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Page 1 Yeung Unit 4: Chemical Equilibrium Text adapted from CK-12.org and Manitoba Education Curriculum and Teaching Name:
Page 1 Yeung
Unit 4: Chemical Equilibrium
Text adapted from CK-12.org and Manitoba Education Curriculum and Teaching
Name:
Page 2 Yeung
Unit 4: Chemical Equilibrium
ü Relate the concept of equilibrium to physical and chemical systems (Conditions necessary to achieve equilibrium) 1 hour
ü Write equilibrium law expressions from balanced chemical equations for heterogeneous and homogeneous systems (1 hour)
ü Use the value of the equilibrium constant to identify how far a system at equilibrium has gone towards completion (0.5 hours)
ü Solve problems involving equilibrium constants (2 hours) ü Use Le Chatelier’s Principle to predict shifts in equilibrium. (Include: temperature changes,
pressure/volume changes, changes in reactant/product concentration, the addition of a catalyst, the addition of an inert gas, and the effects of the various stresses on equilibrium constant.) (2 hours)
ü Perform a lab to demonstrate a) equilibrium constant b) le Chatelier’s principle ü Interpret concentration versus time graphs (1 hour) ü Practical uses of Le Chatelier’s principle (1 hour) ü Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility
(0.5 hours) ü Solve problems involving Ksp (3 hours) common ion problems ü Practical applications of salts with low solubilities (0.5 hours)
Total: 11 hours
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Relate the concept of equilibrium to physical and chemical systems (Conditions necessary to achieve equilibrium) 1 hour
Introduction
Consider this generic reaction: . Based on what we have learned so far, you might assume that the reaction will keep going forward, forming C and D until either A or B (or both) is completely used up. When this is the case, we would say that the reaction “goes to completion.” Reactions that go to completion are referred to as irreversible reactions.
Some reactions, however, are reversible, meaning that products can also react to re-form the reactants. In our example, this would correspond to the reaction . During a reversible reaction, both the forward and backward reactions are happening at the same time.
Blue bottle activity/demonstration
Dissolve 4g of NaOH in 270ml of water in a Erlenmeyer flask. Add 4g of dextrose or glucose and 1ml of Methylene blue to the NaOH solution. Place a stopper tightly and shake vigorously. Observe!
Equilibrium
As we learned earlier, the rate of a reaction depends on the concentration of the reactants. At the very beginning of the reaction , we would not expect the reverse reaction to proceed very quickly. If only a few particles of and have been created in a large flask of and , then it is very unlikely that they will find each other because the concentration of and is just too low. If and cannot find each other and collide with the correct energy and orientation, no reaction will occur.
However, as more and more and are created, it becomes more and more likely that they will find each other and react to re-form and . Conversely, as and are being used up, the forward reaction slows down for the same exact reason. The concentration of and decreases over the course of a reaction because there are less and particles in the same size flask. At some point, the rates for the forward and reverse reactions will be equal, at which point the concentrations will no longer change. If and are being destroyed at the same rate that they are being created, the overall amount should not change over time. At this point, the system is said to be in equilibrium. A qualitative description of this process for the reaction between hydrogen and sulfur to make dihydrogen sulfide is shown below.
Chemists use a double arrow to show that a reaction is in equilibrium.
For the reaction above, the chemical equation would be:
This indicates that both directions of the reaction are occurring. Note that a double-headed arrow ( ) should not be used here because this has a different chemical meaning.
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Dynamic Equilibrium
One characteristic of a system at equilibrium is that the macroscopic properties do not change over time. Macroscopic properties are those that describe the system as a whole. These can be observed and measured without needing to investigate the system in terms of individual molecules. Examples of macroscopic properties are temperature, pressure, and concentration.
One example of a macroscopic property is temperature. Although temperature can be explained on a microscopic level by the speed at which particles are moving around, individual molecules do not have a “temperature.” A glass of water with a constant temperature contains molecules with a large range of different speeds, and the speed of any individual molecule is constantly changing as it collides with other molecules and gains or loses kinetic energy. However, if you average out the properties of all these molecules (
) for an 8 oz glass of water), the overall temperature appears constant, despite continuous changes on the microscopic level.
Concentration is another example of a macroscopic property. It describes the total amount of a substance, but it does not make any statement about what is occurring on a molecular level. When a reaction is at equilibrium, the concentration of each component is constant over time. As we saw before, both the forward and reverse reactions are still taking place, but since they are moving at the same rate, there is no change for the system as a whole. The term dynamic equilibrium refers to a state where no net change is taking place, despite the fact that both reactions are still occurring. Individual molecules are still being formed and broken down, but the system as a whole is not changing over time.
Other Conditions Necessary for Equilibrium
In order for a reaction to reach an equilibrium state, it needs to take place in a closed system. Consider the following reaction:
This equilibrium is established in a closed can of soda because neither the reactants nor the products can
leave the system. However, when you open the soda, one of the products ( ) is free to escape into the
atmosphere. If a molecule of escapes, it is no longer available to collide with the water molecules left behind, so it can no longer participate in the reverse reaction. Therefore, the forward reaction will eventually go
to completion until there is no more available. Note that since is an acid, an open soda will become less acidic over time. Acidity is an important component of taste, so flat soda really does taste different for reasons other than texture.
Another requirement for a system to stay in equilibrium is that the temperature and pressure stay constant. As we learned in the previous chapter, temperature and pressure both have an effect on the rate of a reaction. However, the effect might be greater for the forward reaction than the reverse reaction, or vice versa. For example, in the reaction above, adding heat will favor the forward reaction more than the reverse, resulting in
the production of more . This is true even if the can remains closed. Eventually, a new equilibrium will be established at the new temperature, but while the temperature is changing, the system is no longer in equilibrium.
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The top left graph shows the concentration of the reactants and products in the reaction. Equilibrium is established at the beginning of the plateau. Compared to the reaction rate of the graph (top right), the lines meet because the rates are in equilibrium with each other, thus reaching the same amount. Both the forward and the reverse reactions are operating at the same rate.
