x 10-6
Pryridine
C5H5N
1.4 x 10-9
Aniline
C6H5NH2
4.2 x 10-10
Urea
NH2CONH2
1.5 x 10-14
Example 7:
The base dissociation constant, Kb, for quinine is 3.3 x 10-6. Calculate the [OH-] and the
pH of a 1.7 x 10-3 mol/L solution of quinine.
Solution 7:
Kb quinine = 3.3 x 10-6.
Since the formula of quinine will not affect our calculation, we will use the symbol Q to
represent quinine.
Begin by writing the dissociation equation for quinine.
Q(aq) + H2O(l) ↔ HQ + (aq) + OH-(aq)
Q(aq) + H2O(l)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Kb =
[HQ + ][OH − ]
[Q ]
↔
1.7 x10-3
-x
1.7 x 10-3 –x
3.3 x10-6 =
HQ + (aq)
0.00
+x
+x
+
OH-(aq)
0
+x
+x
( x )( x )
(1.7 × 10−3 − x )
Determine if the value of x is negligible
[Q ] 1.7 × 10−3
=
K b 3.3 × 10−6
> 500
Therefore the value of x is negligible.
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SCH4U – Chemistry
3.3 x 10-6 =
Lesson 19
x2
1.7 × 10−3
x = ± 7.5 x 10-5 (negative value not possible)
x = 7.5 x 10-5
pH = 14.00 – pOH
pH = 9.87
Example 8:
Sodium acetate, CH3COONa is used for developing photographs. Find the value of Kb
for the acetate ion. Then calculate the pH of a solution that contains 12.5g of sodium
acetate dissolved in 1.00L of water.
Solution 8:
Ka for acetic acid = 1.81 x 10-5
Begin by finding Kb using the relationship
Kw = Ka x Kb
1× 10 −14
1.8 × 10 −5
= 5.6 × 10 −10
Kb =
Next calculate the concentration of sodium acetate
12.5g
82.0g / mol
= 0.152mol
η=
Concentration for one litre is 0.152 mol/L
Write out the equation for the dissociation of sodium acetate acting as a base
CH3COO-(aq) + H2O(l) ↔ CH3COOH(aq) + OH-(aq)
[CH3COO-] = 0.152 mol/L
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SCH4U – Chemistry
Lesson 19
CH3COO-(aq) + H2O(l) ↔ CH3COOH(aq) +
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
0.152
-x
0.152 – x
0.00
+x
+x
OH-(aq)
0
+x
+x
[CH3COOH ][OH − ]
Kb =
[CH3COO − ]
Kb =
[ x ][ x ]
[0.152 − x ]
Determine if the value of x is negligible or not
[CH3COO − ]
0.152
=
Kb
5.6 × 10 −10
> 500
Therefore the value of x is negligible.
5.6 x 10-10 =
x2
0.152
x = 9.2 x 10-6 = [OH-] (only positive root possible)
[H3O+] =
Kw
[OH − ]
[H3O+] = 1.1 x 10-9mol/L
pH = -log[H3O+]
pH = 8.96
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SCH4U – Chemistry
Lesson 19
Support Questions
22. An aqueous solution of household ammonia has a molar concentration of 0.105
mol/L. Calculate the pH of the solution.
23. Morphine, C17H19NO3, is a naturally occurring base to control pain. A 4.5 x 10-3 mol/L
solution has a pH of 9.93. Calculate Kb for morphine.
24. An aqueous solution of ammonia has a pH of 10.85. What is the concentration of
the solution?
Buffer Solutions (resist changes in pH)
•
•
•
contain a weak acid/conjugate base or weak base/conjugate acid pair
have a buffer capacity related to its concentration
play an important role in biological systems (homeostasis)
Acid-Base Titration Curves
A graph of the pH of an acid (or base) against the volume of an added base (or acid) is
called an acid-base titration curve.
9 common analytical technique to determine an unknown concentration
9 acid and base are completely reacted at the equivalence point
9 indicator changes colour at chemical end point
9 best results if end point ≈ equivalence point
9 phenolphthalein is often used
Figure 19.2: A sample titration curve
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SCH4U – Chemistry
Lesson 19
Key Question #19
1. Recreate and complete the following table by calculating the missing
values and indicating whether each solution is acidic or basic. (1/2 mark each, 8
marks total)
[H3O+] (mol/L)
4.1 x 10-5
pH
9.42
[OH-] (mol/L)
pOH
Acidic or basic?
6.9 x 10-2
9.2
2. A 0.10 mol/L solution of a weak acid was found to be 4.0% dissociated. Calculate
Ka. (5 marks)
3. Calculate the pH of a 0.10 mol/L solution of ascorbic acid, H2C6H6O6(aq) and the
equilibrium concentration of H2C6H6O6, HC6H6O6-(aq) and C6H6O62-(aq). Ascorbic acid
is a polyprotic acid. (8 marks)
4. What is the value of the base ionization constant, Kb, for the acetate ion, C2H3O2-(aq).
(3 marks)
5. Calculate the pH of a 0.100 mol/L aqueous solution of hydrazine, N2H4(aq), a weak
base. The Kb for hydrazine is 1.7 x 10-6. (5 marks)
Copyright © 2008, Durham Continuing Education
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SCH4U
Grade 12
University Chemistry
Lesson 20 – Solubility Equilibriums
SCH4U – Chemistry
Lesson 20
Lesson 20: Solubility Equilibriums
Solutions have two components: a solute and a solvent. In salt water solution, for
example, salt is the solute and water is the solvent. The ions in the salt can dissociate
into ions. In some solutions, equilibrium can be established in certain solutes that have
low solubility, which is the focus of this unit.
What You Will Learn
After completing this lesson, you will be able to
•
•
•
•
•
•
•
describe, using the concept of equilibrium, the behaviour of ionic solutes in solutions
that are unsaturated, saturated, and supersaturated;
calculate the molar solubility of a pure substance in water or in a solution of a
common ion, given the solubility product constant (Ksp), and vice versa;
predict the formation of precipitates by using the solubility product constant;
solve equilibrium problems involving concentrations of reactants and products and
Ksp,
predict, in qualitative terms, whether a solution of a specific salt will be acidic, basic,
or neutral;
solve problems involving acid-base titration data and the pH at the equivalence
point.
identify effects of solubility on biological systems (e.g., kidney stones, dissolved
gases in the circulatory system of divers, the use of barium sulfate in medical
diagnosis);
Acid-Base Properties of Salt Solutions
The pH of an aqueous salt solution can be predicted using reactions between water and
the dissociated ions of the salt. Ions that do not react with water produce a neutral
solution. Ions that do react with water produce a solution with an excess of H3O+(aq) or
OH-(aq). The extent of the reaction determines the pH of the solution.
