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Chapter 17a
Reaction Rates and Equilibrium
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Chapter 17
Table of Contents
2
Dark Energy and Dark Matter-Not in Textbook
17.1 How Chemical Reactions Occur
17.2 Conditions That Affect Reaction Rates
17.3 The Equilibrium Condition
17.4 Chemical Equilibrium: A Dynamic Condition
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3
Dark Energy
In a 1998 study led by Adam Riess, and a group of scientists from the Mount Stromlo Observatory, which is part of the Australian National University, Harvard University and Johns Hopkins University and Space Telescope Science Institute found, by observing supernovas in distant galaxies that the universe is expanding faster and faster. This violates Newton’s second law of motion that says acceleration is due a force. What force?????
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4
Dark Energy
Basically, dark energy is what they attribute to the accelerated expansion of the universe which means it isn’t dark as to light, but dark as to they know nothing about it. Attempts to explain or measure this energy have largely failed.
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5
Dark Matter
Dark matter was discovered by observing that the arms of spiral move at the same speed in violation of Kepler’s Law of planetary motion. It was explained by some unknown (dark) matter that distorted known gravitational effects. It was later found that the gravity of super massive black holes at the center of galaxies was not enough to hold the galaxies together. Gravity was not enough to hold the small local clusters of galaxies together as well as the super clusters of galaxies. Long filamental lines of matter, evident of some sort of attractive forces, has also been found between the super clusters of galaxies. Dark Matter has been used to explain all of these
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6
New Model of the Cosmos
The new model of the Cosmos puts about 4.6% of the universe being made up of atoms and molecules like what we think we know something about, about 23% is made up by dark matter and the rest (72%) is composed of dark energy.
They say the Universe was different 13.7 billion years ago?????
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Chapter 17
Table of Contents
7
Rates of Chemical Reactions
4 C3H5N3O9 6 N2 + 10 H2O + 12 CO2 + O2 + 5720kJNitroglycerine Exothermic
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Chapter 17
Table of Contents
8
An Example of Reaction Rates
Fast Reaction vs. Slow Reaction
35/97 people died in 1937
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Section 17.1
How Chemical Reactions Occur
Return to TOC
9
Collision Model
• Molecules must collide in order for a reaction to occur.
• Rate depends on concentrations of reactants and temperature.
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Conditions That Affect Reaction Rates
Section 17.2
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10
• Concentration – increases rate because more molecules lead to more collisions.
• Temperature – increases rate. Why?
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Conditions That Affect Reaction Rates
Section 17.2
Return to TOC
11
How to Tame Allergic Reactions
How can you slow down a histamine attack?
Histamine attacks are greater when you are hot. Cooling down affected areas can reduce allergy symptoms.
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Conditions That Affect Reaction Rates
Section 17.2
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12
Activation Energy
• Minimum energy required for a reaction to occur.
Wood must have Ea to light and burn!
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Conditions That Affect Reaction Rates
Section 17.2
Return to TOC
13
What makes Switzerland unique?
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Conditions That Affect Reaction Rates
Section 17.2
Return to TOC
14
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Conditions That Affect Reaction Rates
Section 17.2
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15
Chemical Reactions must go over an energy hill like a mountain (Swiss Alps).
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Conditions That Affect Reaction Rates
Section 17.2
Return to TOC
16
Catalyst
• A substance that speeds up a reaction without being consumed.
• Enzyme – catalyst in a biological system
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Conditions That Affect Reaction Rates
Section 17.2
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17
Catalyst
• A substance that speeds up a reaction without being consumed.
• Enzyme – catalyst in a biological system
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Conditions That Affect Reaction Rates
Section 17.2
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18
Catalyst
• A substance that speeds up a reaction without being consumed.
Chlorofluoro Carbons (CFC’s) are acting as catalysts to decompose the ozone (O3) layer. The ozone layer is formed from cosmic radiation and protects us from UV light.
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Conditions That Affect Reaction Rates
Section 17.2
Return to TOC
19
Depletion is measured by T.O.M.S.
“Total Ozone Mapping Spectrometer”
The below dark shaded are shows the amount of depletion around the Antarctica
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Conditions That Affect Reaction Rates
Section 17.2
Return to TOC
Copyright © Cengage Learning. All rights reserved 20
An Amana refrigerator, one of many appliances that now use HFC-134a. This compound is replacing CFC’s, which lead to the destruction of atmospheric ozone.