Review Questions
1. What does the term dynamic equilibrium mean? 2. List all of the conditions required to reach chemical equilibrium?
a. . b. . c. . d. . e. .
3. Of the following conditions, which is not required for a dynamic equilibrium? a. rate of the forward reaction equals the rate of the reverse reaction. b. reaction occurs in an open system c. reaction occurs at a constant temperature d. reaction occurs in a closed system
4. Which of the following systems, at room temperature and pressure, can be described as a dynamic equilibrium?
a. an open flask containing air, water and water vapor b. a glass of water containing ice cube cubes and cold water c. a closed bottle of soda pop d. an open flask containing solid naphthalene
5. Is each of the following in a state of equilibrium? Explain. a. Ice cubes are melting in a glass of water with a lid on it b. Crystals of potassium dichromate were dissolved in water, and now the water is a uniform
orange color with a small amount of crystal left in the closed container. c. An apple that is left on the counter for a few days, it dries out and turns brown.
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ü Write equilibrium law expressions from balanced chemical equations for heterogeneous and homogeneous systems (1 hour)
ü Use the value of the equilibrium constant to identify how far a system at equilibrium has gone towards completion (0.5 hours)
ü Solve problems involving equilibrium constants (2 hours)
Introduction
You were introduced to two types of equations in the chapter on “Chemical Kinetics.” One consisted of reactions where the overall equation and the reaction mechanism were exactly the same. In those reactions, the reaction occurred with a single collision. The other type of reactions involved a reaction mechanism consisting two or more steps, and the overall equation was the sum of the equations for the reaction mechanism.
For those reactions that consist of a single collision, the reaction rate can be expressed by inserting the concentrations of the reactants into the expression:
, where and are the reactants
For the reactions that have reaction mechanisms of two or more steps, the relationship between the concentration of a particular reactant and the reaction rate is more complex. The rate of these reactions can still be expressed in a similar manner, but which reactants go into the expression and how they are involved must be determined experimentally.
In the previous section, you were introduced to the possibility of a reverse reaction. For the reaction between and yielding and , we have not only the forward reaction:
but we also could have a reverse reaction:
If this reverse reaction occurs, it would be the same simple collision that was involved in the forward reaction. Therefore, the reaction rate of the reverse reaction could be expressed as:
, where the products and have become reactants
When this reaction reaches equilibrium and the two rates become constant and equal,
and
Algebraic manipulation of the expression yields
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Since both and are constants, then dividing one constant by another yields another constant. The new constant is called the equilibrium constant and is symbolized by the capital letter .
The Equilibrium Constant
The equilibrium constant ( or ) represents a mathematical relatioxnship that shows how the concentration of one reaction component is related to the concentration of other reactants and products at
equilibrium. Sometimes you may see the equilibrium constant written with other subscripts, such as ,
, , or . These indicate that the constant describes a specific type of reaction, but they operate exactly the same as any other equilibrium constant.
The equilibrium constant for any reaction can be written by following a few simple rules. We have already learned how to write the expression for a simple generic reaction:
But suppose the reaction was . Writing the equation with a coefficient of 2 in front of the is exactly the same as writing the equation this way:
In this second form of the equation, we have just written twice instead of using a coefficient to show that two molecules are involved in the reaction. If we write the equilibrium constant expression from this last equation, it would look like:
With our knowledge of algebra, we know that writing is the same as writing . Therefore, the equilibrium constant would become
As you can see, the coefficient in the equation has become an exponent in the equilibrium constant. The general rule for writing an equilibrium constant expression is to write the coefficient of each species in the reaction as an exponent in the equilibrium constant expression. So, for the reaction:
where the lower case and are coefficients, the equilibrium constant expression would be:
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Here is an example equation and its equilibrium constant expression.
Do states of matter matter?
There are cases where the concentration of a particular reactant or product does not change when it is used or produced. The concentration of a substance is the moles of the substance divided by the volume (in liters) occupied. In the case of a substance dissolved in a beaker of water, its volume is the same as the volume of the solution. If half of the substance is used up in a reaction, the remaining half is still dissolved in the same volume of water. Therefore, its concentration will be cut in half. In the case of a solid reactant, however, it is not dissolved in the water, so its volume is the volume of the solid itself. If half of the solid is used in a reaction, the amount of moles remaining has been cut in half, but the volume of the solid has also been cut in half. Its concentration, then, stays exactly the same. This is also true of any solid or liquid that is not dissolved in a solution.
If the state of a reactant or product is indicated by or , its concentration can change, but if its state is
indicated by or , its concentration cannot change. Since the concentrations of these substances cannot change, their numerical value in the equilibrium constant expression will be a constant. To simplify the equilibrium expression, these constants can be algebraically combined with to produce a new constant, . To avoid confusion, chemists have decided to always omit solids and liquids from the equilibrium expression, and it is assumed that the constant concentration values for those reaction components are already combined into the equilibrium constant. Here are some examples.
Note the coefficient of 3 becomes an exponent.
Note the two solids are omitted and the coefficients become exponents.
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Note the solids are omitted and the coefficient becomes an exponent.
Note the liquid water is omitted and the denominator is omitted when it is 1.
Mathematics with Equilibrium Expressions
The mathematics involved with equilibrium expressions can range from the straightforward to the complex. For example, look at the sample question below.
Example:
Determine the value of when the equilibrium concentrations are: . Are the reactants or products favored? Explain your answer.
Solution:
Step 1: Write the equilibrium constant expression:
Step 2: Substitute in given values and solve:
The equilibrium constant value is the ratio of the concentrations of the products over the reactants. Therefore, a value of 4.9 for the equilibrium constant indicates that there are more products (numerator) than there are reactants (denominator). If the equilibrium constant is 1 or nearly 1, it indicates that the molarities of the reactants and products are about the same. If the equilibrium constant value is a large number, it indicates that the great majority of the material is in the form of products at equilibrium, and and we say “the products are strongly favored.” If the equilibrium constant is small, it indicates that the reactants are strongly favored.
