15.0 equilibrium
TRANSCRIPT
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Unit D: Equilibrium
Textbook Reference: Chapters 15 and 16
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+ Equilibrium1.1K: Define equilibrium and state the criteria that apply to a chemical system in equilibrium; i.e.,
closed system, constancy of properties, equal rates of forward and reverse reactions
1.2K: Identify, write and interpret chemical equations for systems at equilibrium
1.4K: Define Kc to predict the extent of the reaction and write equilibrium-law expressions for
given chemical equations, using lowest whole-number coefficients
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+Chemical Equilibrium
Not all reactions are quantitative (reactants products) Evidence: For many reactions reactants are present even after the
reaction appears to have stopped
Recall the conditions necessary for a chemical reaction:
Particles must collide with the correct orientation and have sufficient
energy
If product particles can collide effectively also, a reaction is said to
be reversible
Rate of reaction depends on temperature, surface area and
concentration
100%
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+Chemical Equilibrium Consider the following reversible reaction:
The final state of this chemical system can be explained as acompetition between:
We assume this system is closed (so the reactants and productscannot escape) and will eventually reach a:
DYNAMIC EQUILBRIUM
- Opposing changes occur simultaneously at the same rate
The collisions of
reactants to form
products
The collisions of
products to re-form
reactants
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+Modelling Dynamic Equilibrium
Mini Investigation pg. 678
Volume of
Cylinder #1
Volume of
Cylinder #2
25.0 0.0
20.0 5.0
17.0 8.0
14.0 11.0
11.0 14.0
8.0 17.0
5.0 20.0
2.0 23.02.0 23.0
2.0 23.0
2.0 23.0
Assume large strawtransfers 5 mL each time
and the smaller straw
transfers 2 mL each time
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+Chemical Equilibrium
Consider the following hypothetical system:
AB + CD AD + BC forward reaction, thereforeAD + BCAB + CD reverse reaction
Initially, only AB and CD are present. The forward reaction is occurring exclusively at
its highest rate.
As AB and CD react, their concentration decreases. This causes the reaction rate to
decrease as well.
As AD and BC form, the reverse reaction begins to occur slowly.
As AD and BCs concentration increases, the reverse reaction speeds up.
Eventually, both the forward and reverse reaction occur at the same rate =
DYNAMIC EQUILIBRIUM
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+4 Conditions of Dynamic Equilibrium*
1. Can be achieved in all reversible reactions when the rates ofthe forward and reverse reaction become equal
Represented by rather than by
2. All observable properties appear constant (colour, pH, etc)
3. Can only be achieved in a closed system (no exchange of
matter and must have a constant temperature)
4. Equilibrium can be approached from either direction. This
means that the equilibrium concentrations will be the same
regardless if you started with all reactants, all products, or a
mixture of the two
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Types of Equilibrium
1. Phase Equilibrium: a single substance existing in
more than 1 phase
Example: Liquid water in a sealed container with
water vapour in the space above it
Water evaporates until the concentration of water
vapour rises to a maximum and then remains
constant
2. Solubility Equilibrium: a saturated solution
Rate of dissolving = rate of recrystallization
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Types of Equilibrium
3. Chemical Equilibrium reactants and products in a closed system
Example: The Hydrogen-Iodine Equilibrium System
The rate of reaction of the reactants decreases as the number of reactant molecules
decrease. The rate at which the product turns back to reactants increases as the number of
product molecules increases. These two rates become equal at some point, after which thequantity of each will not change.
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+Describing the Position of Equilibrium
1. Percent Yield- the yield of product measured at equilibrium
compared with the maximum possible yield of product.
% yield = product eqm x 100 %
product max
The equilibrium concentration is determined experimentally, the
maximum concentration is determined with stoichiometry
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1. Percent Yield Example
If 2.50 mol of hydrogen gas reacts with 3.0 mol of iodine gas in
a 1.00L vessel, what is the percent yield if 3.90 mol of hydrogen
iodide is present at equilibrium
% yield = product eqm x 100 %
product max
H2(g) + I2(g) 2HI(g)
2.50 mol x (2 mol HI) = 5.0 mol HI1 mol H2(g)
Describing the Position of Equilibrium
% yield = 3.90 mol x 100 %
5.00 mol
= 78%
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2. Using an Equilibrium Constant, (Kc)
This relationship only works if all concentrations are at equilibrium at
a constant temperature in a closed system
Think products over reactants
If the Kc > 1, the equilibrium favours products
If the Kc < 1, the equilibrium favours reactants
Describing the Position of Equilibrium
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2. Using an Equilibrium Constant, (Kc)
Example #1: Write the equilibrium law expression for the
reaction of nitrogen monoxide gas with oxygen gas to form
nitrogen dioxide gas.