Salts that form NEUTRAL Solutions
Strong acids and strong bases dissociate completely in water:
HA(aq) + H2O(aq) Æ H3O+(aq) + A-(aq)
strong acid
•
•
•
•
conj. base (weak)
Anion formed by a strong acid is a much weaker base than water
Salt containing an anion of a strong acid tends to have no effect on pH (ie: Cl-)
Cation formed by a strong base is a much weaker acid than water
Salt containing a cation of a strong base tends to have no effect on pH (ie: Na+)
Copyright © 2008, Durham Continuing Education
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SCH4U – Chemistry
Lesson 20
Salts that form ACIDIC Solutions
Weak bases dissociate very little, therefore, the equilibrium lies to the left:
NH3(aq) + H2O(l)
weak base
NH4+(aq) + H2O(l)
•
'
NH4+(aq) + OH-(aq)
conj. acid (strong)
'
NH3(aq) + H3O+(aq)
Salts of weak bases and strong acids form acidic solutions
i.e.: NH4Cl
Salts that form BASIC Solutions
Weak acids (CH3COOH) form conjugate bases (CH3COO-) that are relatively strong
CH3COO-(aq) + H2O(l)
•
'
CH3COOH(aq) + OH-(aq)
Salts of weak acids and strong bases form basic solutions
i.e.: NaCH3COOH
Salts of Weak Bases and Weak Acids
•
•
Both ions react with water
Use Ka and Kb to determine which ion is stronger
¾ If Ka > Kb, the solution is acidic
¾ If Ka < Kb, the solution is basic
Example 1:
Predict the acid/base property of an aqueous solution of each of the following salts. If
you predict that the solution is not neutral, write the equation for the reaction that
causes the solution to be acidic or basic.
a) ammonia nitrate, NH4NO3
b) sodium chloride, NaCl
c) ammonia hydrogen carbonate, NH3HCO3
Solution 1:
Determine whether the cation is from a strong or weak base, and whether the anion is
from a strong or weak acid.
a) Ammonium, NH4 is from a weak base (ammonia)
Nitrate, NO3 is from a strong acid (nitric acid) ∴ the solution is acidic
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SCH4U – Chemistry
Lesson 20
b) Na is from strong base (NaOH)
Cl is from strong acid (HCl)
Therefore the solution is neutral
c) NH3 is from weak base (NH3)
HCO3 is weak acid (H2CO3)
In order to answer this, we must examine the Ka/Kb values.
Ka for NH4+ = 5.6 x 10-10 (calculate using Kw = Ka x Kb)
Kb for H2CO3 = 2.2 x 10-8 (calculate using Kw = Ka x Kb)
Because the Kb for hydrogen carbonate is larger than the Ka for ammonia, the solution
is basic.
Calculating pH at Equivalence
The equivalence point in a titration occurs when just enough acid and base have been
mixed for a complete reaction to occur, with no excess of either. A chemical pH
indicator can be used instead of a pH meter. A pH indicator is a weak organic acid
which is in equilibrium with its conjugate base. The point at which the indicator changes
colour is called the end-point. You must know the approximate pH of the solution at
equivalence when selecting an appropriate indicator for a particular titration. This can
be done with a titration curve or using calculations and salt information since the pH at
equivalence in a titration is the same as the pH of an aqueous solution of the salt
formed!
RECALL:
ACID + BASE Æ WATER + SALT
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SCH4U – Chemistry
Lesson 20
Example 2:
20 mL of 0.20 mol/L NH3(aq) is titrated against 0.20mol/L HCl(aq). Calculate the pH at
equivalence.
Solution 2:
Write the chemical equation to represent the reaction
NH3(aq) + HCl(aq) Æ NH4Cl(aq)
Moles of ammonia = 0.20 mol/L x 0.020L
= 4.0 x 10-3 mol
= moles HCl (1:1)
4.0 × 10 −3 mol
0.20mol / L
= 0.020L
Volume HCl =
Total volume of solution = 40 mL
The salt, NH4Cl is formed from a weak base which reacts with, and a strong acid, which
does not react with water. The pH of the solution is therefore determined by the extent
of the following reaction
NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
4.0 × 10−3 mol
0.040L
= 0.10mol / L
[NH4Cl] =
Kb ammonia is 1.8 x 10-5
Ka = 5.6 x 10-10 (Using Kw =Ka x Kb)
Next set up your ICE table
NH4+(aq) + H2O(l)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
0.10
-x
0.10 – x
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↔
NH3(aq)
0.00
+x
+x
+
H3O+(aq)
0
+x
+x
Page 49 of 80
SCH4U – Chemistry
Lesson 20
Determine if x is negligible
[NH4 ]
0.10
=
Ka
5.6 × 10−10
> 500, therefore x can be ignored
[NH3 ][H3O + ]
Ka =
[NH4 + ]
5.6 X 10-10 =
( x )( x )
0.10
x = 7.5 x 10-6 mol/L (negative value not valid)
pH = -log[H3O+]
pH = 5.13
Possible indicators you could use would be methyl red or bromocresol green.
Support Questions
25. a) Predict whether a 0.10 mol/L solution of NaNO2(aq) solution will be acidic, basic, or
neutral.
b) Calculate the pH of a 0.10 mol/L solution of NaNO2(aq).
26. If 50.0 mL of 0.10 mol/L hydrobromic acid is titrated with 0.10 mol/L aqueous
ammonia, determine the pH at equivalence.
Solubility Equilibria
Recall:
∆G = ∆H - T∆S
When a salt dissolves:
9 Entropy (∆S) always increases because ions in solution are more disordered
9 Most solids dissolve to a greater extent at higher temperatures
9 If the overall enthalpy change (∆H) is negative, the formation of a solution is
favoured
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SCH4U – Chemistry
Lesson 20
HETEROGENEOUS EQUILIBRIUM: A Solubility System
For solubility systems of sparingly soluble ionic compounds, such as Barium Sulphate
used in GI Tract x-rays, equilibrium exists between the solid ionic compound and its
dissociated ions in solution.
BaSO4(s) ' Ba2+(aq) + SO42- (aq)
The Solubility Product Constant, Ksp
K=
[Ba 2+ (aq ) ][SO4 2− (aq ) ]
Like the equilibrium constant,
Ksp is temperature-dependent.
[BaSO4( s ) ]
K [BaSO4( s ) ] = [Ba 2+ (aq ) ][SO4 2−(aq ) ]
K sp = [Ba 2+ (aq ) ][SO4 2−(aq ) ]
You can use the value of Ksp for a compound to determine the concentration of its ions
in a saturated solution. Table 20.1 below contains select Ksp values. There is a more
detailed list in table 3 located in Appendix A.