Example:
At a given temperature, the reaction produces the following concentrations at equilibrium: . Find .
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Solution:
Step 1: Write the equilibrium expression:
Step 2: Substitute the given values and solve:
Example:
For the same reaction and the same temperature as the previous example, the concentrations of the substances at equilibrium become: . What is the concentration of ?
Solution:
Since it is the same reaction at the same temperature, the value will be the same, 1.34.
Step 1: Rearrange the equilibrium constant expression to solve for the unknown:
Step 2: Substitute and solve:
Equilibrium constant calculation with an unknown using I-C-E tables
Example:
H2 (g) + F2(g) ßà 2HF (g)
1.00 moles of hydrogen and 1.00 moles of fluorine are sealed in a 1.00 L flask at 150°C and allowed to react. At equilibrium, 1.32 moles of HF are present. Calculate the equilibrium constant.
Setup a table with I for Initial, C for change, and E for at equilibrium
Use stoichiometry from the moles of HF to find moles of Hydrogen and Fluorine.
H2 F2 2HF I 1.0 1.0 0.0 C E 1.32
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To calculate the equilibrium constant, plug into the equilibrium formula.
Solving equilibrium constant without an known concentrate at equilibrium
Example 2:
N2(g) + O2(g) 2 NO(g)
The equilibrium constant is 6.76.
If 6.0 moles of nitrogen and oxygen gases are placed in a 1.0 L container, what are the concentrations of all reactants and products at equilibrium?
Because we don’t know the amount at equilibrium, we will use a variable such as “x” for the products at equilibrium. Since there is a coefficient of 2, it will be 2x.
H2 F2 2HF I 1.0 1.0 0.0 C -0.66 -0.66 +1.32 E 0.34 0.34 1.32
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Now, plug into equilibrium formula,
Solve for x:
à à
Now finally, the concentration for NO would be 2 (3.4) = 6.8 mol/l, while for N2 and O2 would be 6.0 – 3.4 = 2.6mol/l.
Example in class: When hydrogen iodide (HI) is placed in a closed container it will decompose into its elements hydrogen and iodine and an equilibrium will be established:
2 HI (g) ó H2(g) + I2(g)
If the initial concentration of HI and the equilibrium constant for the system are known, then we can calculate the equilibrium concentration for HI, H2, and I2. These type of calculations are generally performed using ICE tables (or reaction tables).
Initial concentration of HI: 0.025 mol/L
Kc=2.2x10-3
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Review Questions
1. Why are solids and liquids not included in the equilibrium constant expression? 2. What does the value of mean in terms of the amount of reactants and products? 3. What is the correct equilibrium constant expression for the following
reaction: ?
a.
b.
c.
d. 4. What is the correct equilibrium constant expression for the following
reaction: ?
a.
b.
c.
d.
5. Consider the following equilibrium system: . At a certain temperature, the equilibrium concentrations are as
follows: , and . What is the equilibrium constant for this reaction?
a. b. c. d.
6. For the reaction , the equilibrium constant
was found to be 3.86 at a certain temperature. If is placed in
a container, what is the concentration of at equilibrium?
a.
b.
c. d. not enough information is available
7. Write the equilibrium constant expressions for each of the following equations: a.
b.
c.
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d.
Calculating Equilibrium Constants etc..
8. A mixture at equilibrium at 827°C contains 0.552 moles of CO2, 0.552 moles H2, 0.448 moles CO, and 0.448 moles of H2O in a 1.00 L container. What is the value of the equilibrium constant, Keq?
CO2(g) + H2(g) CO(g) + H2O(g)
9. The equilibrium constant for the reaction 4 H2(g) + CS2(g) CH4(g) + 2 H2S(g)
at 755°C is 0.256. What is the equilibrium concentration of H2S if at equilibrium [CH4] = 0.00108 mol/L, [H2] = 0.316 mol/L, [CS2] = 0.0898 mol/L?
10. Find the value of Keq for the equilibrium system ZnO(s) + CO(g) Zn(s) + CO2(g)
if at equilibrium there are 3.0 moles of CO, 4.0 moles of Zn and 4.0 moles of CO2 in a 500.0 mL container. USE ICE TABLES for the following
11. The decomposition of hydrogen iodide to hydrogen and iodine occurs by the reaction
2 HI(g) H2(g) + I2(g) Hydrogen iodide is placed in a container at 450°C an equilibrium mixture contains 0.50 moles of hydrogen iodide. The equilibrium constant is 0.020 for the reaction. How many moles of iodine and hydrogen iodide are present in the equilibrium mixture?
12. H2(g) + Cl2(g) 2 HCl(g) A student places 2.00 mol H2 and 2.00 mol Cl2 into a 0.500 L container and the reaction is allowed to go to equilibrium at 516°C. If KC is 76.0, what are the equilibrium concentrations of H2, Cl2 and HCl?
13. If Keq = 78.0 for the reaction A(s) + 2 B(g) 2 C(g) and initially there are 5.00 moles of A and 4.84
moles of B in a 2.00 L container, how many moles of B are left at equilibrium?
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Outcome to know: Use Le Chatelier’s Principle to predict shifts in equilibrium. (Include: temperature changes, pressure/volume changes, changes in reactant/product concentration, the addition of a catalyst, the addition of an inert gas, and the effects of the various stresses on equilibrium constant.) (2 hours)
Introduction
When a reaction has reached equilibrium with a given set of conditions, if the conditions are not changed, the reaction will remain at equilibrium forever. The forward and reverse reactions continue at the same equal and opposite rates, and the macroscopic properties remain constant.
It is possible, however, to alter the reaction conditions. For example, you could increase the concentration of one of the products, or decrease the concentration of one of the reactants, or change the temperature. When a change of this type is made in a reaction at equilibrium, the reaction is no longer in equilibrium. When you alter something in a reaction at equilibrium, chemists say that you put stress on the equilibrium. When this occurs, the reaction will no longer be in equilibrium, so the reaction itself will begin changing the concentrations of reactants and products 'until the reaction comes to a new position of equilibrium. How a reaction will change when a stress is applied can be explained and predicted and is the topic of this lesson.
