Chapter 6 Chemical Equilibrium

Chemistry 534
CHAPTER 6
Chemical Equilibrium
The subject of any investigation is called a system. When the study involves experimentation, the
system is said to be real. When the study involves ideas, the system is said to be ideal.
The advantage of an ideal system is that it makes possible "study by analysis". That is, an ideal
system may be broken down into its constituent parts. Each part becomes a system in itself yet
simpler and easier to understand than the whole. On the other hand, in a real system, it is usually
difficult or impossible to remove the various parts without causing the system to either
malfunction or collapse.
▼ TOPIC-1: CHEMICAL EQUILIBRIUM A MATTER OF BALANCE
The phenomenon of chemical equilibrium occurs in a variety of different systems in nature, from
personal biological systems to global ecological systems. In industry, the control of chemical
reactions by the manipulation of their equilibrium conditions is very important in many
processes.
Equilibrium is a balance, a compromise achieved by two opposing tendencies in nature each
attempting to cause change. One compromise is the natural tendency to raise the entropy
(disorder) of a system as much as possible. The other compromise is the natural tendency to
lower the energy of a system as much as possible. Once equilibrium has been reached, the two
tendencies are in a state of balance and the necessity for change no longer exists. In fact, at
equilibrium, there is a natural tendency to resist change.
As long as a system is away from equilibrium, it will experience a tendency to reach equilibrium.
The tendency will be greater the greater the distance from equilibrium. Furthermore, useful work
can only be obtained from a system which is on its way to equilibrium. Once at equilibrium, no
work can be obtained from a system.
Note:
All changes in nature are due to the tendency on the part of different systems to
reach a state of equilibrium.
In this chapter, we learn the principles governing chemical equilibrium. This includes the factors
which influence equilibrium as well as the mathematical laws governing its behavior.
1
EXPERIMENT 1 Consequences of reversibility ✫ A DEMO
Objective: To investigate reversible reactions
In this demonstration, we will use a mechanical model to simulate the behavior of a reversible
reaction.
The mechanical system consists of two 1000 mL beakers (labeled beaker-A and beaker-B), a 250
mL beaker, and a 100 mL beaker. At the start, beaker-A is filled with water. The objective is to
transfer water back and forth between the two 1000 mL beakers via the two smaller beakers.
Note that the 250 mL beaker represents the forward reaction, allowing water to go towards the
right, while the 100 mL beaker represents the reverse reaction, allowing water to go towards the
left.
ã Remember: Since equilibrium is a continuous process, in our mechanical model there
must be a continuous exchange of water between beaker-A and beaker-B.
Also, the exchange can only take place via the two intermediate beakers.
In this particular experiment, due to the size of the containers used, the maximum water which
can be transferred towards the right is 250 mL while the maximum water which can be
transferred towards the left is 100 mL.
We will now verify that, once equilibrium is established, a fixed quantity of water will
continuously be exchanged between beaker-A and beaker-B.
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1) How does the rate of reaction vary as the reaction progresses?
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2) Which graph illustrates the evolution (or progress) of the forward reaction?
3)
What can you say concerning the rate of the reverse reaction during the same period of time.
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4) Which graph illustrates the evolution (or progress) of the reverse reaction?
5) What happens when the system reaches equilibrium?
a) In terms of the macroscopic properties:
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b) In terms of the forward and reverse rates of reaction:
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3
In the last experiment, did you observe that, once equilibrium was reached, the amount of water
transferred left and right remained constant? This resulted in beaker-A and beaker-B both
retaining a constant (although unequal) amount of water.
Note that unlike our mechanical model, in a chemical system the right and left transfers occur
simultaneously in a single container. As such, the mass on each side of the equilibrium remains
constant. This does not mean that the total mass on the left equals the total mass on the right.
What it does mean is that the total mass on the left remains constant and the total mass on the
right remains constant.
þ IMPORTANT
Do not confuse a reaction which goes to completion with a reversible reaction in equilibrium
(which does not go to completion). In a "completion" reaction, the system consists of two
sections analogous to a "before-and-after" situation. The reactants, which are always written on
the left, constitute the "before" situation. The products, which are always written on the right,
constitute the "after" situation. During the "before" situation, some or all of the reactants are
totally used up. In this type of equation, the total mass on the left (mass of the reactants) equals
the total mass on the right (mass of the products). Stated differently, in an ordinary chemical
reaction, the total mass before the reaction equals the total mass after the reaction.
However, an equilibrium reaction does not go to completion. Such a reaction never ends. Both
the substances on the left and the substances on the right are continuously and simultaneously
reacting. This is not a "before-and-after" situation, it is a now situation. Thus, it is incorrect to
assume that the total mass on the left equals the total mass on the right (as is true for reactions
which go to completion). Indeed, in an equilibrium equation, it is quite possible for the mass on
the left to be very little while the mass on the right to be very great (or vice-versa). Of course, the
total mass of the system (the mass on the left plus the mass on the right) remains constant.
To differentiate a reaction which goes to completion from an equilibrium reaction (which never
goes to completion), the following special "arrows" are used:
4
In effect, at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.
This is why neither side gains or loses mass. Furthermore, the continuous forward and reverse
reactions is what causes the equilibrium to be "dynamic".
Here is an illustration of a chemical system in equilibrium:
4 g/s
4 g/s
e word "rate" means something is occurring "per unit time". That is, as time progresses, the
occurrence continues (accumulation occurs). The symbol for the word "rate" is the slash"/".
Thus we may write:
20 $/h
20$ per hour
for rate of pay
4 L/s
4 litres per second
for rate of flow
60 km/h
60 kilometres per hour
for rate of speed
5 mol/s
5 moles per second
for rate of reaction
6) Assuming that the following system is in equilibrium, draw the equation symbol for the
reaction:
X+Y
W+Z
7) Assuming that the following equilibrium system starts from the left, draw the equation
symbol for the reaction:
X+Y
W+Z
8) Assuming that the following equilibrium system starts from the right, draw the equation
symbol for the reaction:
X+Y
W+Z
5
Activity- A Model for Chemical Equilibrium
In this activity you will simulate a system approaching chemical equilibrium. Your task is to determine if the
simulation is a good model for a chemical system at equilibrium.
Apparatus
Two 150 mL beakers, two plastic rulers
Two pieces of glass tubing of different diameters, 10 - 15 cm long
10 mL graduated cylinder, water with blue food coloring, water with yellow food coloring
Method
1.
2.
3.
4.
5.
6.
Put some blue colored water in one beaker and some yellow colored water in the other.
The only restriction is that the total volume of the two is 100 mL.
Work in pairs. Transfer water from B, the blue beaker, using one of the pieces of glass tubing,
to Y. the yellow beaker. At the same time your partner should transfer water from Y to B
using the other piece of glass tubing. Be sure to keep the glass tube vertical at all times.
Use the rulers to record the height of the water in each beaker after every five exchanges.
Continue transferring water until you have three successive readings that are the same.
When the heights remain constant measure the volume transferred by each glass tube.
Repeat steps 1-5 for different volumes of blue and yellow water, but keep the total volume
100 mL.
Observations
Time
(5 exchanges)
Height of
Height of
Water
Water
Blue Beaker
Yellow Beaker
(mm)
(mm)
1
2
3
etc.
Discussion
1. What are the following analogous to in a chemical reaction? The height of the water in the
two beakers: the transfer of water from one beaker to another: two pieces of glass tubing of
different diameters: the coloring in the water; the volume of water transferred: the volume of
water transferred when the height remains constant.
2. What evidence suggests that you reached a point of equilibrium in this activity? You should
be able to think of two key observations.
3. What property of the system determines the final height of the water in the beakers? What
would this be analogous to in a chemical reaction?
4. Plot height of water versus time in units of five exchanges.
(a)
Describe the graph in words.
(b) What is the significance of the horizontal part of the graph?
(c)
What would be the significance of the slope of a tangent drawn at any point on this
graph?
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(d)
(e)
What happens to the slope of the tangent as the system approaches equilibrium? What
does this imply?
The implications of (d) contradict what is actually happening: you can still see water
being transferred from one beaker to another. Explain how it is possible for the graph
to suggest that the reaction has stopped even though you can see it going on.
5. Consider the following gaseous reaction:
CO(g) + Cl2(g) ⇔ COCl2(g)
One mole of Cl2 was added to 2 mol of CO in a 1 L container. The concentration of the
three gases was measured using a spectrophotometer every minute The following data were
recorded:
Time (min)
0
1
2
3
4
5
6
7
(a)
(b)
(c)
(d)
(e)
(f)
[CO] mol/L
2.00
1.76
1.54
1.33
1.26
1.21
1.22
1.21
[Cl2] mol/L
1.00
0.78
0.53
0.31
0.24
0.20
0.21
0.21
[COCl2]
0.00
0.23
0.48
0.67
0.74
0.79
0.79
0.78
Plot a graph of the data.
Explain why the graph has the shape it has.
What is the significance of the horizontal part of the graph?
What would be the significance of the slope of a tangent drawn at any point on this
graph?
