15.0 equilibrium

7/29/2019 15.0 Equilibrium

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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



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+


Example #1: Write the equilibrium law expression for the

reaction of nitrogen monoxide gas with oxygen gas to form

nitrogen dioxide gas.



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+


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



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+


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.


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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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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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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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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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 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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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



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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)



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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?



A A

BB



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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)





SA

SB



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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.





SA

SB



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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?


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:




+ 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

+ Ka Calculations


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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:




+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.



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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



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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.

+ Buffering Capacity


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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)

15.0 equilibrium

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