Le Châtelier’s Principle
In the late 1800s, a chemist by the name of Henry-Louis Le Châtelier was studying stresses that were applied to chemical equilibria. He formulated a principle, Le Châtelier’s Principle, which states that when a stress is applied to a system at equilibrium, the equilibrium will shift in a direction to partially counteract the stress and once again reach equilibrium. For instance, if a stress is applied by increasing the concentration of a reactant, the equilibrium position will shift toward the right and remove that stress by using up some of the reactants. The reverse is also true. If a stress is applied by lowering a reactant concentration, the equilibrium position will shift toward the left, this time producing more reactants to partially counteracting that stress. The same reasoning can be applied when some of the products is increased or decreased.
Le Châtelier's principle does not provide an explanation of what happens on the molecular level to cause the equilibrium shift. Instead, it is simply a quick way to determine which way the reaction will run in response to a stress applied to the system at equilibrium.
Effect of Concentration Changes
Let's use Le Châtelier's principle to explain the effect of concentration changes on an equilibrium system. Consider the generic equation:
At equilibrium, the forward and reverse rates are equal. The concentrations of all reactants and products remain constant, which keeps the rates constant. Suppose we add some additional , thus raising the concentration of without changing anything else in the system. Since the concentration of is larger than it was before, the forward reaction rate will suddenly be higher. The forward rate will now exceed the reverse rate. Now there is a net movement of material from the reactants to the products. As the reaction uses up reactants, the forward rate that was too high slowly decreases while the reverse rate that was too low slowly increases. The two rates are moving toward each other and will eventually become equal again. They do not
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return to their previous rates, but they do become equal at some other value. As a result, the system returns to equilibrium.
Le Châtelier's principle says that when you apply a stress (adding ), the equilibrium system will shift to partially counteract the applied stress. In this case, the reaction shifts toward the products so that and are used up and and are produced. This reduction of the concentration of is counteracting the stress you applied (adding ).
Suppose instead that you removed some instead of adding some. In that case, the concentration of would decrease, and the forward rate would slow down. Once again, the two rates are no longer equal. At the instant you remove , the forward rate decreases, but the reverse rate remains exactly what it was. The reverse rate is now greater than the forward rate, and the equilibrium will shift toward the reactants. As the reaction runs backward, the concentrations of and decrease, slowing the reverse rate, and the concentrations of and increase, raising the forward rate. The rates are again moving toward each other, and the system will again reach equilibrium. The shift of material from products to reactants increases the concentration of , thus counteracting the stress you applied. Le Châtelier's principle again correctly predicts the equilibrium shift.
The effect of concentration on the equilibrium system according to Le Châtelier is as follows: increasing the concentration of a reactant causes the equilibrium to shift to the right, using up reactants and producing more products. Increasing the concentration of a product causes the equilibrium to shift to the left, using up products and producing more reactants. The exact opposite is true when either a reactant or product is removed.
Example:
For the reaction , what would be the effect on the equilibrium system if:
1. increases
2. increases
3. increases
Solution:
1. The equilibrium would shift to the right. would increase, more would be produced (but
that does not increase its concentration since its a solid), and would decrease.
2. The equilibrium would shift to the right. would decrease, more would be produced
(but again no change in concentration), and would increase.
3. The equilibrium would shift left. and ] would increase, and would be used up but not change its concentration.
Example:
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For the reaction , which way will the equilibrium shift if:
1. decreases
2. decreases
3. decreases
Solution:
1. left 2. left 3. right
Example:
Here's a reaction at equilibrium. Note the phases of each reactant and product.
1. Which way will the equilibrium shift if you add some to the system without changing anything else? 2. After has been added and a new equilibrium is reached, how will the new concentration
of compare to the original concentration of ? 3. After has been added and a new equilibrium has been established, how will the new concentration
of compare to the original concentration of ? 4. After has been added and a new equilibrium has been established, how will the new concentration
of compare to the original concentration of ? 5. Which way will the equilibrium shift if you add some to the system without changing anything else? 6. After has been added and a new equilibrium has been established, how will the new concentration
of compare its original concentration? 7. After has been added and a new equilibrium has been established, how will the new concentration
of compare its original concentration? 8. Which way will the equilibrium shift if you add some to the system without changing anything else?
Solution:
1. The equilibrium will shift toward the products. 2. The new concentration of will be higher than the original. 3. The new concentration of will be higher than the original, but lower than the concentration right
after was added. 4. Since is a solid, its concentration will be the same as the original. There will be less of it since some
was used in the equilibrium shift, but the concentration will be the same. 5. The equilibrium will shift toward the reactants. 6. The new concentration of will be lower than the original. 7. The new concentration of will be higher than the original. 8. Since is a solid, adding will not change its concentration and therefore has no effect on the
equilibrium. It is possible that adding some will increase the surface area of and therefore increase the forward reaction rate, but it will also increase the reverse reaction rate by approximately the same amount, hence no shift in equilibrium.
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The Haber Process and Concentration Change Effects
The reaction between nitrogen gas and hydrogen gas can produce ammonia, . However, under normal conditions, this reaction does not produce very much ammonia. Early in the 20th century, the commercial use of this reaction was too expensive because of the low yield.
A German chemist named Fritz Haber applied Le Châtelier’s principle to help solve this problem. Decreasing the concentration of ammonia by immediately removing it from the reaction container causes the equilibrium to shift to the right, so the reaction can continue to produce more product. Haber used Le Châtelier’s principle to solve this problem in other ways as well. These will be discussed in the following sections.
Summary
• Increasing the concentration of a reactant causes the equilibrium to shift to the right, producing more products.
• Increasing the concentration of a product causes the equilibrium to shift to the left, producing more reactants.
• Decreasing the concentration of a reactant causes the equilibrium to shift to the left, using up some products.
• Decreasing the concentration of a product causes the equilibrium to shift to the right, using up some reactants.