What happens to the slope of the tangent as the system approaches equilibrium? What
does this imply?
Use the analogy to explain the changes in the slope of the tangents.
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The experiment you have just performed demonstrates that visible changes occur during a
chemical reaction. For example, the formation of a gas, the disappearance of the tablets,
effervescence occurring at the surface of the liquid, etc. These changes, seen with the naked eye,
are called macroscopic (large scale) changes. Conversely, changes that occur which cannot be
seen with the naked eye are called microscopic (small scale) changes.
For any chemical system to be in equilibrium, the following three conditions are necessary:
ΠThe system must be closed.
• There must be no macroscopic changes (no visible or observable changes).
Ž There must be some reactants and some products present at all times.
Note:
The system is said to be in dynamic equilibrium because there is continuous
microscopic activity. That is, while there are no observable (macroscopic) changes
(such as temperature, pressure, volume, color, etc.), the molecules are constantly in
motion. As a result, there are always some reactants forming some products, and, at
the same time, some products forming some reactants. This is the nature of
"dynamic" equilibrium.
9 ) Tell which of the following are characteristic of an equilibrium system
a)
b)
c)
d)
e)
f)
g)
h)
i)
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A system which is closed.
There is a gas in the system.
The reactants have completely transformed into products.
There is a loss of matter.
The color of the solution remains constant.
The mass of undissolved matter equals the mass of dissolved solid.
The presence of a catalyst.
There are no observable changes.
The solute has completely dissolved.
10) State and explain whether or not each of the following systems are in equilibrium:
a)
A distillation flask contains a constant quantity of alcohol.
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b)
The flame of a Bunsen burner keeps its shape, height, and color for 30 minutes.
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c)
In a water boiler, the temperature and pressure do not vary.
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d)
The water level of Lac Saint-Jean remains unchanged for a week.
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e)
The column of alcohol inside a thermometer remains steady.
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11) Give two examples from every day experience of systems in equilibrium:
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12) To summarize , complete the following phrase by filling in the blanks:
A closed system is in equilibrium if its __________________________________properties
are _______________________ and if both __________________________ and
___________________ are always present. Furthermore, the same equilibrium condition
may be attained by starting with either the __________________________________ or
the ____________________________.
End of experiment
The equilibrium system just studied can attain equilibrium either with the reactants or with the products as the
starting materials. Such reactions are called "reversible" reactions as they simultaneously proceed in both the
forward and reverse directions.
To indicate the fact that a reaction is reversible, a special "two-way" arrow is drawn in the equation as shown below:
CaCl2(aq) + Na2SO4(aq) ⇔ CaSO4(s) + 2 NaCl(aq)
13) Using your own words, define the following reactions:
a)
Forward reaction
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b)
Reverse reaction
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c)
Reversible reaction
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TOPIC-2: UNSTABLE EQUILIBRIUM
To better understand equilibrium, we must understand the factors which influence its operation.
We will investigate four factors and study their effect on chemical equilibrium. The factors are
called "stresses" as they place a "stress" on the equilibrium of a system.
ΠVariation in the concentration of the reactants
• Variation in the pressure of the system
Ž Variation in the temperature of the system
• The presence of a catalyst
1Review of concentration
l that concentration is the amount of matter per unit volume, similar to density.
ever, whereas density refers to the amount of solid matter per unit volume
s per unit volume), concentration refers to the number of particles per unit volume
of liquid.
concentration is the number of particles there are in a specific volume of liquid, the more particles
in the liquid, the greater the concentration (and vice-versa). For example, suppose we have a
tablespoon of salt. If we pour this salt into a glass of water, we will have a specific
concentration of salt water. Now suppose we pour a tablespoon of salt into a huge container of
water. Again we have salt water. This time, however, the concentration is much lower (as there
are fewer particles per unit volume).
ause concentration is mass (moles) per unit volume and the standard volume is the
litre, the unit for concentration is moles/litre or mol/L (or M for short).
14) Suppose a group of 10 people visit the following three locations where, at each location,
they spread out in the space available:
•A football field
‚ An elevator
ƒ A theater
a) At which location is the concentration of people the greatest? _____________
b) At which location is the concentration of people the least? _______________
Note: It is customary to use square brackets "[ ]" to represent the words "the concentration
of". Thus, for example, we may write:
[people] to mean "the concentration of people"
[NaCl] to mean "the concentration of sodium chloride"
[HCl] = 1.0 mol/L to mean "the concentration of HCl is 1.0 mol/L"
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Le Chatelier's Principle
If a system at equilibrium is subjected to a change, reactions occur to counteract that
change and restore the system to a new equilibrium.
Factors that can change the equilibrium state:
1. Concentration: If a reactant is added to an equilibrium system, the forward reaction speeds
up; if a product is added, the reverse reaction speeds up. Eventually, a new equilibrium is
reached.
example: [Co(H2O)6]2+ + 4 Cl- ⇔ [CoCl4]2- + 6 H2O
red ions
blue ions
a)
When more chloride ions are added, there is an increase in the number of successful collisions with the red
ions and so the rate of the forward reaction increases. The solution turns blue as more products are formed.
Very soon, there is an increase in the number of successful collisions between product molecules and the
rate of the reverse reaction increases. A new equilibrium is reached when the rate of the reverse reaction
"catches up" to the rate of the forward reaction. It is called "new" because both rates are faster than the
original rates.
b)
When more water is added, there is an increase in the number of successful collisions with the blue ions and
so the rate of the reverse reaction increases. The solution turns red as more reactants are formed. Very soon,
there is an increase in the number of successful collisions between reactant molecules and the rate of the
forward reaction increases. A new equilibrium is reached when the rate of the forward reaction 'catches up',
with the rate of the reverse. It is called 'new" because both rates are faster than the original rates.
NOTE: At any equilibrium state, there are some red ions and some blue ions present. The colour you see
depends on which ion is present in the greatest concentration. If there is an equal amount of both ions,
the solution is purple.
PROBLEMS:
i. Consider the following equilibrium system: S02(g) + 1/2 O2(g) ⇔ S03(g)
a) If more oxygen is added to the system, in which direction will the equilibrium be shifted?
b) How will this shift affect the concentration of each substance in the system?
ii. Consider the following equilibrium system: 2 H20(g) ⇔ 2 H2(g) + 02(g)
Name 3 concentration changes that would be made that would result in an increase In the amount of
water vapour.
2. Temperature: An increase in temperature will result in an increase in the rate of the
endothermic reaction
example: [Co(H2O)6]2+ + 4 Cl- + heat ⇔ [CoCl4]2- + 6 H2O
red ions
blue ions
a) An increase in temperature increases the rate of the forward reaction The reactants
absorb heat energy and collide more successfully to form products. The solution turns blue.
b) A decrease in temperature slows down the rate of the forward reaction Since the reverse
reaction is still proceeding at its original rate, the solution turns red.
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Experiment 2: Applying Le Chatelier's Principle
In this experiment, reversible reactions will occur; reactants will form products and products will
form reactants. A change in the relative amounts of reactants and products can be observed by
noting colour changes or the formation of a precipitate. The following equilibrium system will be
studied:
2 CrO42- (aq) ⇔ Cr2O72-(aq)
The concentration of each ion present in the system depends on the amount of hydrogen ions
present. The concentration of hydrogen ions can be increased by adding hydrochloric acid (HCl)
and can be decreased by adding sodium hydroxide, a source of OH ions. The OH- ions react with
H+ ions to form water. What effect will this stress (ie. adding or removing H+ ions) have on the
equilibrium system?
Part 1:
Step 1:
Step 2:
Observe the colour of the chromate and dichromate ions.
chromate (CrO42-) ___________ dichromate (Cr2O72-) ____________
Place 10 drops of each solution into separate test tubes. Add NaOH solution,
drop by drop, until a colour change is noted in one of the tubes. Keep these
tubes for step 5.
CrO42-
initial colour
final colour
Cr2O72-
Step 3:
Place 10 drops of each solution into separate test tubes. Add HCl solution,
drop by drop, until a colour change is noted in one of the tubes. Keep these
tubes for step 4.
initial colour
final colour
CrO42Cr2O72-
Step 4:
Add NaOH solution, drop by drop, to the test tubes from step 3 until a colour
change is noted in one of the tubes.
Result: _______________________________
Step 5:
Add HCl solution, drop by drop, to the test tubes from step 2 until a colour
change is noted in one of the tubes.
Result: _______________________________
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15) Did the addition of hydrogen ions from HCl favour the forward or reverse reaction?
_____________________________________________________________________
2 CrO42- (aq) ⇔ Cr2O72-(aq)
16) Add H+ ions to the equation above and balance the other side by adding water molecules.
Explain the effect of adding OH- ions.
_____________________________________________________________________
Part 2:
Step 1: Add 10 drops of dichromate solution to each of 7 separate test tubes. Classify the
solutions in the table below as acid, base, or neutral. Add drops of solution, one to
each of the 7 test tubes containing the dichromate solution until a colour change is
noted.