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Conditions That Affect Reaction Rates
Section 17.2
Return to TOC
21
Use a catalytic converter to convert the polluting exhaust gases of burned lead-free gasoline into harmless gases. Platinum (Pt) is the catalysts used. Only a small amount is needed.
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The Equilibrium Condition
Section 17.3
Return to TOC
Copyright © Cengage Learning. All rights reserved 22
Equilibrium
• The exact balancing of two processes, one of which is the opposite of the other.
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The Equilibrium Condition
Section 17.3
Return to TOC
Copyright © Cengage Learning. All rights reserved 23
Chemical Equilibrium
• A dynamic state where the concentrations of all reactants and products remain constant.
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Section 17.4
Chemical Equilibrium: A Dynamic Condition
Return to TOC
24
Chemical Equilibrium
• On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.
• Macroscopically static • Microscopically dynamic
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Section 17.4
Chemical Equilibrium: A Dynamic Condition
Return to TOC
Copyright © Cengage Learning. All rights reserved 25
The Reaction of H2O and CO to Form CO2 and H2 as Time Passes
Equal numbers of moles of H2O and CO are mixed in a closed container.
The reaction begins to occur, and some products (H2 and CO2) are formed.
The reaction continues as time passes and more reactants are changed to products.
Although time continues to pass, the numbers of reactant and product molecules are the same as in (c). No further changes are seen as time continues to pass. The system has reached equilibrium.
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Section 17.4
Chemical Equilibrium: A Dynamic Condition
Return to TOC
Copyright © Cengage Learning. All rights reserved 26
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Section 17.4
Chemical Equilibrium: A Dynamic Condition
Return to TOC
Copyright © Cengage Learning. All rights reserved 27
Chemical Equilibrium
• Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction.
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Section 17.4
Chemical Equilibrium: A Dynamic Condition
Return to TOC
Copyright © Cengage Learning. All rights reserved 28
Concept Check
Consider an equilibrium mixture in a closed vessel reacting according to the equation:
H2O(g) + CO(g) H2(g) + CO2(g)
You add more H2O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.
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Section 17.4
Chemical Equilibrium: A Dynamic Condition
Return to TOC
Copyright © Cengage Learning. All rights reserved 29
Concept Check
Consider an equilibrium mixture in a closed vessel reacting according to the equation:
H2O(g) + CO(g) H2(g) + CO2(g)
You add more H2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.
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Section 17.4
Chemical Equilibrium: A Dynamic Condition
Return to TOC
30
Reactions Rates and Equilibrium
Chapter 17b
W
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Section 17.4
Chemical Equilibrium: A Dynamic Condition
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31
17.5 The Equilibrium Constant: An Introduction
17.6 Heterogeneous Equilibria
17.7 Le Châtelier’s Principle
17.8 Applications Involving the Equilibrium Constant
17.9 Solubility Equilibria
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Section 17.5
The Equilibrium Constant: An Introduction
Return to TOC
Copyright © Cengage Learning. All rights reserved 32
Consider the following reaction at equilibrium:
jA + kB lC + mD
• A, B, C, and D = chemical species.• Square brackets = concentrations of species at equilibrium.• j, k, l, and m = coefficients in the balanced equation.• K = equilibrium constant (given without units).
j
l
k
m
[B][A]
[D] [C]K =
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Section 17.5
The Equilibrium Constant: An Introduction
Return to TOC
Copyright © Cengage Learning. All rights reserved 33
Example
N2(g) + 3H2(g) 2NH3(g)
2
33
2 2
NH =
N HK
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Section 17.5
The Equilibrium Constant: An Introduction
Return to TOC
Copyright © Cengage Learning. All rights reserved 34
• K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially.
• For a reaction, at a given temperature, there are many equilibrium positions but only one value for K. Equilibrium position is a set of equilibrium
concentrations.
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Section 17.5
The Equilibrium Constant: An Introduction
Return to TOC
Copyright © Cengage Learning. All rights reserved 35
Concept Check
Consider the following equilibrium reaction:HC2H3O2(aq) H+(aq) + C2H3O2
–
(aq)Determine the equilibrium constant expression for the dissociation of acetic acid.
a) b)
c) d)
]OHHC[
]OHC][H[K
232
232
]OHHC[
]OHC[][HK
232
232
]OHC][H[
]OH[HC K
232
232
]H[
]OH[HC K 232
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Section 17.5
The Equilibrium Constant: An Introduction
Return to TOC
Copyright © Cengage Learning. All rights reserved 36
Exercise
For the reaction below, calculate the value of the equilibrium constant, given the equilibrium concentrations.N2O4(g) 2NO2(g)[N2O4] = 0. 055 M [NO2] = 0.060 M
a) K = 0.050b) K = 0.92c) K = 1.1d) K = 0.065
K = (0.060)2/0.055 = 0.065
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Section 17.5
The Equilibrium Constant: An Introduction
Return to TOC
37
What is the equilibrium expression for the following?