Questions
1. What is the effect on the equilibrium position if the [reactants] is increased? 2. What is the effect on the equilibrium position if the [reactants] is decreased?
3. Which of the following will cause a shift in the equilibrium position of the equation: ? i. add ii.
add iii. remove iv. remove a. i and ii only b. ii and iii only c. ii and iv only d. i, ii, iii, and iv
4. Which of the following will cause a shift in the equilibrium position of the
equation: ? i. add ii. add iii.
remove iv. remove a. ii only b. i and iii only c. ii and iv only d. i, ii, iii, and iv only
5. For the reaction: , what would be the effect on the equilibrium if:
a. is added?
b. is removed?
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c. is added?
d. is added?
6. Answer the following questions when is increased in the following system at equilibrium: . a. Write the equilibrium constant expression. b. Which direction will this equilibrium shift?
c. What effect will this stress have on ?
7. For the reaction , what would be the effect on the equilibrium system if:
a. increases? b. mass of decreases?
c. increases?
d. decreases?
Effect of a catalyst
We saw in the previous module that adding a catalyst to a system decreases the activation energy of a reaction. This will cause the rate of a reaction to increase. However, a catalyst lowers the activation energy of BOTH forward and reverse reactions equally.
Therefore, adding a catalyst to a system at equilibrium will NOT affect the equilibrium position. However, if a catalyst is added to a system which is not at equilibrium, the system will reach equilibrium much quicker since forward and reverse reaction rates are increased.
Effect of Changes of Pressure
A second type of stress studied by Le Châtelier was the effect of changing the pressure of a system at equilibrium. Reactants and products that are in solid, liquid, or aqueous states are not compressible, so raising or lowering the pressure on a system that contains no gases will have no effect on the equilibrium. If, however, the reaction at equilibrium contains at least one gas in either the reactants or products, altering the pressure will alter the equilibrium.
To get a better understanding of how pressure changes are affecting gaseous reaction components, it may help to consider the relationship between pressure and concentration for a gas at a given temperature. Recall that for an ideal gas, , where is a constant. Since temperature is also held constant in this case, and can be combined to form a new constant, which we’ll call . Dividing both sides by
volume now gives us . Remember that is in moles and is in liters; moles/liters = molarity. Thus, the pressure of a gas at a constant temperature is directly proportional to its concentration. When we are talking about the partial pressure of a gas, keep in mind that we are essentially talking about its concentration. This will help you to grasp some of the potentially confusing effects pressure can have on the equilibrium position.
Suppose we have the following reaction at equilibrium.
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Since the reaction has at least one gaseous substance involved in the reaction, its equilibrium will be affected by a change in the partial pressure of that gas. In order for this reaction to reach equilibrium, it would have to be in a closed reaction vessel so that the gaseous products do not escape. The gas must stay in contact with the solution for the reverse reaction to occur.
There are at least three ways to increase the pressure in the area above the liquid in the cylinder: 1) add some other gas not involved in the reaction, 2) add some gaseous into the space above the liquid, and 3) push the piston down so the space above the liquid in the cylinder is decreased (see figure below).
1. Increasing the pressure in the cylinder by adding a gas not involved in the reaction will increase the total gas pressure in the cylinder, but it will not affect the partial pressure of , so there will be no effect on the equilibrium of this reaction. This concept might be easier to understand if we think about it in terms of concentration rather than pressure. If we have the same amount of gas in the same amount of space, the concentration of the gas and thus its partial pressure will not change. The fact that some other gas was added does not change either of these key values. This is why only changes to the partial pressure of the reaction component, not the total pressure, are able to affect the equilibrium position.
2. Adding some gaseous to the cylinder is the same as adding a reactant or product, which has already been discussed earlier.
3. Lowering the piston in the cylinder pushes the molecules into a smaller space, increasing both the partial pressure of gaseous and its concentration in
the space above the liquid. Since the concentration of is increased, the reverse reaction rate will increase and the equilibrium will shift toward the reactants.
The reverse of this is also true. If you expand the volume of the cylinder by raising the piston, the partial pressure and concentration of will decrease. When the concentration of decreases, the reverse rate slows and the forward reaction rate will drive the equilibrium toward the products. This equilibrium shift increases the concentration and partial pressure of , once again counteracting the stress you applied.
Increasing Pressure Shifts Equilibrium Toward Fewer Moles of Gas
When you have a reaction at equilibrium with gaseous substances on both sides of the equation, the explanation for what happens is more complicated. Consider the following equation:
In this case, there are gaseous reactants and gaseous products. It should be easy to see that if we reduce the volume above this reaction to half of its previous volume, the partial pressures and the concentrations of these two gases will be doubled. Therefore, both reaction rates will increase because concentrations on each side have been increased. If we assume that we are still dealing with single collision reactions (so that the forward
and reverse rates can be expressed as and ), then we can see that doubling the concentration of B doubles the forward rate and doubling the concentration of D doubles the reverse rate. Both rates will increase, but since they increase by the same factor, the equilibrium will not shift.
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Now consider the reaction below:
This time, we see that there is one gas in the reactants and two gases in the products. If we once again reduce the volume of the gases by half, the partial pressures and the concentrations of the gases will double. Therefore, we double the forward reaction rate because we doubled the concentration of . In terms of the products, we are doubling the concentrations of and . The new reverse rate is:
Therefore, the new reverse rate will be four times the original reverse rate. If we double the forward reaction rate and quadruple the reverse rate, the equilibrium shift will be toward the reactants.
When you increase the pressure (by reducing volume) on a reaction at equilibrium, the equilibrium shift will be toward the side that has fewer moles of gas. A decrease in pressure due to a volume expansion will shift the equilibrium to the side with more moles of gas. Once again, you should note that Le Châtelier's principle predicts this result. If the equilibrium shift is converting moles of gas to of gas, then the shift is reducing the number of moles of gas and the total pressure will decrease. You applied a stress by increasing the pressure, and the equilibrium shift tends to counteract that stress by reducing pressure.