Solution
CH3COOH
HNO3
Ca(OH)2
H2S04
KOH
C2H5OH
NH4OH
Step 2:
Solution
CH3COOH
HNO3
Ca(OH)2
H2S04
KOH
C2H5OH
NH4OH
Type
initial colour
final colour
Repeat step 1 using 7 test tubes containing chromate solution.
Type
initial colour
final colour
17) Which solutions behaved like HCl? _____________________________________________
18) Which solutions behaved like NaOH? ___________________________________________
19) How can you account for the behaviour of C2H5OH? _______________________________
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Part 3:
Step 1: Add 10 drops of chromate solution to a test tube. Add 10 drops of barium nitrate
solution, Ba(N03)2. Observe the formation of the precipitate. Allow it to settle.
Ba2+(aq) + CrO42--(aq) à BaCrO4(s)
Step 2: Repeat step 1 using dichromate solution. Allow the precipitate to settle.
20)
The chromate solution contains both chromate and dichromate ions. The dichromate
solution contains both chromate and dichromate ions. The colour that you see depends on
which ion is present in the greater concentration. Barium nitrate only forms a precipitate
with chromate ions. Analyze your results.
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________________________________________________________________________
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Experiment 3: Equilibrium and temperature
ÄA DEMO
We know that temperature affects the rate of a chemical reaction. Does a change in temperature
also affect the state of equilibrium? Let's find out experimentally using the following equilibrium
system:
2 N02(g) ⇔ N2O4(g)
reddish
colorless
To test for any change in equilibrium, we will set up two systems labeled SYSTEM-1 and
SYSTEM-2. Each system will contain the above reaction in a separate glass tube. One system
will be kept at room temperature, while the other will be subjected first to a temperature of about
80 0C and then to a temperature of about 0 0C.
Recall that a change in temperature changes the volume of a gaseous system. Therefore, to keep
the volume constant (while we change the temperature), the glass tube is sealed at both ends.
Step 1. Submerge a glass tube containing NO2 gas in a beaker of hot water (about 80 0C).
Observe the gaseous reaction.
Step 2. Now replace the beaker of hot water with a beaker of cold water and ice. Again observe
the reaction.
21) What are the equilibrium properties of the system at room temperature?
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22) Has an increase in temperature changed the equilibrium state? Explain:
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23) Has a decrease in temperature changed the equilibrium state? Explain:
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24) Here is the equation of the system we used in terms of energy:
2 N02(g) ⇔ N204(g) + 57.2 kJ
Complete the following phrase:
For this system, a(n) _______________ in temperature favors an exothermic or forward reaction.
Conversely, a(n) _________________ in temperature favors an endothermic or reverse reaction.
ÄEnd of experiment-3
Let's summarize the results of experiment-3.
The equilibrium reaction was: 2 N02(g) ⇔ N204(g) + 57.2 kJ
• The system remained in equilibrium as long as the temperature remained constant.
• Each time the temperature was changed (a stress introduced), the system reacted and a new
equilibrium position was established.
• An increase in temperature upset the equilibrium by increasing the kinetic energy of the
molecules. To re-establish equilibrium, the system partially counteracted or opposed the stress
by consuming some of the added heat energy. It did this by shifting the equilibrium towards the
left so as to convert some of the extra kinetic energy (heat) into enthalpy. The shift lowered the
kinetic energy of the molecules and gradually re-established equilibrium.
• A decrease in temperature upset the equilibrium by decreasing the kinetic energy of the
molecules. To re-establish equilibrium, the system countered the stress by releasing heat
energy. It did this by shifting the equilibrium towards the right so as to release heat. The shift
raised the kinetic energy of the molecules and gradually re-established equilibrium.
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â Remember: For any system m equilibrium, an increase in temperature causes the system to
shift towards the side opposite the energy in the equation (in the direction which
uses up energy). Here are some examples of what happens when we INCREASE
the temperature (the reverse occurs when we decrease the temperature):
Endothermic reaction
A + B + heat ⇔ C + D
System shifts right
Exothermic reaction
A + B ⇔ C + D + heat
System shifts left
EXPERIMENT-4:
Equilibrium and pressure
ÄA DEMO
Objective: To study the effect of pressure on equilibrium
To study the effect of pressure on a system in equilibrium we require a gaseous reaction. We'll
use nitrogen dioxide gas, N02(g), which produces dinitrogen tetroxide, N204(g) with which it
forms an equilibrium system.
To obtain the nitrogen dioxide gas, concentrated nitric acid, HNO3, is made to react with
powdered copper, Cu(s). The reaction produces nitrogen dioxide gas and water vapor.
Cu(s) + 4 HNO3(l) ⇔ 2 NO2(g) + Cu(NO3)2(s) + 2 H20(g)
To remove the water vapor, we pass the mixture of gases through a "drying tube" which
contains calcium chloride, CaCl2. The calcium chloride attracts the water resulting in the
following gaseous equilibrium system:
2 NO2(g) ⇔ N204(g)
reddish
colorless
For convenience, we will contain the system in a syringe. The syringe forms a closed
equilibrium system whose pressure can easily be varied by moving the piston.
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Your teacher or lab technician will prepare the equilibrium system as follows:
Step-1:
Place about 2 g of powdered copper into a 250 mL Erlenmeyer filter flask and seal
the flask with a "twistit" rubber stopper.
Step-2:
Place about 4 g of calcium chloride into a drying tube and connect one end to the
flask. Seal the other end with a rubber cap.
Step-3:
Using a 10 mL syringe, slowly inject 4 mL of concentrated nitric acid, HNO3, into
the Erlenmeyer flask thereby starting the reaction.
ÄNote: Due to its very reactive nature, handle the nitric acid with
caution at all times (avoid inhaling any fumes).
Step-4:
When nitrogen dioxide gas is noticeable in the drying tube (reddish color), draw
about 70 mL of gas into the syringe.
Step-5:
Remove the syringe and immediately seal it (with a rubber stopper or cap).
Ä Note: When the syringe is sealed, pull out the piston slightly so that the
red color of NO2(g) becomes visible.
25) What are the properties of this system in equilibrium?
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Step-6:
Press the piston of the syringe so as to double the pressure.
26) What are the new properties of this system?
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________________________________________________________________________
27) When the pressure was doubled, a stress acted on the system. The system then responded to
the stress (increase in pressure). Did the response by the system favor a forward or reverse
reaction? Explain:
________________________________________________________________________
________________________________________________________________________
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28) Select the correct symbol which represents the temporary reaction as the system adjusts to
the increase in pressure:
2 NO2 (g)
Step-7:
N204(g)
Adjust the piston of the syringe so that the volume is increased.
29) What are the new properties of the system?
_______________________________________________________________________
_______________________________________________________________________
30) Was the response by the system to favor the forward or reverse reaction? Explain:
_______________________________________________________________________
_______________________________________________________________________
31) Select the correct symbol which represents the temporary reaction as the system adjusts to
the last change in pressure:
2 N02(g)
N204(g)
End of experiment-4
Let's summarize the results of experiment-4
The system in equilibrium consisted of the following gaseous reaction:
2 N02(g) ⇔
reddish
1 N204(g)
colorless
C The system remained in equilibrium as long as the pressure remained constant.
CEach time the pressure was changed (a stress introduced), the system reacted and a new
equilibrium position was established.
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• An increase in pressure upset the equilibrium by causing the gas molecules to get closer
together. This caused a temporary shift towards the right (forward reaction). The system shifted
towards the right because, as the equation shows, two N02(g) molecules transform themselves
into one N204(g) molecule. This diminishes the total number of molecules in order to
compensate for the reduced volume.
• A decrease in pressure upset the equilibrium by causing the gas molecules to take up more
space. This caused a temporary shift towards the left (reverse reaction). The system shifted
towards the left because, as the equation shows, one N204(g) molecule transforms itself into two
N20(g) molecules. This increases the total number of molecules in order to compensate for the
larger volume available.
➢ Remember:
For any system in equilibrium containing gases, if the pressure is increased, the system
will shift towards the side having the fewest number of gaseous moles.
Here are some examples of what happens when we INCREASE the pressure (the reverse
occurs when we decrease the pressure):
A(g) + 2B(g) ⇔ C(g) + D(g)
3 gas moles ⇔ 2 gas moles System shifts right
2A(g) + B(g) ⇔ 2C(g) + 3D(g)
3 gas moles ⇔ 5 gas moles System shifts left
3A(s) + 2 B(g) ⇔ 2 C(g) + D(g) 2 gas moles ⇔ 3 gas moles System shifts left
20
A(g) + B(g) ⇔ C(g) + D(g)
2 gas moles ⇔ 2 gas moles No effect
3A(s) + 2 B(s) ⇔ 2C(s) + D(g)
0 gas moles ⇔ 1 gas mole System shifts left
Experiment 5: Equilibrium and a Catalyst
Objective: To investigate the effects of a catalyst on equilibrium
Recall that a catalyst increases the rate of a reaction. Does a catalyst also affect the equilibrium
state of a system? Let's find out.
