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Unit 2: EquilibriumIcons are used to prioritize notes in this section.Make some notes: There are SOME important items on this page that should be copied into your notebook or highlighted in your printed notes.

Copy as we go: There are sample problems on this page. I EXPECT YOU to copy the solutions into a notebook— Even if you have downloaded or printed the notes!!!

Look at This: There are diagrams or charts on this page. Look at them and make sure you understand them. (but you don’t need to copy them)

Extra Information: This page contains background information that you should read, but you don’t need to copy it.

R Review: There is review material on this page. It is up to you to decide if you want to make notes or highlight it, depending on how well you remember it.

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Unit 4(formerly Module 5)

Equilibrium

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Equilibrium• Overview:

Equilibrium is the concept of a system remaining “in balance”. A system in equilibrium does not change at the macroscopic level (the level that we can detect with our senses).True equilibrium should not be confused with homeostasis or “steady-state”, a process by which living organisms attempt to maintain a consistent internal conditions by absorbing or excreting materials from and to their environment.

11.0

4

Equilibrium vs. Homeostasis• Equilibrium usually exists in

a closed system, where materials cannot easily enter or leave.

• Examples: • A reversible chemical reaction

occurs inside a closed container. The reaction appears to have stopped, but at a molecular level changes are still going on.

• A liquid in a sealed bottle does not appear to evaporate.

• Homeostasis usually exists in an open system, where materials can enter and leave.

• Examples: • A dog lives in a kennel. It eats

and excretes roughly equal amounts, and therefore maintains a steady weight and internal conditions.

• A cell in a human body maintains a balance of nutrients.

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The Temporary Nature of Equilibrium

• Although we study equilibrium as if it is an unchanging state, in reality dynamic processes as well as static forces may act upon the equilibrium.

• Eventually something will upset the equilibrium, temporarily or permanently throwing it “out of balance”

• Often a new equilibrium will be re-established after the original equilibrium is upset.

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Examples

• A precariously balanced rock formation can endure in “equilibrium” for centuries. Suddenly the balanced rock falls, temporarily disturbing the equilibrium.

• A reversible chemical reaction inside a beaker has reached a state of equilibrium and appears to have stopped. A researcher adds more of one of the reactants, upsetting the equilibrium. The reaction temporarily resumes until a new equilibrium is established.

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The Old Man of the MountainFor two hundred years a precarious rock formation in New Hampshire was said to resemble the face of an old man. It had become a symbol of New Hampshire, appearing on postcards, road signs and coins

On May 3, 2003 The rock face collapsed. In terms of equilibrium, we could say that this was a static equilibrium that endured for centuries, until it was disturbed by a spring storm. Afterwards a new equilibrium was established, that unfortunately no longer resembled a face.

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9

Chapter 11

Qualitative Aspects of Chemical Equilibrium

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Qualitative Aspects of Equilibrium

• What is an equilibrium? What properties does it have? How can we distinguish dynamic, and static equilibria and tell them apart from simple steady states?

11.1

11

Static Equilibrium• Static equilibrium exists when a

system remains unchanging without any active, dynamic processes involved

• One rock sitting on another is at static equilibrium, even if it is balanced precariously. No dynamic processes are acting on it, so it remains unchanged.• Static equilibrium is a bit boring. We

seldom deal with it in chemistry.

Note: Gravity is not considered to be a

dynamic force.It is a static force.

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Dynamic Equilibrium

• Dynamic equilibrium is the result of two opposing, active processes occurring at the same rate. No visible changes take place, but there are constant changes in the particles at a microscopic level.

• There are several types of dynamic equilibrium of interest to chemists, which are described on the following slides including:• Phase Equilibrium• Solubility Equilibrium• Chemical Equilibrium

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• A dynamic equilibrium is a bit like a hockey game.

• Barring penalties, there is always the same number of players on the ice, but some players are constantly leaving the bench as others return to it.

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11

14

Phase Equilibrium(1st type of dynamic equilibrium)

• Phase equilibrium is a dynamic equilibrium that occurs when a single substance is found in several phases or states within a system as the result of a physical change.

• Example:• In a closed bottle a water may exist as both a liquid

and a gas at the same time (eg. Water vapour above liquid water). As water molecules evaporate from the liquid phase, other water molecules condense from the gaseous phase.

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Solubility Equilibrium(2nd type of dynamic equilibrium)

Solubility Equilibrium occurs when a solute is dissolved in a solvent, and an excess of the solute is in contact with the saturated solution.• Example: If you add too much sugar to a cup

of tea, the tea becomes saturated with sugar and no more appears to dissolve. In fact, some molecules of sugar are dissolving as other molecules recrystallize back into solid sugar.

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Chemical Equilibrium(3rd type of dynamic equilibrium)

• Chemical equilibrium occurs when two opposing chemical reactions occur at the same rate, leaving the composition of the system unchanged.

• Example: Dinitrogen tetroxide (N2O4) and nitrogen dioxide (NO2) can exist in the same container. Each can change into the other, and at equilibrium they do so at the same rate.

N2O4 2 NO2

This is the most important type of equilibrium in chemistry!!

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Irreversible and Reversible Reactions

• Some chemical reactions are easily reversed, like the electrolysis of water.

• Others, such as the burning of wood are impossible to reverse under laboratory conditions.

11.2

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Irreversible Chemical Reactions

An irreversible reaction is a reaction that can only occur in one direction, from reactants to products.• This definition is assumed to refer to reactions

occurring under normal laboratory conditions.• In a lab, it is easy to burn a piece of wood. It is

impossible, under laboratory conditions, to turn ash, smoke, carbon dioxide and water back into wood.

The growth of a tree does allow wood to be produced from materials that might include wood ashes, but growing a tree takes decades, and requires countless changes involving many complex chemical mechanisms and multiple organic catalysts (enzyme systems). Burning is therefore NOT considered reversible!

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Remember!Irreversible = One Way

Reactants ProductsReactants can become products, but products cannot turn back into reactants.

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Reversible Chemical Reactions• Some reactions are easily reversed

using common laboratory procedures. • For example, it is possible to

decompose water into hydrogen and oxygen in a electrolytic cell, and equally possible to synthesize water from hydrogen and oxygen in a fuel cell:

2H2O + electrical energy 2 H2 + O2

2H2 +O2 2H2O + electrical energy

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Remember!Reversible = Both Ways

Reactants ProductsReactants can become products, and products can also turn back into reactants. Also show as: reactants products

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Reversibility and Equilibrium

Only reversible reactions can

produce a true dynamic chemical EQUILIBRIUM

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A Chemical System is at Equilibrium if it meets these Criteria:

1. The System is Closed, for example by being sealed inside a container so material cannot enter or leave.

2. The Change is reversible. The reaction or change can proceed in both direct and reverse directions.

3. There is no Macroscopic Activity. Nothing seems to be happening – the properties of the system are constant.These unchanging properties can include: colour, amount of undissolved solute, concentration, pressure etc.