Describing the Position of Equilibrium
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2. Using an Equilibrium Constant, (Kc)
Note: The Kc value describes the extent of the forward reaction.
Kc reverse = 1 . = The reciprocal value
Kc forward
Example #2: The value of Kc for the formation of HI(g) from H2(g)
and I2(g) is 40, at a given temperature. What is the value of Kc for
the decomposition of HI(g) at the same temperature.
Kc reverse = 1 . = 1 = 0.025
Kc forward 40
Describing the Position of Equilibrium
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2. Using an Equilibrium Constant, (Kc)
Note: For heterogeneous equilibrium systems, DO NOTinclude liquids and solids in the expression. (They are
assumed to have fixed concentrations)
Example #3: Write the equilibrium law expression for the
decomposition of solid ammonium chloride to gaseous ammoniaand gaseous hydrogen chloride
Example #4: Write the equilibrium law expression for the reactionof zinc in copper(II) chloride solution.
Describing the Position of Equilibrium
The solid is
omitted from the
expression
The solids, as well as
spectator ions (Cl-)
are omitted from the
expression
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+Describing the Position of Equilibrium
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+PRACTICE
HW Book pg. 1
Read the notes, and complete Exercises #1-10 at the bottom
HW Book pg. 2
Complete Qs # 1 and 3
Lab Exercise 15.B pg. 686 - complete Analysis
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+ Equilibrium Concentrations2.3K: Calculate equilibrium constants and concentrations for homogeneous when
concentrations at equilibrium are known
initial concentrations and one equilibrium concentration are known
the equilibrium constant and one equilibrium concentration are known.
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+Calculations in Equilibrium Systems
Using the equilibrium law expression to determine whether asystem is at equilibrium:
Substitute in the given concentrations to the equilibrium expression.
If the value is the equilibrium constant, the system is at equilibrium
If the value is larger, this means there are more products that
reactants. To reach equilibrium, the reaction must proceed to the
left (towards the reactants)
If the value is smaller, this means there are more reactants than
products. To reach equilibrium, the reaction must proceed to the
right (towards the products)
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+Calculations in Equilibrium Systems
Example #1: In the following system:
N2(g) + 3H2(g) 2NH3(g)
0.249 mol N2(g), 3.21 X 10-2 molH2(g) and 6.42 X 10
-4 mol NH3(g)
are combined in a 1.00 L vessel at 375oC, Kc = 1.2
Is the system at equilibrium?
Kc = NH32 = (6.42 x 10 -4)2 =
0.0500
N2(g) H2(g)3
(0.249)(3.21 x 10-2
)3
If not, predict the direction in which the reaction must proceed.
Value is below Keq = therefore more reactants than products, so
reaction must proceed to the right
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+Calculations in Equilibrium Systems
Example #2: Find the equilibrium concentration of the ions that
are formed when solid silver chloride is dissolved in water. The
equilibrium constant for this reaction is Kc = 5.4
X 10-4.
AgCl(s) Ag+
(aq) + Cl-(aq)
Kc = Ag+
(aq) Cl-(aq)
5.4 x 10 -4 = x2
0.023 mol/L = x = Ag+(aq)eq = Cl-(aq)
eq
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+ICE Charts and Equilibrium Calculations
STEPS:
Always write out the equilibrium reaction and equilibrium law
expression if not given.
Draw an ICE Chart (Initial, Change in and Equilibrium
concentrations) (I + C = E)
Substitute values where appropriate
Solve for x
Solve for equilibrium concentrations
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+ICE Charts and Equilibrium Calculations
Example #1: Consider the following equilibrium at 100 oC:
N2O4(g) 2 NO2(g)
2.0 mol of N2O4(g) was introduced into an empty 2.0 L bulb. After
equilibrium was established, only 1.6 mol of N2O4(g) remained.
What is the value of Kc?
E: 1.0 x = 0.80 solve for x x = 0.20 2x = 0.40
Solve for Kc = (0.40)2 = 0.20
(0.80)
N2O4(g) 2NO2(g)
I: 1.0 mol/L 0
C: - x + 2x
E: 1.0 x = 0.80 2x
2.0 mol = 1.0 mol/L (I)
2.0L
1.6 mol = 0.8 mol/L (E)
2.0L
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+ ICE Charts and Equilibrium CalculationsExample #2
A 10 L bulb is filled with 4.0 mol of SO2(g), 2.2 mol of O2(g) and 5.6 mol of
SO3(g). The gases then reach equilibrium according to the followingequation:
2SO2(g) + O2(g) 2SO3(g)
At equilibrium, the bulb was found to contain 2.6 mol of SO2(g). Calculate
Kc for this reaction.