Compound
Magnesium sulphate, MgSO4
Lead (II) chloride, PbCl2
Barium fluoride, BaF2
Cadmium carbonate, CdCO3
Copper (II) hydroxide, Cu(OH)2
Silver sulfide, Ag2S
Ksp
5.9 x 10-3
1.7 x 10-5
1.5 x 10-6
1.8 x 10-14
2.2 x 10-20
8 x 10-48
Let’s try a few sample calculations involving Ksp;
Example 3:
Calculate the molar solubility (in mol/L) of a saturated solution of the substance.
Silver chloride, AgCl, has a Ksp = 1.77 x 10-10. Calculate its solubility in moles per liter.
Solution 3:
Begin by writing the dissociation equation for silver chloride:
AgCl(s) ↔ Ag+ (aq) + Cl-(aq)
Then write the Ksp expression for silver chloride
Ksp = [Ag+] [Cl-]
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SCH4U – Chemistry
Lesson 20
This is the equation we must solve. First we put in the Ksp value:
1.77 x 10-10 = [Ag+] [Cl-]
When examining the chemical equation and see that there is a 1:1 ratio between Ag+
and Cl-. We know this from the coefficients (both one) of the balanced equation.
That means that the concentrations of the two ions are EQUAL. We can use the same
unknown (x) to represent both.
Substituting, we get:
1.77 x 10-10 = (x) (x)
Now, take the square root of both sides.
x = 1.33 x 10-5 mol/L
This is the answer because there is a one-to-one relationship between the Ag+
dissolved and the AgCl it came from. So, the molar solubility of AgCl is 1.33 x 10-5
mol/L.
Example 4:
Calculate the molar solubility (in mol/L) of a saturated solution of the substance.
Here are the two substances:
a) Sn(OH)2 Ksp = 5.45 x 10-27
b) Ag2CrO4 Ksp = 1.12 x 10-12
Solution 4:
a) Tin(II) hydroxide
First write the equation for the dissociation:
Sn(OH)2 ↔ Sn2+ + 2 OHand then the Ksp expression:
Ksp = [Sn2+] [OH-]2
The ratio between Sn2+ and OH- is 1:2. That means that however much Sn2+ dissolves,
DOUBLE that amount of OH- dissolves.
One Sn2+ makes two OH-. That means that if 'x' Sn2+ dissolves, then '2x' of the OH- had
to have dissolved.
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SCH4U – Chemistry
Lesson 20
Substituting x into the Ksp equation:
5.45 x 10-27 = (x) (2x)2
4x3 = 5.45 x 10-27
Solving that, we get:
x = 1.11 x 10-9 mol/L
This is the answer because there is a 1:1 ratio between Sn2+ and Sn(OH)2
b) Silver chromate
Dissociation equation
Ag2CrO4 ↔ 2 Ag+ + CrO42-
Ksp = [Ag+]2 [CrO42-] = 1.12 x 10-12
1.12 x 10-12 = (2x)2(x)
**Allow x to equal the concentration of whatever ion has a one in front of it. That ion will
always be in a one-to-one ratio with the solid which is dissolving. So solving for x gives
the molar solubility of the substance.
Solving for x:
4x3 = 1.12 x 10-12
x = 6.54 x 10-5 mol/L
The Common Ion Effect
Adding a common ion to a solution increases the concentration of that ion in solution.
As a result, equilibrium shifts away from the ion (Le Châtelier’s Principle). Since Ksp is
constant at a given temperature, an increase in the concentration of one ion must be
accompanied by a decrease in the concentration of the other ion, achieved by the
formation of a precipitate. A buffer is an example of the common ion effect. Since
buffers are in homogeneous equilibria, the initial concentration of reactants needs to be
considered. Ka is used instead of Ksp.
Example 5:
Silver chloride, AgCl will be dissolved into a solution containing 0.0100mol/L in chloride
ion. What is the solubility of AgCl?
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SCH4U – Chemistry
Lesson 20
Solution 5:
Write the dissociation equation for AgCl:
AgCl(s) ↔ Ag+(aq) + Cl-(aq)
Next, write the Ksp expression:
Ksp = [Ag+] [Cl-]
First substitute in the Ksp value:
1.77 x 10-10 = [Ag+] [Cl-]
Now, we have to reason out the values of the silver and chloride ions. The problem
specifies that [Cl¯] is already 0.0100. I get another 'x' amount from the dissolving AgCl.
Of course, [Ag+] is 'x.'
Substituting into the equation:
1.77 x 10-10 = (x) (0.0100 + x)
This will wind up to be a quadratic equation which is solvable via the quadratic formula.
However, there is a chemical way to solve this problem. We reason that 'x' is a small
number, such that '0.0100 + x' is almost exactly equal to 0.0100. If we were to use
0.0100 rather than '0.0100 + x,' we would get essentially the same answer and do so
much faster. So the problem becomes:
1.77 x 10-10 = (x) (0.0100)
x = 1.77 x 10-8 mol/L
There is another reason why neglecting the 'x' in '0.0100 + x' is possible. Measuring Ksp
values are difficult to do and, hence, have a fair amount of error already built into the
value. So the very slight difference between 'x' and '0.0100 + x' really has no bearing on
the accuracy of the final answer. Why not? Because the Ksp already has significant error
in it to begin with. Our "adding" a bit more error is insignificant compared to the error
already there.
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SCH4U – Chemistry
Lesson 20
Support Questions
27. Calculate the Ksp for magnesium fluoride at 25oC, given a solubility of 0.00172
g/100mL.
28. Calculate the solubility of silver iodide at 25oC. The Ksp of AgI(s) is 1.5 x 10-16 at
25oC.
29. What is the molar solubility of PbCl2(s) in a 0.2 mol/L NaCl(aq) solution?
30. Calculate the solubility of silver chloride in a 0.10 mol/L solution of sodium chloride
at 25oC. At SATP, Ksp AgCl(s) = 1.8 x 10-10.
Predicting the Formation of a Precipitate
The Ion Product, Qsp
• Expression is identical to the solubility product constant, but concentrations are
not necessarily at equilibrium.
MgSO4(s) ' Mg2+(aq) + SO42-(aq)
Ksp = 5.9 X 10-3
Qsp < Ksp
Qsp = Ksp
Qsp = [Mg2+][SO42-]
Qsp > Ksp
yThe system attains equilibrium by moving
to the right, favoring dissociation.
y More solid can dissolve.
y The system is at equilibrium.
y No more solid can dissolve.
y No precipitate forms.
y The system attains equilibrium by moving
to the left, favoring precipitation.
y A precipitate forms until the equilibrium is
reached.
RECALL: General solubility guidelines can be used to help predict whether or not a
precipitate will form. These are summarized below in table 20.1 below.