(NH4)2S2O8(aq) + 2 KI(aq) ⇔ I2(aq) + K2S04(aq) + (NH4)2S04(aq)
Our system consists of equal volumes of ammonium peroxydisulfate, (NH4)2S208(aq) and
potassium iodide, KI(aq). After a short time, the system reaches equilibrium. To find out whether
or not a catalyst affects equilibrium, we divide the mixture into two equal parts. One part will be
left untouched, while copper ions, Cu2+, will be added as a catalyst to the other part.
Step-1: Pour 20 mL of solution-A and 20 mL of solution-B into a 100-mL beaker.
Step-2: Wait approximately 5 minutes to allow the system to reach equilibrium. Ä Note: The
solution will turn yellowish.
Step-3: To the remaining 20 mL of solution-A, add the remaining 20 mL of solution-B. and add
one drop of copper nitrate, Cu(NO3)2, the catalyst. (Note the time it takes for the
reaction to reach equilibrium)
Step-4: Observe the reaction and compare it with the uncatalyzed reaction.
32) Describe the properties of the reaction before the addition of a catalyst:
________________________________________________________________________
________________________________________________________________________
33) Describe the properties of the reaction after the addition of a catalyst:
________________________________________________________________________
________________________________________________________________________
34) Concerning the catalyst:
a)
Did the catalyst affect the equilibrium position of the system? Explain:
________________________________________________________________________
________________________________________________________________________
b)
Did the catalyst affect the rate at which equilibrium is reached? Explain:
________________________________________________________________________
________________________________________________________________________
ÿ End of experiment-5
21
Let's summarize the results of experiment-5 which used the following reaction:
(NH4)2S2O8(aq) + 2 KI(aq) ⇔ I2(aq) + K2S04(aq) + (NH4)2S04(aq)
•
The addition of a catalyst did not affect the equilibrium of the reaction. In effect a catalyst
increased both the forward and the reverse reactions equally. As a result, the system arrived
at equilibrium faster. A catalyst, therefore, brings the system to equilibrium faster but does
not shift the equilibrium. The same equilibrium concentration of reactants and products are
present with or without the catalyst.
•
A catalyst brings the reaction to equilibrium faster by lowering equally the activation energy
for the forward and the reverse reactions.
35) Complete the following by filling-in the blanks:
If an external stress acts on a system in equilibrium, the system has three possibilities of
counteracting the stress. One is to favor the ______________ reaction by shifting the
equilibrium towards the ______________ side of the equilibrium equation, the other is to
favor the ______________ reaction by shifting the equilibrium towards the
______________ side of the equilibrium equation or no change (pressure) if there are an
equal number of moles (for gases only).
Analyzing the results of our experiments on equilibrium, we can make a generalization
which helps predict the shift in equilibrium caused by an external stress. This
generalization was first discovered and stated by Henri Louis Le Chatelier and, because it
is applicable to such a large number of systems, it is called Le Chatelier's Principle.
LE CHATELIER'S PRINCIPLE:
When a system in equilibrium is subjected to a change, processes
occur that tend to counteract the imposed change and the system
reaches a new state of equilibrium.
Henri Louis Le Chatelier was born in Paris in 1850. He was a chemist, a metallurgist and a
teacher at Coll•ge de France in Paris. He contributed significantly towards the development of
plaster and cement. In 1884 Henri Louis Le Chatelier formulated the now famous law bearing
his name.
22
USING LE CHATELIER'S PRINCIPLE TO PREDICT CHANGES IN EQUILIBRIUM
36)
Which reaction, forward or reverse, is favored by the addition of H+(aq) ions to each of the
following systems:
a)
H+(aq) + F-(aq) ⇔ HF(aq)
b) CH3COOH(aq) ⇔ CH3COO-(aq) + H+(aq)
c)
37)
a)
b)
H2C03(aq) ⇔CO32-(aq) + 2 H+(aq)
______________________
______________________
______________________
Here are more systems in equilibrium. For each, one of the species (reactants) has an arrow
above it indicating that its concentration has been decreased. Describe the direction of the
equilibrium shift and the effect which this change has on the concentration of the other
species (reactants).
Þ
NH4+(aq) + OH-(aq) ⇔ H20(l) + NH3(g)
_____________________________________________________________________
_____________________________________________________________________
Þ
CO32-(aq) + NH3(aq) ⇔ NH4+(aq) + OH-(aq)
_____________________________________________________________________
_____________________________________________________________________
Þ
c) HCl(aq) + NH3(aq) ⇔ NH4 (aq) + Cl-(aq)
_____________________________________________________________________
_____________________________________________________________________
+
d)
Þ
N2(g) + 3 H2(g) ⇔ 2 NH3(g)
_____________________________________________________________________
_____________________________________________________________________
38) FeSCN2+(aq) ions are responsible for the reddish color of the following solution:
Fe3+(aq) + SCN-(aq) ⇔ FeSCN2+(aq)
yellow
colourless
red
Explain what happens to the color of the solution if we increase the concentration of the
FeSCN2+(aq) ions:
_______________________________________________________________________
23
_______________________________________________________________________
39) The following system in equilibrium is subjected to an increase in pressure:
2 H2(g) + O2(g) ⇔ 2 H20(g)
a) Which reaction is favored, forward or reverse? Explain:
______________________________________________________________________
______________________________________________________________________
b) What effect does this change have on the quantity of hydrogen molecules in the system?
Explain:
______________________________________________________________________
______________________________________________________________________
40) Suppose we decrease the volume of each of the systems below. Predict in which
direction, forward or reverse, the equilibrium shifts.
41)
a) N2(g) 3 H2(g) ⇔ 2 NH3 (g)
_____________________
b) SO2(g) + H20(l) ⇔ H2S03(aq)
_____________________
c) 4 HCl(g) + 02(g) ⇔ 2 H2O(g) + C12(g)
_____________________
d) CH3COOH(aq) ⇔ H+(aq) + CH3COO-(aq)
_____________________
e) H2(g) + I2(g) ⇔ 2 HI(g)
_____________________
The fermentation of champagne produces carbon dioxide gas, C02(g), in solution. When
the bottle is sealed, a closed system is formed. The fermentation reaction, however,
continues producing carbon dioxide gas in the champagne.
C02(g) in the champagne ⇔ C02(g) in the air inside bottle
When the bottle is opened, explain why bubbles form.
_______________________________________________________________________
_______________________________________________________________________
42) The following aqueous solution is in equilibrium:
2 NaI(aq) + (NH4)2S2O8(aq)
colorless
colorless
⇔
(NH4)2SO4(aq) + I2(aq ) + Na2SO4(aq)
colorless
yellowish colorless
If we add a catalyst to the system, predict the change in color of the solution.
________________________________________________________________________
24
________________________________________________________________________
43) State two ways of shifting the equilibrium in the following reaction to favor the formation
of water vapour.
H20(g) ⇔ H2(g) + 1/2 02(g))H = 241.6 kJ
________________________________________________________________________
________________________________________________________________________
Determining Equilibrium Concentrations
Often, given some initial concentrations, we want to know the final concentrations of the species
at equilibrium. One technique involves the use of a table with the values of the known and
unknown concentrations.
The number of columns in the table depends upon the number of species in the equation. Here is
a typical concentration table for the reaction: A ⇔ B + C
Equation:
Initial
Reaction
Equilibrium
A
⇔
B
0
+
C
0
ÄImportant: Make sure that the equation is balanced.
In the table, the three rows below the equation represent the concentration of the species at three
different stages of the reaction:
Initial is the initial concentration of the species before equilibrium starts.
Reaction is the concentration ratio of the species during the forward reaction
(equilibrium being established, reactants used/products produced)
Equilibrium is the concentration of the species when equilibrium is established.
In using this technique, note that the Initial line represents the situation before the reaction
begins. As such, the concentration of the products in the Initial line is always zero (as shown
above).
Furthermore, the concentrations of the species in the Reaction line are always in proportion with
the coefficients of the equation. The reason for this is because they are used and produced
according to the coefficients of the equation. What this means is that, if we know any one of the
concentrations in the Reaction line, using ratio and proportion we can easily calculate the
concentration of the other species for this line (simply by looking at the coefficients of the
25
balanced equation).
To find the concentration at equilibrium, note that the reactants (the species on the left) are
subtracted (Initial - Reaction), while the products (the species on the right) are added
(Initial + Reaction).
46) In a 1.0 L flask, a chemist places 4.0 mol of substance A and 4.0 mol of substance B When
the system reaches equilibrium, there are 2.0 mol of substance X. Fill in the concentration table
below given the reaction:
A(g) + 2B(g) ⇔ 2X(g) + 4Y(g)
ÄNote: Concentration must be expressed as moles/Litre. In this case, the concentrations are
not given directly. What is given is the number of moles and a volume of one-litre.
Thus, the given concentrations are: 4 mol/L, 4 mol/L and 2.0 mol/L respectively.