4. There is Molecular Activity. Reactions continue at the microscopic or molecular level

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Definitionsof some easily confused terms

• Macroscopic: Occurring at the level we can detect with our senses, as opposed to microscopic. Observable changes.

• Microscopic: Occurring at a level below what we can see. Too small to observe without instruments.

• Dynamic Equilibrium: A balance that involves two opposing active processes that are occurring at the same rate. This contrasts with Static Equilibrium and Steady State (Homeostasis).

• Static Equilibrium: A balance that does not involve active processes.

• Steady State (including Homeostasis): An apparent balance that occurs in an open system. An unchanging set of properties is maintained, but materials enter and leave the system.

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• Page 287• Read all the questions, make sure you

understand them, be prepared to answer them verbally next class.

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Le Châtelier’s Principle

• Henri Louis Le Châtelier (1850-1937) was a French chemist who is most famous for his studies of chemical equilibrium.

• In addition he studied metal alloys and, with his father, was involved in the development of methods of purifying aluminum

11.4

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Equilibrium is not eternal• An equilibrium can exist for a long time, only

to change (be upset) when certain conditions change.

• After it is upset, there is a period of adjustment, then a new equilibrium is established.

Original Equilibrium

New Equilibrium

Equilibrium Upset

Adjusting...

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Some factors that might upset an equilibrium.

• Which of these factors do you think might affect the amount of reactant and product at equilibrium?

• Maybe Temperature?• Maybe Pressure?• Maybe Concentration of reactants and

products? • Maybe Catalyst?

• Most of them do, but one does not.• We’ll see which one doesn’t a bit later!

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• Henri LeChâtelier studied many of these factors to see how they could effect a system at equilibrium.

• He found some factors could favour the direct reaction, increasing the amount of product.

• Others could favour the reverse reaction, increasing the amount of reactant.

• Regardless, eventually an equilibrium was re-established, but with new amounts of product and reactant.

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When an Equilibrium is Upset...

• Henri LeChatelier stated the following generalization:

• “If the conditions of a system in equilibrium change, the system will react to partially oppose this change until it attains a new state of equilibrium.”

• In other words: The system will establish a new equilibrium.

.

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• Will any reversible reaction will eventually reach equilibrium?• Answer: yes, as long as it in a closed system.

• Will the amount of product always equal the amount of reactant at equilibrium? • Short answer: NO! they are not always equal.

• At equilibrium the amount of “reactant” and “product” may vary depending on several factors.

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The Effect of a Catalyst

• Adding a catalyst to a system already at equilibrium will have NO EFFECT.• Adding a catalyst to a system that has not yet

reached equilibrium will cause it to reach equilibrium faster.

• Why? A catalyst increases both forward and backward rates equally, so the final result will be the same, but the process of reaching equilibrium will be faster.

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• Increasing the temperature will favour the endothermic reaction.

• Decreasing the temperature will favour the exothermic reaction.

Effect of Temperature

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Effect of Pressure

• Increasing the pressure may favour the reaction that produces fewer gas particles

• Decreasing the pressure may favour the reaction that produces more gas particles.

• Note: only reactants or products that are in the gaseous state are counted towards the effects of pressure.

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Effects of Concentration

• The effects of concentration of the reactants and products are most important of all. Increasing or decreasing the concentration of Reactants WILL have an effect.

• Remember: Pure solids(s) and liquids(l) do NOT have a variable concentration.

• Before we can see the effects of changing a concentration we should remember what LeChatelier said...

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Remember LeChatelier’s Principle:

• “If the conditions of a system in equilibrium change, the system will react to partially oppose this change until it attains a new state of equilibrium.”

• In other words, any “stress” or change that you make to the system will cause it to react in a way that tries to (partially) undo the change that you made.

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Changes in Concentration• If you increase the concentration of a reactant

(gas or aqueous), the equilibrium will shift to use up some of the reactant you added.

• If you increase a product (gas or aqueous), the equilibrium will shift to reduce the product and make more reactant

H2(g) + I2(g) 2HI(g)

If you increase the concentration of Hydrogen… …The system will react to reduce

the amount of Hydrogen… …by shifting the reaction towards the product

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H2 + I2 2HI

Sudden increase in[H2].

The equilibrium changes: [HI] goes way up, [I2] goes down.

This causes [H2] to adjust towards its original level

The equilibrium changes: [HI] goes way up, [I2] goes down.

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Note:

• Adding more of a undissolvable solid or pure liquid normally has NO EFFECT on a system at equilibrium.

• It can only affect the equilibrium if the concentration changes, and usually only gases and aqueous solutions have variable concentration.

• However:• adding water to an aqueous solution can change its

concentration by dilution.• Adding solid to a solution that is not saturated MIGHT

increase its concentration (if it dissolves!).

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Changes in Temperature• Increasing the temperature causes the

equilibrium to shift in the direction that absorbs heat (the endothermic direction).

• Decreasing the temperature shifts the equilibrium in the exothermic direction.

2SO2 + O2 2SO3

If you increase the temperature… …The system will react to reduce

the temperature……by shifting the reaction towards endothermic side

Heat+

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2SO2 + O2 2SO3 + heat

Sudden increase in temperature.

The endothermic reaction kicks in, getting

rid of some SO3, and creating more SO2 and

more O2 and cooling things off

This causes a lowering of the temperature,

moving it towards its original level

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Changes in Pressure(only affects systems where one or more materials are gases)

• Increasing the pressure causes the equilibrium to shift in the direction that has the fewest gas molecules.

• Decreasing the pressure shifts the equilibrium in the direction that produces more gas molecules.

If you increase the pressure… …the system will create fewer molecules……to reduce the pressure.

8 molecules 4 molecules

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N2 + 3 H2 2 NH3

When the pressure increases, the equilibrium adjusts, making more NH3 (since 2 molecules NH3 are fewer particles than 3 molecules of H2 plus 1 molecule of N2)

Sudden increase in pressure.

Pressure partially adjusts towards the original level..

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Special Note Re. Gases

• Only gases can be affected by pressure.• If only one side of an equation has gases,

then...• Increasing pressure will favour the side with no

gases. • Decreasing the pressure will favour the side with

gases.