Table 20.1: Solubility Guidelines
Solubility Guidelines
1. Soluble cations: NH4+ and group I metals (Li+, Na+, etc.)
2. Soluble anions: polyatomics: ClO4-, NO3-, SO42-, HSO4-, CH3CO2halogens: X=Cl-, Br-, I- but not F3. most other salts are insoluble
4. insoluble exceptions to guideline 2: AgX, PbX2, Hg2X2
XSO4 (X=Ca, Sr, Ba, Pb)
5. soluble exceptions: Ba(OH)2, MS (M=group 2 Metals: Mg+2, Ca+2, etc)
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SCH4U – Chemistry
Lesson 20
Example 6:
Will a precipitate form when 1.0 L of 3.0 x 10-10mol/L Cu(NO3)2 is added to 1.0 L of 2.0 x
10-11M Na2S?
Solution 6:
Always consider both possible precipitates. The first possibility will be the positive ion
from the first salt and the negative ion from the second salt, while the second possibility
will be the positive ion from the second salt and the negative ion from the first salt.
The two possibilities are:
a) CuS
the Solubility Table says “low solubility” (the Cu+2 ion is part of the "All
others" category)
b) NaNO3
the Solubility table says soluble, therefore we can eliminate this
possibility.
Write the dissociation equation for the possible precipitate:
CuS(s)
↔
Cu+2(aq)
+
S-2(aq)
3.0 x 10-10mol/L
2.0 x 10-11mol/L
2.0
2.0
-
Note that both concentration values are divided by two. This is because equal volumes
of the two solutions were combined, which causes the concentrations of all species to
be halved.
Qsp = [Cu+2] [S-2]
Qsp = (1.5 x 10-10) (1.0 x 10-11)
Qsp = 1.5 x 10-21
Compare Qsp to Ksp
From the "Solubility Constant Table" (Table 2 Appendix A) the Ksp for CuS is:
Ksp = 6.0 x 10-37
Therefore: Qsp > Ksp and a precipitate will form
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SCH4U – Chemistry
Lesson 20
Analytical Applications
•
•
•
Fractional precipitation is a process in which ions are selectively precipitated
from solution, leaving other ions behind
Filtration or a centrifuge is used to remove solid precipitates from a mixture
Analysis may be qualitative or quantitative
Support Questions
31. If 100 mL of a .100 mol/L CaCl2(aq) and 100mL of 0.0400mol/L Na2SO4(aq) are mixed
at 20oC, determine whether a precipitate will form. For CaSO4(aq) at 20oC, Ksp is 3.6
x 10-5
Key Question #20
1. a) Predict whether a 0.20 mol/L solution of ammonia chloride, NH4Cl(aq), will be
acidic, basic, or neutral. (3 marks)
b) Calculate the pH of a 0.20 mol/L solution of NH4Cl(aq). (5 marks)
2. a) When 25 mL of 0.10 mol/L HBr is titrated with 0.10 mol/L NaOH (aq), what is the
pH at the equivalence point? (7 marks)
b) Select an appropriate indicator. (2 marks)
3. The maximum solubility of silver cyanide, AgCN, is 1.5 x 10-8 mol/L at 25EC.
Calculate Ksp for silver cyanide. (3 marks)
4. At 25oC, Ksp for PbI2 is 9.8 x 10-9. What is the molar solubility of PbI2 in water at
25oC? (3 marks)
5. Determine the molar solubility of lead (II) iodide, PbI2, in 0.0500 mol/L NaI. (5 marks)
6. A solution contains 0.15 mol/L of NaCl and 0.0034 mol/L Pb(NO3)2. Does a
precipitate form? Include a balanced equation for the formation of the possible
precipitate. Ksp is 1.7 x 10-5. (5 marks)
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Lesson 17
1. What are the conditions necessary for equilibrium?
ÆMust have a closed system.
ÆMust have a constant temperature.
ÆEa must be low enough to allow a reaction.
2. What is a forward reaction versus a reverse reaction?
In a forward reaction, the reactants collide to produce products and it goes from left
to right.
3. What are the characteristics of equilibrium?
ÆForward rate is equal to the reverse rate.
ÆThe concentration of reactants and products are constant. (Not equal!)
ÆMacroscopic properties are constant (color, mass, density, pressure,
concentrations).
4. When ammonia is heated, it decomposes into nitrogen gas and hydrogen gas
according to the following equation:
2NH3(g) ↔ N2(g) + 3H2(g)
When 4.0 mol of NH3(g) is introduced into a 2L container and heated to a particular
temperature, the amount of ammonia changes to 2.0 mol. Determine the
equilibrium concentrations of the other two entities.
[NH3]initial =
4.0mol
=2.0mol/L
2.0L
[NH3]equilibrium =
2.0mol
=1.0 mol/L
2.0L
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
2NH3(g) ↔
2.0
-2x
2.00-2x
N2(g)
0.00
+x
x
+
3H2(g)
0.00
+3x
3x
[NH3(g)] = 2.0 mol/L -2x
= 1.0 mol/L
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2.0mol/l -2x =1.0mol/L
x = 0.5 mol/L
[N2(g)] = x =0.5mol/L
[H2(g)] = 3x
= 3(0.5 mol/L)
= 1.5 mol/L
5. When carbon dioxide is heated in a closed container, it decomposes into carbon
monoxide and oxygen according to the following equilibrium equation:
2 CO2(g) ↔2CO(g) + O2(g)
When 2 mol of CO2(g) is placed in a 5.0L container and heated to a particular
temperature, the equilibrium concentration of CO2(g) is measured to be 0.39mol/L.
Determine the equilibrium concentrations of CO(g) and O2(g).
[CO2(g)]initial =
2.0mol
=0.4 mol/L
5L
[CO2(g)]equilibrium 0.39 mol/L
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
2CO2(g) ↔
0.4
-2x
0.4 mol/L-2x
2CO(g)
0.00
+2x
2x
+
O2(g)
0.00
+x
x
0.39 mol/L = 4 mol/L -2x
x = 0.005 mol/L
[CO]equilibrium = 2(0.005mol/L) = 0.01 mol/L
[O2(g)]equilibrium = 0.005 mol/L
6. At 40oC, 2.0 mol of pure NOCl(g) is introduced into a 2.0L flask. The NOCl(g) partially
decomposes according to the following equilibrium equation:
2NOCl(g) ↔ 2NO(g) + Cl2(g)
At equilibrium, the concentration of NO(g) is 0.032 mol/L. Determine the equilibrium
concentrations of NOCl(g) and Cl2(g) at this temperature.