CH4(g) + 2 H2S(g) <==> CS2(g) + 4 H2(g)
Keq = ----------------[CS2][H2]4
[CH4][H2S]2
H2(g) + I2(g) <==> 2 HI(g)
Keq = ----------[HI]2
[H2][I2]
Fe3+(aq) + SCN-(aq) <==> Fe(SCN)+2(aq)
Keq = ------------------[Fe(SCN)+2]
[Fe3+][SCN-]
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Section 17.6
Heterogeneous Equilibria
Return to TOC
38
Homogeneous Equilibria
• Homogeneous equilibria – involve the same phase:
N2(g) + 3H2(g) 2NH3(g)
HCN(aq) H+(aq) + CN-(aq)
C2H5OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l)
2SO2(g) + O2(g) 2SO3(g)
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Section 17.6
Heterogeneous Equilibria
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39
Heterogeneous Equilibria• Heterogeneous equilibria – involve more than
one phase:
2KClO3(s) 2KCl(s) + 3O2(g)
2H2O(l) 2H2(g) + O2(g)
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Section 17.6
Heterogeneous Equilibria
Return to TOC
40
• The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. The concentrations of pure liquids and solids
are constant.
2KClO3(s) 2KCl(s) + 3O2(g)
H2O(l) H+(aq) + OH-(aq) K=[H+][OH-]
3
2 = OK
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Section 17.6
Heterogeneous Equilibria
Return to TOC
Copyright © Cengage Learning. All rights reserved 41
Concept Check
Determine the equilibrium expression for the reaction:
CaF2(s) Ca2+(aq) + 2F–(aq)
a) b)
c) d)
][CaF
]][F[CaK
2
-2
]CaF[
]F[][CaK
2
-2
2 -
2
[Ca ][2F ]K
[CaF ]
2-2 ]F][[CaK
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Section 17.7
Le Châtelier’s Principle
Return to TOC
42
• If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.
O3 (g) + Cl (g) O2 (g) + OCl(g)
Equilibrium shifts to counter a disturbance.Hills and Valleys!
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Section 17.7
Le Châtelier’s Principle
Return to TOC
Copyright © Cengage Learning. All rights reserved 43
Effect of a Change in Concentration
• When a reactant or product is added the system shifts away from that added component.
• If a reactant or product is removed, the system shifts toward the removed component.
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Section 17.7
Le Châtelier’s Principle
Return to TOC
44
Effect of a Change in Volume
The system is initially at equilibrium.
The piston is pushed in, decreasing the volume and increasing the pressure. The system shifts in the direction that consumes CO2 molecules, lowering the pressure again.
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Section 17.7
Le Châtelier’s Principle
Return to TOC
Copyright © Cengage Learning. All rights reserved 45
Effect of a Change in Volume
• Decreasing the volume
The system shifts in the direction that gives the fewest number of gas molecules.
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Section 17.7
Le Châtelier’s Principle
Return to TOC
46
Effect of a Change in Volume
• Increasing the volume The system shifts in the direction that increases its
pressure or the greatest number of gas molecules.
https://www.youtube.com/watch?v=pnU7ogsgUW8
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Section 17.7
Le Châtelier’s Principle
Return to TOC
Copyright © Cengage Learning. All rights reserved 47
Effect of a Change in Temperature
• The value of K changes with temperature. We can use this to predict the direction of this change.
• Exothermic reaction – produces heat (heat is a product) Adding energy shifts the equilibrium to the left (away
from the heat term).• Endothermic reaction – absorbs energy (heat is a
reactant) Adding energy shifts the equilibrium to the right (away
from the heat term).
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Section 17.7
Le Châtelier’s Principle
Return to TOC
48
Effect of Temperature on Equilibrium
2NO2(g) N2O4(g) + Heat
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Section 17.7
Le Châtelier’s Principle
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Industrial Application-The Manufacture of Ammonia
N2(g) + 3H2(g) 2NH3(g) ΔH = -92.4 kJ mol-1
To increase production how would you manipulate the equilibrium?
