1 unit 2: energy icons are used to prioritize notes in this section. make some notes: there are some...
TRANSCRIPT
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Unit 2: EnergyIcons 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
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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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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
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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
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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 NO2This 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 using electrolysis, and equally possible to synthesize water from hydrogen and oxygen by controlled burning:
2H2O + electrical energy 2 H2 + O2
2H2 +O2 H2O + heat 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.
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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.
• Before we can see the effects of changing a concentration we must examine the mathematics of an equilibrium reaction.
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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,
the equilibrium will shift to use up some of the reactant you added.
• If you increase a product, the equilibrium will shift to reduce the product and make more reactant
H2 + I2 2HI
If you increase the amount 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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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 particles.
• Decreasing the pressure shifts the equilibrium in the direction that produces more gas particles.
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 NH3 takes up less space than H2 + 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
• With each product/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
BA
DC
k
k
][][
][][
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
BA
DC
k
k
][][
][][
ba
dc
c BA
DCK
][][
][][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 BA
DCK
][][
][][
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 BA
DCK
][][
][][
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
*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
OHCN
OHHCNKc
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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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
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The I.C.E. method
• A technique that can help solve some equilibrium problems
• Stands for:• Initial concentration
• Change in concentration
• Equilibrium concentration
• A technique that can help solve some equilibrium problems
• Stands for:• Initial concentration
• Change in concentration
• Equilibrium concentration
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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
(ΔC)
NegativeRatio
(reactant)
Sum Sum Sum
PositiveRatio
(product)
NegativeRatio
(reactant)
Molar ratio
Molar ratio
Molar ratio
Molar ratio
Reactant + Reactant Product + ProductWrite 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.
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 know
The 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.
Add to find the equilibrium values (adding a negative number is like subtracting)
Fill in the missing squares in “C” row. They will be ratios to the one you know
The numbers will be negative for reactants, positive for products.
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.
Use the Equilibrium Concentrations to answer any further questions (such as finding Kc , Ka or Kb)
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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: (convert to moles per litre)
Initial concentrations Ci [NO] = 10 mol in 2 L = 5 mol/LCi [O2] = 8 mol in 2L = 4 mol/L
assume: Ci [NO2] = 0 mol/L
Final Concentration: Cf [NO2] = 8 mol in 2L = 4 mol/L
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NO2 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
(Ci [NO2])
4 mol/L
(Ci[O2])
0 mol/L
(Ci [NO2])
4 mol/L
(Cf [NO2])
+4 mol/L(ΔC [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 know
The 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
16
21
4
][][
][2
2
22
22
ONO
NOKc
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.
Let x represent the change in Hydrogen Concentration
Let: ΔC[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 know
The 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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Continuing the ProblemSet up the Kc formula, substituting in the expression from line “E”
.][
.][
react
prodKc So..
)3)(2(
)2(25
2
xx
x
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
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Using the Quadratic formula
015012521 2 xx Equation to be solved
a
acbbx
2
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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
xCH 22
xCI 32
xCHI 2
29.229.422
HC 33.067.122
HC
33.167.132
IC
34.3)67.1(2 HIC
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
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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.
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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
72
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)
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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)
74
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.
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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.
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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)
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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
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Bronsted-Lowry and conjugates
• One of the results of the Bronsted-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)
Acting Base
ConjugateAcid
ConjugateBase
H+ H+
79
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 acid dissociate differently:• H2SO4(aq) + 2H2O(l) 2H3O+
(aq) + SO4-(aq) (diproteic acid)
• H3PO4 (aq) + 3H2O 3H3O+(aq) + PO4
-(aq) (triproteic acid)
80
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!• 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-)
81
Equivalence of H+ and H3O+
• Some textbooks use H+ instead of 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, and is used in many texts.
[H+]=[H3O+]In most cases, when
water is present
82
• 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 reactions
Tables of Ka values help us compare the “natural strength” of various acids.
Similarity of Ka Kb Kw and Kc
83
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:
][
]][[
2OH
OHHKw
]][[ OHHKw
84
• 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
85
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 neutral water will be close to 6!
• At the freezing point (0°C) it will be about 7.5!
86
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.
87
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
aqaq
a HA
AOHK
][
]][[
)(
)()(
aq
aqaq
a HA
AHK
NOTE: Because of the equivalence of [H+] and [H3O+] in some texts, this would be shown as
88
100][
][%
)(
)(3
aq
aq
HA
OHionization
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:
89
Module 5, Lesson 4Ph, [H3O+] and Ka of Acids
• Strength, concentration and acidity• What are pH and pOH• pH (by the chart)• pH (by calculation)• The I.C.E. method
• For finding concentrations, Ka, Kb or Kc .
90
Strength, Concentration and Degree of Acidity
The strength and concentration of an acid determine its degree of acidity
STRENGTHThe natural strength of an acid, 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 related to its pH
91
What are 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 pH3.
92
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 pOHpOH = - log [OH-]
93
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 6
1 10-1 10-13 13 9 10-9 10-5 5
2 10-2 10-12 12 10 10-10 10-4 4
3 10-3 10-11 11 11 10-11 10-3 3
4 10-4 10-10 10 12 10-12 10-2 2
5 10-5 10-9 9 13 10-13 10-1 1
6 10-6 10-8 8 14 10-14 10-0 0
7 10-7 10-7 7
Acids Bases
*Concentration in mol/L Temperature = 25°C
For simplicity I use [H+] instead of [H3O+]
94
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.
95
(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 S.D.)
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 (I’ve rounded to 3 S.D.)
96
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
97
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
98
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)
99
Unit 4: Chapter 12.2.5Solubility Product Constant
SolubilityCalculating the solubility constant
Examples
100
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
101
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.
102
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 formua:
M
mn
mass (g)
molar mass(g/mol)Moles
103
• The solubility depends on the temperature• 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!
104
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.
• Ksp is calculated in a similar way to other equilibrium constants, but it’s a bit easier!
105
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
106
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
107
• 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
• nBaSO3 = 0.0025 / 217.4 =1.15x10-5 mol
• So... [Ba2+] = [SO32-] = 1.15x10-5 mol/L
• Ksp = [Ba2+][SO32-]
=(1.15x10-5)(1.15x10-5) =1.32x10-10
1 : 1Dissociates equally (1:1)
M
mn
108
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
M
mn
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)
109
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)2(1.3×10-4)
Ksp =8.8×10-12
[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
110
• 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?
111
• 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
112
• 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...
4
104.3 113
x
123 105.8 x3 123 105.8 x
Lmolx /1004.2 4
311 4104.3 x
211 )2(104.3 xx
113
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
114
• 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
115
Exercises on Solubility
• Page 340 # 36 to 39