[NOCl]initial =
2.0mol
= 1.0 mol/L
2L
[NO(g)]equilibrium = 0.032 mol/L
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Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
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2NO(g)
2NOCl(g) ↔
1.0
0.00
-2x
+2x
1.0 mol/L-2x 2x
+
Cl2(g)
0.00
+x
x
0.032 mol/L = 2x
x = 0.016 mol/L
[NOCl]equilibrium = 1.0 mol/L -2(0.016 mol/L)
[NOCl]equilibrium = 0.968 mol/L
[Cl2(g)] = 0.016 mol/L
7. In each process, how does the entropy of the system change?
a) Ice melting –Increase (solid Æ liquid)
b) water vapour condensing -Decrease (gas Æliquid)
c) sugar dissolving in water -Increase (solid Æliquid)
d) HCl(g) + NH3(g) Æ 2NH4Cl(s) -Decrease (gas Æ solid)
e) CaCO3(s) Æ CaO(s) + CO2(g)- Increase (solid Æ solid and gas)
Lesson 18
8. Write the equilibrium expression for each reaction
a) The reaction between nitrogen gas and oxygen gas at high temperatures:
N2(g) + O2(g) ↔ 2NO(g)
[NO ]2
Kc =
[N2 ][O2 ]
b) The reaction between hydrogen gas and oxygen gas to form water vapour:
2H2(g) + O2(g) ↔ 2H2O(g)
Kc =
[H2O ]2
[H2 ]2 [O2 ]
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c) The REDOX equilibrium of iron and iodine ions in aqueous solution:
2Fe3+(aq) + 2I-(aq) ↔ 2Fe2+(aq) + I2(aq)
[Fe 2+ ]2 [I2 ]
Kc =
[Fe3+ ]2 [I ]2
d) The oxidation of ammonia
4NH3(g) + 5O2(g) ↔ 4NO(g) + 6 H2O(g)
[NO ]4 [H2O ]6
Kc =
[NH3 ]4 [O2 ]5
9. At 25oC, the value of Kc for the following reaction is 82.
I2(g) + Cl2 ↔ 2ICl(g)
0.83 mol of I2(g) and 0.83 mol of Cl2(g) are placed in a 1.0L container. What are the
concentrations of the gases at equilibrium?
I2(g)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Kc =
82 =
0.83
-x
0.83-x
+
Cl2 ↔
0.83
-x
0.83- x
2ICl(g)
0.00
+x
2x
[ICl ]2
[I2 ][Cl2 ]
[2 x ]2
[0.83 − x ][0.83 − x ]
9.055 =
2x
[0.83 − x ]
Take the square root of each side
7.516 -9.055 x = 2x
7.516 = 11.055 x
x = 0.680 mol/L
[I2] = [Cl2] = 0.83-.680 = 0.15 mol/L
[ICl] = 1.36 mol/L
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10. At a certain temperature, Kc = 4.0 for the following reaction:
2HF(g)↔H2(g) + F2(g)
A 1.0L reaction vessel contained 0.045 mol of F2(g) at equilibrium. What was the
initial amount of HF in the reaction vessel?
2HF(g)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
↔
H2(g) +
2x
0.090
2x -.090
F2(g)
0.00
+.045
.045
0.00
+.045
.045
(.045)(.045)
(2 x − .090)2
x = 0.056mol / L
4.0 =
The initial amount of [HF] is 2x, thus 2(.056) = 0.11 mol/L
11. Consider the following reaction
SO2(g) + NO2(g) ↔NO(g) + SO3(g)
In a 1.0L container, 0.017 mol of SO2(g) and 0.011 mol of NO2(g) were added. The
value of Kc for the reaction at 200K is 4.8. What is the equilibrium concentration of
SO3(g) at this temperature?
SO2(g)
Initial
concentration
(mol/L)
Change in
concentration
(mol/L)
Equilibrium
concentration
(mol/L)
+
NO2(g) ↔
NO(g)
+
SO3(g)
0.017
0.011
0.00
0.00
-x
-x
+x
+x
.017-x
.011-x
+x
+x
[NO ][SO 3 ]
[SO2 ][NO2 ]
( x )( x )
Kc =
(0.017 − x )(0.011 − x )
Kc =
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Use quadratic equation to solve
3.8 x2 – 0.134x + 0.0008975 = 0
−b ± b 2 − 4ac
x=
2a
X = 0.026 (invalid) or x =0.0089 mol/L
Therefore the equilibrium concentration of SO3 is 0.0089 mol/L
12. Consider the following reaction,
CO(g) + Cl2(g) ↔COCl2(g)
0.055 mol of CO(g) and 0.072 mol of Cl2(g) are placed in a 5.0L container. At 870K,
the equilibrium constant is 0.20. What are the equilibrium concentrations of the
mixture at 870K?
13. The following equation represents the equilibrium reaction for the dissociation of
phosphene gas.
COCl2(g) ↔CO(g) + Cl2(g)
At 100oC, the value of Kc for this reaction is 2.2 x 10-8. The initial concentration of
COCl2 in a closed container at 100oC is 1.5 mol/L. What are the equilibrium
concentrations of CO(g) and Cl2(g)?
Start by dividing the smallest initial concentration by Kc to determine whether you
can ignore the changes in concentration.
Smallest initial concentration
1.5
=
Kc
2.2 × 10−8
>500
Because this value is well above 500, you can ignore the changes in [CO] and [Cl2]
Next set up your ICE table
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COCl2(g)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
↔
1.5
-x
1.5-x ~1.5
CO(g)
0.0
+x
x
+
Cl2(g)
0.00
+x
x
Now set up your equilibrium expression, and substitute in the equilibrium values
from your ICE table.
Kc =
[Cl 2 ][CO ]
[COCl2 ]
2.2 x 10-8 =
( x )( x )
1.5
3.3 x 10-8 = x2
x = 1.8 x 10-4 mol/L = [CO] = [Cl]
14. At 448°C the equilibrium constant, Kc, for the reaction,
H2(g) + I2(g) ↔ 2HI(g)
is 51. Predict how the reaction will proceed to reach equilibrium at 448°C if we start
with 2.0 × 10-2 mol of HI, 1.0 × 10-2 mol of H2, and 3.0 × 10-2 mol of I2 in a 2.0L
container.
The initial concentrations are;
[HI] = 2.0 × 10-2 mol/2.0 L = 1.0 × 10-2 M
[H2] = 1.0 × 10-2 mol/2.0 L = 5.0 × 10-3 M
[I2] = 3.0 × 10-2 mol/2.0 L = 1.5 × 10-2 M
The reaction quotient (Qc) is;
(1.0 × 10−2 )
[HI ] =
= 1.3
Q=
[H2 ][I2 ] ( 5.0 × 10−3 )(1.5 × 10−2 )
2
2
Because Q < Kc, [HI] will need to increase and [H2] and [I2] decrease to reach
equilibrium; the reaction will proceed from left to right.
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15. Consider the following equilibrium:
N2O4(g)
2NO2(g)
H° = 58.0 kJ
In what direction will the equilibrium shift when each of the following changes is
made to a system at equilibrium:
a)
b)
c)
d)
e)
add N2O4
remove NO2
increase the total pressure by adding N2(g)
increase the volume
decrease the temperature?