Kc = (0.70)2 = 48
(0.15)(0.26)2
2SO2(g) O2(g) 2SO3(g)
I: 0.40 0.22 0.56
C: - 2x - x + 2xE: 0.40 2x = 0.26 0.22 - x 0.56 + 2x
E: 0.26 0.15 0.70
4.0 mol = 0.40 mol/L (I)10.0L
2.6 mol = 0.26 mol/L (E)
10.0L
2.2 mol = 0.22 mol/L (I)10.0L 5.6 mol = 0.56 mol/L (I)10.0L
0.40
2x = 0.26X = 0.07
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+PRACTICE
HW Book pg. 3
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+ ICE Charts and Equilibrium Calculations Example #3: Using a perfect square
Given the following reaction:
N2(g) + O2(g) 2NO(g) Kc = 0.00250
Determine the equilibrium concentrations for all species present given that the
initial concentration of each reactant is 0.200 mol/L.
0.00250 = (2x)2 square root both sides 0.005 = 2x = 0.01 0.05x = 2x
(0.200-x)2 0.200 x = 0.01 = 2.05x = 0.00488
N2(g) O2(g) 2NO(g)
I: 0.200 0.200 0
C: - x - x + 2x
E: 0.200 - x 0.200 - x 2x
E: 0.195mo l/L 0.195mo l/L 0.00976mol /L
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+ ICE Charts and Equilibrium Calculations Example #4: Using the Approximation Rule
Calculate the concentration of gases produced when 0.100 mol/L COCl2(g)decomposes into carbon monoxide and chlorine gas. The Kc for this reaction
is 2.2 X 10-10.
2.2 x10-10 = (x)2 Approximation Rule: When the reactants/Kc is larger than 1000, you can
(0.100-x) disregard the change in concentration. So 0.100 x = 0.100
2.2 x 10-10 = x2 = x = 4.69 x 10-6 mol/L
0.100
COCl2(g) CO(g) Cl2(g)
I: 0.100 0 0
C: - x + x + x
E: 0.100 - x x x
E: 0.100 mol /L 4.7 x 10-6mol/L 4.7 x 10-6mol/L
0.100 = >1000
2.2x 10 -10
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+PRACTICE
HW Book pg. 4
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+ Le Chateliers Principle1.3K: Predict, qualitatively, using Le Chateliers principle, shifts in equilibrium caused by
changes in temperature, pressure, volume, concentration or the addition of a catalyst and
describe how these changes affect the equilibrium constant
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+Le Chateliers Principle
Le Chateliers Principle is very useful in predicting how a
system at equilibrium will respond to change.
It states that when a system at equilibrium is disturbed, the
equilibrium shifts in the direction that opposes the change, until
a new equilibrium is reached.
There are three common ways an equilibrium may be
disturbed:
Change in the concentration of one of the reactants or products
Change in the temperature Change in the volume of a container
Addition of a catalyst (less common)
Changes in the temperature, will
change the Kc value. No other
changes will affect this value.
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+Le Chateliers Principle
Effect of Changes in Concentration
If a system at equilibrium is disturbed by the addition of a reactant (or the
removal of a product), Le Chateliers principle predicts that the equilibrium will
shift right.
2N2O(g) + 3O2(g) 4 NO2(g)
If the disturbance is the removal of a reactant (or the addition of a product), Le
Chateliers principle predicts that the equilibrium will shift left.
2N2O(g) + 3O2(g) 4 NO2(g)
Since concentrations of solids are constants and do not appearin expressions
for K, removing or adding some solid does not cause shifts.
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+Concentration Change
Addition of reactant, HF(g)will shift equilibrium to the right
More products will be produced,
and a new equilibrium is
established
Removal of product, HCl(g), willshift the equilibrium to the right
More products will be produced,
and a new equilibrium is
established
If you see spike in either a reactant or
product, which causes a gradual change
in the other entities a substance has
been added or removed.
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+ Le Chateliers Principle
Effect of Changes in Temperature
In endothermic reaction, heat acts like a reactant. Increasing thetemperature shifts the reaction right. Decreasing the temperature, shifts the
reaction left
Heat + 2N2O(g) + 3O2(g) 4 NO2(g)
In exothermic reactions, heat acts like a product. Increasing the temperatureshifts the reaction left. Decreasing the temperature, shifts the reaction right.
4 NO2(g) 2N2O(g) + 3O2(g) + Heat
The equilibrium constant, Kc is temperature dependent
Reaction Type Role of heat Effect of T Effect of T Endothermic Reactants + heat products K K
Exothermic Reactants products + heat K K
Remember K gets bigger
if there are more
products being created(i.e. shifts to the right)
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+Temperature Change
The temperature decreases, at the time indicated by the dotted
line.
This will cause the equilibrium to shift to the right, creating
more products, until a new equilibrium is established.