I2(s) I2(g)Decreased pressure

Increased pressure

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• Page 304, Questions 1 to 4

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Chapter 12

The Quantitative Aspects of Equilibrium

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Chapter 12

In this section we will explore the mathematical aspects of equilibrium, including:

• The Equilibrium Constant (Kc)

• The Equilibrium Law

12.1

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The Equilibrium Constantand Equilibrium Law Expressions

• The equilibrium constant (Kc) is:• A number• derived from an equilibrium law expression.• a ratio between the concentration of products and

the concentration of reactants of a reversible reaction at equilibrium, but…

• With each aqueous or gaseous product and reactant raised to the power of its corresponding coefficient

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Deriving the Equilibrium Law

• The next two slides show how the equilibrium law was derived from the rate law. If you want to understand the relationship between rates and equilibrium you should follow this.

• If you just want to use the equilibrium law to find Kc, you can skip forward three slides.

Warning:

Math Content Ahead

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A Generalized Reversible Reaction with reactants and products

aA + bB ↔ cC + dD reactants products

Forward rate: rdir = kdir [A]a [B]b

Reverse rate: rrev = krev [C]c [D]d

At equilibrium, rdir = rrev

so, through the magic of algebra… ba

dc

rev

dir

BADC

kk

][][][][

Products on top

Reactants on bottom

Forward reaction = dirReverse reaction = rev

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Simplifying the Rate Constant

• The two separate rate constants (kdir and krev) are often replaced by a single equilibrium constant, Kc:

ba

dc

rev

dir

BADC

kk

][][][][

ba

dc

c BADCK

][][][][

Becomes:

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The Equilibrium Law• The lowercase letters represent

coefficients,• the uppercase letters are

chemical formulas, • the square brackets mean

concentration.• Kc is the equilibrium constant

ba

dc

c BADCK

][][][][

For a chemical equation of the type:

aA + bB ↔ cC + dD

“Products” always go on the top!

“Reactants” always go on the bottom!

The equilibrium law expression is:

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Variants of the Equilibrium Constant

keq

Kc

Ka

These symbols are all used forEquilibrium Constants in different text books

or for different types of reaction.

I usually use Kc, since your text book and the study guide both use it. The old textbook used keq. In problems involving acids and bases, Ka and Kb are often used.Ksp is used for problems involving the solubility product.

ba

dc

c BADCK

][][][][

Kb

Ksp

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What does the Equilibrium Constant mean?

• It is a ratio between the amount of reactant and product that exist at equilibrium.

• If Kc is greater than 1, then the direct reaction is favoured, and there is more product than reactant at equilibrium*

• If Kc is less than 1, then the reverse reaction is favoured, and there is more reactant than product at equilibrium*

• If Kc is 0, the reaction is impossible.• if Kc is infinite (∞) the reaction is spontaneous and

irreversible so as much reactant as possible will change to product

• Reactions we call irreversible have high Kc value (>1010)• Reactions that don’t normally occur have Kc values near 0 (<10-10)

*this is a slight over-simplification, since the formula can be complicated, but it is generally true.

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Effect of Temperature on an Equilibrium Constant

• An equilibrium constant relates to concentrations, and only remains constant if the other conditions, such as temperature, remain fixed.

• If the temperature were to change, so would the value of Kc

• How the value of Kc might change depends on the type of reaction (exothermic or endothermic) and how the temperature changed (increased or decreased)

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Table Showing Effects of Temperature on Equilibrium Constants

Type of reaction Temperature change

Favoured Reaction Change in Kc

Exothermic* Increase Reverse () DecreaseExothermic* Decrease Direct () IncreaseEndothermic Increase Direct () IncreaseEndothermic Decrease Reverse () Decrease

*Exothermic means either ΔH < 0 or that Reactants Products + Energy

Notice that any change in temperature that favours the direct reaction will cause the value of Kc to increase.

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Fixed Concentrations• Some materials have “fixed” concentrations, ie.

Their concentrations cannot change in an equilibrium.• Examples:

• An undissolved solid, (it can’t have a concentration unless it dissolves.)

• A pure liquid. (a pure substance always has its maximum concentration)

• These cases, which include all substances with the (s) and (l) phase markers, are not included in equilibrium calculations.

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Example• What is the equilibrium expression for the following

equation: CN1-

(aq) + H2O(l) HCN(aq) + OH1- (aq)

][]][[

1

1

CN

OHHCNKc

]][[]][[

21

1

OHCNOHHCNKc

Answer:

Not this:

Why? Because H2O (liquid water) is a pure substance and therefore has a fixed concentration. Substances with fixed concentration are not included in an equilibrium expression.

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• Copy the following equations, and write the equilibrium law expression for each

1) H2(g) + I2(g) 2 HI(g)

2) 2 BrCl(g) Cl2(g) + Br2(g)

3) CO2(g) + H2O(l) H2CO3(aq)

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Calculating Equilibrium Concentrations

• To calculate the concentrations of reactants and products at equilibrium we sometimes use a table that records the initial concentration, the change in concentration and the final equilibrium concentration of each reactant or product.

• We call such a table an I.C.E. table

12.1.4

The ICE method

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• The I.C.E. method is a technique that can help solve some equilibrium problems

• I.C.E. stands for:

• Initial concentration [A]I• Change in concentration Δ[A]• Equilibrium concentration [A]E

The I.C.E. method

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Identify the molar ratios (based on the coefficients of the equation)Eliminate any unused ratios, such as those based on liquids and undissolved solids.

Reactant 1 Reactant 2(if needed)

Product 1 Product 2(if needed)

I(Initial)

C(Change)

E(Equilibrium)

Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in

Concentration Before

Concentration Before

Concentration Before

Concentration Before

Concentration After

ConcentrationDifference

(Δ[C])

NegativeRatio

(reactant)

Sum Sum Sum

PositiveRatio

(product)

NegativeRatio

(reactant)

Molar ratio

Molar ratio

Molar ratio

Molar ratio

Reactant + Reactant Product + Product

Write the chemical equation of the reaction. Show all reactants and Products

Enter the information you already know into the table. If no values are given for the initial product concentrations, it is usually safe to assume they are zero.Look for a column that you can complete!

Fill in the missing squares in “C” row. They will be ratios to the one you knowThe numbers will be negative for reactants, positive for products.Add to find the equilibrium values (adding a negative number is like subtracting)Use the Equilibrium Concentrations to answer any further questions: such as finding Kc , Ka Kb or Ksp)

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ICE method: Step by StepWrite the chemical equation of the reaction. Show all reactants and Products

Identify the molar ratios (based on the coefficients of the equation)Eliminate any unused ratios, such as those based on liquids and undissolved solids.Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in

Look for a column that you can complete!