Example:
For the reaction , what would be the effect on the equilibrium position when the pressure is increased by reducing the volume?
Solution:
If the pressure increases, the reaction would shift to the side with the least number of moles of gas. Since there are of gaseous reactants and of gaseous products, the equilibrium would shift right, producing more products.
Example:
For the reaction , what would be the effect on the equilibrium system when (1.) the volume increases, and (2.) the volume decreases
Solution:
1. When the volume is increased, pressure decreases, so the equilibrium will shift toward the side with
more moles of gas. Therefore, will decrease, and and will increase. 2. When the volume is decreased, pressure increases, so the equilibrium will shift toward the side with
fewer moles of gas. Therefore, will increase, and and will decrease.
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The Haber Process and the Effect of Pressure Change
Earlier, we discussed the reaction that produces ammonia, :
We also stated that the German chemist Fritz Haber used Le Châtelier’s principle to develop a method that produces more product. Previously, we saw that when ammonia is immediately removed from the reaction container, thus decreasing the concentration, the equilibrium shifts to make up for the stress and produces more ammonia.
Now we can add another improvement: Since there are of gas molecules in the reactants and of gas molecules in the product, increasing the pressure (by decreasing the volume) will shift the equilibrium to the right, producing more ammonia. Later, we will discuss one more factor to complete our discussion of the Haber process.
Summary
• A decrease in volume will cause an increase in pressure, shifting the equilibrium to the side with fewer moles of gas.
• An increase in volume will cause a decrease in pressure, shifting the equilibrium to the side with more moles of gas.
Questions
1. What is the effect on the equilibrium position if the pressure is increased? 2. What is the effect on the equilibrium position if the pressure is decreased? 3. Use Le Châtelier’s Principle to predict what will happen to the following reaction at equilibrium if the
pressure is increased: . Mark all that apply. a. equilibrium position shifts right b. equilibrium position shifts left
c. will decrease
d. will increase 4. Use Le Châtelier’s principle to predict what will happen to the following reaction at equilibrium if the
pressure is decreased: . Mark all that apply. a. equilibrium position will not shift b. equilibrium position shifts left
c. will increase
d. will increase 5. Use Le Châtelier’s principle to predict what will happen to the following reaction at equilibrium if the
volume is decreased: . Mark all that apply. a. equilibrium position shifts right b. equilibrium position shifts left
c. will increase
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d. will decrease
6. For the reaction , what would be the effect on the equilibrium system if the pressure increases (or the volume decreases)?
7. For the reaction , what would be the effect on the equilibrium system if the pressure decreases (or the volume increases)?
8. For the reaction , what would be the effect on the equilibrium system if
a. the pressure increases? b. the volume decreases?
9. For the reaction , what would be the effect on the equilibrium system if
a. the pressure increases? b. the volume decreases?
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Effect of Changing Temperature
Raising the temperature will increase the average speed of the individual particles, thus causing more frequent collisions. Additionally, this increase in energy means that more particles will have the energy necessary to overcome the activation barrier. Overall, a rise in temperature increases both the frequency of collisions and the percentage of successful collisions.
It should be clear that increasing the temperature of the reaction vessel will increase both the forward and reverse reaction rates, but will it increase both rates equally? Let's examine the potential energy diagram of a reaction to see if we can gain any insight there. Here is the potential energy diagram for our usual theoretical reaction:
As you can see, the forward reaction has a small energy barrier while the reverse reaction has a very large energy barrier. With the reactants and products at the same temperature, the forward reaction will be much faster that the reverse reaction if the concentration of reactants is equal to the concentration of products.
For nearly every reaction, either the forward or reverse reaction will require more activation energy than the other. The addition of energy to a reaction will increase both reaction rates, but it increases the rate of the slower reaction more. In an exothermic reaction, the products are lower in energy than the reactants, so they have a higher barrier to climb in order to reach the same transition state. Accordingly, the reverse rate will be slower for an exothermic reaction, so increasing the temperature will shift the equilibrium to the left (reactants). The reverse is true for an endothermic reaction. By looking at a potential energy diagram, you should be able to tell 1) whether the reaction is exothermic or endothermic, 2) whether the forward or reverse reaction would be slower, assuming equal concentrations of reactants and products, and 3) which direction the equilibrium would shift in respond to a change in temperature.
Following the same reasoning as above, we can see that decreasing the temperature of a reaction produces an equilibrium shift in the opposite direction. Cooling an exothermic reaction results in a shift to the right, and cooling an endothermic reaction causes a shift to the left. Le Châtelier's principle correctly predicts the
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equilibrium shift when systems are heated or cooled. An increase in temperature is the same as adding energy to the system. Look at the following reaction:
This could also be written as:
When changing the temperature of a system at equilibrium, energy can be thought of as just another product
or reactant. For this reaction, of energy is produced for every mole of and 2 moles
of that react. Therefore, when the temperature of this system is raised, the effect will be the same as increasing any other product. As the temperature is increased, the equilibrium will shift away from the stress, resulting in more reactants and less products. As you would expect, the reverse would be true if the temperature is decreased.
A summary of the effect temperature has on equilibrium systems is shown in Table below:
The Effect of Temperature on an Endothermic and an Exothermic Equilibrium System
Exothermic Endothermic
Increase Temperature Shifts left, favors reactants Shifts right, favors products
Decrease Temperature Shifts right, favors products Shifts left, favors reactants
Example:
Predict the effect on the equilibrium position if the temperature is increased in each of the following:
1.
2.
Solution:
1. The reaction is endothermic. With an increase in temperature for an endothermic reaction, the reaction will shift right, producing more products.
2. The reaction is exothermic. With an increase in temperature for an exothermic reaction, the reaction will shift left, producing more reactants.
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1. For the following reaction
5 CO(g) + I2O5(s) ↔ I2(g) + 5 CO2(g) ΔHo = -‐1175 kJ
for each change listed, predict the equilibrium shift and the effect on the indicated quantity.