+ energy
If you see a gradual change in the
reactants with an opposite
change in the products, you have
a temperature change going on
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+ Le Chateliers Principle
Effect of Changes in the Volume of the Container
If volume is decreased, pressure increases (Boyles Law Chem 20)
So the reaction will shift in the direction which contains the fewest moles of g as
2N2O(g) + 3O2(g) 4 NO2(g)
If volume is increased, pressure decreases (Boyles Law Chem 20)
So the reaction will shift in the direction which contains the most moles of gas
2N2O(g) + 3O2(g) 4 NO2(g)
If both sides of the equation have the same number of moles of gas, the
change in volume of the container has no effecton the equilibrium.
Pressure 4 moles are
fewer than 5
Pressure5 moles are
more than 5
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+Volume Change
The volume of the container decreased, at the time indicated
by the dotted line. This will cause a pressure increase.
This will cause the equilibrium to shift to the right, the side with
fewer moles of gas, creating more products, until a new
equilibrium is established.
If you see a spike in all of the
entities = P increase, V decrease
If you see a drop in all of the
entities = P decrease, V increase
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+ Le Chateliers Principle
Effect of the Addition of a Catalyst
Catalysts speed up the rate at which equilibrium is obtained, but have no
effect on the magnitude of K. They increase both the forward and backward
rate of reaction.
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+
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+ Identify the nature of the changes imposed on the followingequilibrium system at the four times indicated by coordinates A,
B, C and D
At A, the concentration (or pressure) of every chemical in the system is decreased by
increasing the container volume. Then the equilibrium shifts to the left (the side withmore moles of gas)
At B, the temperature is increased. Then the equilibrium shifts to left.
At C, C2H6(g) is added to the system. Then the equilibrium shifts to the left.
At D, no shift in equilibrium position is apparent; the change imposed must be addition
of a catalyst, or of a substance that is not involved in the equilibrium reaction.
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+Practice
Change Direction of Shift (, ,
or no change)
Effect on quantity Effect (increase, decrease,
or no change)
a) Decrease in volume Kc
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)
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+Practice
Homework Book pg. 5 and 6
Le Chatelier Puzzle
Le Chatelier Lab
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+Acids and Bases - Kw2.1K: Recall the concepts of pH and hydronium ion concentration and pOH and hydroxide ion
concentration, in relation to acids and bases
2.2K: Define Kw to determine pH, pOH, [H3O+] and [OH] of acidic andbasic solutions
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+ The Water Ionization Constant, Kw
Even highly purified water has a very slight conductivity. This is
due to the ionization of some water molecules into hydronium ionsand hydroxide ions.
The water ionization equilibrium relationship is so important, it gets its own special
symbol and name: Ionization Constant for Water, Kw
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+The Water Ionization Constant, Kw
Since the mathematical relationship is simple, we can easilyuse Kw to calculate either the hydronium or hydroxide ion
concentration, if the other concentration is know.
The presence of
substances other than
water decreases the
certainty of the Kw value
to two significant digits;
1.0 x 10 -14
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+Kw Calculations
Example: A 0.15 mol/L solution of hydrochloric acid at 25C is
found to have a hydronium ion concentration of 0.15 mol/L.
Calculate the amount concentration of the hydroxide ions.
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+Kw Calculations
Example #2: Calculate the amount concentration of the
hydronium ion in a 0.25 mol/L solution of barium hydroxide.
Ba2+
= 0.25 mol/L x 2 mol = 0.50 mol/L1 mol
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+Kw Calculations
Example #3: Determine the hydronium ion and hydroxide ion
amount concentration in 500 mL of an aqueous solution for
home soap-making containing 2.6 g of dissolved sodium
hydroxide.
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+Do You Remember? pH and pOH
+
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+Acid Strength as an Equilibrium Position
Do you remember the difference between strong and weak acids?
Strong acids ionize completely (quantitatively), even though this
could be written with a double arrow, it is simpler to use a single arrow
to show the the reaction is >99.9%
Strong Acids: HClO4(aq), HI(aq), HBr(aq) , HCl(aq) , HSO4(aq), HNO3(aq)
Weak acids ionize (react with water) only partially (
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+% Ionization
In a 0.10 mol/L solution of acetic acid, only 1.3% of the
CH3COOH molecules have reacted at equilibrium to formhydronium ions. Calculate the hydronium ion amount
concentration.
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+% Ionization
The pH of 0.10 mol/L methanoic acid solution is 2.38. Calculate
the percent reaction for ionization of methanoic acid.