Equation:
Initial
Reaction
Equilibrium
47)
A(g)
+
2 B(g)
⇔
2 X(g)
+
4Y(g)
In a 10 litre container, a chemist places 6.0 mol of N2(g) and 10 mol of H2(g). At
equilibrium, 4.0 mol of N2(g) remains. Fill in the concentration table below given the
reaction:
N2(g) + 3 H2(g) ⇔ 2 NH3(g)
ÄNote: Again, the concentrations are not given directly. We are given the number of moles
of the reactants and a total volume. From this, we can calculate the concentrations
in mol/L as follows:
For the 6.0 mol of N2(g), we have 6 mol/l0. litres = 0.60 mol/L
For the 10 mol of H2(g), we have 10 mol/l0. litres = 1.0 mol/L
= 0.40 mol/L
For the 4.0 mol N2(g), we have 4 mol/l0. litres
Equation
Initial
Reaction
Equilibrium
ã
26
N2(g)
+
3 H2(g)
⇔
2 NH3(g)
Remember: Concentration must be expressed as moles per litre, mol/L.
48) Five moles of PCl5(g) are sealed in a one litre container. At equilibrium, an analysis shows
that the mixture contains two moles of C12(g). Complete the equilibrium concentration
table below:
PCl5(g) ⇔ PCl3(g) + C12(g)
Equation:
Initial
Reaction
Equilibrium
PCl5(g)
⇔
PCl3(g)
+
C12(g)
49) At room temperature, 0.012 mol of S02(g) and 0.01 mol of 02(g) were placed in a 100 mL
container. At equilibrium, 0.008 mol of S03(g) were formed. Fill-in the equilibrium
concentration table for this reaction:
2 S02(g) + 02(g)
⇔ 2 S03(g)
ÄNote: The volume here is 100 mL (or 0.100 L). Since the unit for concentration is moles
per LITRE, we need to express the given concentrations as mol/L. Thus;
for the 0.012 mol of S02(g), we have 0.012 mol/0.100 litres = 0.120 mol/L
= 0.100 mol/L
for the 0.01 mol of 02(g), we have 0.01 mol/0.100 litres
for the 0.008 mol of S03(g), we have 0.008 mol/0.100 litres = 0.800 mol/L
Equation:
2 S02(g)
+
02(g)
⇔
2 S03(g)
Initial
Reaction
Equilibrium
27
▼ TOPIC-3: THE EQUILIBRIUM CONSTANT
Using Le Chatelier's Principle, we can predict qualitatively the effect of a change (stress) when
one of the three factors disrupts equilibrium:
ΠA change in concentration
• A change in temperature
Ž A change in pressure
It would be useful to predict quantitatively the effects of a stress. For example, while
qualitatively we can predict the shift in equilibrium caused by a change in concentration, it
would be useful to predict the quantitative effect.
Mathematical predictions are derived from mathematical formulas. Thus, we need to develop a
formula (in accordance with Le Chatelier's Principle) to mathematically predict the effects on
equilibrium caused by a specific external stress.
➠ EXPERIMENT-6: Discovering a mathematical formula
Objective:
To discover a mathematical relationship between the concentration
of the species (reactants) and their effects on a system in equilibrium.
The system we'll study is the dissociation of acetic acid (vinegar):
CH3COOH(aq) ⇔ H+(aq) + CH3COO-(aq)
We have two different concentrations of acetic acid (CH3COOH or HCH3COO). For
convenience, they have been labeled SYSTEM-1 and SYSTEM-2..
•
•
SYSTEM-1: CH3COOH 1.0 x 10-1 mol/L
SYSTEM-2: CH3COOH 1.0 x 10-3 mol/L
Record your data in the table below.
28
Step-1:
Record the temperature of the systems (room temperature)
Step-2:
Using pH paper (or a pH meter if available), determine the pH of SYSTEM-1.
Step-3:
Determine the pH of SYSTEM-2.
50) Record your experimental data here:
At Start
System
Concentration
(mol/L)
1
1.0 x 10-1
2
1.0 x 10-3
51)
At Equilibrium
Temperature
(oC)
pH
[H+]
by calculation
Now calculate and fill-in the concentration of hydrogen ions, the [H+] column in the table
above, for SYSTEM-1 and SYSTEM-2. (Convert and record your answers in scientific
notation.)
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
52) Complete the concentration table below for SYSTEM-1 by:
a) Filling-in the Initial line with your experimental data (from previous table).
b) Filling-in the [H+] in the Equilibrium line (with data from previous table).
c) Using ratio and proportion as well as calculation to complete the table.
Equation:
CH3COOH
⇔
H+
+
CH3COOInitial
Reaction
Equilibrium
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
29
53) Complete the concentration table below for SYSTEM-2 (same as above):
Equation:
CH3COOH
⇔
H+
+
CH3COOInitial
Reaction
Equilibrium
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
54) Summarize the Equilibrium concentrations of SYSTEM-1 and SYSTEM-2 here:
System
1
Equilibrium Concentrations (mol/L)
CH3COOH
H+
CH3COO-
2
Now that we have the concentrations at equilibrium for both systems, let's try to discover a
mathematical relationship between concentration and equilibrium.
55)
Using the concentration values from the "summary" table above, calculate the value of the
following mathematical combinations (for both systems):
SYSTEM-1
SYSTEM-2
Π[CH3COOH] x [H+] x [CH3COO-]
__________
__________
• [H+] x [CH3COO-]
[CH3COOH]
__________
__________
___________
___________
Ž [CH3COOH] + [H+] + [CH3COO-]
56)
30
Study the results of your calculations for SYSTEM-1 and SYSTEM-2. How do they
compare with the results of other students?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
57) What general conclusion can you make from this experiment?
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
End of experiment-6
äTHE LAW OF MASS-ACTION
Cato Maximilian Guldberg and his brother-in-law, Peter Waage, studied equilibrium reactions at
Christiana University in Oslo, Norway. They arrived at a conclusion similar to yours. They
concluded that there is a definite mathematical relationship governing the behavior of
equilibrium systems and concentration changes. They called their conclusion the "Law of MassAction". Although mathematically simple to use, in words, the law sounds complicated. It states:
The multiplication of the molar concentration of the
products, raised to the power of their coefficients,
divided by the multiplication of the molar
concentration of the reactants, raised to the power of
their coefficients, results in a constant value provided
the temperature does not change.
The constant is known as the equilibrium constant and is designated by the letter "K" and is
sometimes written as Keq or Kc.
For example:
In the following reaction, which occurs at a temperature of 5000 C, the
calculation of K is:
2NH3(g) ⇔ 3H2(g) + 1N2(g)
K = [H2]3 x [N2]'
[NH3]2
K = 6.02 x 10-2 mol2/L2
Coupled with Le Chatelier's Principle, the Law of Mass-Action is very useful.
31
According to Le Chatelier's Principle, when we modify the conditions of a system in
equilibrium, the system tries to undo the modification as much as possible. Moreover, the Law
of Mass-Action tells us that, in reacting to the modification, the system will adjust the
concentrations of the species in such a way that the value of the equilibrium constant (K) does
not change. In the reaction above, if we change the concentration of any species, the equilibrium
constant remains at 6.02 x 10-2 (provided the temperature is not changed).
* IMPORTANT:
When using the equilibrium formula, ignore the concentration of solids and of water in the
liquid state (as their concentration remains constant).
Consider the dissociation of ammonia:
2 NH3(g) ⇔ 3 H2(g) + 1 N2(g)
The Law of Mass-Action tells us that we may change the concentration of any species without
changing the value of the equilibrium constant. Thus, for example, we may add hydrogen gas,
H2(g), to this system. As we do, the system responds by reducing the concentration of hydrogen
gas. It does this by shifting the equilibrium towards the left (thereby consuming more hydrogen
gas). Moreover, the adjustment takes place in such a way that the value of the equilibrium
constant remains "constant".
58) State the expression of the equilibrium constant for each system below:
a) CO(g) + 2 H2(g) ⇔ CH3OH(g)
K = ___________________
b) H2(g) + C12(g) ⇔ 2 HCl(g)
K = ___________________
c) 2 N02(g) ⇔ 2 NO(g) + 02(g)
K = ___________________
d) 02(g) ⇔ 02(aq) + 12.5 kJ
K = ___________________
e) 4 HCl(g) + 02(g) ⇔ 2 H20(g) + 2 C12(g)
K = ___________________
Since there are different types of equilibrium systems, it is convenient to group them into
families. To differentiate the equilibrium constants of one family from another, it is customary to
use a subscript.
32
Here are some examples of equilibrium families and the designation of their equilibrium
constants:
CHEMICAL EQUATION
CH3COOH(aq) ⇔ CH3COO-(aq) + H+(aq)
HF(aq) ⇔ F-(aq) + H+(aq)
HNO2 (aq ) ⇔ NO2- (aq ) + H+(aq)
BaSO4(s) ⇔ Ba2+(aq) + SO42-(aq)
Pbl2(s) ⇔ Pb2+(aq) + 2 I-(aq)
AgCl(s) ⇔ Ag+(aq) + Cl-(aq)
H2O(l) ⇔ H+(aq) + OH-(aq)
N2(g) + 3 H2(g) ⇔ 2 NH3(g)
H2(g) + I2(g) ⇔ 2 HI(g)
PCl5(g) ⇔ PCl3(g) +C12(g)
02(g) ⇔ O2(aq)
Constant
Ka
COMMENT
Dissociation of substances
which liberate H+(aq) ions
_________ (acids).
Ksp
Equilibrium between a
solid and its ions in an
aqueous solution.