Change
Direction
of Shift
( → ; ← ; or no change)
Effect on
Quantity
Effect
(increase, decrease,
or no change)
(a) decrease in volume amount of I2O5
(b) raise temperature amount of CO(g)
(c) addition of I2O5(s) amount of CO(g)
(d) addition of CO2(g) amount of I2O5(s)
(e) removal of I2(g) amount of CO2(g)
_____________________________________________________________________________ 2. Consider the following equilibrium system in a closed container:
Ni(s) + 4 CO(g) ↔ Ni(CO)4(g) ΔHo = -‐ 161 kJ
In which direction will the equilibrium shift in response to each change, and what will be the effect on the indicated quantity?
Change
Direction
of Shift
( → ; ← ; or no change)
Effect on
Quantity
Effect
(increase, decrease,
or no change)
(a) add Ni(s) Ni(CO)4(g)
(b) raise temperature Kc
(c) add CO(g) amount of Ni(s)
(d) remove Ni(CO)4(g) CO(g)
(e) decrease in volume Ni(CO)4(g)
(f) lower temperature CO(g)
(g) remove CO(g) Kc
Answers: # 1(a) ←, no change; (b) ←, increase; (c) no change, no change; (d) ←, increase; (e) →, increase.
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Answers: # 2(a) no change, no change; (b) ←, decrease; (c) →, decrease; (d) →, decrease; (e) →, increase; (f) →, decrease; (g) ←, no change.
Introduction to Solubility Equilibrium
We will now switch our focus in dissolving salts and looking at the ions that are partially dissolved in solution. Since equilibrium constant expressions omit solids, the equilibrium constant expressions for the dissolution of a salt will have a denominator of 1, so no denominator will be shown. Here are some examples with their equilibrium constant expressions.
In this final section of the chapter, we will look at solubility equilibria.
Salts in an Equilibrium System
Consider the dissociation reaction for copper(II) hydroxide:
The equilibrium constant for this reaction is very low with a value of . A very small value
for means that the reactant (the solid salt) is heavily favored, so very few and ions
will actually be formed when is allowed to dissolve in water. In other words, the solution will become saturated very quickly.
This type of equilibrium will be established for any saturated solution that also contains some of the solid salt. When writing the equilibrium expression for a dissociation reaction, the equilibrium constant is often given the
subscript “sp,” which stands for solubility product. The solubility product constant works the same as any other equilibrium constant, but because calculations for this type of reaction are often quite similar, they are frequently differentiated in this way.
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Example:
Tooth enamel is composed of the compound hydroxyapatite, , which has a value
of . Write the dissociation equation and comment on the value of the equilibrium constant.
Solution:
Ca5(PO4)3OH(s) ßà Ca5(PO4)3+ (aq) + OH-
(aq)
The equilibrium constant is very small, so the reactants are heavily favored. This means that tooth enamel will not dissolve readily in water (definitely a good thing).
Example: Write the dissociation reaction and the solubility product constant expression for each of the following solids.
1.
2.
Solution:
1.
2.
Calculating Ksp From Solubility
The of a slightly soluble salt can be calculated from its solubility. Solubilities are usually given in grams/liter, or occasionally moles/liter. If the solubility is given in moles/liter, you can follow the same process as demonstrated below, except you would skip the first step.
Given the solubility of copper(I) bromide ( ), the molarity of a saturated solution can be determined. From that value, we can find the concentrations of the individual ions.
In this case, one ion and one ion is formed from each formula unit of , so
when of dissolves, the concentration of the ions in solution will
be and .
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The expression for is . Substituting the calculated values into
the expression will give:
.
The process is only slightly more complicated for salts that dissociate into more than two ions. Consider the
salt calcium phosphate Ca3(PO4)2. The solubility of calcium phosphate is . First, we
convert the solubility in to .
Then, using the dissolving equation, we determine the molarity of the ions in solution.
For every of calcium phosphate that dissolves, there will be 3 times as many calcium ions and 2 times as many phosphate ions in solution. Therefore,
We finish the calculation by writing the expression and substituting the molarities of the ions into the expression and calculate.
Calculating Solubility From Ksp
Not only can be calculated from the solubility, but the solubility can be calculated from the value.
Given the for ( ), we can use the dissociation equation and some algebra to calculate the concentrations of each ion in solution.
If we allow to represent the moles of that dissolves in one liter of
water, and , since only one of each ion is produced per formula unit of
. We can then write the expression for , set it equal to the given value, substitute the assigned variables into the equation, and solve.
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Because of how we defined , we can see that the molarity of will be the same so the solubility
of is .
This problem becomes slightly more difficult when the salt generates more than two ions. Let's try lead(II)
fluoride. Here is the dissolving equation, the expression, and the value for :
Once again, we let represent the solubility of . Therefore, the concentration of lead ions in solution will be and the concentration of fluoride ions in solution will be . Substitute these
variables into the expression and solve for .
Calculating Ion Concentrations from Ksp
The Ksp of magnesium hydroxide is 8.9 x 10-12. What will be the equilibrium concentrations of the dissolved ions in a saturated solution of Mg(OH)2?
Solution:
We are given the Ksp but no equilibrium concentrations or anything else. So we will need to set up a ICE table.
1st step: We need to find the molar ratios, so establish your balanced equation
Mg(OH)2 à Mg2+(aq) + 2OH-
(aq)
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2nd step: We will set Mg2+ as our x and 2OH- as 2x
3rd step: Use the equilibrium to solve for unknown. Make sure the coefficient for OH is your exponent Ksp = [Mg2+][OH-]2
Ksp = x (2x) 2
Ksp = 4x2
8.9 x 10-12 = 4x2 Isolate for x
x = 1.3 x 10-4
[Mg2+] = x = 1.3×10–4 mol/L [OH–] = 2x = 2.6×10–4 mol/L
Lesson Summary
• Equilibrium constants for slightly soluble salts are called solubility product constants.