+
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+Practice
HW Book pg. 9 and 10
Pg. 716 #1, 3, 5 Check answers in back of book
Pg. 718 #7 Check answers in back of book
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+ Bronsted-Lowry Acids and Bases1.5K: Describe BrnstedLowry acids as proton donors and bases as proton acceptors
1.6K: Write BrnstedLowry equations, including indicators, and predict whether reactants or
products are favoured for acid-base equilibrium reactions for monoprotic and polyproticacids and bases
1.7K: Identify conjugate pairs and amphiprotic substances
+
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+Bronsted-Lowry Acid-Base Concept
Focuses on the role of the chemical species in a reaction rather
than on the acidic or basic properties of their aqueous
solutions.
Bronsted Lowry Definition for an Acid:proton donor
Bronsted Lowry Definition for an Base:proton acceptor
+
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+Bronsted-Lowry Acid-Base Concept
Bronsted-Lowry Reaction Equation: is an equation written to show
an acid-base reaction involving the transfer of a proton from oneentity (an acid) to another (a base)
Bronsted-Lowry Neutralization: is a competition for protons that
results in a proton transfer from the strongest acid present to the
strongest base present.
The Bronsted-Lowry concept does away with defining a substance as
being an acid or base. Only an entitythat is involved in a proton
transfer in a reaction can be defined as an acid or base and only fora particular reaction
Remember this is just a theoretical definition, not a theory, because it does
not explain why a proton is donated or accepted, and cannot predict which
reaction will occur for a given entity in any new situation.
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+Bronsted-Lowry Acid-Base Concept
Protons may be gained in a reaction with one entity, but lost in a
reaction with another entity.
The empirical term, amphoteric, refers to a chemical substance with the
ability to react as either an acid or base.
The theoritical term, amphiprotic, describes an entity (ion or molecule)
having the ability to either accept or donate a proton.
Example: When bicarbonate ions are in aqueous solution, some react with
the water molecules by acting as an acid, and some react by acting as a
base. Kc values given for these reactions, show that one predominates.
The number of ions acting as a base is over 2000x more than the number
reacting as an acid. = BASIC Solution
+
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+Bronsted-Lowry Acid Base Concept
The Amphoteric Nature of Baking Soda It can partially neutralize a strong acid, but also a strong base
Practice: pg. 724 #2-6
+ Conjugate Acids and Bases
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+ Conjugate Acids and Bases
In an acid-base reaction, there will always be two acids and two
bases.
The original acidon the left and the acid on the right createdby
adding a proton.
The original base on the left and the base on the right createdby
removing a proton
A pair of substances with formulas that differ only by a proton is
called a conjugate acid-base pair
This analogy shows why
acetic acid is a weak acid.
The proton is more strongly
attracted to the acetic acid
molecule than it is to the
water.
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+Acids/Bases
What about strong acids:
When HCl reacts with water, the water wins the competition
against the Cl- for the proton. This is why at equilibrium,
essentially all of the HCl molecules have lost protons to water.
NOTE: The stronger the base, the more it attracts a proton(proton acceptor). The stronger the acid, the less it attracts its
own proton (proton donor)
>99%
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+Conjugate Acids and Bases
RULE: The stronger the base, the more it attracts a proton(proton acceptor). The stronger the acid, the less it attracts its
own proton (proton donor)
What does this mean about their conjugate pair??
The stronger an acid, the weaker is its conjugate base.
If you are good at donating a proton, this means the conjugate base
is not good at competing for it (weak attraction for protons)
The stronger a base, the weaker is its conjugate acid.
If you are good at accepting a proton, this means the conjugate acidis not good at giving it up (strong attraction for protons).
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+
+
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+Acid-Base Reactions
What is happening?
Collisions are constantly occurring. Each time, a proton is transferred
to a stronger proton attractor.
Theoretically, it could transfer several times (each time to a stronger
proton attractor.) But once, it is transferred to the strongest basepresent, the proton will remain there as nothing outcompetes it.
Likewise, once the strongest acid has given up its proton, its
conjugate base cannot gain one back (as it is the weakest proton
attractor in the whole system)
So what does this mean?
Proton transfer occurs between the strongest acid and strongest
base. All other transfers are negligible so are ignored.
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+Predicting Acid-Base Reactions
+ Predicting Acid Base Reactions
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+ Predicting Acid-Base Reactions
1) List all entities present initially (ions, atoms, molecules, H2O(l)) as
they exist in aqueous solution.
H3O+
(aq) is the SA that can exist. If a stronger acid is dissolved, it reacts instantly and
completely with water to form H3O+
(aq). So all strong acids are written as H3O+
(aq)
OH-(aq) is the SB that can exist. If a stronger base is dissolved, it reacts instantly and
completely with water to form OH-(aq). The only example of this: soluble ionic oxides,
write the cation and the oxide ion is written as OH-(aq)
No entity can react as
a base if it is weaker
than water.