Kw
Dissociation of water
K
Other types of equilibrium
systems.
__________ ___________________________
59) State the expression of the equilibrium constant for each system below:
a) AgC1(s) ⇔ Ag+(aq) + Cl-(aq)
Ksp =
________________________
b) H20(1) ⇔ H+(aq) + OH-(aq)
Kw
=
________________________
c) 2 N02(g) ⇔ 2 NO(g) + 02(g)
K
=
_________________________
d) 02(g) ⇔ 02(aq) + 12.51 kJ
K
=
__________________
60) The equilibrium system below occurs at a temperature of 250 0C. Given the equilibrium
concentrations, find the equilibrium constant (K).
PC15 (g) ⇔ PCl3(g) + C12(g)
[PCl5 (g)] = 0.0175 mol/L
[PCl3(g)] = 0.0267 mol/L
[C12(g)] = 0.0267 mol/L
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
33
61) Acetic acid dissolves in water according to the following equation:
CH3COOH(aq) ⇔ H+(aq) + CH3COO-(aq)
Calculate the value of the equilibrium constant, at 25 oC, given the concentrations of the
species at equilibrium to be:
[CH3COOH(aq)]
= 0.20 mol/L
= 0.0019 mol/L
[CH3COO (aq)]
+
[H (aq)]
= 0.0019 mol/L
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
62) The temperature of a saturated solution of strontium chromate, SrCrO4(s), is 25 oC. Calculate
the value of the equilibrium constant if the concentration of the species at equilibrium are:
[Sr2+(aq)] = 0.0060 mol/L
[CrO42-(aq)] = 0.0060 mol/L
The equation of the reaction is: SrCrO4(s) ⇔ Sr2+(aq) + CrO42-(aq)
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
63) A saturated solution of lead iodide, PbI2, is represented by this equation:
PbI2(s) ⇔ Pb2+(aq) + 2 I-(aq)
Calculate the equilibrium constant for this reaction, given that at 25 0C the concentrations of
the products are:
[Pb2+(aq)] = 0.00l3 mol/L
[I-(aq)] = 0.0026 mol/L
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
34
64) An equilibrium system contains, at 448 0C, 0.25 moles of H2(g), 0.25 moles of I2(g) and
1.75 moles of HI(g). Calculate the equilibrium constant at this temperature knowing that the
volume of the system is 0.50 litres.
H2 (g) + I2(g) ⇔ 2 HI(g) + heat
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
65) A 10 litre sealed flask contains 0.40 moles of nitrogen gas (N2), 3.6 moles of hydrogen gas
(H2), and 0.070 moles of ammonia gas (NH3) at a temperature of 350 0C. Determine the
equilibrium constant for this system.
Reminder:
When calculating the equilibrium constant, concentration is expressed as mol/L.
N2(g) + 3 H2(g) ⇔ 2 NH3(g)
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
66) At 25 0C, the concentrations of sulfate ions (S042-) and barium ions (Ba2+) are
1.02 x 10-5 mol/L. Calculate the equilibrium constant for this system.
BaSO4(s) ⇔ Ba2+(aq) + SO42-(aq)
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
67) Contrary to what one might think, pure water does not contain only H20(1) molecules.
Actually, at 25 0C, for every litre of water, there are 1.0 x 10-7 moles of water molecules
which dissociate according to this equation:
H20(l) ⇔ H+(aq) + OH-(aq)
Calculate the value of the equilibrium constant for this system.
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
35
t TOPIC 4: THE CONSTANT RULES
The reactions representing the dissociation of acids in a solution form a particular family of
equilibrium systems whose constant is known as the "acidity constant", Ka . We can represent
these reactions with the following general equation:
HB(aq) ⇔ H+ (aq) + B- (aq)
where B represents an anion (negative ion).
Let's study the acidity constant (Ka) of two different acids.
EXPERIMENT 7: Is Ka A Characteristic Property?
Objective: To determine and compare the acidity constants (Ka) of two different acids.
68) Outline an experimental procedure to determine the acidity constants of 1.00 x 10-3 M acetic
acid (CH3COOH) and 1.00 x 10-3 M Lactic acid (CH3CHOHCO2H).
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
69) Record your experimental results:
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
70) Determine the value of the equilibrium constants:
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
71) Classify the acids in increasing order of their acidity constant, Ka.
Œ
•
36
ACID
Ka CONSTANT
72) Which acid has dissociated the most, thereby providing the greatest number of H+ (aq)
ions? Explain.
______________________________________________________________________________
______________________________________________________________________________
73) Define the expression, degree of dissociation of an acid.
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
ÄEnd of experiment 7
74) Describe the relationship between:
a) The acidity constant (Ka) and the concentration of H+(aq) ions.
_________________________________________________________________________
_________________________________________________________________________
b) The acidity constant (Ka) and the degree of dissociation.
_________________________________________________________________________
_________________________________________________________________________
75) Which of the following acids has the greatest tendency to dissociate in an aqueous
solution? Explain:
•HCHO2
‚ HNO2
ƒ HCN
Ka = 2.l x l0 -4
Ka = 4.5 x l0-4
Ka = 7.0 x l0-10
_________________________________________________________________________
_________________________________________________________________________
76) A strong acid is a substance which, in an aqueous solution, highly dissociates to produce
H+(aq) ions. Which of the following acids is the strongest? Explain:
• HCIO2
‚ HClCH2CO2
ƒ HC1O
Ka = l.2 x l0-2
Ka = l.4 x l0-3
Ka = l.0 x 10-8
________________________________
________________________________
________________________________
77) Which acid produces a solution with the highest [H+] level? The highest pH level?
Explain:
• HF
Ka = 6.7 x l0-4
‚ HCN
Ka = 7.0 x l0-10
_________________________________________________________________________
37
Review (Acids and Bases)
.wACIDS-BASES
An acid is a compound which when dissolved in water produces a solution that:
• Conducts electricity
‚ Reacts with metals (such as Zn and Mg) to give off hydrogen gas (H2)
ƒ Changes blue litmus paper to red
„ Tastes sour (example; vinegar)
A base is a compound which when dissolved in water produces a solution that:
• Conducts electricity
‚ Reacts with acids to neutralize its properties
ƒ Changes red litmus paper to blue ( F Hint: remember Base Blue)
„ Tastes bitter and feels slippery (example: javel water)
wWATER
In pure water, the concentration of hydrogen ions equals the concentration of hydroxide ions.
That is, [H+] = [OH-] and the water is said to be neutral.
ÄNote: It has been found experimentally that Kw = [H+] x [OH-] = 1.0 x l0-14
Therefore; [H+] = 1.0 x 10 -7 mol/L
[OH-] = 1.0 x l0-7 mol/L
.wNEUTRALIZATION REACTION
A reaction between an acid and a base neutralizes each other forming a salt and water.
ACID + BASE à SALT + WATER
Example:
HCl + NaOH à NaCl + H20
ÄNote: All of the hydrogen ions of the acid (H+) combine with all of the hydroxide
ions (OH-) of the base.
Net ionic equation:
38
H+(aq) + OH-(aq) à H2O(l)
Review (continued)
• pH SCALE (by definition pH = -log[H+]
A scale used to conveniently specify how acidic or basic a solution is:
Neutral
Acid
Base
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Vinegar (2.5)
Blood (7.4)
• TITRATION
This is an experimental technique to find the concentration of an unknown solution by reaction
with a known solution. The process involves neutralizing the solution (adding an acid to a basic
solution or vice-versa). By measuring the amount of acid (or base) added whose concentration
we know, we can find the concentration of the unknown solution.
Remember:
ACID
BASE
(H+ concentration)(volume) = (OH- concentration)(volume)
ÄNote: Be careful about the particular acid or base used. Write the dissociation
equations of both the acid and the base in order to determine the correct H+
and OH- concentrations.
If the pH is less than 7, the solution is acidic. If the pH is greater than 7, the solution is basic.
A pH of 7 is a neutral solution.
Some common acids:
HCl (hydrochloric acid, also known as muriatic acid))
H2S04 (sulfuric acid)
HN03 (nitric acid)
HC2H302 or CH3COOH (acetic acid or vinegar)
wBUFFERS
Buffers are mixtures of chemicals which make a solution resist a change in its pH. Solutions
which have a resistance to changes in their pH value because of the presence of buffers are called
buffer solutions.
39
•
ACID-BASE INDICATORS
Modern chemistry labs have pH meters which conveniently measure the pH of a solution.
Traditionally, however, pH papers are used. When pH paper is dipped into a solution, the wetted
portion of the paper changes color. By color comparison, the pH value can be determined.