Enter the information you already know into the table. If concentrations are not given as mol/L you may have to convert them. If not told otherwise, you may assume that the initial product concentrations are zero.

Fill in the missing squares in “C” row. They will be ratios to the one you knowThe numbers will be negative for reactants, positive for products.Add to find the equilibrium values (adding a negative number is like subtracting)

Use the Equilibrium Concentrations to answer any further questions (such as finding Kc , Ka or Kb)

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Copy this work!Sample Problem

(see p. 316 and 317 in Text Book)

At a given temperature, 10 moles of nitrogen oxide (NO) and 8 moles of Oxygen (O2) are placed in a 2 litre container. After a given period of time, the following equilibrium is obtained:

2 NO(g) + O2(g) 2NO2(g)

Once equilibrium is attained, 8 moles of NO2 have been produced. Calculate the equilibrium constant.Preliminary work: It is necessary to convert to moles per litre!

Initial concentrations [NO]I = 10 mol in 2 L = 5 mol/L

[O2]I = 8 mol in 2L = 4 mol/Lassume: [NO2] I= 0 mol in 2L = 0 mol/L

Final Concentration: [NO2]E = 8 mol in 2L = 4 mol/L

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NO O2(g) NO2

I(Initial C)

C(Change)

E(Equilibrium)

Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in

5 mol/L

[NO]i

4 mol/L

i[O2]

0 mol/L

[NO2]i

4 mol/L

[NO2]e

+4 mol/LΔ [NO2]

-4 mol/L(-ratio)

1 mol/L 2 mol/L

-2 mol/L(-ratiot)

2 1 2

2 NO(g) + O2(g) 2NO2(g)

Write the chemical equation of the reaction. Show all reactants and ProductsIdentify the molar ratios (based on the coefficients of the equation)Enter the information you already know into the tableLook for a column that you can complete!Fill in the missing squares in “C” row. They will be ratios to the one you knowThe numbers will be negative for reactants, positive for products.Add to find the equilibrium values (adding a negative number is like subtracting)

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Finishing the ProblemNow you need to find the Kc value. To do this, use the values from the “E” line of your table (the equilibrium concentrations) and substitute them into the Kc formula:

82

1621

4][][

][2

2

22

22

ONONOKc

The value of the equilibrium constant is 8(although you do not need to give the units of an equilibrium constant, since our concentrations were in moles/Litre, and our time is assumed to be in seconds it would be safe to call it mol/(L∙s))

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A Tougher Sample Problem(see p. 316 and 317 in Text Book)

At 1100K the equilibrium constant of the following reaction is 25H2(g) + I2(g) 2HI(g)

2 moles of hydrogen and 3 moles of iodine are placed in a 1 L container. What is the concentration of each substance when the reaction attains equilibrium..

This time we have no numbers for equilibrium concentrations, we will have to use algebraic variable for some of the values.We can use x to represent the change in Hydrogen Concentration

Let: Δ[H2] = -x (why negative? Because hydrogen is a reactant!)

Initial concentrations: [H2] = 2 mol/L, [I2] = 3 mol/L

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H2 I2(g) HI

I(Initial C)

C(Change)

E(Equilibrium)

Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in

2 mol/L 3 mol/L 0 mol/L

2x mol/L

+2x mol/L-x mol/L

2-x mol/L 3-x mol/L

-x mol/L

1 1 2

H2(g) + I2(g) 2HI(g)

Write the chemical equation of the reaction. Show all reactants and ProductsIdentify the molar ratios (based on the coefficients of the equation)Enter the information you already know into the tableLook for a column that you can complete!Fill in the missing squares in “C” row. They will be ratios to the one you knowThe numbers will be negative for reactants, positive for products.Add to find the equilibrium values (adding a negative number is like subtracting)

69

Continuing the ProblemSet up the Kc formula, substituting in the expression from line “E”

.][

.][reactprodKc So..

)3)(2()2(25

2

xxx

This equation can be rearranged in the following steps:2)2()3)(2(25 xxx

22 4)56(25 xxx 22 425125150 xxx

015012521 2 xx Carried over to next slide

70

Using the Quadratic formula

015012521 2 xx Equation to be solved

aacbbx

242

Quadratic formula, where: a=21, b=125, c=150

)21(2)150)(21(4)125()125( 2

x substitute

29.4x 67.1xTwo solutions to formula

But only one of the solutions will give a real answer!

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Choosing the Best Answers29.4x 67.1xTwo solutions to formula

Try finding the correct concentrations using the solutions

xH E 2][ 2

xI E 3][ 2

xHI E 2][

29.229.42][ 2 H 33.067.12][ 2 H

33.167.13][ 2 I

34.3)67.1(2][ HI

At equilibrium: The concentration of hydrogen will be 0.33 mol/L, The concentration of iodine will be 1.33 mol/L and The concentration of hydrogen iodide will be 3.34 mol/L

Concentration cannot be negative!

Solve using 4.29 Solve using 1.67

72

Simplifying I.C.E. tables(the 5% rule)

Sometimes, when doing an ICE table you may have to subtract a very small value from a relatively large value, for example 2.0 mol/L – 1.0x10-4 mol/L. In this case, don’t bother doing the subtraction, since by the time you change it to show significant digits, the result will be the same: 2.0 mol/L – 0.0001 mol/L = 1.9999 mol/L ≈ 2.0 mol/L.

In fact, as a rule of thumb, if the number you subtract is less than 5% of the original number, you can skip the subtraction.

73

• Page 318 • Questions #1 to 8, (Equilibrium Law and Kc)• Questions #19 to 21, (I.C.E. method)

• Optional Assignments:• Link to another worksheet

74

Acids and Bases• The concept of acids and bases has been with us for a long

time, but there has always been difficulty with the exact definitions.

• In this section we will look at four theories about acids and bases (Don’t worry, you only need to remember 2 of them)

• We will find out the difference between “strength”, “concentration” and “acidity” of an acid.

• We will also explore the concepts of pH and pOH a bit deeper than you did in grade 9 or 10.