Answer:
a) The system will adjust, so as to decrease the concentration of the added N2O4;
the equilibrium consequently shifts to the right, in the direction of products.
b) The system will adjust to this change by shifting to the side that produces more
NO2; thus, the equilibrium shifts to the right.
c) Adding N2 will increase the total pressure of the system, but N2 is not involved in
the reaction. The partial pressures of NO2 and N2O4 are unchanged, and there is no
shift in the position of the equilibrium.
d) The system will shift in the direction that occupies a larger volume (more gas
molecules); thus, the equilibrium shifts to the right.
e) The reaction is endothermic; therefore, we can imagine heat as a reagent on the
reactant side of the equation. Decreasing the temperature will shift the equilibrium in
the direction that produces heat, and so the equilibrium shifts to the left, toward the
formation of more N2O4. Note that only this last change also affects the value of the
equilibrium constant, K.
Lesson 19
16. Calculate the concentration of hydronium ions in 4.5 mol/L HCl
HCl(aq) + H2O(l) Æ H3O+(aq) + Cl-(aq)
Hydrochloric acid is a strong acid that dissociates completely
[H3O+] = 4.5 mol/L (1:1 ratio)
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17. Calculate the concentration of hydroxide ions in 3.1 mol/L KOH.
KOH(aq) Æ K+(aq) + OH-(aq)
Potassium hydroxide is a strong base that dissociates completely
[OH-] = 3.1 mol/L (1:1)
18. Phenol C6H5OH is used as a disinfectant. An aqueous solution of phenol was found
to have a pH of 4.72. Is phenol acidic, neutral or basic? Calculate [H3O+] [OH-] and
pOH of the solution.
This solution is acidic (pH <7)
pOH = 14 - 4.72 = 9.28
[H3O+] = 10-4.72
[H3O+] = 1.9 x 10-5 mol/L
[OH-] = 10-9.28
[OH-] = 5.24 x 10-10 mol/L
19. Calculate the pH of a sample of vinegar that contains 0.083 mol/L acetic acid. What
is the percent dissociation of the vinegar?
Since acetic acid is a weak acid, and does not ionize completely, we have to set up
an ICE Table to determine the hydronium ion concentration.
First write the equation for the dissociation of acetic acid, HCH3COOH
HCH3COOH(aq) + H2O(l) ↔ H3O+(aq) + HCH3COO-(aq)
Ka = 1.8 x 10-5 mol/L
HCH3COOH(aq) +H2O(l) ↔ H3O+(aq) + HCH3COO-(aq)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
•
0.083
-x
0.083-x
0.00
+x
+x
0.00
+x
+x
Note the water is not included in the equilibrium expression since it is in liquid
form.
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Ka =
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[HCH3COO − ][H3O + ]
[HCH3COOH ]
1.8 x10-5 =
( x )( x )
(0.083 − x )
**The value for Ka for acetic acid (found in table 1, appendix A) is 1.8 x 10-5**
Next check to see if the amount of acid that dissociates is negligible compared to
initial concentration of the acid.
[HCH3COOH ]
0.083
=
Ka
1.8 × 10−5
> 500
Since this value is > 500, the amount that dissociate is negligible compared with
initial concentration. In other words we can disregard –x in (0.083- x).
Solve the Ka expression above
1.8 x10-5 =
( x )( x )
(0.083)
x = 1.22 x 10-3 mol/L = [H3O+]
pH = -log[H3O+]
pH = 2.91
% dissociation = 1.5%
20. A chemist finds that 0.03% of hypochlorous acid molecules are dissociated in a 0.40
mol/L solution of the acid. What is the value of Ka for the acid?
HOCl(aq) + H2O(l) ↔ OCl-(aq) + H3O+(aq)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
HOCl(aq) +
0.40
-1.2 x 10-4
.399
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H2O(l) ↔ OCl-(aq)
0.00
+1.2 x 10-4
+1.2 x 10-4
+
H3O+(aq)
0.00
+1.2 x 10-4
+1.2 x 10-4
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Ka =
[OCl − ][H3O + ]
[HOCl ]
Ka =
(1.2 × 10 −4 )(1.2 × 10 −4 )
(.399)
Ka = 3.6 x 10-8
21. Adipic acid is a diprotic acid used to manufacture nylon. Its formula can be
shortened to H2Ad. The acid dissociation constants for adipic acid are Ka1 = 3.71 x
10-5 and Ka2=3.87 x 10-6. What is the pH of a 0.085 mol/L solution of adipic acid?
H2Ad(aq) + H2O(aq) ↔ HAd-(aq) + H3O+(aq)
Given Ka1 for H2Ad= 3.71 x 10-5, Ka2 for HAd- = 3.87 x10-6
H2Ad(aq) + H2O(l) ↔
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Ka =
0.085
-x
0.085-x
HAd-(aq)
0.00
+x
+x
H3O+(aq)
+
0.00
+x
+x
[HAd − ][H3O + ]
[H2 Ad ]
3.71 x10-5 =
( x )( x )
(0.085 − x )
Next determine if the amount of dissociation of phosphoric acid is negligible or not.
[H2 Ad ]
0.085
=
Ka
3.71 × 10 −5
> 500
Therefore, x is probably negligible, so our solution becomes much simpler.
x2
0.085
x = 1.78 × 10 −3
3.71 x10 −5 =
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Now write the expression for the dissociation of HAdHAd-(aq) + H2O(l) ↔ Ad2-(aq) + H3O+(aq)
HAd-(aq) + H2O(l) ↔
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Ka =
1.78 x 10-3
-x
1.78 x 10-3 -x
Ad2-(aq)
0.00
+x
+x
+
H3O+(aq)
1.78 x 10-3
+x
1.78 x 10-3 +x
[ Ad 2− ][H3O + ]
[HAd − ]
( x )(1.78 × 10−3 + x )
3.87 x 10 =
(1.78 × 10 −3 − x )
-6
Now check to see if the value of x is negligible.
[HAd − ] 1.78 × 10 −3
=
Ka
3.87 × 10 −6
> 500
Therefore the value of x is negligible.
x(1.78 × 10 −3 )
1.78 × 10 −3
= 3.87 × 10 −6
3.87 × 10 −6 =
pH = -log[1.78 x 10-3]
pH =2.74
22. An aqueous solution of household ammonia has a molar concentration of 0.105
mol/L. Calculate the pH of the solution.
NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
Write the expression for the dissociation of ammonia:
Kb =
[OH ][NH 4 + ]
[NH3 ]
From Table 2, Appendix A, Kb = 1.8 x 10-5
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Set up ICE table
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
NH3(aq) + H2O(l)
0.105
-x
0.105 -x
↔ NH4+(aq)
0.00
+x
+x
+
OH-(aq)
0.00
+x
+x
Check to see if x is negligible
[NH3 ]
0.105
=
Kb
1.8 × 10 −5
> 500
Since this value is greater than 500, the amount that dissociates can be neglected.