Certain liquids are also used as indicators.
ÄNote: When the pH is specified as a whole number, the [H+] equals
one times ten to the negative power of the pH value.
For example, if the pH = 4, then [H+] = 1 x l0-4 mol/L. However, if the pH
is a fractional number, for example, 6.4, then you can use a scientific
calculator to obtain the [H+] like this:
Once you have calculated the [H+] using one of the methods above, convert your answer to
scientific notation. For example, if the pH is 4.2, [H+] = 6.3 x l0-5 M.
Press: 1 X
40
10
YX 4.2 +/- =
----------------DISPLAY----------------0.000063095 = 6.3 x 10-5
t TOPIC 5: ACID-BASE TITRATION
Recall that a neutralization reaction is one in which a base and an acid neutralize each other
forming a salt and water. What exactly happens during the process of neutralization? Let's find
out with an experiment.
We will analyze the behavior of an acid solution by the gradual addition of a basic solution.
Recall that during a neutralization reaction, the H+(aq) ions of the acid react with the OH-(aq)
ions of the base to form water.
H+(aq) + OH-(aq) ➝ H20(l)
wEXPERIMENT-8:
Objective:
The acid- base competition
To study acid-base reactions.
This experiment consists of two parts:
• PART-A: Neutralization of a strong acid
•PART-B: Titration techniques
• PART-A: Neutralization of a strong acid
We will perform the neutralization of hydrochloric acid, HCl(aq), by gradually adding (from a
burette) sodium hydroxide, NaOH. Note that both the acid and the base have a concentration of
0.10 mol/L.
Step-1: Using a funnel, fill a 50 mL burette with 0.10 M NaOH
ÄNote: For convenience, add slightly more NaOH than required then carefully drain the excess
into a "waste" beaker to obtain the desired level.
Step-2: As accurately as possible using a pipette, pour 10.0 mL of HCl into a 50 mL
Erlenmeyer flask.
ÄHint: If a pipette is not available, measure the HCl using another burette or
a 10 mL graduated cylinder
Step 3: Add three drops of phenolphthalein to the acid solution and stir gently.
(The phenolphthalein will be used as a pH indicator)
41
Step-4: From the burette, you will add 1.0 mL portions of NaOH (one at a time) into the
HCl flask. (You will be adding 20 separate portions in all stirring gently between
each portion.)
Step-5: After each addition of NaOH, dip a stirring rod into the flask. Wet a pH indicator
paper and check the pH of the solution. Record the pH value and your observation
in the table below. (If available, use a pH meter)
Ä Note: Rinse the pH meter with distilled water after each use.
Step-6: Stop the neutralization when 20.0 mL of NaOH have been added.
Vol. of HCl (mL)
Total Vol. of NaOH
added (mL)
pH
Observations
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
78) In general, how does the pH of an acid vary with the addition of a base?
__________________________________________________________________________
79) In your experiment, how did the pH vary as you added the NaOH solution?
__________________________________________________________________________
42
80) Using your experimental data, trace the graph of the pH (y-axis) as a function of the volume
(x-axis) of the base added.
ÄNote: For an acid to be neutralized by a base (of the same concentration), it is
necessary that the quantity of the acid equals the quantity of the base. We can
identify this "equivalence" from the graph as it represents that section having
the steepest slope and is known as the equivalence point. At this point, the
solution loses its acidic and basic properties as the number of H+(aq) ions
equals the number of OH-(aq) ions.
81) From the graph, determine the pH at the equivalence point.
_________________
82) What is the pH value for the turning point in your experiment? _________________
43
• PART-B: Titration techniques
A titration is an accurate method of determining the concentration of an unknown solution using
a solution of known concentration. In a titration, we carefully determine the equivalence point.
Once we have determined the equivalence point, we can calculate the concentration of the
unknown solution.
To perform the titration, follow the steps below. When finished, record the volume and
concentration of the base added. Ü Be prepared to be patient!
Step 1:
Step 2:
Step 3:
Step 4:
Place 10.0 mL of an HCl acid solution whose concentration is unknown into an
Erlenmeyer flask.
Add three drops of phenolphthalein to the acid solution and stir gently.
(The phenolphthalein will be used as a pH indicator)
Adjust the level of NaOH in the burette to the nearest whole number.
Using the burette, add the NaOH base to the solution in the Erlenmeyer
flask by portions of about 1 mL.
ÄNote: When you notice the solution has a tendency of remaining pink in
color, you are approaching the equivalence point. When this happens,
rather than use 1 mL portions, add the NaOH drop by drop (so as not to
over do it).
IMPORTANT: As you stir the solution, the pink color should disappear. If the color is very
pink, you have passed the end point and must re-do the titration.
Step 5: Continue adding "drops" of NaOH (stirring gently each time).
The "end point" is reached when the entire solution remains slightly pink.
ÄHint: Be sure to record the volume of NaOH added. This way, should you add
too much and need to start over, you will know how much is "too much".
83) Record your data here:
Volume of NaOH:
__________________
Concentration of NaOH
__________________
84) Determine the number of moles of NaOH:
___________________________________________________________________________
___________________________________________________________________________
44
85) At the equivalence point:
The number of moles of base = the number of moles of acid = ____________________
86) Concerning the unknown acid, from your titration:
a) How many moles of acid were you given?
______________________
b) What was the concentration of this acid?
______________________
(See below)
ÄEnd of experiment-8
Review of concentration calculations
Note that we can mathematically find the concentration of an acid or a base. In doing so, four
variables are involved: the concentration and volume of the acid (CA and VA), and the
concentration and volume of the base (CB and VB).
Knowing that Concentration = moles/Volume, moles = Concentration x Volume
And, since moles of acid (H+) = moles of base (OH-), we have:
cone. of acid x volume of acid = conc. of base x volume of base
CA x VA = CB x VB
or:
[H+] x VA = [OH-] x VB
[H+] = [OH-] x VB
VA
= (0.5)(20) = 0.2 mol/L
(50)
Now using the equation for the dissociation of H2S04, we get:
1H2S04 ➝ 2H+ + SO42x
0.2
2x = (1)(0.2)
x = [H2S04] = 0.1 mol/L
45
t TOPIC 6: ENRICHMENT
An example calculating the acidity constant (Ka) for hydrogen sulfide acid will familiarize you
with the technique used to solve such problems.
Sample Problem
Suppose we have a solution of hydrogen sulfide acid, H2S, whose concentration is
1.0 x 10-1 mol/L. We determine experimentally that, at equilibrium, [H+] ions is
1.0 x l0-4 mol/L. What is the value of the acidity constant of H2S?
H2S(aq) ⇔ H+(aq) + HS-(aq)
Our technique in finding the equilibrium constant involves creating an equilibrium concentration
table illustrating the three stages of equilibrium. After filling-in the table, we use the Law of
Mass Action to find the value of the acidity constant (Ka).
Recall that in the concentration table, the balanced equation is written on the first line, the Initial
line represents the concentration of the system before the reaction starts, the Reaction line
represents the concentration of the reactants used and the products produced (equilibrium being
established), and the Equilibrium line represents the concentrations of the substances at
equilibrium. Like this:
Equation:
H2S (aq)
⇔
H+ (aq)
+
HS- (aq)
Initial
Reaction
-
+
+
Equilibrium
From the given concentrations, we fill-in the table accordingly. In this case, we are given that at
the start of the reaction, the concentration of H2S is 1.0 x 10-1 mol/L and that at equilibrium, the
concentration of H+ is 1.0 x 10-4 mol/L. Thus, we have:
H+ (aq)
H2S (aq)
Initial
1.0 x 10-1
0
Reaction
-
+
Equilibrium
46
⇔
Equation:
1.0 x 10
+
HS- (aq)
0
+
-4
We can now complete the H+ column this way:
Initial =
0
Reaction = + ?
Equilibrium = 1.0 x 10-4
Our table now looks like this:
Equation:
H2S (aq)
Initial
1.0 x 10-1
Reaction
⇔
H+ (aq)
HS- (aq)
+
0
0
+ 1.0 x 10-4
-
+
1.0 x 10-4
Equilibrium
Remember that the concentrations on the Reaction line are in proportion with the coefficients of
the equation. This means that if we know any concentration on this line, we can easily find the
rest of the concentration (on the Reaction line).