For this reason, the
conjugate bases of the
strong acids are not
considered bases in
aqueous solutions
+ Predicting Acid Base Reactions
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+ Predicting Acid-Base Reactions
1) List all entities present initially (ions, atoms, molecules, H2O(l)) as
they exist in aqueous solution.
Example: What will be the predominant reaction if spilled drain cleaner
(sodium hydroxide) solution is neutralized by vinegar?
List entities present:
Na+(aq) OH-(aq) CH3COOH(aq) H2O(l)
+ Predicting Acid Base Reactions
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+ Predicting Acid-Base Reactions
2) Identify and list all possible aqueous acids and bases, using
the Bronsted-Lowry definitions.
Use the entity lists of the Relative Strengths of Acids and Bases (in your data
booklet). Amphiprotic entities are labeled fro both possibilities. Conjugate
bases on SAs are not included. Metal ions are treated as spectators.
Example: What will be the predominant reaction if spilled drain cleaner(sodium hydroxide) solution is neutralized by vinegar?
List entities present:
Na+(aq) OH-(aq) CH3COOH(aq) H2O(l)
A A
BB
+ Predicting Acid Base Reactions
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+ Predicting Acid-Base Reactions
3) Identify the strongest acid and the strongest base present,
using the table of Relative Strengths of Acids and Bases.
Use the order of the entities in the Relative Strengths of Acids and Bases
table to identify the SA (the highest one on the table) and the SB (the lowest
one on the table)
Example: What will be the predominant reaction if spilled drain cleaner
(sodium hydroxide) solution is neutralized by vinegar?
List entities present:
Na+(aq) OH-(aq) CH3COOH(aq) H2O(l)
SA
SB
+ Predicting Acid Base Reactions
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+ Predicting Acid-Base Reactions
4) Write an equation showing a transfer of one proton from the
SA to the SB, and predict the conjugate base and the conjugate
acid to be the products.
Example: What will be the predominant reaction if spilled drain cleaner
(sodium hydroxide) solution is neutralized by vinegar?
List entities present:
Na+(aq) OH-(aq) CH3COOH(aq) H2O(l)
SA
SB
+ Predicting Acid Base Reactions
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+ Predicting Acid-Base Reactions
5) Predict the approximate position of equilibrium,
using the following Learning Tip
Example: What will be the predominant reaction if spilled
drain cleaner (sodium hydroxide) solution is neutralized by
vinegar?
Na+(aq) OH-(aq) CH3COOH(aq) H2O(l)
SA
SB
The reaction of H3O+
(aq) and OH-(aq) is always
quantitative (100%) so a single arrow is used
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+Practice
Pg. 731 #9,11,13,15 due tomorrow
Start HW Book pg. 16
Tomorrow:
Table Building and Thought Lab HW Book pg. 15 and 16
+ Table Building
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+ Table Building Lab Exercise 16.D
If a reaction is >50%, it favours products and the SA is above the SB.
If a reaction is
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+ Table Building Lab Exercise 16.D
+
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+Thought Lab
Homework Book pg. 13 and 14
5 Step Method Practice
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+Acids and Bases Ka2.2K: Define Ka , Kb and use these to determine pH, pOH, [H3O
+] and [OH] of acidic and
basic solutions
2.3K: Calculate equilibrium constants and concentrations for Bronsted-Lowry acid base systemswhen:
concentrations at equilibrium are known
initial concentrations and one equilibrium concentration are known
the equilibrium constant and one equilibrium concentration are known.
+ Th A id I i ti C t t K
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The Acid Ionization Constant, Ka
One important way that chemists communicate the strength of
any weak acid is by using the equilibrium constant expressionfor the ionization of weak acids. (Ka)
Look at your Relative Strengths of Acids and Bases Table:
All of the strong acids have a Ka value listed as very large;
remember they ionize quantitatively, so the actual acid speciespresent is H3O+
(aq)
All other acids are weak and vary greatly in the extent of reaction at
equilibrium.
We use the equilibrium law to write the formula for Ka
We omit liquid water because we
assume its value is essentially
constant for dilute solutions
Ka Values have only two sig
digs because they are
somewhat inaccurate because
we calculate them based on
assumptions.
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Ka Calculations
Two common Ka calculations:
Calculating Ka from empirical data (Examples #1-3)
Using Ka to predict hydronium ion concentration from an initial
concentration of weak acid. (Example #4 and 5)
+ Ka Calculations
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Ka Calculations
Example #1: The pH of a 1.00 mol/L solution of acetic acid is
carefully measured to be 2.38 at SATP. What is the value of Ka
for acetic acid?
Regardless of size, Ka values are usually expressed in scientific notation = 1.7 x 10-5
1.00mol/L 0.0042 mol/L = 0.9958
(rounds to 1.00 precision rule)
Change in concentration is negligible in
this casebut not always
+ Ka Calculations
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Ka Calculations
Example #2: A student measures the pH of a 0.25 mol/L
solution of carbonic acid to be 3.48. Calculate the Ka for
carbonic acid from this evidence.