• We will explore the mathematical relationships between concentration and pH

12.2

75

Properties of Acids & Bases• Acids are solutions that:

• Turn litmus red, but leave phenolphthalein clear.• React with active metals to give off H2 gas• React with carbonate salts to give off CO2 gas• Taste sour (if safe to taste)• Have low pH numbers (below 7)

• Bases are solutions that:• Turn litmus blue, and turn phenolphthalein purple.• Taste bitter (if safe to taste)• Seldom react with metals or carbonates• Have High pH numbers (above 7)• Emulsify fats and oils

76

Theories of Acids and BasesHypothesis#1: Lavoisier’s Mistake

(optional item)

Lavoisier (1776) dealt mostly with strong oxyacids, like HNO3 and H2SO4. He claimed that:

• An acid is a substance that contains oxygen.• This hypothesis is now known to be completely wrong,

but it did give oxygen its name: “oxy” (meaning acid)+ “gen” (former or creator)

77

Theories of Acids and Bases(Hypothesis#2: Arrhenius’ Theory)

Arrhenius (c. 1884) claimed that:• An acid is a substance that dissociates in water

to produce H+ ions• A base is a substance that dissociates in water

to produce OH- ionsArrhenius’ theory is still used to this day as a “simplified” way of explaining acids and bases. It explains most of the properties of acids... Why their formulas usually begin with H, why they give off hydrogen when reacting with metals, etc. But this theory does not account for acidic and basic salts– substances that act like acids and bases, but don’t have an H or OH in their formula, or for anhydrous acids, or for reactions that occur outside of water.

78

Theories of Acids and Bases(Hypothesis#3: Brønsted-Lowry)

Brønsted and Lowry (1923) proposed:• An acid is a substance from which a proton

(H+) can be removed. An acid is a proton donor.

• A base is a substance that can cause a proton to be removed from an acid. A base is a proton acceptor.

Although a bit complicated for explaining simple acid/base reactions, this is the main theory in use today.

79

Brønsted-Lowry and Conjugates• One of the results of the Brønsted-Lowry theory

is that in a reaction, each acid has a corresponding base and each base has a corresponding acid (called their conjugates)

HCl(aq) + H2O(l) H3O+(aq) + Cl-

(aq)

Acid (strong)

Base(weak)

ConjugateBase of HCl

ConjugateAcid of H2O

H+ H+

Becomes

Becomes

80

Simple way of finding conjugates• The formula of a conjugate base is the formula of the acid

with one H removed, and more negative charge.• The conjugate base of HBr is Br-

• The conjugate base of HCN is CN-

• The conjugate base of water (H2O) is OH-

• The first conjugate base of H3PO4 is H2PO4-

• H3PO4 can have other conjugates, such as HPO42-, PO4

3-

• The formula of a conjugate acid is the usually the negative ion, with an H added in front of it and one step more positive:• The conjugate acid of F- is HF• The conjugate acid of water (H2O) is H3O+

81

Theories of Acids and Bases(Hypothesis#4: Lewis Acids)

Gilbert Lewis (c.1940) proposed:• An acid is a substance that can accept an electron

pair to form a covalent bond.• A base is a substance that can donate a pair of

electrons to form a covalent bond.

This theory is not explained in the new textbook, and will probably not be required for examinations. It is given here as optional enrichment material. The Lewis theory is the only one that can explain the properties of acidic & basic salts.

(optional item)

82

Comparing the HypothesesTheory An Acid is... A Base is... Details

Lavoisier A substance with oxygen

Incorrect hypothesis

Arrhenius A substance that dissociates in water to produce H+ ions

A substance that dissociates in water to produce OH- ions.

All acids have HAll bases have OH

Bronsted-Lowry

A substance from which an H+ ion can be removed (a proton donor)

A substance which can remove the H+ from an acid (a proton acceptor)

All acids have HBases can have any negative ion.

Lewis A substance that can accept a pair of electrons to form a covalent bond

A substance that can donate a pair of electrons to form a covalent bond

Acids may have H+ or some other + ions, a base may have OH- or some other – ions

Incorrect Hypothesis

Optional Hypothesis

83

Assignment• Find the Conjugate base of each of these acids:

1) HCl2) HNO3

3) HBrFind the conjugate acid of each of these bases:1) CN-

2) HCO3-

3) H2O

84

Dissociation

• Both theories of acids tell us that acids are compounds that “dissociate” in water to give off H+ ions (protons), which immediately attach to water molecules to become H3O+ ions

• Eg. HCl added to water breaks up into:• H+ and Cl- ions, the H+ ions join H2O as shown by:

• HCl(aq) + H2O(l) H3O+(aq) + Cl-

(aq) (monoproteic acid)

• Different acids can dissociate differently:• H2SO4(aq) + 2H2O(l) 2H3O+

(aq) + SO4-(aq) (diproteic acid)

• H3PO4 (aq) + 3H2O 3H3O+(aq) + PO4

-(aq) (triproteic acid)

85

Dissociation of Water• In distilled water, at any given moment about

2 molecules out of every billion (0.00000002%) exist as ions: H+ and OH-

• This means there is an equilibrium reaction occurring.

H2O H+ + OH- • Since so few of the molecules are ionized, the K value of this

equilibrium must be tiny! (see calculation 3 slides from now)

• This also means that water can act as either a very weak acid or a very weak base! (it can give off H+ or OH-)

86

Equivalence of H+ and H3O+

• Some textbooks use H+ and other books use H3O+ to represent the hydrogen ion content of acids.• Although H3O+ better represents the actual state of

the ions in a water solution, H+ is simpler to write.• As long as the solvent is water, the two are the same

[H+]=[H3O+]In most cases, when

water is present

87

• Ka is called the acidity constant

• Kb is called the basicity constant

• Kw is the ionization constant of water• but all three are calculated the same way as Kc and are used

for similar purposes.

We usually use Ka or Kb in acid-base reactionsTables of Ka values help us compare the “natural strength” of

various acids.

Similarity of Ka Kb Kw and Kc

88

The Ionization Constant of Water (Kw)

• Water is H2O, but at any given temperature a tiny fraction (≈ 2/ 109) of water molecules dissociate into ions:

• H2O(l) H+(aq) + OH-

(aq)

• Kw represents a ratio between the ionized and unionized molecules:

][]][[

2OHOHHKw

]][[ OHHKw

89

• Because of the equivalence of H+ and H3O+ in water solutions, you could also write:

• At room temperature [H3O+] and [OH-] in pure water are both about 1.0×10-7 mol/LSo...

• The value of Kw varies a bit with temperature (see p 328), but has an average value of 1.0 × 10-14 at 25°C.

]][[ 3 OHOHKw

CatKw 25100.1 14

]10][10[ 77 wK

90

Effect of Temperature on Kw

• At lower temperatures, Kw becomes smaller. Less water molecules dissociate when it is cold

• At higher temperatures, Kw becomes larger. More water molecules dissociate when it is hot.• Kw affects the pH of water. We say that the pH of

pure water is 7, but in reality it varies by temperature:

• At the boiling point (100°C) , the pH of pure water will be close to 6!

• At the freezing point (0°C) it will be about 7.5!

91

The Equilibrium Constant of an Acid

• The acidity constant of an acid (Ka) represents the fraction of the acid which will dissociate and release H+ (ie. H3O+) ions.