Review Questions
1. What is the solubility product constant? Give an example. 2. Why is solubility considered a special case for chemical equilibria? 3. Nickel hydroxide is a slightly soluble salt. Its dissociation reaction is represented
as: . Which of the following represents the solubility
product constant expression, ?
a.
b.
c.
d.
4. The for is . What is at equilibrium?
a.
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b.
c. d. not enough information is given
5. The for is . What is at equilibrium?
a.
b.
c.
d. 6. Magnesium hydroxide is the acive component in milk of magnesia, a suspension used to cure
indigestion. It has an equilibrium constant of . Write the dissociation equation and comment on the value of the equilibrium constant.
7. Write the dissociation reactions for the following salts as well as the expressions. a. calcium fluoride b. chromium(II) carbonate c. arsenic(III) sulfide
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The Common Ion Effect
When an ionic compound dissolves in pure water, the initial concentration of each ion is zero. However, if an ionic compound dissolves in a solution that has an ion in common with the compound, this is not the case.
Even though the starting concentrations may not be zero, the product of the ions must still equal the solubility product constant.
For example, how would the solubility of silver chloride in pure water be compared to dissolving silver chloride in tap water?
Let's write out the dissociation equation:
AgCl(s) Ag+(aq) + Cl–(aq)
Tap water often has chlorine added to kill bacteria. The chlorine exists as chloride ions, so when we dissolve silver chloride in tap water, chloride ions are present. The silver chloride and the tap water have the chloride ions in common. According to Le Chatelier's Principle, adding more chloride ions to a saturated solution of silver chloride would cause the equilibrium position to shift to the left to use up the excess product. This would result in more solid formed, and a decreased solubility. This decreased solubility due to the presence of a common ion is called the common ion effect.
Le Chatelier's Principle predicts that the solubility of an ionic solid in a solution containing a common ion decreases its solubility. Let's see if this is supported by the calculations.
Example 1. Determine the solubility of silver chloride in pure water and in a solution of 0.10 mol/L sodium chloride. The Ksp of AgCl is 1.7 x 10-10.
Solution
Step 1: Solubility of AgCl in pure water.
In order to calculate the solubility of AgCl, we must need the concentrations for the ions. Since we don’t have the exact concentrations for the ions, x can be used.
AgCl(s) Ag+(aq) + Cl–(aq)
Equilibrium x x
Since [AgCl] = x, the solubility of AgCl in pure water is 1.3×10–5mol/L.
Step 2: Set up ICE table for the sodium chloride solution.
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There won’t be any concentration of Silver ions because there weren’t anything in the solution while there is 0.10 concentration of the Cl- from the AgCl.
AgCl(s) Ag+(aq) + Cl–(aq)
I 0 0.10
C + x + x
E x 0.10 + x
Step 3: Substitute values into the solubility product expression.
When we substitute the equilibrium values into the equilibrium law, we will get a quadratic.
Since Ksp is small, very little AgCl dissociates. We can assume that the value of x is insignificant compared to 0.10. That means 0.10 + x is approximately equal to 0.10.
The solubility of AgCl is 1.7×10–9 mol/L in 0.10 mol/L NaCl and 1.3×10–5 mol/L in pure water.
The solubility decreases, just as we (and Le Chatelier) predicted.
Example 2. The Ksp of lead (II) chloride, PbCl2, is 1.6×10–5. What is the solubility of lead (II) chloride in a 0.10 mol/L solution of magnesium chloride, MgCl2?
Solution
Step 1: Determine the concentration of the common ion, Cl–.
MgCl2(s) → Mg2+(aq) + 2 Cl–(aq)
From the dissociation of magnesium chloride,
[Cl–] = 2 x [MgCl2] = 0.20 mol/L
A note of caution. The concentration of the chloride ion is doubled because it's source is the MgCl2 NOT the PbCl2.
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Step 2: Set up an ICE table for PbCl2.
PbCl2(s) Pb2+(aq) + 2 Cl–(aq)
I 0 0.20
C + x + 2x
E x 0.20 + 2x
Step 3: Substitute values into the solubility product expression.
Since Ksp is small, we will assume the value of 2x is insignificant. Ksp and the concentration of the ions are related so if Ksp is small, the concentration is small.
Don't forget to square the [Cl–]!
The solubility of PbCl2 is 4.0×10–4 mol/L in 0.10 mol/L MgCl2.
Example 3. The Ksp of lead (II) chloride is 1.6×10–5. What is the solubility of lead (II) chloride in a 0.10 mol/L solution of lead (II) nitrate, Pb(NO3)2?
Solution
Step 1: Determine the concentration of the common ion, Pb2+
Pb(NO3)2(s) → Pb2+(aq) + 2 NO3–(aq)
From the dissociation of lead (II) nitrate,
[Pb2+] = [Pb(NO3)2] = 0.10 mol/L
Step 2: Set up an ICE table for PbCl2. Setup ICE table for the species with the Ksp.
PbCl2(s) Pb2+(aq) + 2 Cl–(aq)
I 0.10 0
C + x + 2x
E 0.10 + x 2x
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Step 3: Substitute values into the solubility product expression.
Since Ksp is small, we will assume that the value of x is insignificant compared to 0.10.
The solubility of PbCl2 is 6.3×10–3 mol/L in 0.10 mol/L Pb(NO3)2.
Answer the following questions. Be sure to show all your work.
1. Silver iodide, AgI, has a solubility product of 8.5×10–17. What is the solubility, in moles per Litre, of AgI in
a. pure water b. 0.010 mol/L HI c. 0.010 mol/L MgI2 d. 0.010 mol/L AgNO3
2. Magnesium fluoride, MgF2, has a solubility product of 8.0×10–8. Calculate the solubility, in mol/L, of
magnesium fluoride in
a. pure water b. 0.50 mol/L NaF c. 0.50 mol/L MgCl2
3. Gold (III) chloride, AuCl3, has a Ksp of 3.2 ×10–25. Calculate its solubility, in mol/L, in
a. pure water b. 0.20 mol/L HCl c. 0.20 mol/L MgCl2 d. 0.20 mol/L Au(NO3)3