( x )( x )
0.105
x = 1.37 × 10 −3
1.8 × 10−5 =
pOH = -log [1.37 × 10-3] = 2.86
pH = 14.00 - 2.86
= 11.14
23. Morphine, C17H19NO3, is a naturally occurring base to control pain. A 4.5 x 10-3 mol/L
solution has a pH of 9.93. Calculate Kb for morphine.
C17H19NO3(aq) + H2O(l) ↔ C17H19NO3H+(aq) + OH-(aq)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Kb =
C17H19NO3(aq) + H2O(l) ↔ C17H19NO3H+(aq) + OH-(aq)
4.5 x 10-3
0.00
0.00
-x
+x
+x
+x
4.5 x 10-3 -x
+x
[OH − ][C17H19NO3H + ]
[C17H19NO3 ]
pOH = 14.00 − 9.93 = 4.07
[OH-] = 10-4.07
[OH-] = 8.5 x 10-5mol/L (this is also the value of x)
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Solve for Kb
(8.5 × 10 −5 )2
Kb =
4.4 × 10 −3
= 1.6 × 10 −6
24. An aqueous solution of ammonia has a pH of 10.85. What is the concentration of
the solution?
NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
pOH = 14 – 10.85
pOH = 3.15
[OH-] = 10-3.15
[OH-] = 7.1 x 10-4 mol/L
Set up an ICE table
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Kb =
NH3(aq) + H2O(l)
x
-7.1 x 10-4
x - 7.1 x 10-4
↔ NH4+(aq)
0.00
+7.1 x 10-4
7.1 x 10-4
+
OH-(aq)
0.00
+7.1 x 10-4
7.1 x 10-4
[NH4 + ][OH − ]
[NH3 ]
We can disregard the amount of ammonia that dissociates (check using [NH3]/Kb)
(7.1× 10−4 )2
1.8 x 10 =
[NH3 ]
-5
[NH3] = 2.8 x 10-2 mol/L
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Lesson 20
25. a) Predict whether a 0.10 mol/L solution of NaNO2(aq) solution will be acidic, basic, or
neutral.
Write the dissociation equation for sodium nitrite
NaNO2(s) Æ Na+(aq) + NO2-(aq)
Na will have no effect on the pH
NO2- is the conjugate base of the weak acid, HNO2(aq)
Therefore the solution will be basic
b) Calculate the pH of a 0.10 mol/L solution of NaNO2(aq).
NO2(aq) + H2O(l) ↔ OH-(aq) + HNO2(aq)
Kb =
[OH −][HNO2 ]
[NO2 − ]
Calculate the Kb using Kw = KaKb
Kb = 1.4 x 10-11
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
NO2(aq) + H2O(l) ↔ OH-(aq) + HNO2(aq)
.10
0.00
-x
+x
.10-x
x
0.00
+x
x
(In this case, we will disregard the amount of NO2 dissociated since .10/Kb >500)
1.4 x 10-11 =
x2
0.10
x = 1.2 x 10-6 = [OH-]
pOH = -log[1.2 x 10-6]
pOH = 5.92
pH = 8.08 (14 – pOH)
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26. 50.0 mL of 0.10 mol/L hydrobromic acid is titrated with 0.10 mol/L aqueous
ammonia. Determine the pH at equivalence.
NH3(aq) + HBr(aq) → NH4Br(aq)
Moles of NH3(aq) = 0.20 mol/L × 0.020 L
= 4.0 × 10−3 mol
[NH3] = [HBr] (1:1), the volume of aqueous ammonia added must be the same as the
volume of hydrobromic acid (50 mL).
Therefore, [NH4Br] = 0.050 mol/L
The titration results in an aqueous solution of ammonium bromide, NH4Br(aq).
NH4+(aq) is the conjugate acid of a weak base, so it reacts with water. Br−(aq) is the
conjugate base of a strong acid, so it does not react with water. The solution will be
acidic.
NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
Kb for NH3(aq) is 1.8 × 10−5 .
Ka can be calculated using the relationship KaKb = Kw.
Ka = 5.6 × 10−10
Set up an ICE table
NH4+(aq) + H2O(l) →
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
0.050
-x
0.050-x
NH3(aq) +
H3O+(aq)
0.00
+x
x
0.00
+x
x
Check to see is the amount of ammonium that dissociates is negligible.
[NH4 + ]
.050
=
Ka
5.6 × 10−10
> 500
Therefore x can be ignored
Ka =
[NH3 ][H3O + ]
[NH4 + ]
5.6 × 10-10=
( x )( x )
0.050
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x = 5.3 × 10-6 mol/L
x = [H3O+] = 5.3 × 10−6 mol/L
pH = −log[H3O+]
= −log(5.3 × 10−6)
= 5.28
27. Calculate the Ksp for magnesium fluoride at 25oC, given a solubility of 0.00172
g/100mL.
MgF2(s) Æ Mg2+(aq) + 2F-(aq)
Ksp = [Mg2+][F-]2
Calculate the concentration of MgF2
[MgF2] = 0.00172g/100mL x 1mol/62.31g x 1000mL/L
[MgF2] = 2.8 x 10-4 mol/L
MgF2(s)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Mg2+(aq) +
Æ
0
+x
x
2F-(aq)
0
+2x
2x
x = 2.8 x 10-4 mol/L
Ksp = (2.8 x 10-4)(2 x 2.8 x 10-4)2
Ksp = 8.4 x 10-11
28. Calculate the solubility of silver iodide at 25oC. The Ksp of AgI(s) is 1.5 x 10-16 at
25oC.
AgI(s) ↔ Ag+(aq) + I-(aq)
1.5 x 10-16 = x2
x = 1.2 x 10-8 mol/L
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29. What is the molar solubility of PbCl(s) in a 0.2 mol/L NaCl(aq) solution?
NaCl(s) Æ Na+(aq) + Cl-(aq)
In addition, we are adding chlorine ions (common ion) from PbCl2
PbCl2(s) ↔Pb2+(aq) + 2Cl-(aq) Ksp = 1.7 x 10-5
Next set up your ice table
Pb2+(aq) +
PbCl2(s) ↔
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
0
+x
x
2Cl-(aq)
.20
+2x
.2 + 2x
Ksp = [Pb2+][Cl-]2
1.7 x10-5 = x (0.2 + 2x)2
Since PbCl2 has a very low solubility, we can make the assumption the 0.2 + 2 x ~
0.2
Solve for x
x = 4.2 x 10-4
30. Calculate the solubility of silver chloride in a 0.10 mol/L solution of sodium chloride
at 25oC. At SATP, Ksp AgCl(s) = 1.8 x 10-10.