• It is one of two factors that affect what the pH of an acid solution will be. • (the other factor is the concentration of the acid)

• Acids with low Ka values are considered naturally “weak” acids. Acids with high Ka values are considered naturally “strong” acids.

92

If we consider an acid to have the formula HA then its dissociation can be represented by HA(aq) + H2O(l) H3O+

(aq) + A-(aq)

In this case, we can calculate the Ka value this way:

][]][[

)(

)()(3

aq

aqaqa HA

AOHK

][]][[

)(

)()(

aq

aqaqa HA

AHK

NOTE: Because of the equivalence of [H+] and [H3O+] in some texts, this could be shown as

93

100][

][%)(

)(3

aq

aq

HAOHionization

The Ionization Percentage of an Acid

If you ever want to work out the percentage of an acid that is ionized (like if you had too much time on your hands, or were bored or something) you can use this formula:

94

• Copy the problem and solve:Johnny dissolves 3.4 mols of dried weak acid

powder in 500 mL of water. HA(aq) + H2O(l) H3O+

(aq) + A-(aq)

When he measures the acidity he discovers that the acid has 4.5x10-2 mol/L of H3O+ ions.

What is the Ka of this acid?

What is the percentage ionization of this acid?

95

pH, [H3O+] and Ka of Acids

There are three different factors that affect the apparent strength of an acid. In this section we will discover some of the ways of measuring acids and bases, including:

• Strength vs. concentration vs. acidity• What are pH and pOH?• pH (by the chart)• pH (by calculation)• Applying the I.C.E. Method to Acids and Bases

• For finding [H+], [OH-]concentrations, Ka, Kb.

12.

96

Measures of an Acid

• There are three different ways of measuring acids:1) The natural strength Ka

2) The concentration mol/L3) The degree of Acidity pH

97

Natural Strength of an Acid• A “strong” acid is one that dissociates completely

(100%) when dissolved in water. Strong acids have Ka values close to infinity (huge numbers > 1010). Strong acids dissociate irreversibly.Typical Reaction: HA + H2O H3O+ + A-

• A “weak” acid is one that does NOT dissociate completely when dissolved in water. Weak acids have small Ka values, usually less than 1. Weak acids dissociate reversibly, creating a possible equilibrium.Typical reaction: HA + H2O H3O+ + A-

98

“Strong” Acids and “Weak” AcidsAcids shown in red, Conjugate Base in blue

Strong Acid Name Reaction Ka (>>1)

Hydrochloric Acid

HCl(aq) +H2O H3O+ + Cl- ∞

Hydrobromic Acid

HBr(aq) +H2O H3O+ + Br- ∞

Hydroiotic Acid HI(aq) +H2O H3O+ + I- ∞Nitric Acid HNO3(aq) +H2O H3O+ + NO3

- ∞Sulphuric Acid H2SO4+2H2O 2H3O+ + SO4

- ∞Weak Acid Name Reaction Ka (<1)

Iodic HIO3+H2O H3O++IO3- 1.7x10-1

Oxalic Acid H2C2O4+H2O H3O++HC2O4 5.8x10-2

Formic Acid HCOOH +H2O H3O++HCOO- 1.8x10-4

Acetic Acid CH3COOH +H2O H3O+

+CH3COO-1.7x10-5

Citric Acid H3C6H5O7+H2O H3O++H2C6H5O7 3.2x10-7

Hydrogen Peroxide

H2O2 + H2O H3O+ + HO2- 2.4x10-12

Water H2O + H2O H3O+ OH- 1.0x10-14

99

Concentration of an Acid• The concentration of an acid is the number of

moles of an acidic substance that have been dissolved in a volume of water. This is determined by the concentration formula:

6 mol/L is considered a very concentrated acid 0.1 mol/L is considered a fairly dilute acid

VnC Where:

n = number of moles of acidic solute V= volume of solution in Litres

100

Strength, Concentration and Acidity

The strength and concentration of an acid together determine its degree of acidity

STRENGTHThe natural strength of an

acidic compound, determined by its Ka value

ConcentrationThe number of moles per

litre of the acidic compound in solution.

Degree of AcidityThe effective strength of an acid,

based on its [H3O+] concentration and Usually recorded as its pH

101

Degree of Acidity

The degree of acidity is the EFFECTIVE strength of the acid, that is, how effective the acid is at reacting with other materials.

Acidity can be measured using:a) the H3O+ concentration, or more commonly

b) a measure called the pH

102

Who invented pH and pOH?(optional background information)

The degree of acidity of a solution is directly related to its [H3O+] concentration. Unfortunately this is often a small number expressed in scientific notation, such as 1.3x10-5 mol/L or 2.3x10-13 mol/L. Not easy numbers to remember, compare or write.

In 1909 Søren Sørensen suggested an easier way to record the degree of acidity of solutions. It was a logarithmic scale called pH. Each level of the pH scale represents a 10 fold difference in H3O+ (or H+) concentration.

ie. A acid of pH 2 has 10 times more [H+] than one that’s pH3.

103

pH and pOHpH is a measure of the degree of acidity,

given by the following formula:pH = - log [H3O+]

Or: (since [H3O+] is equivalent to [H+] in water solutions)

pH = - log [H+]Although less used, there is also a measure

of the degree of alkalinity, called pOH:pOH = - log [OH-]

104

pH and Concentration of H+ ionsFor solutions of an exact pH you may use the chart below:

pH [H+]* [OH-]*

pOH

pH [H+]*

[OH-]* pOH

0 10 0 10-14 14 8 10-8 10-6 61 10-1 10-13 13 9 10-9 10-5 52 10-2 10-12 12 10 10-10 10-4 43 10-3 10-11 11 11 10-11 10-3 34 10-4 10-10 10 12 10-12 10-2 25 10-5 10-9 9 13 10-13 10-1 16 10-6 10-8 8 14 10-14 10-0 07 10-7 10-7 7

Acids Bases

*Concentration in mol/L Temperature = 25°C

For simplicity I used [H+] instead of [H3O+]

105

What about “in-between” pH values?

• The formula for pH is: • pH = -log [H+]• On your calculator you must find out how to calculate the

negative logarithm of a number!

• The formula for [H+] concentration is:• [H+]=10 –pH or… [H+]=log-1(-pH)

• Sometimes log-1 is called antilog: [H+] =antilog (-pH)• Sometimes log-1 is called inverse log: [H+] =invlog (-pH)• Sometimes log-1 is called 10x: [H+] =10(-pH)

• On your calculator find out raise 10 to a negative number or how to calculate the inverse logarithm (antilog) of a negative number.