+ Ka Calculations
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a
Example #3: The pH of a 0.400 mol/L solution of sulfurous acid
is measured to be 1.17. Calculate the Ka for sulfurous acid
from this evidence.
According to the equilibrium law, the
Ka for sulfurous acid is1.4 x 10-2
+ Ka Calculations
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Ka Calculations Example #4: Predict the hydronium ion concentration and pH
for a 0.200 mol/L aqueous solution of methanoic acid.
1.8 x 10-4
= x2
x = 0.006 = H3O+
(aq) concentration
(0.200)
Approximation Rule:0.200 = >1000
1.8 x 10 -4
So (0.200-x) = 0.200
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a Ca cu at o s Example #5: Predict the hydronium ion concentration and pH
for a 0.500 mol/L aqueous solution of hydrocyanic acid.
6.2 x 10-10
= x2
x = 1.8 x 10-5
= H3O+
(aq) concentration
(0.500)
Approximation Rule:
0.500 = >10006.2 x 10 -10
So (0.500-x) = 0.500
+P ti K
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Practice Ka
HW Book pg. 15 Ka and Strength of Acids
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+Acids and Bases Kb2.2K: Define Ka , Kb and use these to determine pH, pOH, [H3O
+] and [OH] of acidic and
basic solutions
2.3K: Calculate equilibrium constants and concentrations for Bronsted-Lowry acid base systemswhen:
concentrations at equilibrium are known
initial concentrations and one equilibrium concentration are known
the equilibrium constant and one equilibrium concentration are known.
+Base Strength & Ionization Constant K
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Base Strength & Ionization Constant, Kb
Ionic hydroxides, such as NaOH(aq) or Ca(OH)2(aq), are
assumed to dissociate completely upon dissolving.
Finding the hydroxide ion concentration does not involve any
reaction with water
Example: Find the hydroxide ion concentration of a 0.064 mol/Lsolution of barium hydroxide
+Base Strength & Ionization Constant K
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Base Strength & Ionization Constant, Kb
We can communicate the strength of weak bases, with the
equilibrium constant for its reaction with water.
The equilibrium constant is called the base ionization constant,
Kb
Two common Kb calculations:
Calculating Kb from empirical data (Example #1)
Using Kb to predict the concentration of hydroxide ions when the initial
concentration is known (Example #2)
+K C l l ti
We will use the same method as Ka calculations, but there is
usually one extra step because pH values need to be
t d t fi d h d id i t ti
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Kb Calculations
Example #1: A student measures the pH of a 0.250 mol/L
solution of aqueous ammonia and finds it to 11.32. Calculatethe Kb for ammonia
converted to find hydroxide ion concentrations
14 = pH + pOH
pOH = 2.68
10-2.68 = 0.0021 = OH-(aq)
Kb for ammonia is 1.8 x 10-5
Remember Kb has
only 2 sig digs
+P ti
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Practice
Pg. 746 #11-13 Check answers in back of book
+ Calculating OH- from Kb
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g b First problem, the data booklet has Ka values not Kb values.
What do we do?
Kw = KaKb So Kb = Kw/Ka remember Kw = 1.00 x 10-14
Example #1: Solid sodium benzoate forms a basic solution.
Determine the Kb for the weak base present.
+ Calculating OH- from Kb
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g b Example #2: Find the hydroxide ion amount concentration, pOH, pH and the
percent reaction (ionization) of a 1.20 mol/L solution of baking soda.
Baking soda = NaHCO3(s) Na+
(aq) + HCO3-(aq)
For HCO3-(aq), the conjugate acid is H2CO3(aq) whose Ka is = 4.5 x 10
-7
Approximation Rule:
1.20 = >1000
2.2 x 10 -8
So (1.20-x) = 01.20
2.2 x 10-8 = x2 . x = 1.6 x 10-4 = OH-(aq)
2.2x 10-8
+ Calculating OH- from Kb
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g b Example #2: Find the hydroxide ion amount concentration, pOH, pH and the
percent reaction (ionization) of a 1.20 mol/L solution of baking soda.
2.2 x 10-8 = x2 . x = 1.6 x 10-4 = OH-(aq)
2.2x 10-8
+Amphoteric Species
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Amphoteric Species
If an entity can react as either a Bronsted-Lowry acid or base, how
do you know which will be the predominant reaction?
Find the Ka value, calculate the Kb value, which ever is larger wins!
Example: Which reaction predominates when NaHSO3(s) is dissolvedin water to produce HSO3
-(aq) solution? Will the solution be acidic or
basic?