In this case, the concentrations are in the ratio of 1 to 1 to 1 Thus, we can fill-in these values in
the table:
Equation:
H2S (aq)
Initial
1.0 x 10-1
Reaction
- 1.0 x 10
⇔
H+ (aq)
+
0
-4
HS- (aq)
0
-4
+ 1.0 x 10-4
+ 1.0 x 10
1.0 x 10-4
Equilibrium
We can now fill in the Equilibrium line by subtraction in the H2S column and by addition in the
HS- column so that our completed table looks like this:
Equation:
H2S (aq)
Initial
1.0 x 10-1
⇔
- 1.0 x 10
Equilibrium
9.9 x 10-2
+
0
-4
Reaction
H+ (aq)
HS- (aq)
0
-4
+ 1.0 x 10
+ 1.0 x 10-4
1.0 x 10-4
1.0 x 10-4
Using the concentrations in the Equilibrium line, we solve for Ka :
Ka = [H+(aq)] x [HS-(aq)]
[H2S(aq)]
=
(l.0 x l0-4) x (l.0 x l04) = 1.0 x10-7
(9.99 x 10-2)
47
87) In a 4.0 L sealed container, we place 4.0 mol of dinitrogen tetroxide, N204(g). The gas
partially dissociates and establishes equilibrium at a temperature of 27 0C. If, at equilibrium,
there is 1.6 mol of nitrogen dioxide N02(g), calculate the equilibrium constant for this
system.
a) Determination of concentrations:
ÄNote: Be sure to specify the given concentrations as moles/litre. This means the
volume of the container must be one litre. Otherwise, if the given
volume is greater than one litre, divide the number of moles by the
number of litres, if the given volume is specified in mL, divide the number
of moles by 1000.
Equation:
Initial:
Reaction:
Equilibrium:
N204(g)
⇔
2 N02(g)
b) Determination of the equilibrium constant:
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
88) At 25 0C, NOBr(g) partially decomposes. Knowing that initially there were
3.0 mol of NOBr(g) and that at equilibrium there remains 2.0 mol, determine the value of the
equilibrium constant. The volume of the system is 1.0 L.
Equation:
2 NOBr(g)
⇔ 2 NO(g)
+ Br2(g)
Initial:
Reaction:
Equilibrium:
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
_____________________________________________________
48
89 At 25 0C, a saturated solution is obtained by adding 2.0 mol of calcium sulfate (CaSO4) to 1.0
L of water. Upon analysis, it is determined that the concentration of the calcium ions (Ca2+) is
7.81 x l0-3 mol/L. Determine the equilibrium constant (also known as the solubility constant,
Ksp).
CaSO4(s) ⇔ Ca2+(aq) + SO42-(aq)
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
90) In a given solution at 25 0C, the concentration of H+(aq) ions is 2.5 x 10-4 mol/L. What is the
concentration of the OH-(aq) ions if the equilibrium constant is 1.0 x 10-14?
H20(1) ⇔ H+(aq) + OH-(aq)
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
49
91) Acetic acid (CH3COOH) dissociates partially in water at a temperature of 25 0C. At
equilibrium, the concentration of the acid is 0.996 mol/L. What is the concentration of the
H+(aq) ions in this solution if the equilibrium constant is 1.8 x 10-5 ?
CH3COOH(aq) ⇔ H+(aq) + CH3COO-(aq)
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
92) In a 1.0 L balloon, we place 5.0 mol of nitrogen dioxide gas (NO2). At equilibrium, we find
1.5 mol of oxygen gas (O2). What is the value of the equilibrium constant for this system?
2 N02(g) ⇔ 2 NO(g) + O2(g)
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
50
◆ Supplementary Problems
Part 1: Neutralisation Problems
1. Exactly 12.0 mL of 0.024 M NaOH are required to neutralize 20.0 mL of HCl solution.
What is the concentration of HCl?
2. What is the concentration of acetic acid in vinegar when 32.5 mL of 0.56 M NaOH are
required to neutralize 15.0 mL of vinegar?
3. A sample of Saniflush containing 81.0 g of NaHSO4 is dissolved in 4.0 L of water and
50.0
mL of it is titrated with 2.0 M KOH. . What is the volume of KOH required for a
complete reaction?
4. An effluent from a research station was suspected of containing phosphoric acid H3P04
To determine the acid concentration, it was titrated with 1.0 M NaOH. If 17.3 mL of NaOH
were required to remove the first H+ in a 10.0 mL sample of the acid, what is the molarity of
the acid?
5. A 25.0 mL sample of muriatic acid (slightly impure HCl sold as a cleanser) reacted
exactly with 45.0 mL of NaOH. If the NaOH had a pH of 12.0, find the pH of the acid.
6. Oxalic acid, H2C204, is a solid acid that is slightly soluble in water. When 6.25 g of acid
are dissolved in water containing the indicator phenol red, the solution turns yellow. If 32.2
mL of KOH are required to change the indicator to red at the equivalence point, what is the
concentration of the KOH ?
Part 2: Le Chatelier's Principle Problems
7. Acetic acid is a weak carboxylic acid which ionizes in water as follows:
CH3COOH ⇔ CH3COO- + H+
If one drop of HCl is added to this system:
a) in which direction will the equilibrium shift?
b) what effect will this have on the concentration of each substance in the system?
8. If one drop of NaOH is added to the system in question 7,
a) in which direction will the equilibrium shift?
b) what effect will this have on the concentration of each substance in the system?
9. In large, public swimming pools, Cl2 (g) can be added to the water to produce
hypochlorous acid, HOCl, which prevents the growth of algae and bacteria. The
equilibrium system is:
Cl2(g) + H2O(l) ⇔ Cl- (aq) + H+ (aq) + HOCl (aq)
a) If the pH increases, what happens to the [H+] in the system?
b) If the pH increases, in which direction will the equilibrium shift?
51
10. Tooth decay results in the dissolving of tooth enamel, Ca5(PO4)3OH , in the mouth. The
following equilibrium is set up:
Ca5(PO4)3OH(s) ⇔ 5 Ca2+ (aq)
+
3 PO43-
+ OH- (aq)
When sugar ferments on teeth, H+ ions are added to the system.
a) In which direction does this change cause the equilibrium to shift?
b) What effect does this shift have on the tooth enamel?
11. Chicken eggs are formed by the precipitation of calcium carbonate, CaCO3. Chickens, like
dogs, do not perspire and, therefore, in hot weather they must resort to panting". This releases
water and carbon dioxide from their bodies. Using the following sequence of steps:
1.
H20(l) + C02(g) ⇔ CO2(aq) ⇔ H2CO3(aq) ⇔ H+ (aq) + HCO3-(aq)
2.
HCO3- (aq) ⇔ H+(aq) + CO32-(aq)
3.
Ca2+(aq) + CO3 2-(aq) ⇔ CaCO3(s)
a) explain why chickens lay eggs with thinner shells in summer. (hint: When they pant, which
way does the equilibrium shift?)
b) explain why giving the chickens soda water (carbonated water) to drink solves the whole
problem.
Part 3: Equilibrium Constant Problems
12. a) Write the equilibrium equation for the ionization of one proton (H+) from H2SeO3(aq).
b) Write the equilibrium constant expression.
13. The pH of a 0.1 M solution of cyanic acid (HCNO) is 4.0.
a) Write the equation for the ionization of cyanic acid
b) Write the Ka expression and calculate the value of Ka.
14. The Ka for hypoiodous acid (HOI) is 2.5 x 10-11. What would you expect the [H+] of
a 0.010 M solution of this acid to be?
15. A 0.100 M solution of HOBr has a pH of 4.843. Find Ka.
16. A 0.25 M solution of benzoic acid (C6H5COOH) has a [H+] of 4.0 x 10-3 M. Find Ka.
17. Ka = 1.3 x 10-2
Ka = 6.7 x 10-4
a) Which acid is the strongest?
b) Which acid has the highest pH?
52
Ka = 6.6 x 10-10
18. The initial concentration of an aspirin solution (HC9H704) is 1.368 x 10-3 M. At equilibrium,
the pH is 3.28. Find Ka.
PART 4: pH Problems
19. Solution X has a pH of 2 and solution Y has a pH of 4. Which of the following statements is
true?
a) The hydrogen ion concentration of X is half that of Y.
b) The hydrogen ion concentration of X is twice that of Y.
c) The hydrogen ion concentration of X is 100 times that of Y.
d) The hydrogen ion concentration of X is 1/100 that of Y.
20.
If the pH value of an electrolyte changes from 4 to 3, what has happened to the hydrogen
ion concentration and the hydroxide ion concentration?
21. a) What is the pH of a 0.01 M nitric acid solution? Nitric acid is a strong acid (ie. 100%
dissociated).
b) What is the pH of a 0.01 M solution of hydrofluoric acid that is 8% ionized?
22. Calculate the pH of a 0.015 M solution of potassium hydroxide, a strong base.
23. A solution of hydrobromic acid has a pH of 4.84. Find the concentration of hydrogen ions.
24. a) Complete the following table:
pH
[H+]
4.26
?
?
6.25 x 10-3 M
2.95
?
?
1.80 x 10-1 M
b) List the acids in order from strongest to weakest.
25. Vinegar is about 100 times more acidic than distilled water. What is the approximate pH of
vinegar?
26. Ammonia is about 100 times more basic than distilled water. What is the approximate pH of
ammonia?
27. If the pH of a solution decreases by a factor of 1, the concentration of hydrogen ions
___________ by a factor of _________.
28. A strong acid is one which has a _______ pH value and a ____ Ka value.
29. A weak acid is one which has a ________ pH value and a ________ Ka value.
30. In pure water, the [H+] = [OH-] = 1.0 x 10-7M. The pH value = -log[H+] = _______ ?
53