106

(TI 83 instructions)Type: log (1.40 x 10 ^ (-) 8) Enter

Question: Find the pH if the H3O+ concentration is 1.40x10-8 mol/L

(–) log ^ (–) Enter

The answer should be:pH=7.85… (I’ve rounded to 3 Sig.Fig.)

Solution: pH = – log(1.40 x 10 – 8)

Question: Find the H3O+ concentration if the pH is 4.30

(TI-83 instructions)10 ^ (-) 4.3^ (–)

Solution: [H+] =10 – 4.30

Enter

The answer should be:pH=5.01… x 10-5 mol/L(I’ve rounded to 3 Sig.Fig.)

107

Using the Windows Calculator• Switch to scientific view To find pH, use: 1.4 Exp 8 =

To find [H3O+] use: 4.30 =or: 10 4.30 =

Finding the pH if H3O+ concentration is 1.40x10-8 mol/L

Finding [H3O+] concentration if pH is 4.30

7.85…

5.01…x10-5

108

Try: (1.4 Exp 8 +/- ) log +/- =

Or: (1.4 Exp 8 +/- ) log +/- =

Traditional Scientific CalculatorsQuestion: Find the pH if the H3O+ concentration is 1.40x10-8 mol/LSolution: pH = – log(1.40 x 10 – 8)

log +/-Exp +/-

EE +/- log +/-

Question: Find the H+ concentration if the pH is 4.30

Try: 10 yx 4.30 +/- =Or: 4.30 +/- Inv log =Or 4.30 +/- 2ndF 10x =Or: 10 ^ (-) 4.3

Solution: [H+] = 10 - 4.30

7.85...

5.01... ×10-5

109

1. Find the [H+] and [OH-] of the following pH solutions without using a calculator:• A) pH=4 B) pH=12 C) pH=9 D) pH=7 E) pH=8

2. Find the pH of the following solutions without a calculator:• A) [H+]=1.0x10-8 mol/L B) [OH-]=1.0x10-3 mol/L

• C) [H+]=1.0x10-5 mol/L D) [OH-]=1.0x10-9 mol/L

3. Find the [H+] of these solutions using the 10x or antilog function of your calculator: [H+]=10-pH

• A) pH=3.7 B) pH=9.8 C) pH=6.2 D) pH=4.0 E) pH=7.1

4. Find the pH of the following solutions using the log function of your calculator: pH= - log [H+].• A) [H+]=4.5x10-3 mol/L B) [H+]=3.4x10-8 mol/L

• C) [H+]=3.0x10-7mol/L D) [H+]=2.5x10-2 mol/L

110

Problem 5.

5. Calculate the Ka value of a monoproteic, weak acid if a 0.10 mol/L (initially) solution has a pH of 5.5 when it reaches equilibrium.• Assume: HA H+ + A-

• Hint: You must find the concentration of [H+] ions before you do the problem.

2 sig. digits

2 sig. digits

111

1. Answers• A) 1x10-4, 1x10-10 mol/L B) 1x10-12, 1x10-2 mol/L • C) 1x10-9, 1x10-5 mol/L D) 1x10-7, 1x10-7 mol/L • E) 1x10-8, 1x10-6 mol/L

2. Answers:• A) pH=8 B) pOH-=3 pH= 11• C) pH=5 D) pOH-=9 pH=5

3. Answers in mol/L• A) 2.0x10-4 mol/L B) 1.6x10-10 mol/L C) 6.3x10-7 mol/L • D) 1.0x10-4 mol/L E) 7.9x10-8 mol/L

4. Answers: pH= - log [H+].• A) pH= 2.3 B) pH=7.5• C) pH=6.5 D) pH=1.6

Solutions to Problems 1- 4

112

Solution to Problem 5

• Calculate the Ka value of a monoproteic weak acid if a 0.1 mol/L (initially) solution has a pH of 5.5 when it reaches equilibrium.• Assume: HA H+ + A-

• Hint: You must find the concentration of [H+] ions before you do the problem.

• Find the [H+]: • [H+] = log-1 (-5.5) or [H+] = 10 -5.5

• [H+] = 3.162x10-6 mol/L• For an extremely accurate solution, use the I.C.E. method to

find the equilibrium concentrations… (continued on next slide)

• For a quicker solution, using the 5% rule, see later slides

113

HA H+ A-

I (initial) 0.10 0 0

C (change)

E (equil.) 3.163x10-6

+3.163x10-6

+3.163x10-6

-3.163x10-6

3.163x10-69.9997x10-2

HA H+ + A-

1 : 1 : 1

Ka = [H+][A-] = (3.163x10-6 x 3.163x10-6) mol/L =1.0004 x10-10

[HA] 9.9997x10-2 mol/L ≈ 1.0x10-10

Round to 2 Sig. Digits

114

The Short-cut:An alternative solution using 5% rule

Ka = [H+][A-] = (3.163x10-6 x 3.163x10-6) mol/L =1.0004 x10-10

[HA] mol/L

= 1.0x10-10

HA H+ A-

I (initial) 0.10 0 0

C (change)

E (equil.) 3.163x10-6

3.163x10-6

assume

Since 3.163 x 10-6 is less than 5% of 0.1, we can use the 5% rule and

simply write 0.1, instead of subtracting (0.1 – 0.000003163)

and using an ICE table

After rounding to the correct number of significant digits, our

answer is the same as doing it the hard way!

0.100.10

115

A Half-Evil Example

Calculate the pH of an aqueous solution of formic acid HCOOH at 0.20 mol/L if its acidity constant is 1.8x10-4.The equation of this reaction is as follows

HCOOH(aq) + H2O(l) H3O+(aq) + HCOO-

(aq)

See the solution on page 333(I told you, it’s only half-evil)

116

Assignments

Page 339, Questions 4 to 23

117

Unit 4: Chapter 12.2.5Solubility Product Constant

SolubilityCalculating the solubility constant

Examples

118

Solubility

• A saturated solution contains:• Dissolved solute (usually ions) in the solution• Some non-dissolved solute (crystal)at the bottom

Dissolved ions

Solid solute crystals

119

In a saturated solution...

• There is an equilibrium situation:Solute(s) ion+

(aq) + ion–(aq)

Some of the solute dissociates (dissolves into ions) and at the same time, some of the ions crystallize back into solid.