Ka = 6.3 x 10-8 Kb = Kw = 1.0 x 10-14 = 7.1 x 10-13
Ka1.4 x 10-2
The Kavalue far exceeds the Kbvalue, so an aqueous solution of this
substance will be acidic because the hydrogen sulfite ion will react
predominately as a Bron sted-Lowry acid.
+Practice
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Practice
Pg. 750 #1-6
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+Acids and Bases - Buffers1.8K: Define a buffer as relatively large amounts of a weak acid or base and its conjugate in
equilibrium that maintain a relatively constant pH when small amounts of acid or base are
added.
+ Chem 20 Review:
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A graph showing the continuous change of pH during an acid-base titration,
which continues until the titrant is in great excess, is called a pH curve
Endpoint refers to the point in a titration analysis where the addition of titrant
is stopped. The endpoint is defined (empirically) by the observed colourchange of an indicator.
Equivalence Point refers to the point in any chemical reaction where
chemically equivalent amounts of the reactants have combined. This point is
determined by stoichiometry.
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+Titration Analysis
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Titration Analysis
Chem 20 Review: Selecting proper indicators
Alizarin yellow is not a suitable indicator because it
will change colour long before the equivalence point
of this strong acid-strong base reaction, which
theoretically has a pH of exactly 7.
Orange IV is also unsuitable; its colour change
would occur too late.
The pH at the middle of the colour change range for
bromothymol blue is 6.8, which very closely
matches the equivalence point pH; so bromothymol
blue should give accurate results.
+Acid Base Indicators
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Any acid base indicator is really two entities for which we use the same name: a
Bronsted-Lowry conjugate acid-base pair.
At lease one of the entities is visibly coloured, so you can tell simply by looking
when it forms or is consumed. Examples include:
Phenolphthalein: conjugate acid is colourless, conjugate base is bright pink
Bromothymol Blue: conjugate acid is yellow, conjugate base is blue, and when they are in equal
quantities (appear green to the human eye)
Litmus Paper red (HIn) to blue (In-)
We will use the designation HIn for the conjugate acid and In- for the conjugate base as
their actual formulas can be very complex.
Summary:An indicator is a conjugate weak acid-weak base pair formed when an
indicator dye dissolves in water.
+Practice:
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Practice:
Homework Book pg. 17 (first half)
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+ Polyprotic Entities
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A pH curve for the addition of NaOH to a sample of H3
PO4(aq)
displays only two rapid
changes in pH even though H3PO4(aq) is triprotic.
This is because only two of the transfers are quantitative. The third reaction never goes
to completion, but instead establishes an equilibrium.
General Rule: Only quantitative reactions produce detectable equivalence points in an
acid-base titration.
+pH Curves
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pH Curves
+Practice
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Practice
Homework Book pg. 17 (2nd half)
+ General Rule
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Strong Acid to
Weak Base:
pH at
equivalence
point is always
lower than 7
Strong Base to
Weak Acid:
pH atequivalence
point is always
higher than 7
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+ pH Curve Shape
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p p
SA-SB: water is the only acid or base present = neutral solution
SA-WB: a weak acid (NH4+)is present along with water, at the equivalence
point, so the solution is acidic (pH < 7)
WA-SB: a weak base (CH3COO-) is present along with water, at the
equivalence point, so the solution is basic (pH > 7)
WA-WB: do not have detectable equivalence points, because the reactions
are usually not quantitative.
+Practice:
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Practice:
Pg. 762 #11-14
+ Buffers
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A bufferis a relatively large amount of any weak acid and its conjugate
base, in the same solution. In equilibrium, they maintain a relatively
constant pH when small amounts of acid or base are added.
I.e. The addition of a small amount of base
produces more acetate ions. The very small
change in the acid-base conjugate pair ration
and the complete consumption of the OH-
explains why the pH change is very slight
The addition of a small amount of acid
produces more acetic acid. The very small
change in the acid-base conjugate pair ratio
and the complete consumption of the H3O+explains why the pH change is very slight
+Buffer Example: Blood Plasma
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Buffer Example: Blood Plasma
Blood plasma has a remarkable buffering ability, as shown by
the following table.
This is very useful, as a change of more than 0.4 pH units, can
be lethal. If the blood were not buffered, the acid absorbed
from a glass of orange juice would likely be fatal.
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The limit of the ability of a buffer to maintain a pH level.
When one of the entities of the conjugate acid-base pair reacts
with an added reagent and is completely consumed, the
buffering fails and the pH changes dramatically.
All of the CH3COOH(aq) is used up, OH-
additions will now cause the pH to
drastically increase
All of the CH3COO-(aq) is used up, H3O
+
additions will now cause the pH to
drastically decrease
+Practice
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Practice
Pg. 766 #16-21
Homework Book pg. 18 (Acid/Base Stoichiometry Review)