2. Lower Temperature
1. Lower Volume
3. Remove Product
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Section 17.7
Le Châtelier’s Principle
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Copyright © Cengage Learning. All rights reserved 50
Concept Check
Consider the reaction: 2CO2(g) 2CO(g) + O2(g)
How many of the following changes would lead to a shift in the equilibrium position towards the reactant?
I. The removal of CO gas.
II. The addition of O2 gas.
III. The removal of CO2 gas.
IV. Increasing the pressure in the reaction by decreasing the volume of the container.
a) 1
b) 2
c) 3
d) 4
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Section 17.7
Le Châtelier’s Principle
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Copyright © Cengage Learning. All rights reserved 51
Concept Check
One method for the production of hydrogen gas can be described by the following endothermic reaction:
CH4(g) + H2O(g) CO(g) + 3H2(g)
How many of the following changes would decrease the amount of hydrogen gas (H2) produced?
I. H2O(g) is added to the reaction vessel.II. The volume of the container is doubled.III. CH4(g) is removed from the reaction vessel.IV. The temperature is increased in the reaction
vessel.
a) 1
b) 2
c) 3
d) 4
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Section 17.8
Applications Involving the Equilibrium Constant
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52
• A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right. Reaction goes essentially to completion.
The Extent of a Reaction
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Section 17.8
Applications Involving the Equilibrium Constant
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53
• A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left. Reaction does not occur to any
significant extent.
The Extent of a Reaction
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Section 17.8
Applications Involving the Equilibrium Constant
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54
• The value of K for a system can be calculated from a known set of equilibrium concentrations.
• Unknown equilibrium concentrations can be calculated if the value of K and the remaining equilibrium concentrations are known.
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Section 17.8
Applications Involving the Equilibrium Constant
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Copyright © Cengage Learning. All rights reserved 55
Concept Check
If the equilibrium lies to the right, the value for K is __________.
large (or >1)
If the equilibrium lies to the left, the value for K is ___________.
small (or <1)
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Section 17.8
Applications Involving the Equilibrium Constant
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56
Concept Check
At a given temperature, K = 50. for the reaction:H2(g) + I2(g) 2HI(g)
Calculate the equilibrium concentration of H2 given:[I2] = 1.5 × 10–2 M and [HI] = 5.0 × 10–1 M
a) 1.5 × 10–2 Mb) 3.0 × 10–2 Mc) 5.0 × 10–1 Md) 3.3 × 10–1 M
K = (HI)2/(H2)(I2)
50 = (5.0 × 10–1)2/(H2)(1.5 × 10–2)
(H2) = 3.3 × 10–1 M
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Section 17.9
Solubility Equilibria
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57
• The equilibrium conditions also applies to a saturated solution containing excess solid, MX(s). Ksp = [M+][X] = solubility product constant
The value of the Ksp can be calculated from
the measured solubility of MX(s).
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Section 17.9
Solubility Equilibria
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Copyright © Cengage Learning. All rights reserved 58
Solubility Equilibria
• Solubility product (Ksp) – equilibrium constant; has only one value for a given solid at a given temperature.
• Solubility – an equilibrium position.
Bi2S3(s) 2Bi3+(aq) + 3S2–(aq)
2 33+ 2sp = Bi S K
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Section 17.9
Solubility Equilibria
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59
Concept Check
If a saturated solution of PbCl2 is prepared by dissolving some of the salt in distilled water and the concentration of Pb2+ is determined to be 1.6 × 10–2 M, what is the value of Ksp?
a) 2.6 × 10–4
b) 2.0 × 10–4 PbCl2 Pb2+ + 2 Cl-
c) 3.2 × 10–2 x x 2xd) 1.6 × 10–5
Ksp = [Pb2+] [Cl–]2 = (x)(2x)2=(1.6 × 10–2) (3.2 × 10–2)2 = 1.6 × 10–5
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Section 17.9
Solubility Equilibria
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60
Concept Check
Calculate the solubility of silver chloride in water.
Ksp = 1.6 × 10–10
a) 1.3 × 10–5 M
b) 1.6 × 10–10 M AgCl(s) Ag+ + Cl-
c) 3.2 × 10–10 M x x x
d) 8.0 × 10–11 MKsp = [Ag+][Cl–]
1.6 × 10–10 = (x)(x) = x2
x = 1.3 × 10–5 M