Solid Solute

Dissolved Ions

120

Solubility• The solubility of a substance is the maximum

amount of the substance that dissolves in a given volume of the solvent• Usually given in g/L or sometimes in g/100mL• For calculation purposes you should use molar

solubility (mol/L)• To convert g/100 mL to g/L, multiply by 10• To convert g/L to mol/L, change grams to moles using

the mole formula:

Mmn

mass (g)

molar mass(g/mol)Moles

121

• The solubility of a substance partly depends on the temperature of the water you are dissolving it in.• Standard tables give solubility at 25°C

• Solids generally have a higher solubility at higher temperatures

• Gases generally have a lower solubility at higher temperatures.• There are a few exceptions to these

generalizations!

122

What Is Ksp

• The solubility product constant is a number used to compare the solubilities of different solutes.

• The higher the Ksp, the more of the solute can be dissolved before the solution becomes saturated. Low Ksp values mean very little of the substance will dissolve.

• Ksp is calculated in a similar way to other equilibrium constants, but it’s a bit easier!

123

“Soluble” and “Insoluble” Substances

• A “soluble” substance is a substance with a high Ksp value. At room temperature, a large amount of the substance can be dissolved in water.

• A truly insoluble substance doesn’t exist, but any substance with a very low Ksp is said to be “insoluble”, since so little of it will dissolve in water that it can barely be measured.

See table 8.11 on page 423 to find out what common substance will be “soluble” and “insoluble”

124

General Formula for Ksp

• For a solute that dissociates like this:XmYn(s) mX+

(aq) +nY-(aq)

• The Ksp can be calculated by this way:Ksp = [X+]m[Y-]n

125

Calculating Ksp

• Example 1:• BaSO4(s) Ba2+

(aq) + SO42-

(aq)

• At equilibrium there will be a certain concentration of Ba2+ ions and SO4

2- ions

Ksp = [Ba2+][SO42-]

• Example 2:• CaCl2(s) Ca2+

(aq) + 2Cl-(aq)

• At equilibrium there will be a certain concentration of Ca2+ ions and Cl- ions

Ksp = [Ca2+][Cl-]2

Eliminate solid.

Eliminate solid.

1 : 1 : 1

1 : 1 : 2

126

• Ksp = [Ba2+][SO32-]

• If the solubility of BaSO3 is 0.0025g/L, what is the Ksp of this compound?

• Since it dissociates equally, the concentration of both ions will be equal to the moles of BaSO3 that dissolved (0.0025 g/L).

• We have to convert that to mol/L• MBaSO3 = 137.3 + 32.1 + 3(16.0) = 217.4 g/mol

• nBaSO3 = 0.0025 g/L / 217.4 g/mol =1.15x10-5 mol

• So... [Ba2+] = [SO32-] = 1.15x10-5 mol/L

• Ksp = [Ba2+][SO32-]

=(1.15x10-5)(1.15x10-5) = 1.32x10-10 mol2/L2

1 : 1Dissociates equally (1:1)

Mmn

127

Example• The solubility of silver carbonate (Ag2CO3) is

3.6x10-3 g/100mL at 25C. Calculate the value of the solubility product constant of silver carbonate.

Data:Solubility = 3.6×10-3 g/100mL

Molar Solubility =?[+ ions] = ?[– ions] = ?Ksp= ?

we want the solubility in g/L, so multiply by 10...3.6×10-3 g/100mL = 3.6×10-2 g/L

Now we need it in mol/L, so...n= 3.6×10-2 g /L

275.8 g/moln= 1.3×10-4 mol/L

Molar solubility = 1.3x10- 4 mol/L

Mmn

M = 2(107.8)+12+3(16)Values from periodic table and formula of Ag2CO3.

Carry over to the next slide.

Use solubility:

m = g (per litre)

128

Data:Solubility = 3.6x10-3 g/100mLMolar Solubility= 1.3x10-4 mol/L

[CO32-]

[Ag+]

Ksp=?

Equation:

Ag2CO3(s) 2 Ag+ + CO32-

Ksp = [Ag+]2 [CO32-]

[CO32-]=1.3×10- 4 mol/L

[Ag+] = 2.6×10- 4 mol/L

Ksp=(2.6×10- 4 mol/L)2(1.3×10-4 mol/L)

Ksp =8.8×10-12 mol3/L3

[Ag+] is double [CO32-]. Multiply by 2

1 : 2 : 1

= 1.3x10- 4 mol/L= 2.6x10- 4 mol/L

The molar concentration of CO3 2-(aq)

in solution will equal the molar solubility!

Info carried over from previous slide

129

• At the annual chemistry Christmas party, a careless chemist spills a whole bottle of calcium fluoride, CaF2, into a 2 litre punch bowl filled with fruit punch. Most of the calcium fluoride dissolves, but a little powder settles to the bottom of the punch bowl.

• The Ksp of calcium fluoride is 3.4*10-11

• CaF2 dissociates: CaF2(s) Ca2+(aq) + 2F-

(aq)

• The lethal dose of fluoride ions is 1g.• Will all of the eight chemists at the party die if each

drinks a cup of the punch? (1 cup ≈ 250mL or ¼ L.) • Would one chemist die if he drank all the punch?

130

• Ksp = [Ca2+] [F-]2

• The dissociation formula is:CaF2(s) Ca2+

(aq) + 2F-(aq)

• So there will be twice as much F- produced as Ca2+

• Let’s choose a variable to represent the concentration of Ca2+ ions produced at equilibrium. Double that variable to represent the F- ion concentration.Let x = [Ca2+] and 2x = [F-]Note: The units of x will be mol/L

131

• Ksp = [Ca2+] [F-]2

• Simplify x(2x)2

• Reverse the equation and divide by 4 to isolate the x3

• Simplify:• Cube root of both sides...• ...Gives us “x”, but we

aren’t finished...

4104.3 11

3

x

123 105.8 x3 1233 105.8 x

Lmolx /1004.2 4

311 4104.3 x

211 )2(104.3 xx

132

But x represents [Ca2+], and since the [F-] is twice as high:

So to change this to grams per litre we must multiply by the molar mass of F-, which is 19.0 g/mol (note: this is not F2)

A cupful (250mL) is one quarter of this, so...

Lmolx /1004.2 4

LmolF /1008.41004.22][ 44

LgF /1076.70.191008.4][ 34

cupgF /1094.1][ 3

133

• Each cup contains only 0.00194 g of fluoride ions, so nobody would die from drinking one cup.

• If somebody drank the whole punchbowl (2 litres = 8 cups) he would still only get 0.0155 g of fluoride ions. Still well below the deadly limit. The chemists would survive and have sparkling white teeth!

• Note: Although this dosage might not be lethal it could have long-term side effects. Never drink contaminated punch! The maximum recommended concentration of fluoride in water is 1 mg/L, about one eighth what is in the punch.

cupgF /1094.1][ 3

134

Exercises on Solubility

• Page 340 # 36 to 39