Initial [Cl-] = 0.10 mol/L
Ksp = [Ag+][Cl-] = 1.8 x 10-10
x = solubility of AgCl(s) = [Ag+] = [Cl-]
AgCl(s) ↔
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Ag+(aq) +
0
+x
x
Cl-(aq)
.10
+x
.1 + x
0.10 + x ~ 0.10
1.8 x 10-10 = x (.1)
x = 1.8 x 10-9
The solubility of AgCl(s) is 1.8 x 10-9mol/L
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31. If 100 mL of a .100 mol/L CaCl2(aq) and 100mL of 0.0400mol/L Na2SO4(aq) are mixed
at 20oC, determine whether a precipitate will form. For CaSO4(aq) at 20oC, Ksp is 3.6 x
10-5.
Begin by writing the equation for the reaction. Use the solubility table to determine
solid formed.
CaCl2(aq) + Na2SO4 Æ CaSO4(s) + NaCl(aq)
Calculate the concentration of Calcium and sulphate ions after mixing)
[Ca2+] = 0.100 mol/L x 100mL/200mL = 0.0500 mol/L
[SO42-] = 0.0400 mol/L x 100mL/200mL = 0.0200 mol/L
CaSO4 ↔ Ca2+(aq) + SO42-(aq) , Ksp is 3.6 x 10-5
Q for CaSO4 = [Ca2+][SO42-(aq)]
= (0.0500)(0.0200)
Q = 1.00 x 10-3
Q > Ksp
Therefore a precipitate will form.
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APPENDIX A
Table 1: Acid/Base Constants at 25oC
Acid
Acetic acid
Formula
CH3CO2H
Acetylsalicylic acid (aspirin) HC9H7O4
Ka1
Ka2
Ka3
3.0x10-13
1.8x10-5
3.0x10-4
Aluminum ion
Al(H2O)43+
1.2x10-5
Arsenic acid
H3AsO4
2.5x10-4
5.6x10-8
Ascorbic acid
H2C6H6O6
7.9x10-5
1.6x10-12
Benzoic acid
C6H5COOH 6.3x10-5
Carbonic acid
H2CO3
4.2x10-7
Ferric ion
Fe(H2O)63+
4.0x10-3
Formic acid
HCO2H
1.8x10-4
Hydrocyanic acid
HCN
4.0x10-10
Hydrofluoric acid
HF
7.2x10-4
Hydrogen peroxide
H 2O 2
2.4x10-12
Hydrosulfuric acid
H2S
1.0x10-7
Hypochlorous acid
HClO
3.5x10-8
Nitrous acid
HNO2
4.5x10-4
Oxalic acid
H2C2O4
5.9x10-2
Phenol
C6H5OH
1.0x10-10
Phosphoric acid
H3PO4
7.5x10-3
6.3x10-8
Sulphuric acid
H2SO4
very large
1.2x10-2
Sulphurous acid
H2SO3
1.7x10-2
6.4x10-8
Zinc ion
Zn(H2O)42+
2.5x10-10
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4.8x10-11
1.0x10-19
6.4x10-5
3.6x10-13
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Formula
Kb
Ammonia
NH3
1.8x10-5
Aniline
C6H5NH2
7.4x10-10
Caffeine
C8H10N4O2
4.1x10-4
Codeine
C18H21O3N
8.9x10-7
Diethylamine
(C2H5)2NH
6.9x10-4
Dimethylamine
(CH3)2NH
5.9x10-4
Ethylamine
C2H5NH2
4.3x10-4
Hydroxylamine
NH2OH
9.1x10-9
Isoquinoline
C9H7N
2.5x10-9
Methylamine
CH3NH2
4.2x10-4
Morphine
C17H19O3N
7.4x10-7
Piperidine
C5H11N
1.3x10-3
Pyridine
C5H5N
1.5x10-9
Quinoline
C9H7N
6.3x10-10
Triethanolamine
C6H15O3N
5.8x10-7
Triethylamine
(C2H5)3N
5.2x10-4
Trimethylamine
(CH3)3N
6.3x10-5
Urea
N2H4CO
1.5x10-14
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Table 2: Table of Solubility Product Constants (Ksp at 25o C)
Bromides
Carbonates
Chlorides
Chromates
Cyanides
Fluorides
Hydroxides
Iodides
PbBr2
AgBr
BaCO3
CaCO3
CoCO3
CuCO3
FeCO3
PbCO3
MgCO3
MnCO3
NiCO3
Ag2CO3
ZnCO3
PbCl2
AgCl
BaCrO4
CaCrO4
PbCrO4
Ag2CrO4
Ni(CN)2
AgCN
Zn(CN)2
BaF2
CaF2
PbF2
MgF2
AgOH
Al(OH)3
Ca(OH)2
Cr(OH)3
Co(OH)2
Cu(OH)2
Fe(OH)2
Fe(OH)3
Pb(OH)2
Mg(OH)2
Mn(OH)2
Ni(OH)2
Zn(OH)2
PbI2
AgI
6.3 x 10-6
3.3 x 10-13
8.1 x 10-9
3.8 x 10-9
8.0 x 10-13
2.5 x 10-10
3.5 x 10-11
1.5 x 10-13
4.0 x 10-5
1.8 x 10-11
6.6 x 10-9
8.1 x 10-12
1.5 x 10-11
1.7 x 10-5
1.8 x 10-10
2.0 x 10-10
7.1 x 10-4
1.8 x 10-14
9.0 x 10-12
3.0 x 10-23
1.2 x 10-16
8.0 x 10-12
1.7 x 10-6
3.9 x 10-11
3.7 x 10-8
6.4 x 10-9
2.0 x 10-8
1.9 x 10-33
7.9 x 10-6
6.7 x 10-31
2.5 x 10-16
1.6 x 10-19
7.9 x 10-15
6.3 x 10-38
2.8 x 10-16
1.5 x 10-11
4.6 x 10-14
2.8 x 10-16
4.5 x 10-17
8.7 x 10-9
1.5 x 10-16
Oxalates
Phosphates
Sulphates
Sulphides
Sulphates
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BaC2O4
CaC2O4
MgC2O4
AlP04
Ba3(P04)2
Ca3(P04)2
CrP04
Pb3(P04)2
Ag3P04
Zn3(P04)2
BaS04
CaS04
PbS04
Ag2S04
CaS
CoS
CuS
FeS
Fe2S3
PbS
MnS
NiS
Ag2S
ZnS
BaS03
CaS03
Ag2S03
1.1 x 10-7
2.3 x 10-9
8.6 x 10-5
1.3 x 10-20
1.3 x 10-29
1.0 x 10-25
2.4 x 10-23
3.0 x 10-44
1.3 x 10-20
9.1 x 10-33
1.1 x 10-10
2.4 x 10-5
1.8 x 10-8
1.7 x 10-5
8 x 10-6
5.9 x 10-21
7.9 x 10-37
4.9 x 10-18
1.4 x 10-88
3.2 x 10-28
5.1 x 10-15
3.0 x 10-21
1.0 x 10-49
2.0 x 10-25
8.0 x 10-7
1.3 x 10-8
1.5 x 10-14
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